text:
\section{Introduction}

We give a characterization of spaces that can be mapped onto the Baire space by a continuous quasi-open bijection, see Proposition~\ref{another_desc} and Definition~\ref{def_of_pi_sp}. We call such spaces $\pi\mathsurround=0pt$\emph{-spaces}. The class of $\pi\mathsurround=0pt$\emph{-spaces} is countably productive and the complement of a $\sigma\mathsurround=0pt$-compact subset of a $\pi\mathsurround=0pt$-space is also a $\pi\mathsurround=0pt$-space, see Corollaries~\ref{count_prod} and~\ref{sigma_comp}.

The class of $\pi\mathsurround=0pt$-spaces is similar to the class of spaces with a Lusin $\pi\mathsurround=0pt$-base~\cite{patrakeev2015metrizable}:
Every space with a Lusin $\pi\mathsurround=0pt$-base is a $\pi\mathsurround=0pt$-space and every second-countable $\pi\mathsurround=0pt$-space has a Lusin $\pi\mathsurround=0pt$-base, see  Remark~\ref{pi.base} and  Theorem~\ref{when_pi_has_lpb}.
The Baire space and the Sorgenfrey line (and its finite and countable powers) have a Lusin $\pi\mathsurround=0pt$-base~\cite{patrakeev2015metrizable,patrakeev2019property}, so these spaces are $\pi\mathsurround=0pt$-spaces.

Every space that has a Lusin $\pi$-base can be mapped onto every nonemty Polish space by a continuous open mapping~\cite{patrakeev2015metrizable}. In Theorem~\ref{example} we show that there exists a $\pi\mathsurround=0pt$-space without this property.

In Theorem~\ref{bla_main_theorem} we give a characterization of continuous open images of $\pi\mathsurround=0pt$-spaces: For a nonempty topological space ${X}$, the following are equivalent:
\begin{itemize}
  \item[\ding{226}\,] ${X}$ is a continuous open image of a $\pi$-space;
  \item[\ding{226}\,] ${X}$ is a continuous open image of a space that can be mapped onto a Polish space by a continuous quasi-open bijection;
  \item[\ding{226}\,] ${X}$ is a Choquet space of countable $\pi$-weight and of cardinality not greater than continuum.
\end{itemize}

The above theorem implies a characterization of the class of Hausdorff compact continuous open images of $\pi\mathsurround=0pt$-spaces: it is the class of Hausdorff compact spaces of countable $\pi$-weight and of cardinality not greater than continuum, see Corollary~\ref{corr.1}.
And also it implies a characterization of the class of second-countable continuous open images of $\pi\mathsurround=0pt$-spaces: it is the class of second-countable Choquet spaces of cardinality not greater than continuum, see Corollary~\ref{corr.2}.

The class of second-countable continuous open images of $\pi\mathsurround=0pt$-spaces coincides with the class of second-countable continuous open images of  spaces with a Lusin $\pi\mathsurround=0pt$-base, see Corollary~\ref{corr.2}. The question whether the class of all continuous open images of $\pi\mathsurround=0pt$-spaces coincides with the class of all continuous open images of spaces with a Lusin $\pi\mathsurround=0pt$-base remains open, see Question~\ref{quest}.

\section{Notation and terminology}

We use terminology from~\cite{topenc} and~\cite{kunen2014set}. A \emph{space} is a topological space. We also use the following notation.

\begin{nota}\label{not01}%
  The symbol $\coloneq$ means ``equals by definition''\textup{;}
  the symbol ${\colon}{\longleftrightarrow}$ is used to show that the expression on the left side is an abbreviation for the expression on the right side\textup{;}
  \begin{itemize}
  \item [\ding{46}\,]
 $\mathsurround=0pt
 \omega$ $\coloneq$ the set of finite ordinals $=$ the set of natural numbers, so $0=\varnothing\in\omega$ and ${n}=\{0,\ldots,{n}-1\}$ for all ${n}\in\omega;$
  \item [\ding{46}\,]
 $\mathsurround=0pt
 {s}\,$ is a \textit{sequence}$\ {\colon}{\longleftrightarrow}\
 {s}$ is a function whose domain is a finite ordinal or is $\omega$;
\item [\ding{46}\,]
 if ${s}$ is sequence, then

 $\mathsurround=0pt
 \mathsf{length}({s})\:\coloneq$ the domain of ${s}$;
 \item [\ding{46}\,]
   $\mathsurround=0pt
   \langle{s}_0,\ldots,{s}_{{n}-1}\rangle$ $\coloneq$
   the sequence ${s}$ such that  $\mathsf{length}({s})={n}\in\omega$ and  ${s}({i})={s}_{i}$ for all ${i}\in{n};$
 \item [\ding{46}\,]
   $\mathsurround=0pt
   \langle\rangle$ $\coloneq$ the sequence of length {0};
 \item [\ding{46}\,]
   if ${s}=\langle{s}_0,\ldots,{s}_{{n}-1}\rangle$, then

   $\mathsurround=0pt
   {s}\hspace{0.5pt}^{\frown}{x}
   \:\coloneq\:\langle{s}_0,\ldots,{s}_{{n}-1},{x}\rangle$;
 \item [\ding{46}\,]
   if ${s}=\langle{s}_0,\ldots,{s}_{{n}-1}\rangle$ and  ${t}=\langle{t}_0,\ldots,{t}_{{m}-1}\rangle$, then

   $\mathsurround=0pt
   {s}\:{\hat{}}\:{t}
   \:\coloneq\:\langle{s}_0,\ldots,{s}_{{n}-1},{t}_0, \ldots, {t}_{{m}-1}\rangle$;
\item [\ding{46}\,]
 $\mathsurround=0pt
 {f}{\upharpoonright}\hspace{0.5pt}{A}$ $\coloneq$ the restriction of the function ${f}$ to the set ${A};$

 \item [\ding{46}\,] ${g}\circ {f}$ is the composition of functions ${g}$ and ${f}$ (that is, ${g}$ after ${f}$);
  \item [\ding{46}\,]
 $\mathsurround=0pt
 {A}\subset{B}
 \ {\colon}{\longleftrightarrow}\
 {A}\subseteq{B}\enskip\text{and}\enskip{A}\neq{B};$
  \item [\ding{46}\,]
 if ${s}$ and ${t}$ are sequences, then

 $\mathsurround=0pt
 {s}\sqsubseteq{t}
 \ {\colon}{\longleftrightarrow}\
 {s}={t}{\upharpoonright}\hspace{0.5pt}\mathsf{length}({s})\quad$

 (actually, ${s}\sqsubseteq{t}\leftrightarrow{s}\subseteq{t}$);

  \item [\ding{46}\,]
 $\mathsurround=0pt
 {}^{B}\!{A}$ $\coloneq$ the set of functions from ${B}$ to ${A};$

 in particular, ${}^{0}\hspace{-1pt}{A}=\big\{\langle\rangle\big\};$
\item [\ding{46}\,]
 $\mathsurround=0pt
 {}^{{<}\omega}\hspace{-1pt}{A}
 \:\coloneq\:\bigcup_{{n}\in\omega}{}^{{n}}\hspace{-1pt}{A}\,=\,$
 the set of finite sequences in ${A};$
\item [\ding{46}\,]
 if ${p}$ is a point in a space with topology $\tau$, then

 $\mathsurround=0pt
 \tau({p})$ $\coloneq$ $\{{U}\in\tau:{p}\in{U}\}$ $=$
 the set of open neighbourhoods of ${p}$;

  \item [\ding{46}\,]
 $\mathsurround=0pt
 \gamma\,$ is a $\pi\mathsurround=0pt$-\textit{net} for a space ${X}
 \ {\colon}{\longleftrightarrow}\ $ all elements of $\gamma$ are nonempty and for each nonempty open ${U}\subseteq{X}$, there is ${G}\in\gamma$ such that ${G}\subseteq{U}$;
  \item [\ding{46}\,]
 $\mathsurround=0pt
 \gamma\,$ is a $\pi\mathsurround=0pt$-\textit{base} for a space ${X}
 \ {\colon}{\longleftrightarrow}\ \gamma$ is a $\pi\mathsurround=0pt$-net for ${X}$ and all elements of $\gamma$ are open;
  \item [\ding{46}\,]
  $\mathsurround=0pt
  \tau_{\omega^\omega}$ $\coloneq$  the Tychonoff product topology on the set ${}^{\omega\hspace{-1pt}}\omega$, where $\omega$ carries the discrete topology;
  \item [\ding{46}\,]
  $\mathsurround=0pt
  {\omega^{\omega}}$ $\coloneq$ the \emph{Baire space} $=$ the space $\langle{}^{\omega\hspace{-1pt}}\omega,\tau_{\omega^\omega}\rangle$.

  \end{itemize}
\end{nota}

Recall that, in \cite{kechris2012classical}, a \emph{Souslin scheme}  is an indexed family $\langle V_a \rangle_{a \in {}^{{<}\omega}\hspace{-1pt}\omega}$ of sets.

\begin{deff}
 Let ${\bf V} = \langle V_a \rangle_{a \in {}^{{<}\omega}\hspace{-1pt}\omega}$ be a Souslin scheme, $\ll{X},\tau\rangle$ be a space, and $p \in {}^\omega\hspace{-1pt}\omega$. Then

 \begin{itemize}
 \item [\ding{46}\,]  ${\bf V}$  {\it covers} ${X}\ {\colon}{\longleftrightarrow}\  V_{\langle \rangle} = X$ and $V_a = \bigcup_{n \in \omega} V_{a\hspace{0.5pt}^{\frown}n}$ for all $a \in {}^{{<}\omega}\hspace{-1pt}\omega$;
 \item [\ding{46}\,]  ${\bf V}$  {\it partitions} ${X}\ {\colon}{\longleftrightarrow}\  {V}$ covers ${X}$ and  $V_{a\hspace{0.5pt}^{\frown}{n}} \cap V_{a\hspace{0.5pt}^{\frown}{m}} = \emptyset$ for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$ and ${n} \neq {m} \in \omega$;
 \item [\ding{46}\,] the {\it fruit} of ${p}$ in ${\bf {V}}$, $\fruit{p}{{\bf V}}$, is the set $\bigcap_{n \in \omega}V_{p \upharpoonright n}$;
 \item [\ding{46}\,]  ${\bf V}$ is {\it complete} $\ {\colon}{\longleftrightarrow}\ $  $\fruit{q}{{\bf V}} \neq \emptyset$ for all $q \in {}^\omega\hspace{-1pt}\omega$;
 \item [\ding{46}\,]  ${\bf V}$ has {\it strict branches} $\ {\colon}{\longleftrightarrow}\ |\fruit{q}{{\bf V}}|=1$ for all $q \in {}^\omega\hspace{-1pt}\omega$;
 \item[\ding{46}\,] ${\bf {V}}$ is {\it open} on $\ll{X},\tau\rangle\ {\colon}{\longleftrightarrow}\ {V}_{a}\in\tau$  for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$.
 \end{itemize}
\end{deff}

\begin{deff}\label{def.luz.p.base}
 A \textit{Lusin $\hspace{1pt}\mathsurround=0pt\pi$-base} for a space $\ll{X},\tau\rangle$ is an open Souslin scheme $\langle V_a \rangle_{a \in {}^{{<}\omega}\hspace{-1pt}\omega}$ on $X$ that partitions ${X}$, has strict branches, and such that
  \begin{itemize}
 \item[\ding{46}\,]
 $\mathsurround=0pt
   \forall{x}\in{X}\enskip\forall{U}\in\tau({x})$

   $\mathsurround=0pt
   \exists {a}\in{}^{{<}\omega}\hspace{-1pt}\omega\enskip
   \exists{n}\in\omega\ $
   \begin{itemize}
   \item[\ding{226}\,]
  $\mathsurround=0pt
  {x} \in {V}_{a}\enskip$ and
   \item[\ding{226}\,]
  $\mathsurround=0pt
  \bigcup_{{i}\geqslant{n}}{V}_{{a}\hspace{0.5pt}^{\frown}{i}}\subseteq {U}.$
   \end{itemize}
  \end{itemize}
\end{deff}

\begin{nota}\label{not.S.N}\mbox{\ }

  \begin{itemize}

  \item [\ding{46}\,]
  $\mathsurround=0pt
  \mathbf{S}$ $\coloneq$ the \textit{standard Lusin scheme} $\coloneq$ the Souslin scheme $\langle {S}_{a}\rangle_{{a}\in{}^{{<}\omega\hspace{-1pt}}\omega}$ such that

  $\mathsurround=0pt
  {S}_{a}=\{{p}\in{}^{\omega\hspace{-1pt}}\omega:{a}\sqsubseteq{p}\}$
  for all ${a}\in{}^{{<}\omega\hspace{-1pt}}\omega.$
  \end{itemize}
\end{nota}

\begin{rema}\label{rem.baire.space}
  \begin{itemize}
  \item [(a)]
 The family $\{{S}_{a}:{a}\in{}^{{<}\omega\hspace{-1pt}}\omega\}$ is a base for the Baire space.
  \item [(b)]
 The standard Lusin scheme is a Lusin $\pi\mathsurround=0pt$-base for the Baire space.
  \end{itemize}
\end{rema}

\section{$\pi$-spaces}

\begin{deff} \label{def_of_pi_sp}
 A space ${X}$ is a $\pi$-\textit{space} iff there exists an open Souslin scheme $\langle V_a \rangle_{a \in {}^{{<}\omega}\hspace{-1pt}\omega}$ on $X$ that partitions ${X}$, has strict branches, and such that the family $\{{V}_{a} : {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\}$ is a $\pi$-base for ${X}$.
\end{deff}

\begin{rema}\label{pi.base}
  If  $\langle V_a \rangle_{a \in {}^{{<}\omega}\hspace{-1pt}\omega}$ is a Lusin $\pi\mathsurround=0pt$-base for a space ${X},$ then the family $\{{V}_{a}:{a}\in{}^{{<}\omega\hspace{-1pt}}\omega\}$ is a $\pi\mathsurround=0pt$-base for ${X}$.

  It follows that every space with a Lusin $\pi$-base is a $\pi$-space.\hfill$\qed$%
\end{rema}

Recall that a continuous map is \textit{quasi-open} iff the image of every nonempty open set has  nonempty interior.

\begin{lemm} \label{abt_qs_opn}
 For spaces $\ll {X}, \tau \rangle$ and $\ll {Y}, \sigma \rangle$ the following are equivalent:
 \begin{itemize}
  \item[\textup{(a)}] There exists a continuous quasi-open bijection ${f}\colon \ll {X}, \tau \rangle \to \ll {Y}, \sigma \rangle$.
  \item[\textup{(b)}] There exists a topology $\rho \supseteq \sigma$ on ${Y}$ such that
   \begin{itemize}
   \item[\ding{226}\,] $\sigma \setminus \{\varnothing\}$ is a $\pi$-base for $\ll Y, \rho \rangle$ and
   \item[\ding{226}\,] $\ll {Y}, \rho \rangle$ is homeomorphic to $\ll {X}, \tau \rangle$.  \end{itemize}
 \end{itemize}
\end{lemm}

\begin{proof}
 (a) $\Rightarrow$ (b). It is not hard to see that the topology $\rho := \{{f}[{U}] : {U} \in \tau\}$ satisfies all requirements.

 (b) $\Rightarrow$ (a). Take a homeomorphism ${f}\colon \ll {X}, \tau \rangle \to \ll {Y}, \rho \rangle$. Then  ${f}\colon \ll {X}, \tau \rangle \to \ll {Y}, \sigma \rangle$ is a continuous quasi-open bijection.
\end{proof}

\begin{prop} \label{another_desc}
 For a space ${X}$ the following are equivalent\textup{:}
 \begin{itemize}
  \item[\textup{(a)}] ${X}$ is a $\pi$-space.
  \item[\textup{(b)}] There exists a continuous quasi-open bijection ${f}\colon{X}\to\omega^\omega$.
  \item[\textup{(c)}] There exists a topology $\rho\supseteq\tau_{\omega^\omega}$ on ${}^\omega\hspace{-1pt}\omega$ such that
  \begin{itemize}
  \item[\ding{226}\,] $\tau_{\omega^\omega} \setminus \{\varnothing\}$ is a $\pi$-base for $\ll {}^\omega\hspace{-1pt}\omega, \rho \rangle$ and
  \item[\ding{226}\,] $\ll {}^\omega\hspace{-1pt}\omega, \rho \rangle$ is homeomorphic to ${X}$.  \end{itemize}
 \end{itemize}
\end{prop}

\begin{proof}
 (b) $\Leftrightarrow$ (c) follows from Lemma \ref{abt_qs_opn}.

 (a) $\Rightarrow$ (b). Suppose that $\mathbf{V}=\ll {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ is an open Souslin scheme on $X$ that partitions ${X}$ and has strict branches, and such that the family $\{{V}_{a} : {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\}$ is a $\pi$-base for ${X}$. Then for every ${x}\in {X}$,  there is a unique branch ${f}({x}) \in {}^\omega\hspace{-1pt}\omega$  such that $\{{x}\} = \fruit{{f}({x})}{{\bf {V}}}$.
 This gives a mapping  ${f}\colon{X}\to\omega^\omega$, and this mapping is bijective.

 Note that ${f}[{{V}_{a}}] = {S}_{a}$ for all ${a}\in{}^{{<}\omega}\hspace{-1pt}\omega$.
 Then ${X}$ has a $\pi$-base $\{{V}_{a}:{a}\in{}^{{<}\omega}\hspace{-1pt}\omega\}$, whose images $\{{S}_{a}:{a}\in{}^{{<}\omega}\hspace{-1pt}\omega\}$ form a base for $\omega^\omega$ (see Remark~\ref{rem.baire.space}(a)), so the mapping $f$ is quasi-open.
 And also $\omega^\omega$ has a base $\{{S}_{a}:{a}\in{}^{{<}\omega}\hspace{-1pt}\omega\}$ whose  preimages $\{{V}_{a}:{a}\in{}^{{<}\omega}\hspace{-1pt}\omega\}$ are open, so the mapping $f$ is continuous.

 (b) $\Rightarrow$ (a). Suppose that there exists a continuous quasi-open bijection ${f}\colon{X}\to\omega^\omega$. Let   ${V}_{a} \coloneq {f}^{-1}[{S}_{a}]$ for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$. Then $\ll {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ is an open Souslin scheme on ${X}$, it partitions ${X}$ and has strict branches, and the family $\{{V}_{a}:{a}\in{}^{{<}\omega}\hspace{-1pt}\omega\}$ is a $\pi$-base for ${X}$.
\end{proof}

Since finite and countable powers of the Baire space are homeomorphic to the Baire space, equivalency of (a) and (c) in Proposition~\ref{another_desc} implies the following.

\begin{corr} \label{count_prod}
Finite and countable products of $\pi\mathsurround=0pt$-spaces are also $\pi\mathsurround=0pt$-spaces. \hfill$\qed$%
\end{corr}

Since the complement of a $\sigma\mathsurround=0pt$-compact subset in the Baire space is homeomorphic to the Baire space~\cite[Theorems 3.11 and 7.7]{kechris2012classical} and since a nonempty open set in the Baire space cannot be covered by a $\sigma\mathsurround=0pt$-compact set~\cite[Theorem 7.7]{kechris2012classical}, equivalency of (a) and (c) in Proposition~\ref{another_desc} implies another property of $\pi\mathsurround=0pt$-spaces.

\begin{corr} \label{sigma_comp}
The complement of a $\sigma\mathsurround=0pt$-compact subset of a $\pi\mathsurround=0pt$-space is a $\pi\mathsurround=0pt$-space. \hfill$\qed$%
\end{corr}

\begin{teo} \label{when_pi_has_lpb}
 Every second-countable $\pi$-space has a Lusin $\pi$-base.
\end{teo}

\begin{proof}
 Let ${X}$ be a  second-countable $\pi$-space. By Proposition~\ref{another_desc}, there is a topology $\rho\supseteq\tau_{\omega^\omega}$ on ${}^\omega\hspace{-1pt}\omega$ such that
  \begin{itemize}
  \item[\ding{226}\,] $\tau_{\omega^\omega} \setminus \{\varnothing\}$ is a $\pi$-base for $\ll {}^\omega\hspace{-1pt}\omega, \rho \rangle$ and
  \item[\ding{226}\,] $\ll {}^\omega\hspace{-1pt}\omega, \rho \rangle$ is homeomorphic to ${X}$.  \end{itemize}

Suppose that $\{{B}_{k} : {k} \text{ is odd}\}$ is a countable base for $\ll {}^\omega\hspace{-1pt}\omega , \rho \rangle$. We build a  Souslin scheme ${\bf {V}} = \ll {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ on ${}^\omega\hspace{-1pt}\omega $ that partitions ${}^\omega\hspace{-1pt}\omega $ and such that:
 \begin{itemize}
   \item[(a)] $\forall {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\ [{V}_{a} \in \tau_{\omega^\omega} \setminus \{\varnothing\}]$;
   \item[(b)] $\forall {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\ [$ if $\mathsf{length}({a})$ is odd, then there exists ${d}\in{}^{{<}\omega}\hspace{-1pt}\omega$ such that ${V}_{a} = {S}_{d}$ and $\mathsf{length}({d}) \geq \mathsf{length}({a})]$;
   \item[(c)] $\forall {k} \in \omega$ and $\forall {a} \in {}^{k} \omega\ [$ if ${k}$ is odd and ${V}_{a} \cap {B}_{k} \neq \varnothing$, then  $\bigcup_{{n} \geq 1} {V}_{{a}^\frown{n}} \subseteq {B}_{k}]$.
 \end{itemize}

It is easy to check that ${\bf {V}}$ is a Lusin $\pi$-base for $\ll {}^\omega\hspace{-1pt}\omega, \rho \rangle$.

We build ${\bf {V}}$ by recursion on $\mathsf{length}({a})$. Let ${V}_{\ll\rangle} := {}^\omega\hspace{-1pt}\omega $. Suppose we have constructed ${V}_{a}$ for all ${a}\in{}^{{<}\omega}\hspace{-1pt}\omega$ with $\mathsf{length}({a}) \leq {k}$. Let ${a} \in {}^{k}\omega$; we will define ${V}_{{a}^{\frown}{n}}$ for all ${n} \in \omega$.

Suppose that ${k}$ is even. By condition (a), the set  ${V}_{a}$ is nonempty and open in the Baire space.
If ${V}_{a}={}^\omega\hspace{-1pt}\omega$, then put ${C}\coloneq\{\ll\rangle\}$. If ${V}_{a}\neq{}^\omega\hspace{-1pt}\omega$, then put
$$
{C}\coloneq\{{c}
\in{}^{{<}\omega}\hspace{-1pt}\omega:{S}_{c}\subseteq{V}_{a}\text{ and } {S}_{c\upharpoonright (\mathsf{length}(c)-1)} \not \subseteq {V}_{a}\}.
$$
The family $\{{S}_{c}: {c} \in {C}\}$  is disjoint and ${V}_{a} = \bigcup_{{c} \in {C}}{S}_{c}$.
Let $${D}\coloneq\{{c}\:{\hat{}}\:{d}:{c}\in{C}\text{ and }{d}\in{}^{k+1}\omega\}.$$
The family $\{{S}_{d}: {d} \in {D}\}$ is  disjoint and ${V}_{a}= \bigcup_{{d} \in {D}}{S}_{d}$. Also the set ${D}$ is infinite and  $\mathsf{length}({d})\geq \mathsf{length}({a})+1$ for all ${d}\in{D}$. Now we can define the sets ${V}_{{a}^{\frown}{n}}$, ${n}\in\omega$, in such a way that $\{{V}_{{a}^{\frown}{n}}:{n}\in\omega\}=\{{S}_{d}:{d}\in{D}\}$ and all ${V}_{{a}^{\frown}{n}}$ are different.

Suppose that ${k}$ is odd. If ${V}_{a}\cap{B}_{k} = \varnothing$, then we take ${V}_{{a}^{\frown}{n}}$, ${n}\in\omega$, as in the previous case. Suppose that ${V}_{a}\cap{B}_{k}\neq\varnothing$. Since $\tau_{\omega^\omega}\setminus\{\varnothing\}$ is a $\pi$-base for $\ll{}^\omega\hspace{-1pt}\omega,\rho\rangle$, there is ${c} \in {}^{{<}\omega}\hspace{-1pt}\omega$ such that ${S}_{c}\subset {V}_{a}\cap{B}_{k}$. Let ${V}_{{a}^{\frown}0} := {V}_{a} \setminus {S}_{c}
\neq\varnothing$ and let ${V}_{{a}^{\frown}{n}} := {S}_{{c}^{\frown}({n}-1)}$ for all ${n}\geq 1$.
\end{proof}

From Theorem \ref{when_pi_has_lpb} and \cite[Theorem 3.7]{patrakeev2015metrizable} it follows that

\begin{corr}
If ${X}$ is a second-countable $\pi$-space, then there exists an open continuous map from ${X}$ onto every nonempty Polish space. \hfill$\qed$%
\end{corr}

\begin{teo} \label{example}
 There exists a $\pi$-space that can not be mapped onto the Baire space by a continuous open map. It follows that there exists a $\pi$-space without a Lusin $\pi$-base.
\end{teo}

\begin{proof}
Let $N$ be the family of nowhere dense subsets of the Baire space. Set ${B} := \{{U} \setminus {A} : {U} \in \tau_{\omega^\omega} \text{ and } {A} \in {N}\}$. Since ${N}$ is closed under finite unions, it is easy to see that ${B}$ is closed under finite intersections. Also $B$ covers ${}^\omega\hspace{-1pt}\omega$, so $B$ is a base for some topology on ${}^\omega\hspace{-1pt}\omega$, which we denote by $\tau_{N}$.

Let us prove that
 $$
  \ll {}^\omega\hspace{-1pt}\omega, \tau_{N} \rangle \text{ is a } \pi \text{-space}.
 $$
Using Proposotion \ref{another_desc} it is enough to show that $\tau_{\omega^\omega} \setminus \{\varnothing\}$ is a $\pi$-base for $\ll {}^\omega\hspace{-1pt}\omega, \tau_{N} \rangle$. Since $\varnothing \in {N}$, it follows that $\tau_{\omega^\omega} \subseteq \tau_{N}$. Take a nonempty ${W} \in \tau_{N}$. There exist ${U} \in \tau_{\omega^\omega} \setminus \{\varnothing\}$ and ${A} \in {N}$ such that ${U} \setminus {A} \subseteq {W}$. Since ${A}$ is nowhere dense in $\omega^\omega$, there exists ${V} \in \tau_{\omega^\omega} \setminus \{\varnothing\}$ such that ${V} \subseteq {U} \setminus {A}$. It follows that ${V} \subseteq {W}$.

 Since $\tau_{\omega^\omega} \setminus \{\varnothing\}$ is a $\pi$-base for $\ll {}^\omega\hspace{-1pt}\omega, \tau_{N} \rangle$, it is easy to see that
 \begin{equation} \label{nwd}
  \forall {B} \subseteq {}^\omega\hspace{-1pt}\omega\ [\text{if } {B} \text{ is nowhere dense in } \ll {}^\omega\hspace{-1pt}\omega, \tau_{N} \rangle \text{, then } {B} \in {N}].
 \end{equation}

 Now we prove that
 $$
  \text{there is no continuous open map from } \ll {}^\omega\hspace{-1pt}\omega, \tau_{N} \rangle \text{ onto the Baire space}.
 $$
 Suppose on the contrary that $f : \ll {}^\omega\hspace{-1pt}\omega, \tau_{N} \rangle \to \omega^\omega$ is a continuous open surjection. Let ${A}$ be a nowhere dense and not closed subset of the Baire space. Since $f$ is open, it follows that ${f}^{-1}[{A}]$ is nowhere dense in $\ll {}^\omega\hspace{-1pt}\omega, \tau_{N} \rangle$. Then  ${f}^{-1}[{A}] \in {N}$ by  (\ref{nwd}), so ${}^\omega\hspace{-1pt}\omega \setminus {f}^{-1}[{A}] \in \tau_{N}$, and hence $f\big[{}^\omega\hspace{-1pt}\omega \setminus {f}^{-1}[{A}]\big]$ is open in  $\omega^\omega$. But $f\big[{}^\omega\hspace{-1pt}\omega \setminus {f}^{-1}[{A}]\big]$ equals ${}^\omega\hspace{-1pt}\omega\setminus{A}$, so $A$ is closed in $\omega^\omega$, a contradiction.
\end{proof}

Note that the example of a $\pi\mathsurround=0pt$-space in this theorem is not a $T_3$-space.

\section{Description of open images of $\pi$-spaces} \label{sec.4}

Recall that the \emph{Choquet game} on a nonempty space $X$ is defined as follows: Two players, I and II, alternately choose nonempty open sets\medskip

I $\ \ \ \ \ \ U_0\qquad\ \ \ \ U_1\qquad\qquad\ldots$

II $\qquad\ \ \ \ \ V_0\qquad\ \ \ \ \  V_1\qquad\qquad\ldots$\medskip\\
such that $U_0\supseteq V_0\supseteq U_1\supseteq V_1 \ldots\ .$ Player~II wins the run  $\big\ll\ll{U}_{0},{V}_{0}\rangle,\ll{U}_{1},{V}_{1}\rangle,\dots\big\rangle$ of Choquet game on $X$ iff $\bigcap_{n}V_n\neq\varnothing;$ otherwise player~I wins this run.
A nonempty space $X$ is called a \emph{Choquet space} iff player~II has a winning strategy in the Choquet game on $X$. More precise definitions of this notions can be found in~\cite{kechris2012classical}.

In this section we will prove the following theorem:

\begin{teo} \label{bla_main_theorem}
 Let ${X}$ be a nonempty space. Then the following are equivalent\textup{:}
 \begin{itemize}
  \item[\textup{(a)}] ${X}$ is a continuous open image of a $\pi$-space.
  \item[\textup{(b)}] ${X}$ is a continuous open image of a space that can be mapped onto a Polish space by a continuous quasi-open bijection.
  \item[\textup{(c)}] ${X}$ is a Choquet space of countable $\pi$-weight and of cardinality not greater than continuum.
 \end{itemize}
\end{teo}

\begin{corr}\label{corr.1}
A Hausdorff compact space is a continuous open image of a $\pi$-space if and only if it has a countable $\pi$-base and its cardinality is not greater than continuum. \hfill \qed
\end{corr}

\begin{corr}\label{corr.2}
A second-countable space is a continuous open image of a $\pi$-space if and only if it is a continuous open image of a space with a Lusin $\pi$-base if and only if it is a Choquet space of cardinality not greater than continuum.
\end{corr}

\begin{proof}
    We only need to prove that if a second-countable space ${Y}$ is a continuous open image of a $\pi$-space, then it is a continuous open image of a space with a Lusin $\pi$-base. Let $f$ be a continuous open map from a $\pi$-space $\ll {X}, \tau \rangle$ onto ${Y}$, $B$ a countable base for ${Y}$, and ${\bf {V}} = \langle {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ an open Souslin scheme on $\ll {X}, \tau \rangle$ that partitions ${X}$, has strict branches, and such that the family $\{{V}_{a} : {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\}$ is a $\pi$-base for $\ll {X}, \tau \rangle$. Consider the topology $\rho$ on ${X}$ generated by the subbase $\{{V}_{a} : {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\} \cup \{{f}^{-1}[{U}] : {U} \in {B}\}$. It is easy to see that $\ll {X}, \rho \rangle$ is a second-countable $\pi$-space and $f \colon \ll {X}, \rho \rangle \to {Y}$ is a continuous open surjection. From Theorem \ref{when_pi_has_lpb} it follows that $\ll {X}, \rho \rangle$ has a Lusin $\pi$-base.
\end{proof}

\begin{ques}\label{quest}
Do the class of continuous open images of $\pi$-spaces equals the class of continuous open images of spaces with a Lusin $\pi$-base?
\end{ques}

\begin{nota}\label{not.S.N}\mbox{\ } Let ${\bf {V}} = \langle {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ be a Souslin scheme. Then
  \begin{itemize}
 \item [\ding{46}\,] $\mathsf{flesh}({\bf {V}}) = \bigcup_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega} {V}_{a}$;
 \item [\ding{46}\,]  $\mathsf{branches}_{\bf {V}} ({x}) := \{q \in {}^\omega\hspace{-1pt}\omega : x \in \fruit{q}{{\bf V}}\}$.
  \end{itemize}
\end{nota}

\begin{deff}
 A {\it $\pi$-net} Souslin scheme on a space ${X}$ is a Souslin scheme ${\bf {V}} = \langle {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ such that $\mathsf{flesh}({\bf {V}})\subseteq{X}$ and the family $\{{V}_{b}: {b} \sqsupseteq {a}\}$ is a $\pi$-net for the subspace ${V}_{a}$ of ${X}$ for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$.

 A {\it $\pi$-base} Souslin scheme on a space ${X}$ is an open $\pi$-net Souslin scheme on ${X}$.
\end{deff}

\begin{deff}
 A \emph{selector} on a Souslin scheme ${\bf {V}}$ is a surjection  ${f}\colon{}^\omega\hspace{-1pt}\omega\to\mathsf{flesh}({\bf {V}})$ such that for all ${x} \in \mathsf{flesh}({\bf {V}})$, the preimage ${f}^{-1}(x)$ is a dense subset of the subspace $\mathsf{branches}_{\bf {V}}({x})$ of the Baire space.
\end{deff}

If a Souslin scheme ${\bf {V}}$ has  strict branches and covers a set ${X}$, then the function ${f} \colon {}^\omega\hspace{-1pt}\omega \to {X}$ such that $\{{f}({p})\} = \fruit{p}{{\bf {V}}}$ is a selector on ${\bf {V}}$.

A less trivial example of a selector can be obtained as follows. Let ${f}$ be a continuous surjection from the Baire space onto a space ${X}$. Let ${V}_{a} \coloneq {f}[{S}_{a}]$ for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$. Then ${f}$ is a selector on $\ll {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$.

\begin{lemm} \label{main_property_of_fiber_scheme}
 Let ${\bf {V}} = \langle {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ be a Souslin scheme that covers $\mathsf{flesh}({\bf {V}})$ and let ${f}$ be a selector on ${\bf {V}}$. Then ${f}[{S}_{a}] = {V}_{a}$ for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$.
\end{lemm}

\begin{proof}
 Let ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$. We  prove two inclusions.

 For $\subseteq$: Let ${p} \in {S}_{a}$. Since ${f}$ is a selector on ${\bf {V}}$ we have ${p} \in \mathsf{branches}_{\bf {V}}({f}({p}))$; that is, ${f}({p}) \in \fruit{p}{{\bf {V}}}$. Also we have $\fruit{p}{{\bf {V}}} \subseteq {V}_{{p} \upharpoonright \mathsf{length}({a})}$ and ${p} \upharpoonright \mathsf{length}({a}) = {a}$, therefore ${f}({p}) \in {V}_{a}$.

 For $\supseteq$: Let ${x} \in {V}_{a}$. Since ${\bf {V}}$  covers $\mathsf{flesh}({\bf {V}})$, it follows that ${S}_{a} \cap \mathsf{branches}_{\bf {V}}({x}) \neq \varnothing$. Since ${f}^{-1}({x})$ is dense in $\mathsf{branches}_{\bf {V}}({x})$, there exists ${q} \in ({S}_{a} \cap \mathsf{branches}_{\bf {V}}({x})) \cap {f}^{-1}({x})$, so ${x} = {f}({q}) \in {f}[{S}_{a}]$.
\end{proof}

\begin{deff}
 Let $\langle {X}, \tau \rangle$ be a space, ${\bf {V}}$ a Souslin scheme that covers ${X}$, and ${f}$ a selector on ${\bf {V}}$. Then $\sigma_{\tau, {f}}$ is the topology on ${}^\omega\hspace{-1pt}\omega$ generated by the subbase $\{{f}^{-1}[{U}]: {U} \in \tau\}\cup \{{S}_{a}: {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\}$.
\end{deff}

\begin{rema}\label{rema.sigma.tau.f}
 Let $\langle {X}, \tau \rangle$ be a space, ${\bf {V}}$ a Souslin scheme that covers ${X}$, and ${f}$ a selector on ${\bf {V}}$. Then  the family $\{{f}^{-1}[{U}]\cap {S}_{a}:{U} \in \tau,\ {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\}$ is a base for the topology $\sigma_{\tau,{f}}$.  \hfill \qed
\end{rema}

\begin{lemm} \label{surjection}
 Let ${f} \colon {A} \to {X}$ be a surjection, ${S} \subseteq {A}$, ${V} \subseteq {X}$, and ${f}[{S}] = {V}$. Then ${f}\big[{f}^{-1}[{U}] \cap{S}\big] = {U} \cap {V}$ for all ${U} \subseteq {X}$. \hfill \qed
\end{lemm}

\begin{lemm} \label{fXV_is_open}
 Let $\langle {X}, \tau \rangle$ be a space, ${\bf {V}}$ an open Souslin scheme on $\langle {X}, \tau \rangle$ that covers ${X}$, and ${f}$ a selector on ${\bf {V}}$. Then ${f} \colon \langle {}^\omega\hspace{-1pt}\omega, \sigma_{\tau, {f}} \rangle \to \langle {X}, \tau \rangle$ is a continuous open surjection.
\end{lemm}

\begin{proof}
 It is easy to see that ${f}$ is a continuous surjection. By Remark~\ref{rema.sigma.tau.f}, $\{{f}^{-1}[{U}]\cap {S}_{a}:{U} \in \tau,\ {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\}$ is a base for  $\sigma_{\tau,{f}}$, so it is enough to prove that ${f}\big[{f}^{-1}[{U}]\cap {S}_{a}\big]$ is open for all ${U} \in \tau$ and ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$. It is true, because  ${f}\big[{f}^{-1}[{U}]\cap {S}_{a}\big] = {U} \cap {V}_{a}$ by Lemmas \ref{surjection} and  \ref{main_property_of_fiber_scheme}.
\end{proof}

\begin{lemm} \label{weqw}
  Let $\langle {X}, \tau \rangle$ be a space, ${\bf {V}}$ a Souslin scheme that covers ${X}$, and ${f}$ a selector on ${\bf {V}}$. Then the weight of the space $\langle {}^\omega\hspace{-1pt}\omega, \sigma_{\tau, {f}}\rangle$ is less than or equal to the weight of $\langle {X}, \tau \rangle$. \hfill \qed
\end{lemm}

\begin{lemm} \label{sigma_tauV_pispace}
 Let $\langle {X}, \tau \rangle$ be a space, ${\bf {V}} = \ll {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ a $\pi$-net Souslin scheme on $\ll {X}, \tau \rangle$  that covers ${X}$, and ${f}$ a selector on ${\bf {V}}$. Then $\langle {}^\omega\hspace{-1pt}\omega,  \sigma_{\tau, {f}}\rangle$ is a $\pi$-space.
\end{lemm}

\begin{proof}
 We prove that condition (c) in Proposition~\ref{another_desc} is satisfied for $\rho=\sigma_{\tau,f}$. Obviously, $\sigma_{\tau,{f}}\supseteq\tau_{\omega^\omega}$. By Remark~\ref{rema.sigma.tau.f}, $\{{f}^{-1}[{U}]\cap {S}_{a}:{U} \in \tau,\ {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\}$ is a base for  $\sigma_{\tau,{f}}$, so it is enough to prove that for all ${U} \in \tau$ and ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$ with nonempty ${f}^{-1}[{U}]\cap {S}_{a}$, there exists ${b} \in {}^{{<}\omega}\hspace{-1pt}\omega$ such that ${S}_{b} \subseteq {f}^{-1}[{U}]\cap {S}_{a}$.

 From Lemmas \ref{surjection} and  \ref{main_property_of_fiber_scheme} it follows that $f\big[{f}^{-1}[{U}]\cap {S}_{a}\big] = {U} \cap {V}_{a}$,  so ${U} \cap {V}_{a}$ is nonempty open set in the subspace ${V}_{a}$ of $\ll{X},\tau\rangle$. Then  there exists ${b} \in {}^{{<}\omega}\hspace{-1pt}\omega$ such that ${b} \sqsupseteq {a}$ and ${V}_{b} \subseteq {U} \cap {V}_{a}$ because ${\bf {V}}$ is a $\pi$-net Souslin scheme on $\ll {X}, \tau \rangle$. From Lemma \ref{main_property_of_fiber_scheme} it follows that ${S}_{b} \subseteq {f}^{-1}[{V}_{b}] \subseteq {f}^{-1}[{U}]$, and since ${b} \sqsupseteq {a}$, we see that ${S}_{b} \subseteq {S}_{a}$. Therefore ${S}_{b} \subseteq {f}^{-1}[{U}]\cap {S}_{a}$.
\end{proof}

\begin{lemm} \label{lemm_perm}
 Let ${\bf {V}} = \langle {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ be a Souslin scheme and let ${g} \colon \omega \to \omega$ be a surjection. Then
 \begin{itemize}

\item[\textup{(a)}] $\{{g}\circ({a}\:{\hat{}}\:{c}):{c}\in{}^{{<}\omega}\hspace{-1pt}\omega\}=\{({g}\circ{a})\:{\hat{}}\:{c}:{c}\in{}^{{<}\omega}\hspace{-1pt}\omega\}$ for all ${a}\in{}^{{<}\omega}\hspace{-1pt}\omega$;

\item[\textup{(b)}]  $\bigcup_{{n} \in \omega}V_{({g}\circ {a})\hspace{0.5pt}^{\frown}{n}} = \bigcup_{{n} \in \omega}{V}_{{g}\circ ({a}\hspace{0.5pt}^{\frown}{n})}$ for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$;

\item[\textup{(c)}]  $\bigcap_{{n} \in \omega} {V}_{{g}\circ ({q}{\upharpoonright}\hspace{0.5pt}{n})} = \bigcap_{{n} \in \omega} {V}_{({g}\circ {q}) {\upharpoonright} {n}}$ for all ${q} \in {}^\omega\hspace{-1pt}\omega$.\hfill \qed
 \end{itemize}
\end{lemm}

\begin{lemm} \label{blabla}
  Let ${\bf {V}} = \langle {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ be a Souslin scheme and ${x} \in \mathsf{flesh}({\bf {V}})$. Suppose that for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$, if ${V}_{a}\ni{x} $, then there are ${n} \neq {m} \in \omega$ such that $ {V}_{{a}\hspace{0.5pt}^{\frown}{n}}\ni{x}$ and ${V}_{{a}\hspace{0.5pt}^{\frown}{m}}\ni{x}$. Then $\mathsf{branches}_{\bf {V}}({x})$ is a dense-in-itself subspace of the Baire space.\hfill \qed
\end{lemm}

\begin{nota}
 Suppose that ${\bf {V}} = \langle {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ is a Souslin scheme and ${g} \colon \omega \to \omega$. Then ${\bf {V}}^{{g}} = \langle {V}^{{g}}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ is a Souslin scheme such that ${V}^{{g}}_{a} \coloneq {V}_{{g} \circ {a}}$ for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$.
\end{nota}

\begin{lemm}\label{prop_of_Pfv}
 Let ${\bf {V}} = \langle {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ be a Souslin scheme, ${g} \colon \omega \rightarrow \omega$ a surjection, and ${X}$ a space. Then:
 \begin{itemize}
  \item[\textup{(a)}]  If $\,{\bf {V}}\!$ covers ${X}$, then ${\bf {V}}^{{g}}$ covers ${X}$.

  \item[\textup{(b)}]  If $\,{\bf {V}}\!$ is complete, then ${\bf {V}}^{{g}}$ is complete.

  \item[\textup{(c)}]  If $\,{\bf {V}}\!$ is open on ${X}$, then ${\bf {V}}^{{g}}$ is open on ${X}$.

  \item[\textup{(d)}]  If $\,{\bf {V}}\!$ covers ${X}$ and $|{g}^{-1}({n})| \geq 2$ for all ${n} \in \omega$, then $\mathsf{branches}_{{\bf {V}}^{{g}}}({x})$ is a dense-in-itself subspace of the Baire space for all ${x} \in {X}$.

  \item[\textup{(e)}] If $\,{\bf {V}}\!$ is a $\pi$-net Souslin scheme on ${X}$, then ${\bf {V}}^{{g}}$ is a $\pi$-net Souslin scheme on ${X}$.
 \end{itemize}
\end{lemm}

\begin{proof}
 For (a), let ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$. From (b) of Lemma \ref{lemm_perm} it follows that ${V}^{{g}}_{a} = {V}_{{g}\circ {a}} = \bigcup_{{n} \in \omega}{V}_{
 ({g}\circ {a})\hspace{0.5pt}^{\frown}{n}} = \bigcup_{{n} \in \omega}{V}_{{g}\circ ({a}\hspace{0.5pt}^{\frown}{n})} = \bigcup_{{n} \in \omega}{V}^{g}_{{a}\hspace{0.5pt}^{\frown}{n}}$.

 For (b), let ${q} \in {}^\omega\hspace{-1pt}\omega$. From (c) of Lemma \ref{lemm_perm} it follows that
 $$
 \fruit{q}{{\bf V}^{g}} = \bigcap_{{n} \in \omega} {V}^{g}_{{q} {\upharpoonright} {n}} = \bigcap_{{n} \in \omega} {V}_{{g}\circ ({q} {\upharpoonright} {n})} = \bigcap_{{n} \in \omega} {V}_{({g}\circ {q}) {\upharpoonright} {n}} =
 \fruit{{g}\circ{q}}{{\bf V}}\neq \emptyset.
 $$

 (c) is obvious. For (d), let ${x} \in {X}$. Using Lemma \ref{blabla} it is enough to show that for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$, if ${V}^{g}_{a}\ni{x}$, then there are ${n} \neq {m} \in \omega$ such that ${V}^{g}_{{a}\hspace{0.5pt}^{\frown}{n}}\ni{x}$ and ${V}^{g}_{{a}\hspace{0.5pt}^{\frown}{m}}\ni{x}$. Take ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$ such that ${V}^{g}_{a}\ni{x}$. By (a), ${\bf {V}}^{g}$  covers ${X}$, so there exists ${n} \in \omega$ such that ${x} \in {V}^{g}_{{a}\hspace{0.5pt}^{\frown}{n}}$. Take ${m} \in \omega \setminus \{{n}\}$ such that ${g}({m}) = {g}({n})$. Then ${x} \in {V}^{g}_{{a}\hspace{0.5pt}^{\frown}{n}} = {V}_{{g}\circ ({a}\hspace{0.5pt}^{\frown}{n})} = {V}_{{g}\circ ({a}\hspace{0.5pt}^{\frown}{m})} = {V}^{g}_{{a}\hspace{0.5pt}^{\frown}{m}}$.

 For (e), let ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$. We must show that  $\{{V}^{g}_{b}:{b}\sqsupseteq{a}\}$ is a $\pi$-net for the subspace ${V}^{g}_{a}$ of ${X}$. From (a) of Lemma \ref{lemm_perm} it follows that
 $$
 \{{V}^{g}_{b}:{b}\sqsupseteq{a}\}=
 \{{V}^{g}_{{a}\:{\hat{}}\:{c}}:{c}\in {}^{{<}\omega}\hspace{-1pt}\omega\}=
 \{{V}_{{g}\circ ({a}\:{\hat{}}\:{c})}:{c}\in{}^{{<}\omega}\hspace{-1pt}\omega\} =
 $$
 $$
 =\{{V}_{({g}\circ {a})\:{\hat{}}\:{c}}:{c}\in{}^{{<}\omega}\hspace{-1pt}\omega\} =
 \{{V}_{b}:{b}\sqsupseteq{g}\circ{a}\}.
 $$
 Since ${\bf {V}}$ is a $\pi$-net Souslin scheme on ${X}$, we see that $\{{V}_{b}:{b}\sqsupseteq{g}\circ{a}\}$ is a $\pi$-net for the subspace ${V}_{{g}\circ {a}}={V}^{g}_{a}$ of ${X}$.
\end{proof}

\begin{lemm} \label{fiber_scheme}
 Let ${\bf {V}}$ be a complete Souslin scheme that covers $\mathsf{flesh}({\bf {V}})$ and  $|\mathsf{flesh}({\bf {V}})| \leq 2^{\aleph_0}$. Suppose that  $\mathsf{branches}_{\bf {V}}({x})$ is a dense-in-itself subspace of the Baire space for all ${x} \in \mathsf{flesh}({\bf {V}})$. Then ${\bf {V}}$ has a selector.
\end{lemm}

\begin{proof}
 Note that, since $\mathbf{V}$ covers $\mathsf{flesh}({\bf {V}})$, $\mathsf{branches}_{\bf {V}}({x})$ is nonempty for all ${x} \in \mathsf{flesh}({\bf {V}})$.

 Note also that if ${x}\in\mathsf{flesh}({\bf {V}})$ and ${U}$ is nonempty open set in the subspace $\mathsf{branches}_{\bf {V}}({x})$ of the Baire space, then ${U}$ has cardinality $2^{\aleph_0}$. Indeed, $U$ is open in $\mathsf{branches}_{\bf {V}}({x})$ and $\mathsf{branches}_{\bf {V}}({x})$ is closed in $\omega^\omega$, so ${U}$ is Polish. Also ${U}$ is dense-in-itself because $\mathsf{branches}_{\bf {V}}({x})$ is dense-in-itself. Therefore $|{U}|=2^{\aleph_0}$.

 Now, by transfinite recursion on $\mathsf{flesh}({\bf {V}})$ well-ordered in the type of its cardinality, it is easy to build an indexed family $\langle {Q}_{x} \rangle_{{x} \in \mathsf{flesh}({\bf {V}})}$ such that
 \begin{itemize}
 \item[\ding{226}\,] ${Q}_{x}$ is a dense subset of $\mathsf{branches}_{\bf {V}}({x})$ for all ${x} \in \mathsf{flesh}({\bf {V}})$ and
  \item[\ding{226}\,] ${Q}_{x} \cap {Q}_{y} = \emptyset$ for all ${x} \neq {y} \in \mathsf{flesh}({\bf {V}})$.
 \end{itemize}

 Now we can construct a selector ${f}\colon{}^\omega\hspace{-1pt}\omega\to\mathsf{flesh}({\bf {V}})$ on ${\bf {V}}$. If ${p}
 \in{Q}_{x}$ for some ${x}\in\mathsf{flesh}({\bf {V}})$, then set ${f}({p})\coloneq{x}$.  If ${p}\nin\bigcup_{{x}\in\mathsf{flesh}({\bf {V}})}{Q}_{x}$, then choose ${f}({p})\in\fruit{p}{{\bf {V}}}$ arbitrarily. It is easy to see that ${f}$ is a selector on ${\bf {V}}$.

\end{proof}

\begin{lemm} \label{pibasesch_fibpibasesch}
 Let ${X}$ be a space and $|{X}| \leq 2^{\aleph_0}$. Suppose that there exists a complete $\pi$-base Souslin scheme on ${X}$ that covers ${X}$. Then there exists a complete $\pi$-base Souslin scheme on ${X}$ that covers ${X}$ and has a selector.
\end{lemm}

\begin{proof}
 Let ${\bf {V}}$ be a complete $\pi$-base Souslin scheme on ${X}$ that covers ${X}$. Consider a surjection ${g}\colon \omega \to \omega$ such that $|{g}^{-1}({n})| \geq 2$ for all ${n} \in \omega$. From Lemma \ref{prop_of_Pfv} it follows that ${\bf {V}}^{{g}}$ is a complete $\pi$-base Souslin scheme on ${X}$ that covers ${X}$, and that ${\bf {V}}^{{g}}$ meets the premisses of Lemma \ref{fiber_scheme}. Therefore ${\bf {V}}^{{g}}$ has a selector.
\end{proof}

\begin{lemm} \label{suff}
 Suppose that there exists a $\pi$-base Souslin scheme on a space ${X}$ that covers ${X}$ and has a selector. Then ${X}$ is a continuous open image of a $\pi$-space.
\end{lemm}

\begin{proof}
 This lemma follows from Lemmas \ref{fXV_is_open} and \ref{sigma_tauV_pispace}.
\end{proof}

\begin{lemm} \label{desc_op_im}
 A space ${X}$ is a continuous open image of a $\pi$-space if and only if $|{X}| \leq 2^{\aleph_0}$ and there exists a complete $\pi$-base Souslin scheme on ${X}$ that covers ${X}$.
\end{lemm}

\begin{proof}
For $\Rightarrow$: Let ${f}$ be a continuous open surjection from a $\pi$-space ${Y}$ onto ${X}$. Let $\mathbf{V}=\ll {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ be an open Souslin scheme on $Y$ that partitions ${Y}$, has strict branches, and such that the family $\{{V}_{a} : {a} \in {}^{{<}\omega}\hspace{-1pt}\omega\}$ is a $\pi$-base for ${Y}$. It is easy to see that $\ll {f}[{V}_{a}] \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ is a complete  $\pi$-base Souslin scheme on ${X}$ that covers ${X}$.

For $\Leftarrow$: From Lemma \ref{pibasesch_fibpibasesch} it follows that there exists $\pi$-base Souslin scheme on ${X}$ that covers ${X}$ and has a selector. So ${X}$ is a continuous open image of a $\pi$-space by Lemma \ref{suff}.
\end{proof}

Let ${X}$ be a space and ${s}=\big\ll\ll{U}_{0},{V}_{0}\rangle,\dots,\ll{U}_{n},{V}_{n}\rangle\big\rangle$ a sequence of moves in the Choquet game on ${X}$. For a  positive ${k}\leqslant{n}$, we call a pair $\ll{U}_{k},{V}_{k}\rangle$ of the sequence ${s}$ \emph{redundant} iff ${V}_{k}={U}_{k}={V}_{{k}-1}$.
Also we call the pair $\ll{U}_0,{V}_0\rangle$ \emph{redundant} iff ${V}_{0}={U}_{0}={X}$. Let ${f}({s})$ be the sequence that is obtained by removing all redundant pairs from ${s}$. For example, if ${Z}\subset{Y}\subset{X}$, then
$$
{f}\big(\big\ll \ll{X},{X}\rangle,\ll{X},{X}\rangle,\ll{Y},{Y}\rangle,\ll{Y},{Y}\rangle,\ll{Z},{Z}\rangle \big\rangle\big) = \big\ll\ll{Y},{Y}\rangle,\ll{Z},{Z}\rangle\big\rangle.
$$

Let $\Gamma$ be a strategy for player II in the Choquet game on a space ${X}$. The \emph{modification} $\Gamma'$ of the strategy $\Gamma$ prescribes player II to play as follows. Let
$$
{s}=\big\ll\ll{U}_{0},{V}_{0}\rangle,\dots,\ll{U}_{{n}},{V}_{{n}}\rangle\big\rangle
$$
be a sequence of previous moves.
If in the $({n}{+}1\hspace{-1pt})\mathsurround=0pt$th move player I plays ${U}_{{n}+1}={V}_{{n}}$, then $\Gamma'({s},{U}_{{n}+1})\coloneq{V}_{n}$; that is, $\Gamma'$ tells player II to reply with the set ${V}_{{n}+1}\coloneq{V}_{n}$.
If player I plays ${U}_{{n}+1}\neq{V}_{n}$, then $\Gamma'({s},{U}_{{n}+1})\coloneq\Gamma\big({f}({s}),{U}_{{n}+1}
\big)$.
Also we set $\Gamma'(\ll\rangle,{X})\coloneq{X}$ and $\Gamma'(\ll\rangle,{U}_{0})\coloneq\Gamma(\ll\rangle,{U}_{0})$ for  ${U}_{0}\neq{X}$.

\begin{lemm}\label{modified}
 If $\,\Gamma\!$ is a winning strategy for player II in the Choquet game on a space ${X}$, then its modification $\,\Gamma'\!$ is also winning.\hfill\qed
\end{lemm}

\begin{proof}[Proof of Theorem~\ref{bla_main_theorem}]
This theorem says that for every nonempty space ${X}$, the following are equivalent:
 \begin{itemize}
  \item[\textup{(a)}] ${X}$ is a continuous open image of a $\pi$-space.
  \item[\textup{(b)}] ${X}$ is a continuous open image of a space that can be mapped onto a Polish space by a continuous quasi-open bijection.
  \item[\textup{(c)}] ${X}$ is a Choquet space of countable $\pi$-weight and of cardinality not greater than continuum.
 \end{itemize}

(a)$\Rightarrow$(b) follows from Proposition \ref{another_desc}.

(b)$\Rightarrow$(c). Suppose that ${X}$ is a continuous open image of a space ${Y}$ that can be mapped onto a Polish space by a continuous quasi-open bijection. It is easy to show that ${Y}$ is a Choquet space of countable $\pi$-weight and of cardinality not greater than continuum. Then ${X}$ also possesses these properties.

(c)$\Rightarrow$(a).  Using Lemma \ref{desc_op_im}, it is enough to find a complete  $\pi$-base Souslin scheme on ${X}$ that covers ${X}$.
Let $\Gamma$ be a winning strategy for player II in the Choquet game on ${X}$.

We will build Sousin schemes ${\bf {U}} = \ll {U}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ and ${\bf {V}} = \ll {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ such that
\begin{itemize}
 \item[\ding{226}\,]  ${\bf {V}}$ covers ${X}$;
 \item[\ding{226}\,]  ${V}_{a}$ is nonempty and open for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$;
 \item[\ding{226}\,]  $\{{V}_{a^{\frown}{m}} : {m} \in \omega\}$ is a $\pi$-base for the subspace ${V}_{a}$ of ${X}$ for all ${a} \in {}^{{<}\omega}\hspace{-1pt}\omega$;
 \item[\ding{226}\,]  for every branch ${p} \in {}^\omega\hspace{-1pt}\omega$, the sequence $\big\ll \ll{U}_{p{\upharpoonright}\hspace{0.5pt}{0}}, {V}_{p{\upharpoonright}\hspace{0.5pt}{0}}\rangle, \ll{U}_{p{\upharpoonright}\hspace{0.5pt}{1}}, {V}_{p{\upharpoonright}\hspace{0.5pt}{1}}\rangle, \dots \big\rangle$ is a run of the Choquet game on ${X}$ in which player II plays according to the modified strategy $\Gamma'$.
\end{itemize}
Lemma~\ref{modified} says that $\Gamma'$ is a winning strategy for player II in the Choquet game on ${X}$, so the scheme ${\bf {V}}$ is  complete. Also it is easy to see that ${\bf {V}}$ is a $\pi$-base Souslin scheme on ${X}$ that covers ${X}$.

To compete the proof, it remains to construct ${\bf {U}} = \ll {U}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$ and ${\bf {V}} = \ll {V}_{a} \rangle_{{a} \in {}^{{<}\omega}\hspace{-1pt}\omega}$; we do it by recursion on $\mathsf{length}({a})$.
Put ${U}_{\ll \rangle} \coloneq {X}$ and
${V}_{\ll \rangle} := \Gamma'({\ll\rangle},{U}_{\ll \rangle})$. Note that
${V}_{\ll \rangle}={X}$.

Suppose that ${U}_{a}$ and ${V}_{a}$ have been chosen; we will chose ${U}_{a^{\frown}{m}}$ and ${V}_{a^{\frown}{m}}$ for all ${m}\in\omega$.
Consider a sequence $\ll {W}_{m} \rangle_{{m} \in \omega}$ such that  ${W}_{0} = {V}_{a}$ and $\{{W}_{m} : {m} \in \omega\}$ is a $\pi$-base for the subspace ${V}_{a}$ of ${X}$. For all ${m}\in\omega$, we put ${U}_{a^{\frown}{m}} \coloneq {W}_{m}$ and
$$
{V}_{a^{\frown}{m}} := \Gamma'\big(\big\ll\ll{U}_{{a}{\upharpoonright}\hspace{0.5pt}{0}}, {V}_{{a}{\upharpoonright}\hspace{0.5pt}{0}}\rangle,\ll{U}_{{a}{\upharpoonright}\hspace{0.5pt}{1}},{V}_{{a}{\upharpoonright}\hspace{0.5pt}{1}}\rangle, \dots, \ll{U}_{{a}},{V}_{{a}}\rangle \big\rangle,{U}_{a^{\frown}{m}}\big).
$$
\end{proof}


text:
Israeli Knesset passes 'Jewish nation-state' bill
Filed under: Archived,Benjamin Netanyahu,Crime and law,Israel,Jerusalem,Judaism,Knesset,Likud,Middle East,Pages with defaulting non-local links,Politics and conflicts,Religion — admin @ 5:00 am
22 July 2018: Israeli Knesset passes 'Jewish nation-state' bill
21 July 2018: Indian Supreme Court: unconstitutional to bar women of certain age group from entering Sabarimala temple
19 July 2018: After signing peace declarations, Eritrea reopens embassy in Ethiopian capital Addis Ababa
7 July 2018: France: Rainbow pedestrian crossings 'will be permanent!', Paris mayor says
1 July 2018: India: Madhya Pradesh police arrests second accused of gang-raping minor girl in Mandsaur district
On Thursday, the Israeli parliament, the Knesset, passed the "Jewish nation-state" bill granting Israel the status of "national home of the Jewish people". Israeli Prime Minister Benjamin Netanyahu called it "a defining moment in the annals of Zionism and the history of the state of Israel".
The bill was passed at 3 AM with 62–55 majority. Two lawmakers abstained. The bill declared Hebrew as the only official language and demoted Arabic from national language to a "special status" language. Netanyahu said, "An absolute majority wants to ensure our state's Jewish characters for generations to come." Speaker of the Knesset Yuli Edelstein said this bill, which received just 51.66% votes in favour out of the full Knesset of 120, was "one of the most important laws ever to be passed by the Knesset".
The Knesset began the voting on Wednesday. Previously, the bill had clauses declaring the state as a Jewish-only community as well as direction for judiciary courts to consider Jewish values over democratic values. Both of the clauses were challenged by the Israeli President Reuven Rivlin and the Attorney General. The latter clause about Jewish values for courts was dropped in May; the other clause was eventually dropped, and the final draft of the bill was approved on Monday. The bill was first introduced in 2011 by the right-wing Likud party.
The latest draft of the bill said the Hebrew calendar is to be the official calendar of the state. It also read, "Israel is the historical homeland of the Jewish people, in which the State of Israel was established" and "the state views the development of Jewish settlement as a national value and will act to encourage and promote its establishment and consolidation." Israel is home to about 1.8 million Palestinians, who account for roughly 20% of the national population. The bill also declared Jerusalem, in its entirety, as the capital of Israel.
The bill received negative reception from a number of people. The Arab Joint List multi-party alliance called it "anti-democratic, colonialist, racist, and with clear characteristics of apartheid." Calling it a "shameful night", Meretz Party's Tamar Zandberg said it was a "debased and tainted law". The Joint List's Ahmad Tibi questioned, "Why are you afraid of the Arabic language?".
Last week, Netanyahu said, "We will keep ensuring civil rights in Israel's democracy but the majority also has the rights and the majority decides". Jeremy Ben-Ami, the president of J Street, a US-based Jewish Organisation, said the bill "was born in sin, its only purpose is to send a message to the Arab community, the LGBT community and other minorities in Israel, that they are not and never will be equal citizens. Two months ago we celebrated the 70th anniversary of the Israeli Declaration of Independence, where it was written that the State of Israel 'will ensure complete equality of social and political rights to all its inhabitants irrespective of religion, race or gender.' Today Netanyahu's government is trying to ignore those words and the values that they represent."
"Israel passes controversial 'Jewish nation-state' law" — Al Jazeera, July 19, 2018
"Netanyahu cheers Jewish state law as a 'pivotal moment' in Zionist history" — The Times of Israel, July 19, 2018
Raoul Wootliff. "Final text of Jewish nation-state law, approved by the Knesset early on July 19" — The Times of Israel, July 19, 2018
Jonathan Liz and Nou Landau. "Israel Passes Controversial Nation-state Law Defining Country as Jewish National Homeland" — Haartez, July 19, 2018
Gill Hoffman. "Controversial Jewish-Nation Bill Passes Into Law" — The Jerusalem Post, July 19, 2018
Noga Tarnopolsky. "Israeli parliament passes bill defining country as the nation-state of the Jewish people" — Los Angeles Times, July 18, 2018
"Israeli Knesset to vote on Jewish Nation-State Bill" — Al Jazeera, July 18, 2018
Comments Off on Israeli Knesset passes 'Jewish nation-state' bill
Arabs react to Israeli debate on Jerusalem\'s Holy sites
Filed under: Archived,Islam,Israel,Israel/Palestine Conflict,Jordan,Judaism,Knesset,Middle East,Palestine,Religion — admin @ 5:00 am
Arabs react to Israeli debate on Jerusalem's Holy sites
1 July 2015: Palestinian prisoner Khader Adnan stops hunger strike, due to be released
29 May 2015: Tony Blair resigns as envoy of Quartet to Middle East
22 May 2015: FIFA president 'on a mission of peace'
29 April 2015: UN publishes report on 2014 Gaza conflict
18 March 2015: Netanyahu's Likud party wins election in Israel
Location of Palestine
In Jewish tradition, the rock upon which Abraham built his altar for the sacrifice of Isaac (Dome of the Rock interior).
Image: OSU Special Collections & Archives.
Islamic and Jordanian leaders responded yesterday to a debate conducted by the Israeli parliament, the Knesset, addressing current Arab control over holy sites in the Old City of Jerusalem including Al-Aqsa Mosque and the Dome of the Rock.
Last week, members of the Jordanian Parliament condemned the plan for a debate in the Knesset on the issue, but yesterday after the debate in Israel almost 50 of them had petitioned the Jordanian government to stop the enforcement of the 1994 Israel–Jordan peace treaty. The treaty allowed for the exchange of ambassadors and on that point, the Jordanian parliament voted on the recall of the nation's ambassador to Israel. Jordan's Prime Minister Abdullah Nsur said, in remarks to Qatari newspaper Al-Watan, any Israeli action affecting Jordanian control of the Holy sites would violate the treaty and Jordan could review whether or not the treaty had been invalidated. Meanwhile, the Organisation of Islamic Cooperation criticized the Knesset debate as a provocation.
Before the Israeli debate Tuesday, Palestinian protesters took to one of the Gates of the Temple Mount in the Old City, used by Jews, armed with firecrackers and rocks and were met by Israeli police.
On Tuesday, members of the Israeli Knesset began a debate on an issue it had never before broached. MK Moshe Feiglin of Likud started the debate about one of Jerusalem's holiest sites, The Temple Mount. Feiglin spoke to freedom of religion and the right of Israelis to pray on the site controlled by the Arabs while opponents addressed their fears the sensitive topic could harm ongoing peace talks. He said, "Any terror organization can raise its flag there, [but] there can be no trace of the Israeli flag. Only Jews are forbidden to pray at this place."
During the debate, Labor MK Eitan Cabel reacted to Feiglin's motion, "Why stop at the Temple Mount? If you are the new messiah of our times, what about" — he named various specifics of the holy promise of the lands of the Greater Israel — "Why leave them behind?" A "Greater Israel" promised by God is generally considered off limits in civic discourse inside Israel. The Knesset did not vote on the issue.
Al-Aqsa Mosque ranks just behind Mecca and Medina as the holiest of Islamic sites. Jews consider the Temple Mount to be their holiest site.
The Al-Aqsa Mosque from the Western Wall Plaza.
Image: Wilson44691.
Southern aerial view of the Temple Mount in the Old City of Jerusalem.
Image: Godot13.
The portal of the Al-Aqsa Mosque from the north.
Al-Fakhriyyah Minaret & Islamic Museum.
The Al-Aqsa Mosque in Jerusalem.
Image: Jane Fresco.
The Dome of the Rock (close).
Image: بلال الدويك.
The outdoor Islamic Pulpit of the Cadi Borhan-ed-din Jerusalem Holy Land at the Al-Aqsa Mosque, Jerusalem.
Image: Snapshots of The Past.
Al Aqsa Mosque from the Dome of the Rock.
Image: MatthiasKabel.
The Dome of the Chain (Arabic: قبة السلسلة, Qubbat al-Silsila) in front of The Dome of the Rock.
Southern Wall of the Temple Mount in Jerusalem.
Image: Wknight94.
"Jordan PM warns peace deal with Israel may be reviewed" — The Daily Star (Lebanon), February 26, 2014
Mohammad Tayseer, Donna Abu-Nasr. "Jordan Lawmakers Seek to Expel Israeli Ambassador Over Holy Site" — Bloomberg News, February 26, 2014
"Muslim body slams debate on Jewish prayer at Al-Aqsa mosque" — Raidió Teilifís Éireann, February 26, 2014
"Jordan MPs urge scrapping peace treaty with Israel" — Press TV, February 26, 2014
Gil Ronen. "MK Cabel Asks Feiglin: Are You the Messiah?" — Arutz Sheva, February 26, 2014
Adiv Sterman, Gavriel Fiske. "Debate on Temple Mount worship rocks Knesset" — The Times of Israel, February 25, 2014
Comments Off on Arabs react to Israeli debate on Jerusalem\'s Holy sites
Israelis re-elect Netanyahu, centre-left rises
Filed under: Archived,Barack Obama,Benjamin Netanyahu,Israel,Knesset,Likud,Middle East,Pages with defaulting non-local links,Palestine,Palestine Liberation Organization,Tom Morris (Wikinewsie) — admin @ 5:00 am
Location of Israel
Binyamin Netanyahu
Image: US State Dept.
Binyamin Netanyahu is set to be re-elected as prime minister of Israel despite losses suffered by his party, Likud. An alliance of centre-left parties has increased their share of the votes.
Likud now have 31 seats in the Knesset. Netanyahu has stated he seeks to form "as broad a coalition as possible". The newly-formed Yesh Atid party led by former television presenter Yair Lapid is set to play an important part in negotiations over the formation of a coalition government.
As a result of Yesh Atid winning 19 seats in the Knesset, Lapid has been approached by both Netanyahu to form a governing party and by Shelly Yachimovich from the left-wing Israeli Labor Party to join an opposition alliance. Yesh Atid supports changing the law to remove the exemptions ultra-Orthodox Jews have from mandatory military service, a move which along with economic reforms to support the middle class they dubbed "sharing the burden".
Yachimovich invited Yesh Atid "to form a coalition of moderate, social, peace advocate and centrist forces without Benjamin Netanyahu as PM". Yesh Atid have rejected that offer, with Yair Lapid saying that they will not join an "obstructive bloc".
The election sees the highest number of women elected to the Knesset, 26 out of 120.
Jay Carney, the White House press spokesman, reiterated American support for Israel: "The United States remains committed, as it has been for a long time, to working with the parties to press forward the goal of a two-state solution. That has not changed, and it will not change." Carney also laid out the White House view on how to deal with Palestine: "We believe that what needs to take place is direct negotiations between the two parties (Israelis and Palestinians) that addresses the final-status issues and results in a two-state solution". The New York Times noted that a newly-formed coalition government may improve the relationship between Netanyahu and US president Barack Obama.
Dan Shapiro, the US Ambassador to Israel, took to Twitter to congratulate Israelis on the conclusion of their voting process: "Mazal tov to the Israeli people on their just completed elections! Always inspiring to see democracy in action! […] Fascinating election from a US perspective. Every democratic system is different. Important thing is the people have their say." Shapiro said the US "looks forward to working closely with next government of Israel."
Palestine Liberation Organization official Hanan Ashrawi stated: "I am not going to say that now the chances of peace are going to be drastically improved or that we are going to see a sort of left-wing coalition and a peace camp that will take over and produce instant peace. You are not going to have a savior, suddenly producing instant peace."
Harriet Sherwood. "Israel election result hands rising star Yair Lapid a pivotal role" — The Guardian, January 23, 2013
"Israel centrists gain, but Netanyahu squeaks in" — United Press International, January 23, 2013
Lahav Harkov. "Record number of female, religious MKs in Knesset" — The Jerusalem Post, January 23, 2013
Yoav Zitun and Tzvika Brot. "Lapid says 'no' to obstructive bloc" — Ynetnews, January 23, 2013
Moran Azulay. "Yachimovich to Lapid: Don't join PM's gov't" — Ynetnews, January 23, 2013
"Yair Lapid 'rules out forming an anti-Netanyahu bloc'" — BBC News Online, January 23, 2013
Mark Landler. "Netanyahu-Obama Ties May Thaw After Israel Election" — The New York Times, January 23, 2013
"White House calls for resumption of peace talks" — The Jerusalem Post, January 23, 2013
"US 'looking forward' to work with new Israeli gov't" — January 23, 2013
Karin Laub (Associated Press). "Palestinian surprise, skepticism after Israel vote" — The Seattle Times, January 23, 2013
Comments Off on Israelis re-elect Netanyahu, centre-left rises
World leaders react to Obama win
Filed under: 2012 United States presidential election,Africa,Archived,Barack Obama,Benjamin Netanyahu,David Cameron,Europe,European Commission,European Union,France,Germany,Israel,Kenya,Knesset,Likud,Middle East,Nigeria,North America,Pages with categorizable local links,Pages with defaulting non-local links,Palestine,Politics and conflicts,South Africa,Syria,Tom Morris (Wikinewsie),United Kingdom,United States — admin @ 5:00 am
World leaders react to Obama win – Wikinews, the free news source
Image: Pete Souza.
2012 U.S. Presidential Election stories
7 November 2012: World leaders react to Obama win
7 November 2012: United States re-elects Barack Obama
7 November 2012: Australian Broadcasting Corporation plans to call California for Obama before polls close
5 November 2012: Wikinews interviews former New Mexico governor Gary Johnson, presidential nominee of the Libertarian Party
5 November 2012: On the campaign trail, October 2012
Leaders from around the world have congratulated re-elected U.S. President Barack Obama on his success at the ballot box.
Israeli leader Benjamin Netanyahu stated that he "will continue to work with President Obama to preserve the strategic interests of Israel's citizens" and described relations between the U.S. and Israel as "rock solid". Some others within the ruling Likud party were not so optimistic about the relationship, with Knesset member Danny Danon criticising Obama for "ill-advised" policies towards Israel and hoping Obama "resets his course relating to Israel and our region for the next four years".
A statement on behalf of Palestinian leader Mahmoud Abbas expressed hope that Obama "continues his efforts to achieve peace in the Middle East".
The British Prime Minister David Cameron took time from his tour of the Middle East to congratulate Obama and listed as shared priorities a trade deal between the U.S. and the United Kingdom as well as solving the crisis in Syria. Ed Miliband, leader of the opposition Labour Party, stated on Twitter that Obama's success in the election was "based on building fairer economy and optimism about what politics can achieve."
French leader François Hollande said in choosing Obama over Mitt Romney, voters in the United States had voted for "an America of openness and solidarity, fully engaged on the international scene and conscious of the challenges facing our planet: peace, economy and the environment". He listed economic growth, unemployment, and the Middle East as challenges the two politicians would have to face. German leader Angela Merkel stated she "deeply appreciated" meetings and conversations with Obama on issues including "overcoming the global financial and economic crisis".
Herman Van Rompuy, President of the European Council, and Jose Manuel Barroso, President of the European Commission, issued a joint statement which described the U.S. as "a key strategic partner of the European Union" and hoped to meet Obama "at an early date in order to reconfirm our priorities and provide renewed impetus to our joint action".
Across Africa, Obama received congratulations from South African President Jacob Zuma, Nigerian President Goodluck Jonathan, and Raila Odinga, the Prime Minister of Kenya.
"United States re-elects Barack Obama" — Wikinews, November 7, 2012
"Barack Obama elected 44th President of the United States" — Wikinews, November 5, 2008
Joshua Mitnick. "Israel moves quickly to congratulate Obama" — The Christian Science Monitor, November 7, 2012
"Obama's re-election: The world reacts" — BBC News Online, November 7, 2012
"David Cameron congratulates Barack Obama on re-election victory" — BBC News Online, November 7, 2012
Angelique Chrisafis. "France's reaction to Obama's election victory" — The Guardian, November 7, 2012
Comments Off on World leaders react to Obama win
Attorney general drops case against Israeli participants in Freedom Flotilla
Filed under: Archived,Crime and law,Israel,Israel/Palestine Conflict,Knesset,Middle East,Pages with defaulting non-local links,Palestine,Tom Morris (Wikinewsie),Transport — admin @ 5:00 am
Other stories from Israel
23 November 2012: Terror suspects arrested in connection with bombing public bus in Israel
21 November 2012: Israeli airstrikes damage more offices housing international journalists in Gaza City
20 November 2012: Israel sets 36 hour ultimatum for Hamas
19 November 2012: Israeli Defense Force admits to targeting media center in Gaza City airstrike
18 November 2012: Journalists in Gaza City injured in Israeli airstrike
To write, edit, start or view other articles on Israel, see the Israel Portal
Israeli Defense Forces boarding one of the ships bound for Gaza.
Image: Israeli Defense Forces.
Citizens in Israel who participated in the 2010 'Freedom Flotilla' to Gaza will not be prosecuted, according to a statement released today by the Attorney General of Israel, Yehuda Weinstein.
Weinstein said the cases had been closed due to "evidentiary and legal difficulties" related to "proving the basis of the office as well as legal questions regarding the jurisdiction of Israeli law on crimes committed abroad."
Haneen Zoabi, an Israeli Arab and member of the Knesset for the Balad party, was amongst the Israeli citizens who participated in the flotilla. She responded to the dropping of the cases by saying that it "proves what I said all along. My participation in the flotilla was a political act and entirely legal."
Anastassia Michaeli, a Knesset member for the nationalist party Yisrael Beiteinu, said the decision to drop the cases was "a boost and reward for terror in Israel and around the globe".
Gaza Freedom Flotilla
Tomer Zarchin. "AG closes case against Israeli participants of 2010 Gaza flotilla" — Ha'aretz, December 22, 2011
Joanna Paraszczuk. "A-G closes case against Israeli flotilla members" — Jerusalem Post, December 22, 2011
What do you think of this page?
Please take a moment to rate this page below. Your feedback is valuable and helps us improve our website.
Reliability : ExcellentHighFairLowPoor(unsure) Completeness : ExcellentHighFairLowPoor(unsure) Neutrality : ExcellentHighFairLowPoor(unsure) Presentation : ExcellentHighFairLowPoor(unsure)
Comments Off on Attorney general drops case against Israeli participants in Freedom Flotilla
Israel marks Holocaust Memorial Day
Filed under: AutoArchived,Benjamin Netanyahu,History,Israel,Knesset,Middle East,Museums,North America,United States — admin @ 5:00 am
Israel marks Holocaust Memorial Day – Wikinews, the free news source
The Yad Vashem Holocaust memorial museum in Jerusalem.
Yom HaShoa, the Holocaust Memorial Day was marked in Israel yesterday.
In Israel, a two-minute siren brought life to a standstill at 10 a.m. as Israelis observed an annual moment of silence in remembrance of the victims of the Nazis and their collaborators. Pedestrians stood, buses and cars pulled over on streets.
The sirens were followed by services at the Knesset and Yad Vashem Holocaust Martyrs' and Heroes' Remembrance Authority. At the Knesset, Prime Minister Ehud Olmert read the names of the members of the Richter family, relatives of his wife Aliza who were killed in the Holocaust whilst Benjamin Netanyahu, opposition leader, read the names of his wife Sara's relatives. Vice Prime Minister, Shimon Peres spoke of losing his grandfather, whilst Social Affairs Minister Isaac Herzog spoke of his father's cousin who was caught trying to cross the French border and was sent to Auschwitz. Additionally, memorial services were held throughout the day at all schools and institutions.
Virginia Tech Shootings
The spirit of the day was further saddened late on in the day with the news that Professor Liviu Librescu, an Israeli Holocaust Survivor was killed in the Virginia Tech massacre. It was reported that he was murdered while holding off the gunman at his lecture hall entrance so his students could escape through the window.
"Silence, memorial services mark day" — Haaretz, April 17, 2007
"Israeli killed in Virginia massacre" — YNET, April 17, 2007
Comments Off on Israel marks Holocaust Memorial Day
Current Israel Hezbollah conflict being compared with \"Operation Grapes of Wrath\"
Filed under: 2006 Israel-Lebanon conflict,AutoArchived,Israel,Knesset,Lebanon,Middle East,Politics and conflicts — admin @ 5:00 am
Current Israel Hezbollah conflict being compared with "Operation Grapes of Wrath"
2006 Lebanon War
Israel says cluster bomb use was legal
French Prime Minister Dominique de Villepin visits Beirut
Belgian peacekeepers wounded in Lebanon
Studies find radioactive material at Israel bomb site in Lebanon
Israel to end Lebanon blockade
Turkish Parliament approves sending troops to Lebanon
On Sunday, Dr. Azmi Bishara, leader of the Balad Arab nationalist political party in Israel and member of the Israeli Knesset (Israeli Parliament) compared the current military actions by Israel unfavorably with similar previous conflicts in Lebanon.
In an article for the Arab Times Bishara said; "The only difference between today and the earlier bombardments — the 'Day of Reckoning' and 'Grapes of Wrath' between 1993 and 1996 — is that Syrian forces are no longer present in Lebanon."
American Spencer Witt, who is filing daily reports from Lebanon, compared the difficulty average citizens are having today with those experienced during past operations; "Etienne was a freshman in high school during Operation Grapes of Wrath — 16 days of Israeli strikes that resulted in the destruction of Lebanese power plants and left much of the country in the dark. And now this." Witt said.
Israeli Prime Minister Ehud Olmert has said that the killing of 8 Israeli soldiers and the abduction of another two soldiers by Hezbollah militants on July 12 was an act of war; "The murderous attack this morning was not a terror attack but an act of war," Olmert said. Lebanon has Hezbollah members in its parliament and "must bear full responsibility. Israel's response will be restrained but very, very, very painful." Olmert said on July 12, 2006.
Olmert told Israeli legislators today that; "We want a two-kilometer (1.2-mile) space from the border in which it will not be possible to fire rockets toward soldiers and civilians houses and in which there will not be contact with military border patrols."
Israeli soldiers previously patrolled a "security zone" during Israel's 18-year occupation of south Lebanon. Olmert said today; "We do not have any intention of returning to the security zone but want to create an area where there will be no Hezbollah." and he repeated Israel's call for an international force (being referred to as "muscular" or "robust") to be put on the Israel-Lebanon border.
"We need international intervention forces that have military capabilities and ability to respond and enforce, and not forces similar to UNIFIL." he said, referring to U.N. peacekeepers. "I don't know when there will be a cease-fire. This is difficult, but we will stand up to this. … We want to end the violent operation as quickly as possible but we won't end it before we believe we can achieve results that justify the price we have paid."
Dr. Bishara lambasted every aspect of Israel's position in this current conflict; "Any comparison between Israeli Prime Minister Ehud Olmert's and Hezbollah chief Nasrallah's political rhetoric must conclude that the latter is the more rational. His speeches are more consistent with the facts and rely less than Olmert's on religious expressions and allusions…Israeli politicians have no cultural or moral edge over resistance leaders. The latter are far less attached to Iran than the former are to the US…The people who unleashed the brutal war against Lebanon are neither intelligent nor courageous. Quite the opposite; they are mediocrities, cowards and opportunists, but they happen to have military superiority. And they possess the keys to the machinery of a state, a real state, one that is secure in its identity, that has clear national security goals and channels of national mobilization, as opposed to a long deferred project for statehood and a states built on the fragmentation of national identity."
Regarding the current conflict, Israeli Prime Minister Ehud Olmert told an Israeli parliament committee Wednesday that Israel would not reoccupy any part of southern Lebanon.
However, speaking on Israeli army radio, Justice Minister Haim Ramon – a close confidant of Israeli Prime Minister Ehud Olmert – said "everyone understands that a victory for Hezbollah is a victory for world terror". He said that in order to prevent casualties amongst Israeli soldiers battling Hezbollah militants in southern Lebanon, villages should be flattened by the Israeli air force before ground troops move in.
"All those now in south Lebanon are terrorists who are related in some way to Hezbollah," Mr Ramon said.
Mr Ramon's call for the use of greater firepower came as the Israeli cabinet was set to decide whether to broaden its military offensive.
"World 'backs Lebanon offensive'" — BBC News Online, July 27, 2006
Sam F. Ghattas. "Report: 12 Israeli troops die in fighting" — AP, July 26, 2006
Spencer Witte. "The New Beirut Nightlife" — Fox News, July 23, 2006
Dr. Azmi Bishara. "The Hiroshima of Our Times" — Arab News, July 22, 2006
Julie Stahl. "Israel Sees Hizballah Provocations As 'Act of War'" — Crosswalk.com, July 12, 2006
Laurie Copans, AP. "Israel seeks 1.2-mile-wide 'security zone'" — Yahoo!, July 12, 2006
Comments Off on Current Israel Hezbollah conflict being compared with \"Operation Grapes of Wrath\"
Exit polls suggest Kadima victory in Israeli elections
Filed under: Ariel Sharon,AutoArchived,Ehud Olmert,Elections,Israel,Knesset,Likud,Middle East,Politics and conflicts,West Bank — admin @ 5:00 am
Early exit polls for the Israeli parliamentary elections suggest that the newly formed Kadima party of Prime Minister Ariel Sharon and Acting PM Ehud Olmert managed to capture between 28 and 32 seats in the 17th Knesset with Labor coming in second with 20 to 22 seats ahead of Yisrael Beiteinu with 12 to 14 seats. Likud got 11 or 12 seats, the National Religious Party took 8 or 9 seats and Shas 10 or 11 seats. The Pensioners party performed better than expected, capturing 6 to 8 seats, United Torah Judaism got 5 to 6 seats and Meretz around 5 seats. Arab parties are expected to get 7 or 8 seats.
The results mean that Olmert could probably form a center-left coaltion to implement his unilateral disengagement plan for the West Bank.
"Kadima has won today. The next prime minister is Ehud Olmert," said Roni Bar-On, a Member of the Knesset for Kadima. His party did worse than expected from opinion polls, which saw them at gaining about 40 seats.
"This a crisis unlike any that has ever hit Likud," said Dan Naveh, a senior MK of the party referring to Likud's fall from being one of the dominating Israeli parties in the last decades to finishing only 4th.
Correspondents for the BBC stress that Israeli exit polls are considered to be relatively unreliable, and that it may take some time to get definitive results.
Israel legislative election, 2006
"Exit polls show center-left bloc winning 62-66 seats" — Haaretz, March 28, 2006
"Kadima MKs hail Ehud Olmert as next the prime minister" — Haaretz, March 28, 2006
"Polls close in Israeli election" — BBC News Online, March 28, 2006
Comments Off on Exit polls suggest Kadima victory in Israeli elections
The Tel Aviv Magistrates Court sentences Omri Sharon
Filed under: Archived,Ariel Sharon,Crime and law,Israel,Knesset,Middle East,Tel Aviv — admin @ 5:00 am
On February 14, 2006, the Tel Aviv Magistrates Court sentenced Omri Sharon, son of the comatosed Prime Minister Ariel Sharon, to a nine month prison term, a nine month suspended sentence, and a 300,000 Shekel (USD 63,500) fine. Omri's lawyers have already stated they will appeal a "exceptionally harsh verdict". Ariel Sharon has not been charged so far.
Judge Beckenstein decided to delay his prison sentence six months due to the current health of his father. Beckenstein was quoted as saying: "This is a swamp of political corruption and it must be dried up." In addition Gavriel Manor, an accomplice of Omri, was sentenced to a nine month suspended sentence for his involvement.
Omri was indicted on August 28, 2005 over a fundraising scandal relating to his father's 1999 election. At the time Omri was exempt from prosecution due to parliamentary immunity but decided to stand trial.
According to Israeli prosecutors Omri received over $1,300,000 in contributions and donations from Israeli and other companies between July 1999 and February 2000. This violated the amount allowed by party finance laws. He was then charged of investing the funds into a corporation called Annex Research.
On November 14 Omri Sharon plead guilty in a plea bargain with prosecutors. He later resigned from the Israeli Knesset on January 3.
Vered Luvitch. "PM's son going to jail" — Ynetnews, February 14, 2006
"Sharon's son gets jail time in Israel bribery case" — Reuters, February 14, 2006
"Sharon son jailed for corruption" — BBC News Online, February 14, 2006
Comments Off on The Tel Aviv Magistrates Court sentences Omri Sharon
Former German president Johannes Rau dies at 75
Filed under: Archived,Berlin,Europe,Germany,Knesset,Moshe Katsav,Obituaries — admin @ 5:00 am
Johannes Rau 1931-2006
Former German president Johannes Rau has died on Friday morning in his home in Berlin at the age of 75, his office announced. Rau had been president of Germany from 1999 till 2004. He had a heart disease for a long time and had to undergo twice a surgery after leaving office. He is survived by his wife and their three children.
Rau was born on January 16, 1931 in Wuppertal. He first joined the Gesamtdeutsche Volkspartei in 1952. After the party had been dissolved, he followed his political mentor Gustav Heinemann to the Social Democratic Party of Germany (SPD). He was elected member of the state legislature of North Rhine-Westphalia in 1958 and became governor of the state in 1978. He was an unsuccessful candidate for the chancellorship in 1987. He resigned as governor in 1998 when his party promised to nominate him for the presidential elections. During his presidency, Rau became the first German head of state to speak in the Israeli Knesset.
His successor, president Horst Köhler, ordered that a state funeral for Rau be held on February 7. He said Germany "has lost a distinctive politician, a great president and a man who made the world more humane."
Chancellor Angela Merkel said Germany "has lost an exceptional personality".
The governor of North Rhine-Westphalia, Jürgen Rüttgers, called Rau a "great governor" to whom the state and its people owe a great deal.
SPD chair Matthias Platzeck said the Social Democrats lost "an irreplaceable friend, who was one of the most outstanding and popular politicians of our country. […] We will miss him."
Israeli president Moshe Katsav called Rau a "great friend of the Jewish people and of the state of Israel".
"Johannes Rau ist tot" — Der Spiegel, January 27, 2005
"Altbundespräsident Rau gestorben" — AP, January 27, 2005
Horst Köhler. "Bundespräsident Horst Köhler zum Tod von Bundespräsident Johannes Rau" — Bundespräsidialamt, January 27, 2005
""Wir werden ihn vermissen"" — Der Spiegel, January 27, 2005
"Germany's ex-president Rau dies" — BBC News, January 27, 2005
Comments Off on Former German president Johannes Rau dies at 75

redpajama_set_name: RedPajamaCommonCrawl


text:
Home-Working With Poor Internet?
Data Protection, IT Support, Password management, Security
Caught out by a phishing email
Sigh.. we're always trying to reinforce internet safety to our customers, but I'm not too proud to admit that this weekend I got caught out with a phishing email. I...
Data Protection, IT Support, News, Password management, Security
Someone tried to hack us…
Over the past couple of weeks we have noticed that there have been multiple attempts to hack into our website from an IP address located in China, despite the fact...
Oliver Jarvis
iT news, News, Password management, Security, Support
Twitter users urged to change password
Twitter has discovered a bug in it's system that resulted in user passwords being stored in plain text on an internal system – whilst no data breach has been reported,...
Email, Password management, Security
Cyber Attack on Parliament – Emails Compromised
Parliament' s IT system was hacked last Friday, compromising the email accounts of almost 1% of the 9,000 users of the system according to the latest reports. Remote Access Disabled...
Internet, Password management, Security, Virgin
Virgin Media Advises Customers To Change Default Passwords
Virgin Media has advised its 800,000 customers to change their passwords to reduce the risk of hacking after finding that many customers were still using risky default network and router passwords....
IT Support, Password management, Security
Insight – what we use for: passwords
We are bombarded with different products or services to manage our lives, it can be difficult working out which one is the best solution for your needs. We spend quite...
News, Password management, Security
Facebook AND Google Victims Of Massive Phishing Scam
Details have emerged of a phishing scam which took place from 2013 until 2015, allegedly run by one 48-year-old man who claimed both Google and Facebook as victims to the...
info@globe2.net
© Globe2 Ltd.
Trading Terms | Privacy Policy | Code of Practice


I used to have a love affair with Marriott, but then, the inevitable happened: A fancy little boutique turned my head and I left for another brand.
But, then, this: the view that I woke up to every day.
OK, really, I could have been sleeping on sandpaper and showering in a plastic tub and, with that view, I would have been happy. Luckily for us, though, the Stanton underwent a renovation not too long ago that gave the rooms a fresh, modern feel with marble bathrooms, lush beds and yes, a view of the ocean from many rooms (Miami's city view, known for its striking and cutting edge architecture, is no slouch either).
Add to the that the direct beach access, a full service pool area, a park next door with a Citibike station, a dog-friendly policy (with a $150 cleaning fee per stay) and a location on Ocean Drive south of Fifth Street right at the edge of all the action, and boom: the Stanton is the perfect hotel in what might be the most perfect vacation spot in America.
Oh, and those 'loosely interpreted' hotel quality standards that can overshadow South Beach hotel experiences? Not at the Stanton—it's a Marriott. You know what to expect: great service, friendly staff, great quality and the feeling of being well taken care of. And the Stanton's location might be the best in South Beach—Ocean Drive between 1st and 2nd: right on the beach, just a short walk to the fun on Ocean Drive (between 5th and 10th), to shopping on Collins and a quick cab or bike ride to the scene-y area of the Lincoln Road and the hotels, restaurants and night clubs north of 16th Street.
When we arrived at the Stanton Marriott we pulled into the circular driveway where our car was immediately unloaded by the bell staff, and climbed a short flight of steps to the demure multi-level lobby accented by soaring ceilings and Art Deco details. We checked in at the front desk and set off to explore. Up another short staircase is another small lobby area and a long corridor that leads past the spa and gift shop to the restaurant where, during the day, you can enjoy a meal or cocktails with an ocean view. Just outside the corridor is an expansive terrace with several seating areas and cabanas that line the outside wall; guests can rent one for the day or have an outdoor massage. At the east end of the terrace is the outdoor restaurant and bar area that overlook the pool and beach.
From the lobby we took the elevator to the mezzanine level—technically the fourth floor— to our room. While we had booked a room with two queen beds, the hotel upgraded us to a suite for our stay, which was to say the least, delightful (Marriott rewards members take note: This is an upgrade worth planning for!). Since we visited in December right before the holiday rush, the hotel wasn't fully booked and they were gracious to upgrade us. Our room, which had a sizable balcony with a table and chairs, looked north over the hotel's terrace, pool and the park beyond it, all framed by the Atlantic Ocean and palm trees. Again, that view.
Looking at the rooms on the north side of the hotel, it looked like most had a balcony (which has a relatively high glass wall), something we always look for when traveling with kids and a dog. It's like having another room to spill over into.
Our room was beyond comfortable: Flat screen TVs, a modern bathroom and for a suite, enough doors to give everyone a little privacy. The living room had a pull-out bed but it was a twin, so housekeeping brought a rollaway bed, too.
But this is Miami Beach—who wants to hang out in the room?!? Our routine is usually to get up early, stake out poolside chairs by leaving our books on them, get a Starbucks (conveniently located in the hotel at the street level), then when the sun is out and has warmed the pool, take up our spots by the pool.
We are SO RUDE. Or were. The Stanton has a great solution: They politely request that we not hold chairs. What that means is that no one else can, either, and even though by our third day the hotel was busy with a wedding and holiday travelers, there was never an issue with getting a great seat beside one of the hotel's two pools.
And, not everyone wanted to go to the pool (just me and our dog Eli, really): the beach beckoned. Down a short flight of stairs at the end of the pool deck is a walkway out to the beach where the Stanton has a towel and wait station; towels are complimentary, beach chairs can be rented for $13 per day, and a kind and lovely waiter will bring you lunch, snacks, piña coladas or other goodies and bill the charge to your room. Simply civilized.
But back to the pool. Not that I don't 'do' the beach, but Eli is a sun god and likes to soak up the rays; the beach is strictly no dogs allowed. So, I asked if I could bring Eli to the pool with me. It was a bit of a puzzling question; I guess most guests don't ask to bring their dog to the pool, so the pool attendant went off to check, and quickly came back with pretty much what we heard from all hotel staffers during our stay: Of course!
Everyone who walked by gave Eli a pat on the head as he took his spot on the lounge chair next to me and quietly enjoyed the sun. Not a water-lover, he just lounged while we took a dip in the heated pool—yes heated; December temperatures can get a bit frosty at night, so discovering that the pool was a bath-like temperature was delightful, even though it was 78 degrees every day we were there.
Here's the other reason Eli loved the Stanton (and so did we): the park right next door, complete with a playground for young kids and doggy-bag dispensers for when your four-legged companion does his business. On the street right in front of the park is a CitiBike station (which had bikes for us every time we went to rent them). There are locations down the block and a block away, too, and the app and website tell you which stations have bikes and how many. Just pop your credit card in and away you go. Rates are $4 for 30 minutes or $6 for an hour. We rented bikes and rode up to Lincoln Road where we docked them and went shopping.
Then, here's the main reason our girls loved the Stanton (besides the beach and pool and beautiful view): Dolcezza, located just outside the lobby on Ocean Drive (there's a Deco Drive Cigar store too). Dolcezza specializes in hand made chocolate, sweets and gelato, but don't walk in unless you plan to give in to it's intoxication. The gelato is simply divine and the chocolates are wonderful. The kids gave it a yums-up.
Rates for a double room while we were there (shoulder season) ran about $275 a night and about double that for a suite. During the high season (Winter), rates start at about $450 a night and suites are about double that, too.
It's worth noting that while the hotel is fully accessible, there are a lot of levels, so this is something to take into consideration for someone on wheels (or guests with short legs), though it's a short walk from the front door to the elevators and pool. And while hotel is family friendly and there is a great playground right next door, there is not a kids club or other kid-focused features (unless you count the beach, pool and a great vacation spot to be a kid-friendly feature, which we do).
Disclosure: We were guests of the Stanton for two nights of our four night stay; opinions expressed are all my own.


LeVario Blue is a dynamic husband and wife acoustic duo which is comprised of DeQuan & Charie Jones. The combination of smooth rhythmic arrangements layered with the silky smooth vocals of Charie, provides clients with an emotion filled experience that you won't soon forget. LeVario Blue can be booked as an acoustic duo for a more personal intimate feel, or with a full band which includes, keys, bass, guitar and drums. For more than 6 years we have been able to perform for countless people in various settings from weddings and special ceremonies, to full out concerts and outdoor events. When we aren't performing you can often find us volunteering to sing in hospitals and retirement homes.
LeVario Blue has a very unique sound and our song selections are carefully planned out to provide the ultimate entertainment experience for our listeners. We strongly believe that music has the power to transcend all barriers and take listeners on a journey that will leave them feeling inspired and hopefully wanting to hear more. Charie's voice is incredibly smooth and with her beautiful tone she's able to deliver songs with ease and comfort. She has traveled as far as the Czech Republic singing background vocals for various artist and she's often called in as a studio session singer because of her ability to learn and deliver songs effectively.
Please note LeVario Blue will also travel to Colfax, High Point, Wallburg, Winston Salem, Greensboro, Oak Ridge, Jamestown, Rural Hall, Bethania, Walkertown, Belews Creek, Stokesdale, Summerfield, Thomasville, Trinity, Germanton, Pleasant Garden, Sophia, Walnut Cove, Clemmons.


The U.S. Court of Appeals for the Ninth Circuit recently upheld the convictions of two ministers of the Hawaii Cannabis Ministry who admitted using and distributing large quantities of cannabis in accordance with their religious beliefs. The Hawaii Cannabis Ministry was founded in 2000 in the City of Hilo, Hawaii as "a community wherein Cannabis could be celebrated as a sacrament." According to one of the ministers, "[t]he consumption, possession, cultivation, and distribution of cannabis are essential and necessary components of the THC ministry." The Cannabis Ministry distributes marijuana both to its members and to medical marijuana users.
The Ministry would provide its members with "a broad menu of cannabis products to pick up and to take away with them: 'packets,' 'live plants,' 'clones,' 'seeds,' 'candy,' 'brownies and chocolate chip cookies all with cannabis,' 'holy anointing oil,' and 'tinctures.'" In return for the cannabis products, the Cannabis Ministry members would provide a "donation" in some amount.
According to the Ninth Circuit, the Ministry "wrote down a handful of rules nominally designed to ensure that cannabis went out only to Ministry members or medical marijuana users. But in practice these rules were little more than parchment barriers." The Ministry did not confirm that people seeking marijuana were who their Ministry ID cards identified them to be. Further, the Ministry "never advised people who came through the express service that there were restrictions on what 'members' could do with the sacrament," such as use it only for religious purposes.
In 2010, a grand jury indicted the Ministry's reverends and associates with several crimes, including violation of the Controlled Substances Act (CSA). Two ministers pled guilty to one count of conspiracy to manufacture and distribute marijuana. One minister was sentenced to 60 months in prison followed by 4 years of supervised release. The other minister was sentenced to 27 months in prison and 3 years of supervised release.
The ministers challenged their convictions under the Religious Freedom Restoration Act (RFRA), which requires that when government substantially burdens religious exercise it must justify the burden through a compelling government interest advanced in the least restrictive means possible.
The Ninth Circuit found that the government had a compelling interest in "mitigating the risk that cannabis from the Ministry will be diverted to recreational users" and that "mandating [the ministers'] full compliance with the CSA would help to advance this compelling interest to a meaningful degree." The Ninth Circuit specifically pointed to the black market for cannabis in the city of Hilo, as well as the risk that cannabis would end up in the hands of minors. The ministers argued that recent Department of Justice memoranda regarding the CSA with respect to marijuana show that the DOJ has abandoned marijuana enforcement efforts, but the Ninth Circuit rejected this argument. The Ninth Circuit further found that the government satisfied the least restrictive means requirement, and rejected the ministers' claims to the contrary.
The Ninth Circuit's decision in United States of America v. Christie is available here.
*RLUIPA-Defense.com trivia: The Hawaiian word for cannabis is "pakalolo" (read New York Times article here). Hawaii has the highest proportion of production to use in the country (see DrugScience.org report, available here).
Original photo by Frances Ellen. Some rights reserved.


text:
New Waterproof Flyers 8.5" x 11"
25 50 75 100 125 150 200 250 300 400 500 750 1000 1500 2000 2500
8 Mil 10 Mil 14 Mil
25 / 8 Mil - $0.00 USD 25 / 10 Mil - $0.00 USD 25 / 14 Mil - $0.00 USD 50 / 8 Mil - $0.00 USD 50 / 10 Mil - $0.00 USD 50 / 14 Mil - $0.00 USD 75 / 8 Mil - $0.00 USD 75 / 10 Mil - $0.00 USD 75 / 14 Mil - $0.00 USD 100 / 8 Mil - $0.00 USD 100 / 10 Mil - $0.00 USD 100 / 14 Mil - $0.00 USD 125 / 8 Mil - $0.00 USD 125 / 10 Mil - $0.00 USD 125 / 14 Mil - $0.00 USD 150 / 8 Mil - $0.00 USD 150 / 10 Mil - $0.00 USD 150 / 14 Mil - $0.00 USD 200 / 8 Mil - $0.00 USD 200 / 10 Mil - $0.00 USD 200 / 14 Mil - $0.00 USD 250 / 8 Mil - $0.00 USD 250 / 10 Mil - $0.00 USD 250 / 14 Mil - $0.00 USD 300 / 8 Mil - $0.00 USD 300 / 10 Mil - $0.00 USD 300 / 14 Mil - $0.00 USD 400 / 8 Mil - $0.00 USD 400 / 10 Mil - $0.00 USD 400 / 14 Mil - $0.00 USD 500 / 8 Mil - $0.00 USD 500 / 10 Mil - $0.00 USD 500 / 14 Mil - $0.00 USD 750 / 8 Mil - $0.00 USD 750 / 10 Mil - $0.00 USD 750 / 14 Mil - $0.00 USD 1000 / 8 Mil - $0.00 USD 1000 / 10 Mil - $0.00 USD 1000 / 14 Mil - $0.00 USD 1500 / 8 Mil - $0.00 USD 1500 / 10 Mil - $0.00 USD 1500 / 14 Mil - $0.00 USD 2000 / 8 Mil - $0.00 USD 2000 / 10 Mil - $0.00 USD 2000 / 14 Mil - $0.00 USD 2500 / 8 Mil - $0.00 USD 2500 / 10 Mil - $0.00 USD 2500 / 14 Mil - $0.00 USD
Posters include free color printing, rounded corners and our ink-protective nano coating for outdoor use! Plus get free 2-day shipping!
Join countless others by using premium waterproof flyer paper from TerraSlate. Our premium line of flyers are printed on either our 8 Mil / 80lb or our 14 Mil 170lb ultra heavy-stock waterproof stock that gives flyers extra rigidity and stability. Vibrant color reproduction and high contrast give your flyers the ability to stand out and out-last rest.
Perfect for posting outdoors on walls, sign posts, telephone poles, car windows and more. TerraSlate's 100% waterproof and tear-proof substrate ensures your flyers will stand up to the elements including heavy rain, direct sunlight and high wind.
Keep your posters intact and vibrant in the harshest conditions. They come printed on one or on both sides in full color at no extra cost and ensured not to tear or incur water damage for the lifetime of the flyer.
For sale flyers
Concert flyers
Sale flyers
Informational flyers
Maps & directional flyers
Garage sale flyers
Lost dog flyers


text:
Q: Setting adapter to spinner scrolls recycler view to top I have a spinner and a recycler view in one layout. In my activity's onResume, I'm setting an adapter to the spinner and trying to scroll recycler view to the last visible item that I have saved in the activity's onPause. The problem is that setting adapter to spinner, scrolls recycler view to top! I've tested if I don't set the adapter there would be no problem. Here's my code:
override fun onPause() {
      super.onPause()
      currentVisiblePosition = (recyclerView.layoutManager as LinearLayoutManager).findFirstCompletelyVisibleItemPosition()
}

override fun onResume() {
      super.onResume()
      val list = ArrayList<String>()
      list.add("item1")
      list.add("item2")
      spinner.adapter = ArrayAdapter<String>(this,android.R.layout.simple_list_item_1,list)
      (recyclerView.layoutManager as LinearLayoutManager).scrollToPosition(currentVisiblePosition)
}

Do you guys know where the problem is?

redpajama_set_name: RedPajamaStackExchange


The US Air Force is looking into making bombs out of anti-matter. I want a key ring with a speck of it! >One millionth of a gram of positrons contain as much energy as 37.8 kilograms (83 pounds) of TNT, according to Edwards' March speech. A simple calculation, then, shows that about 50-millionths of a gram could generate a blast equal to the explosion (roughly 4,000 pounds of TNT, according to the FBI) at the Alfred P. Murrah Federal Building in Oklahoma City in 1995.
Sort of interesting, but I suspect that the main point of the Air Force's research leans more towards antimatter-catalyzed fission then pure "Star Trek"-grade antimatter weapons. An impressively small number of anti-protons can be used to drastically lower the critical mass needed to get a nuclear explosion. This may or may not be useful as a weapon (does anyone actually want grenade-sized nukes?), but it's fantastically useful as a spacecraft propulsion mechanism. The "orion drive" as it tends to be called in science fiction basically consists of a heavy spacecraft, a thick blast plate, and a bunch of small nukes. Each nuke is thrown out behind the ship and explodes. The blast plate absorbs the blast, and the ship goes flying forward. Repeat a few thousand times, and you're moving along at a really impressive speed.
Obviously, this works better in science fiction then reality. Niven and Pournelle used it to great effect in Footfall, where they nuked my hometown.
There are tons of engineering and political problems with any sort of nuclear-weapon driven spaceship, but one of the more fascinating ones is that the "classical" version scales up, but doesn't scale down. You can build immensely huge spaceships with it, but you can't efficiently build anything much smaller then a flying aircraft carrier, because you can't build efficient nukes with a small enough yield. That's where the antimatter comes in–if you can build efficient nukes with really small yields, then this might become practical. It doesn't solve the political problems, which are probably insolvable, but it brings us closer to being able to efficiently launch large masses into space.