text:
\section{Introduction}
All topological spaces  will be assumed to be Tychonoff.  For a
space $X$ and its compactification $bX$ the complement $bX\setminus
X$ is called a \emph{remainder} of $X$.

The study of the interplay between covering properties of Tychonoff
spaces and those of their remainders takes its origin in the seminal
work  of Henriksen and Isbell \cite{HenIsb57}.
 Being highly homogeneous objects,
topological groups restrict the properties of their remainders in a
special, strong way, unlike general topological spaces. The study of
this phenomenon in our context was started by
 Arhangel'skii  \cite{Arh08, Arh09} and further pursued in his
 joint works with  Choban, van Mill, and others, see \cite{ArhCho21, ArhMil13, ArhMil14,ArhMil15} and
 the references therein.

This motivated the study in \cite{BelTokZdo16} of topological groups
whose remainders have combinatorial covering properties
 which lie between the $\sigma$-compactness and
Lindel\"of property. Recall from \cite{COC1} that a space $X$ is
said to be \emph{Menger} (or has the \emph{Menger property}) if  for
any sequence $\la \U_n: n\in\w\ra$ of open covers of $X$ one can
pick finite sets $\mathcal V_n\subset \mathcal U_n$ in such a way
that $\{\bigcup\mathcal V_n:n\in\omega\}$ is a cover of $X$. A
family $\{W_n:n\in\w\}$ of subsets of $X$ is called an
\emph{$\w$-cover} of $X$, if for every $F\in [X]^{<\w}$, the  set
$\{n\in\w:F\subset W_n\}$ is infinite. The property of
\emph{Scheepers} is defined in the same way as the Menger property,
the only difference being that we additionally demand that
$\{\bigcup\mathcal V_n:n\in\omega\}$ is a $\w$-cover of $X$. It is
immediate that
$$\sigma\mbox{-compact} \Rightarrow
\mbox{Scheepers} \Rightarrow \mbox{Menger} \Rightarrow
\mbox{Lindel\"of}.$$ Through a sequence of reductions it was proved
in \cite{BelTokZdo16} that there exists a Scheepers ultrafilter  if
and only if there exists a topological group $G$ such that $\beta
G\setminus G$ is Scheepers and not $\sigma$-compact if and only if
there exists a topological group  $G$ such that all finite powers of
$\beta G\setminus G$ are Menger   and are not $\sigma$-compact.
Here,  $\mathcal P(\w)$ is as usually identified with the Cantor
space $2^\w$ via characteristic functions, and subsets of $\mathcal
P(\w)$ are considered with the subspace topology. Thus the existence
of a topological group $G$ such that $\beta G\setminus G$ is
Scheepers (resp. has all finite powers Menger) and not
$\sigma$-compact is independent from ZFC: Such a group exists under
$\mathfrak d=\mathfrak c$, and its existence yields $P$-points, see
\cite{BelTokZdo16} and the references therein. Furthermore, it was
proved in \cite{BelTokZdo16} that the existence of a topological
group $G$ such that $(\beta G\setminus G)^2$ is
 Menger but not $\sigma$-compact  is independent from ZFC as well.

Since the same approach does not allow to solve the question whether
consistently every Menger remainder of a topological group is
$\sigma$-compact\footnote{This question remains open.
}, it was asked in \cite{BelTokZdo16} whether the Scheepers and
Menger properties can be distinguished by remainders of topological
groups at all, and whether a Menger remainder of a topological group
can have a non-Menger square. In case of a negative answer outright
in ZFC this would give the consistency of all Menger remainders of
topological groups being $\sigma$-compact, simply by applying the
aforementioned results. However, we provide here two alternative
proofs that both of these questions have consistently
 the affirmative answer, by constructing counterexamples using different approaches,
  see Theorems~\ref{0main}, \ref{0main_f}, and
 Corollary~\ref{cor01}. Our topological groups are actually non-meager $P$-filters
 on $\w$, hence metrizable, $0$-dimensional,  totally bounded, and hereditarily Baire by \cite{Mar98}.
 Since the Menger property is preserved by finite products of
 metrizable spaces in the Miller model \cite{Zdo18} and coincides
 with the Scheepers property under $\hot u<\hot g$ by \cite{Zdo05}
 (this inequality holds
in the Miller model by \cite[Theorem~2]{BlaLaf89} combined with the
results of \cite{BlaShe89}),
 filters like in Theorems~\ref{0main} and \ref{0main_f}
 cannot be obtained in ZFC. Theorem~\ref{0main_density_z} is a
 variation of Theorem~\ref{0main_f} motivated by the question
 whether the density one filter can be diagonalized by a (proper)
 poset adding no dominating reals.

As discussed above, Theorem~\ref{0main} (stating that there exists a
filter $\F$ such that among other properties, both $\F$ and $\F^+$
are Menger, and hence $\F$ cannot be meager by
\cite[Prop.~2]{Zdo05}), cannot be proved in ZFC. In
section~\ref{imp_res} we investigate how far the assumptions on $\F$
can be weakened so that it is still impossible to get such a filter
in ZFC. In this context let us recall  that there are Menger
non-meager filters in ZFC, see \cite{RepZdoZha14}. On the other
hand, the existence of a filter $\F$ in ZFC such that $\F^+$ is
Menger is  unknown and constructing such a filter without additional
set-theoretic assumptions (if it is possible at all) would be
extremely difficult: If $\F^+$ is Menger then $\F$ is non-meager and
$P$, see Corollary~\ref{obs_obv}, and it is a famous
 open problem to construct a non-meager $P$-filter in ZFC. In Theorem~\ref{0_anti_main}
we show that consistently there are no Menger filters $\F$ such that
$\F^+$ is Menger as well.

Let us note that  the properties of Menger, Scheepers,  having
Menger square, etc., are preserved by perfect maps in both
directions.  This implies that if one of the remainders of a space
$X$ has one of these covering properties, then all other also have
it, see the beginning of \cite[\S~3]{BelTokZdo16} for more detailed
explanations.

All undefined topological notions can be found in \cite{Eng}. For
the definitions and basic properties of cardinal characteristics
used in this paper we refer the reader to \cite{Bla10}.

\section{Main result}

We need to recall some standard as well as introduce some ad-hoc
notation and terminology. A family $\HH\subset[\w]^\w$ is called a
\emph{semifilter} if for every $H\in\HH,$ $n\in\w$, and $X\supset
H\setminus n$ we have $X\in\HH$. For a set $\F\subset \mathcal
P(\w)$ and $n\in\w$ we denote by $\la\F\ra_n$ the set
$\{\bigcap\F':\F'\in [\F]^{\leq n}\}$, and $\la\F\ra$ stands for
$$\{X\subset\w:\exists n\exists Y\in\la\F\ra_n (Y\setminus n\subset X)\}.$$
$\F$ is said to be \emph{centered} if $\la\F\ra_n\subset [\w]^\w$
for all $n\in\w$, i.e., if the intersection of any finite subfamily
of $\F$ is infinite. In this case, $\la\F\ra$ is the smallest free
filter on $\w$ containing $\F$. Let us note that  if $\F$ is compact
then so is $\la\F\ra_n$ for any $n\in\w$, being a continuous image
of $\F^n$.  Similarly, by  $\la\F\ra_s$ we shall denote the smallest
semifilter containing $\F$, i.e.,
$$\la\F\ra_s=\{X\subset\w\: :\: \exists F\in\F\exists n\in\w(F\setminus n\subset X)\}.$$
For a family $\sfF$ of subsets of $\mathcal P(\w)$ (i.e.,
$\sfF\subset\mathcal P(\mathcal P(\w))$) we define the
\emph{$\la\ra_*$-saturation} of $\sfF$ as the smallest subfamily
$\sfF_1\supset\sfF$ of $\mathcal P(\mathcal P(\w))$ such that
$\la\bigcup\sfF'\ra_n \in\sfF_1$ for any $\sfF'\in [\sfF_1]^{<\w}$
and $n\in\w$. It is  easy to write the $\la\ra_*$-saturation of a
family $\sfF$ in a precise way,
 thus proving that the $\la\ra_*$-saturation of an infinite $\sfF$ has the same cardinality as $\sfF$,
 and it consists of compact subsets of $\la\bigcup\sfF\ra$ if each $\F\in\sfF$ is compact.
We say that $\sfF$ is \emph{$\la\ra_*$-saturated} if it coincides
with its $\la\ra_*$-saturation.

Given families $\F,\HH$ of subsets of $\w$, we denote by
$\F\wedge\HH$ the family $\{F\cap H:F\in\F, H\in\HH\}$. Again, if
$\F,\HH$ are compact, then so is $\F\wedge\HH$. For
$\sfF,\sfH\subset\mathcal P(\w)$ we define the
\emph{$(\sfF,\wedge)$-saturation} of $\sfH$ as the smallest
subfamily $\sfH_1\supset\sfH$ of $\mathcal P(\mathcal P(\w))$ such
that $\F\wedge\HH\in\sfH_1$ for any $\F\in\sfF$ and $\HH\in\sfH_1$.
Again, it is straightforward to write the $(\sfF,\wedge)$-saturation
of $\sfH$ in the precise way, thus proving that it is a subfamily of
$\mathcal P(\la\bigcup\sfF\ra\wedge\bigcup\sfH)$ of size at most
$\max\{\w,|\sfF|,|\sfH|\}$, and it consists of compact sets if so do
$\sfF$ and $\sfH$. We call $\sfH$ to be
\emph{$(\sfF,\wedge)$-saturated} if it coincides with its
$(\sfF,\wedge)$-saturation.

The next easy auxiliary fact is very similar to \cite[Prop.~7]{GuzHruMar17}
and can be  derived from the latter one in a rather straightforward way.
However, for reader's convenience we give a direct proof.

\begin{lemma}\label{bidi}
Let $\sfF$ be a family of compact subsets of $[\w]^\w$ of size
$|\sfF|<\min\{\hot d,\hot r\}$ and $\phi:\w\to\w$  a finite-to-one
surjection. Then there exists  $Z\subset\w$ such that
$$|\phi^{-1}[Z]\cap F|=|(\w\setminus \phi^{-1}[Z])\cap F|=\w$$
for all $F\in\bigcup\sfF$.
\end{lemma}
\begin{proof}
Let us first assume that $\phi$ is monotone and consider
the strictly increasing number sequence
$\la k_n:n\in\w\ra$   with $k_0=0$ such that
$\phi^{-1}(n)=[k_n,k_{n+1})$ for all $n\in\w$.

For every $\F\in\sfF$ let $h_\F:\w\to\w$ be a strictly increasing function such that
$[h_\F(n), h_{\F}(n+1))\cap F\neq\emptyset$ for all $F\in\F$ and $n\in\w$. Since $|\sfF|<\hot d$, there
exists a strictly increasing $h\in\w^\w$ such that
$$ I_\F:=\{ n\in\w\: : \: (|[h(n),h(n+1))\cap h_\F[\w]|\geq 2)\}$$
is infinite  for all $\F\in\sfF$. Without loss of generality we may assume that
$h[\w]\subset \{k_n:n\in\w\}$, and hence for every (infinite) $I\subset\w$ there exists (an infinite) $Z\subset\w$
such that $\bigcup_{n\in I}[h(n),h(n+1))=\phi^{-1}[Z]$.

 Since $|\F|<\hot r$ there exists $I\subset\w$ such that
 $|I\cap I_\F|=|I_\F\setminus I|=\w$ for all $\F\in\sfF$.
 We claim that
 $$ \big|\bigcup_{n\in I}[h(n),h(n+1))\cap F\big|=\big|\bigcup_{n\in \w\setminus I}[h(n),h(n+1))\cap F\big|=\w $$
 for all $F\in\bigcup\sfF$. Indeed, let us find $\F\in\sfF$ containing $F$ and  pick $n\in I\cap I_\F$.
 Then there exists $m\in\w$ such that
 $[h(n),h(n+1))\supset [h_\F(m),h_\F(m+1))$,
 and hence
 $$\emptyset\neq F\cap [h_\F(m),h_\F(m+1))\subset F\cap [h(n),h(n+1))\subset F\cap\bigcup_{n\in I}[h(n),h(n+1))\cap F.$$
The case of $\w\setminus I$ is analogous.
\smallskip

Now suppose that $\phi$ is arbitrary
 finite-to-one surjection  from $\w$ to $\w$.
Fix a bijection $\theta:\w\to\w$ such that
$\phi\circ\theta$ is a monotone surjection. It follows from the above that there exists
$Z\subset\w$ such that
$$\big|(\phi\circ\theta)^{-1}[Z]\cap \theta^{-1}[F]\big|=\big|(\w\setminus (\phi\circ\theta)^{-1}[Z])\cap \theta^{-1}[F]\big|=\w$$
for all $F\in\bigcup\sfF$, i.e.,
$$\big|\theta^{-1}[\phi^{-1}[Z]]\cap \theta^{-1}[F]\big|=\big|(\w\setminus \theta^{-1}[\phi^{-1}[Z]])\cap \theta^{-1}[F]\big|=\w$$
and thus also
$$|\phi^{-1}[Z]\cap F|=|(\w\setminus \phi^{-1}[Z])\cap F|=\w$$
for all $F\in\bigcup\sfF$ because $\theta$ is a bijection.
\end{proof}

A semifilter $\F$ is called a \emph{$P$-semifilter} if for every
sequence $\la F_n:n\in\w\ra$ of elements of $\F$ there exists a
sequence $\la K_n:n\in\w\ra$ such that $K_n\in [F_n]^{<\w}$ for all
$n\in\w$ and $\bigcup_{n\in\w}K_n\in\F$. If
$\F^+=\{X\subset\w:\forall F\in\F(X\cap F\neq\emptyset)\}$ is a
$P$-semifilter, then we also say that $\F$ is a
\emph{$P^+$-semifilter}. Each semifilter $\F$ on $\w$ gives rise to
the semifilter $\F^{(<\w)}$ on $[\w]^{<\w}\setminus\{\emptyset\}$
generated by the family $\{[F]^{<\w}\setminus\{\emptyset\}\: :\:
F\in\F \}$. If $\F$ is a filter we shall call $\F^+$ the \emph{coideal} of $\F$.

Next, we put together several  known facts about Menger semifilters
established in \cite{ChoRepZdo15} and \cite{GuzHruMar17} and get a
potentially useful characterization.

\begin{theorem}\label{basics}
Let $\F$ be a semifilter on $\w$. Then the following statements are
equivalent:
\begin{itemize}
\item[$(1)$]  $\F$ is Menger;
\item[$(2)$] For every sequence $\la\K_n:n\in\w\ra$ of compact subsets of $\F^+$
there exists an increasing sequence $\la m_n:n\in\w\ra\in\w^\w$ with
the property $\bigcup_{n\in\w}(K_n\cap m_n)\in\F^+$ for any $\la
K_n:n\in\w\ra\in\prod_{n\in\w}\K_n$;
\item[$(3)$] For every sequence $\la\K_n:n\in\w\ra$ of compact subsets of $\F^+$
there exists an increasing sequence $\la m_n:n\in\w\ra\in\w^\w$ with
the property $\bigcup_{n\in\w}\big(K_n\cap
[m_{n-1},m_n)\big)\in\F^+$ for\footnote{We set here $m_{-1}=0$.} any
$\la K_n:n\in\w\ra\in\prod_{n\in\w}\K_n$; and
\item[$(4)$]  $\F^{(<\w)}$ is a $P^+$-semifilter.
\end{itemize}
\end{theorem}
\begin{proof}
The equivalence $(1)\Leftrightarrow (4)$ was established in
\cite[Claim~2.4]{ChoRepZdo15}, and its proof works verbatim also for
semifilters. The implication $(1)\to (3)$ was obtained in
\cite[Prop.~3.4]{ChoRepZdo15}, its proof again works for semifilters
without any changes, while $(3)\to (2)$ is straightforward. Thus it
remains to prove $(2)\to (4)$. Let $\la E_n:n\in\w\ra$ be a sequence
of elements of $(\F^{(<\w)})^+$. For each $n$ set
$\K_n=\{X\subset\w:\forall e\in E_n (X\cap e\neq\emptyset)\}$ and
note that $\K_n\subset\F^+$: If $K\cap F=\emptyset$ for some $K\in
\K_n$ and $F\in\F$, then there is no $e\in E_n\cap [F]^{<\w}$, which
contradicts our choice of $E_n$. Let $\la m_n:n\in\w\ra$ be as in
$(2)$. We claim that $\bigcup_{n\in\w} E_n\cap\mathcal
P(m_n)\in(\F^{(<\w)})^+$, which will imply $(4)$. Indeed, otherwise
there exists $F\in\F$ such that $e\setminus F\neq\emptyset$ for each
$e\in E_n\cap\mathcal P(m_n)$ and $n\in\w$. Thus
$$K_n:=\bigcup\{e\setminus F:e\in E_n, e\subset m_n\}\cup(\w\setminus m_n)\in\K_n$$
for all $n$. However,
\begin{eqnarray*}
F\cap \bigcup_{n\in\w}(K_n\cap m_n)=F\cap
\bigcup_{n\in\w}\bigcup\{e\setminus F:e\in E_n, e\subset m_n\}=\\
=\bigcup_{n\in\w} (F\cap \bigcup\{e\setminus F:e\in E_n, e\subset
m_n\})=\emptyset,
\end{eqnarray*}
and hence $\bigcup\{K_n\cap m_n:n\in\w\}\not\in\F^+$, which
contradicts $(2)$.
\end{proof}

\begin{corollary} \label{obs_obv}
Each Menger semifilter is $P^+$.
\end{corollary}

Theorem~\ref{basics} will be crucial for the proof of the following
fact, which is the main result of the paper.

\begin{theorem}\label{0main}
($\hot r=\hot d=\hot c$). There exists a  filter $\G$ on $\w$ such that
\begin{itemize}
\item[$(1)$] $\G$ and $\G^+$ are Menger;
\item[$(2)$] For every finite-to-one surjection $\phi:\w\to\w$ there exists $X\subset\w$ such that
$\phi^{-1}[X],\phi^{-1}[\w\setminus X]\in\G^+$.
\end{itemize}
\end{theorem}
\begin{proof}
Let us fix the following enumerations:
\begin{itemize}
\item $\{\la\K^\alpha_n:n\in\w\ra:\alpha<\hot c\}$=: the family of all sequences of compact subsets of
$[\w]^\w$;
\item $\{\phi_\alpha:\alpha<\hot c\}$=: the family of all finite-to-one surjections from $\w\to\w$.
\end{itemize}
By recursion over $\alpha<\hot c$ we shall construct a sequence
$\la\la \sfF_\alpha,\sfH_\alpha\ra:\alpha<\hot c\ra$ such that
\begin{itemize}
\item[$(a)$] $\sfF_\alpha,\sfH_\alpha$ are families of compact subsets of $[\w]^\w$
of size $<\hot c$   with $\sfF_0=\sfH_0=\{\{\w\}\}$;
\item[$(b)$] $\bigcup\sfF_\alpha$ is centered and $\sfF_\alpha$ is $\la\ra_*$-saturated;
\item[$(c)$] $\sfF_\alpha\subset\sfF_{\alpha'}\cap\sfH_{\alpha'}$ and $\sfH_\alpha\subset\sfH_{\alpha'}$
 for any $\alpha\leq\alpha'$;
\item[$(d)$] $\bigcup\sfH_\alpha\subset \la\bigcup\sfF_\alpha\ra^+$,
and $\sfH_\alpha$ is $(\sfF_\alpha,\wedge)$-saturated;
\item[$(e)$] If
$\bigcup_{n\in\w}\K^\alpha_n \subset \la\bigcup\sfF_\alpha\ra$,
 then there exists an increasing
number sequence
$\la m^\alpha_n:n\in\w \ra$ such that
$\sfF_{\alpha+1}\ni \K_\alpha$,
where $\K_\alpha=\big\{\bigcup_{n\in\w}(K_n\cap m^\alpha_n):\la K_n:n\in\w\ra\in\prod_{n\in\w}\K^\alpha_n \big\};$
\item[$(f)$] If
$\bigcup_{n\in\w}\K^\alpha_n \not\subset\la\bigcup\sfF_\alpha\ra$,
 then there exists $K_\alpha\in \bigcup_{n\in\w}\K^\alpha_n$ such that
 $\{\w\setminus K_\alpha\}\in\sfH_{\alpha+1}$;
 \item[$(g)$]
 If
$\bigcup_{n\in\w}\K^\alpha_n \subset\la\bigcup\sfH_\alpha\ra_s$,
then there exists an increasing
number sequence
$\la l^\alpha_n:n\in\w \ra$ such that
$\sfH_{\alpha+1}\ni \mathcal L_\alpha$,
where $\mathcal L_\alpha=\big\{\bigcup_{n\in\w}(K_n\cap l^\alpha_n):\la K_n:n\in\w\ra\in\prod_{n\in\w}\K^\alpha_n \big\};$
\item[$(h)$]
 If
$\bigcup_{n\in\w}\K^\alpha_n \not\subset\la\bigcup\sfH_\alpha\ra_s$,
 then there exists $L_\alpha\in \bigcup_{n\in\w}\K^\alpha_n$ such that
 $\{\w\setminus L_\alpha\}\in\sfF_{\alpha+1}$;
 \item[$(j)$] There exists  $Z_\alpha\subset\w$
 such that $\{\phi_\alpha^{-1}[Z_\alpha],\phi_\alpha^{-1}[\w\setminus Z_\alpha]\}\in\sfH_{\alpha+1}$.
\end{itemize}
First, let us assume that we have constructed a sequence $\la\la
\sfF_\alpha,\sfH_\alpha\ra:\alpha<\hot c\ra$ satisfying $(a)$-$(j)$.
We claim that
$$\G:=\big\la\bigcup\{\bigcup\sfF_\alpha:\alpha<\hot c\}\big\ra$$ is as required. It follows that $\G^+$ equals $$\G_1:=\big\la\bigcup\{\bigcup\sfH_\alpha:\alpha<\hot c\}\big\ra_s.$$
Indeed, $\G_1\subset\G^+$
follows directly from $(d)$. To prove that $\G^+\subset\G_1$
let us fix any
$X\not\in\G_1$. Let $\alpha<\hot c$ be such that  $\K^\alpha_n=\{X\}$ for every $n\in\w$
and note that $\la\K^\alpha_n:n\in\w\ra$ satisfies the premises of $(h)$. Therefore there exists
$L_\alpha\in\bigcup_{n\in\w}\K^\alpha_n=\{X\}$ (i.e., $L_\alpha=X$) such that $\{\w\setminus L_\alpha\}\in\sfF_{\alpha+1}$,
and hence $\w\setminus X\in\G$ which yields $X\not\in\G^+$.

Next, we shall establish that both $\G$ and $\G^+$ are Menger.
Let  $\la\K_n:n\in\w\ra$ be a sequence of compact subsets of $\G$ and
$\alpha$ be such that $\K^\alpha=\K^\alpha_n$ for all $n\in\w$. It follows that
$\bigcup_{n\in\w}\K^\alpha_n\subset\la\bigcup\sfF_\alpha\ra$. Indeed, otherwise
by $(f)$ there exists $K_\alpha\in\bigcup_{n\in\w}\K^\alpha_n$ such that $\{\w\setminus K_\alpha\}\in\sfH_{\alpha+1}$,
and therefore $\w\setminus K_\alpha\in\G^+$, which contradicts $K_\alpha\in\G$. Thus
$\la\K^\alpha_n:n\in\w\ra$ fulfills the premises of $(e)$, which yields
 an increasing sequence $\la m_n:n\in\w\ra\in\w^\w$
with the property $\bigcup_{n\in\w}(K_n\cap m_n)\in\G$ for any $\la
K_n:n\in\w\ra\in\prod_{n\in\w}\K^\alpha_n.$ Applying
Theorem~\ref{basics}, we conclude that $\G^+$ is Menger.

To see  that also $\G$ is Menger
let us consider a sequence  $\la\K_n:n\in\w\ra$   of compact subsets of $\G^+$ and find
$\alpha$  such that $\K^\alpha=\K^\alpha_n$ for all $n\in\w$. It follows that
$\bigcup_{n\in\w}\K^\alpha_n\subset\la\bigcup\sfH_\alpha\ra_s$. Indeed, otherwise
by $(h)$ there exists $L_\alpha\in\bigcup_{n\in\w}\K^\alpha_n$ such that $\{\w\setminus L_\alpha\}\in\sfF_{\alpha+1}$,
and therefore $\w\setminus L_\alpha\in\G$, which contradicts $L_\alpha\in\G^+$. Thus
$\la\K^\alpha_n:n\in\w\ra$ fulfills the premises of $(g)$, which yields
 an increasing sequence $\la l_n:n\in\w\ra\in\w^\w$
with the property $\bigcup_{n\in\w}(K_n\cap l_n)\in\G^+$ for any
$\la K_n:n\in\w\ra\in\prod_{n\in\w}\K^\alpha_n.$ Applying
Theorem~\ref{basics}, we conclude that $\G$ is Menger.
\smallskip

Finally, we show how to construct a sequence as above
 satisfying $(a)$-$(j)$. Limit stages are straightforward, so
 suppose that
we have already constructed $\la\la \sfF_\beta,\sfH_\beta\ra:\beta\leq\alpha\ra$ satisfying $(e)$-$(j)$
for all $\beta<\alpha$, $(a),(b)$ and $(d)$ for all $\beta\leq\alpha$, and
$(c)$ for all $\beta\leq\beta'\leq\alpha$. Two cases are possible.
\smallskip

1). $\bigcup_{n\in\w}\K^\alpha_n \subset \la\bigcup\sfF_\alpha\ra$. Then
for every $\HH\in \sfH_\alpha$   we can find an increasing sequence $\la m^{\HH}_n:n\in\w\ra\in\w^\w$
such that $$ K\cap H\cap (m^{\HH}_n\setminus n) \neq\emptyset $$
 for every $K\in\la\K^\alpha_{n}\ra_n$,
$H\in \HH$, and $n\in\w$. Such an $m^{\HH}_n$ exists because of the compactness of the involved sets
and $\HH\subset\bigcup\sfH_\alpha\subset \la\bigcup\sfF_\alpha\ra^+$.
Since $|\sfH_\alpha|<\hot d$, there exists  an increasing sequence $\la m^\alpha_n:n\in\w\ra\in\w^\w$
such that $|\{n\in\w:m^{\HH}_n\leq m^\alpha_n\}|=\w$ for all $\HH\in \sfH_\alpha$.
Set
$$\K_\alpha=\Big\{\bigcup_{n\in\w}(K_n\cap m^\alpha_n):\la K_n:n\in\w\ra\in\prod_{n\in\w}\K^\alpha_n \Big\},$$
and $\sfF_{\alpha+1}$ to be the $\la\ra_*$-saturation of
$\sfF_\alpha \cup \{\K_\alpha\}.$
Since $|\sfF_{\alpha+1}|<\hot c$,
Lemma~\ref{bidi} yields  $Z_\alpha\subset\w$
 such that $\{\phi_\alpha^{-1}[Z_\alpha],\phi_\alpha^{-1}[\w\setminus Z_\alpha]\}\in \la\bigcup\sfF_{\alpha+1}\ra^+$.
 We define $\sfH_{\alpha+1}$ to be
 the $(\sfF_{\alpha+1},\wedge)$-saturation of
 $$\sfH_{\alpha}\cup\big\{ \{\phi_\alpha^{-1}[Z_\alpha],\w\setminus\phi_\alpha^{-1}[Z_\alpha]\}\big\}.$$
We are left with the task of checking that conditions $(a)$-$(j)$
are satisfied. Indeed, $(a)$, $(c)$, and $(e)$-$(j)$ hold
immediately by the construction, in case of $(f)$ and $(h)$ because
of the premises being violated.

Proof of $(b)$, $(d)$ for $\alpha+1$: \ By $(b)$ and $(d)$ for
$\alpha$ it suffices to prove that $\la\K_\alpha\ra_k\wedge
\HH\subset[\w]^\w$  for any
 $\HH\in\sfH_\alpha$ and $k\in\w$.
Let us fix
$$\big\{ \la K^i_n:n\in\w\ra :i\leq k\big\}\subset\prod_{n\in\w}\K^\alpha_n $$
  and   $H\in\HH$, and find
 $n\geq k $  such that $m^{\HH}_n\leq m^\alpha_n$.
 Then $\bigcap_{i\leq k}K^i_n\in\la\K^\alpha_n\ra_n$ and therefore
 \begin{eqnarray*}
 \emptyset\neq\bigcap_{i\leq k}K^i_n\cap H\cap (m^{\HH}_n\setminus n)\subset \bigcap_{i\leq k}K^i_n\cap H\cap (m^{\alpha}_n\setminus n)\subset \\
 \subset\Big(\bigcap_{i\leq k}\bigcup\{K^i_n\cap m^\alpha_n:n\in\w\}\cap H\Big)\setminus n,
 \end{eqnarray*}
 which proves $(b)$ and $(d)$ for $\alpha+1$ and thus completes this case.
 \smallskip

 2).  $\bigcup_{n\in\w}\K^\alpha_n \not\subset \la\bigcup\sfF_\alpha\ra$.

  Two subcases of  $2)$ are possible.

 $2_0)$.   $\bigcup_{n\in\w}\K^\alpha_n \not \subset \la\bigcup\sfH_\alpha\ra_s$.
It follows that there exists $L_\alpha\in
\bigcup_{n\in\w}\K^\alpha_n$ such that
 $\w\setminus L_\alpha\in\la\bigcup\sfH_{\alpha}\ra_s^+\subset\la\bigcup\sfF_\alpha\ra^+$,
 which allows us to define
 $\sfF_{\alpha+1}$ as the $\la\ra_*$-saturation of $\sfF_\alpha\cup\big\{ \{\w\setminus L_\alpha\}\big\} $.
 Lemma~\ref{bidi} yields  $Z_\alpha\subset\w$
 such that $\{\phi_\alpha^{-1}[Z_\alpha],\phi_\alpha^{-1}[\w\setminus Z_\alpha]\}\subset \la\bigcup\sfF_{\alpha+1}\ra^+$.
Let   $\sfH_{\alpha+1}$ be the $(\sfF_{\alpha+1},\wedge)$-saturation of
  $$\sfH_{\alpha}\cup\{\{\phi_\alpha^{-1}[Z_\alpha],\w\setminus\phi_\alpha^{-1}[Z_\alpha]\}\}.$$
 Conditions  $(a)$-$(c)$, and $(e)$-$(j)$ hold for $\alpha+1$
immediately by the construction, in case of $(e)$ and $(g)$ because of the premises being violated,
and for $(f)$ we can simply take $K_\alpha$ to be $L_\alpha$.
Regarding $(d)$ for $\alpha+1$, by $(b)$ and $(d)$ for $\alpha$ it suffices to prove that
 $F_0\cap F_1\cap(\w\setminus L_\alpha)\cap X$ is infinite for any $F_0,F_1\in\bigcup\sfF_\alpha$ and $X\in\{\phi_\alpha^{-1}[Z_\alpha],\w\setminus\phi_\alpha^{-1}[Z_\alpha]\}$, which again has been guaranteed in the course of the construction above.

$2_1)$. $\bigcup_{n\in\w}\K^\alpha_n  \subset \la\bigcup\sfH_\alpha\ra_s$.
In this case we set $\sfF_{\alpha+1}=\sfF_\alpha$.

For every $\F\in \sfF_{\alpha+1}$   we can find an increasing sequence $\la l^{\F}_n:n\in\w\ra\in\w^\w$
such that $$ K\cap F\cap (l^{\F}_n\setminus n) \neq\emptyset $$
 for every $K\in \K^\alpha_n$
and $F\in \F$. Such an $l^{\F}_n$ exists because of the compactness of the involved sets
and $\K^\alpha_n\subset\la\bigcup\sfH_\alpha\ra_s\subset \la\bigcup\sfF_\alpha\ra^+$.
Since $|\sfF_\alpha|<\hot d$, there exists  an increasing sequence $\la l^\alpha_n:n\in\w\ra\in\w^\w$
such that $|\{n\in\w:l^{\F}_n\leq l^\alpha_n\}|=\w$ for all $\F\in \sfF_{\alpha+1}$.
Set
$$\mathcal L_\alpha=\Big\{\bigcup_{n\in\w}(K_n\cap l^\alpha_n):\la K_n:n\in\w\ra\in\prod_{n\in\w}\K^\alpha_n \Big\}.$$
Lemma~\ref{bidi} yields  $Z_\alpha\subset\w$
 such that $\{\phi_\alpha^{-1}[Z_\alpha],\phi_\alpha^{-1}[\w\setminus Z_\alpha]\}\subset \la\bigcup\sfF_{\alpha+1}\ra^+$.
From $\bigcup_{n\in\w}\K^\alpha_n  \not\subset \la\bigcup\sfF_\alpha\ra$
it follows that
  there exists $K_\alpha\in \bigcup_{n\in\w}\K^\alpha_n$ such that
 $\{\w\setminus K_\alpha\}\in \la\bigcup\sfF_\alpha\ra^+$.
Finally, we define $\sfH_{\alpha+1}$ to be
 the $(\sfF_{\alpha+1},\wedge)$-saturation of
 $$\sfH_{\alpha}\cup \big\{ \{\phi_\alpha^{-1}[Z_\alpha], \{\w\setminus\phi_\alpha^{-1}[Z_\alpha],\w\setminus K_\alpha\}\big\}\cup \{\mathcal L_\alpha\}.$$
It is straightforward  to check that conditions $(a)$-$(j)$ are
satisfied, the only slightly non-trivial step  being $\mathcal
L_\alpha\subset\la\bigcup\sfF_{\alpha+1}\ra^+$, which can be proved
analogously  to (but more easily) ``$\la\K_\alpha\ra_k\wedge
\HH\subset[\w]^\w$ for any
 $\HH\in\sfH_\alpha$ and $k\in\w$'' in case $1).$

This concludes our proof, since all possible cases have been
considered.
\end{proof}

In the proof of the following corollary we shall use the classical
result of Hurewicz \cite{Hur27} (see also \cite[Theorem~4.3]{COC2})
stating that $X\subset\mathcal P(\w)$ is Menger if and only if
$f[X]$ is not dominating for any continuous $f:X\to\w^\w$.

\begin{corollary} \label{cor01}
If a semifilter $\HH$ satisfies condition $(2)$ of
Theorem~\ref{0main}, then $\HH^2$ is not Menger and $\HH$ is not
Scheepers. In particular, there is no continuous surjection from
$\HH$ onto $\HH^2$.
\end{corollary}
\begin{proof}
Suppose that $\HH^2$ is Menger, then so is $\HH\times(\sim\HH)$, where
$\sim\HH=\{\w\setminus H:H\in\HH\}$, because $\sim\HH$ is homeomorphic to $\HH$.
Therefore
$\mathcal X:=\HH\times(\sim\HH)\cap\{\la X,X\ra:X\subset\w\}$ is also Menger being a closed subspace
of  $\HH\times(\sim\HH)$, and hence so is  $\HH\cap(\sim\HH)$ as the projection of $\mathcal X$ to
the first (as well as the second) coordinate. Let us note that $\HH\cap(\sim\HH)$ consists of infinite co-infinite subsets of $\w$, and hence the map $h:\HH\cap(\sim\HH)\to\w^\w$,
$h(X)=\{n\in X: (n+1)\not\in X\}$, must have a non-dominating range.

On the other hand, given a strictly increasing $x\in\w^\w$ with $x(0)=0$, let us consider the
monotone surjection $\phi:\w\to\w$ such that $\phi^{-1}(n)=[x(n),x(n+1))$ for all $n$.
It follows that there exists $Z\subset\w$ such that $\phi^{-1}[Z]\in \HH\cap(\sim\HH)$,
and it is easy to see that $h[\phi^{-1}[Z]]\subset x[\w]$. Thus for every $x$ as above there exists
$X\in \HH\cap(\sim\HH)$ with $h(X)$ contained in the range of $x$, which clearly yields
a dominating continuous range of $\HH\cap(\sim\HH)$ and thus leads to a contradiction.

Now, suppose that $\HH$ is Scheepers and consider the clopen cover
$\mathsf O=\{\mathcal O_k:k\in\w\}$ of $\HH$, where $\mathcal
O_k=\{X\subset\w:k\in X\}$. By \cite[Theorem~2]{BabKocSch04} (see
the equivalence of items 1 and 4 there) there exists a disjoint
sequence $\la\mathsf O_n:n\in\w\ra$ of finite subsets of $\mathsf O$
such that for every finite $\HH'\subset \HH$ there exists $n$ with
$\HH'\subset\bigcup\mathsf O_n$. Let $s_n\in [\w]^{<\w}$ be such
that $\mathsf O_n=\{\mathcal O_k:k\in s_n\}$ and consider the
finite-to-one surjection $\phi:\w\to\w$ such that $\phi^{-1}(n)=s_n$
for all $n\in\w$. It follows from the above that for every finite
$\HH'\subset\HH$ there exists $n\in\w$ such that $H\cap
s_n\neq\emptyset$ for all $H\in\HH'$.

On the other hand, using Theorem~\ref{0main}(2) pick $Z\subset\w$ be such that $\HH':=\{\phi^{-1}[Z],\w\setminus\phi^{-1}[Z]\}\subset\HH$
and note that there is no $n\in\w$ such that both sets
$$ \phi^{-1}[Z] \cap s_n=\phi^{-1}[Z]\cap\phi^{-1}(n) \mbox{ \ and \ } (\w\setminus\phi^{-1}[Z])\cap s_n= (\w\setminus\phi^{-1}[Z])\cap \phi^{-1}(n)$$
are non-empty, a contradiction.
\end{proof}

Let us note that the proof of Corollary~\ref{cor01} could be
actually extracted from that of \cite[Lemma~3.1]{BelTokZdo16}, but
we have nonetheless presented it for reader's convenience. Since
every Menger subspace of $\mathcal P(\w)$ has Menger square and is
Scheepers in the Miller model, Theorem~\ref{0main} cannot be proved
in ZFC, as follows from Corollary~\ref{cor01}.

Since each filter has a structure of a topological group,
Theorem~\ref{0main} combined with Corollary~\ref{cor01} answers
\cite[Question~1.8]{BelTokZdo16} in the affirmative, the coideal
$\mathcal G^+$ being the needed counterexample. Another motivation
for Corollary~\ref{cor01} comes from \cite{MedZdo16} where it was
proved that each filter on $\w$ is homeomorphic to its square.
According to \cite[Prop.~8]{MedZdo16}, this result fails for
semifilters, and the counterexample is a Borel comeager semifilter.
However, until now no  semifilter
$\F$ such that both $\F$ and $\F^+$ are non-meager, which in addition  is not homeomorphic to its square, was known, and
Theorem~\ref{0main} combined with Corollary~\ref{cor01}  gives a
consistent example of coideal like that.

For curiosity we exclude below one more possibility for spaces
related to filters $\G$ satisfying Theorem~\ref{0main} to be
homeomorphic.

\begin{corollary} \label{cor02}
If a filter  $\G$ satisfies  Theorem~\ref{0main}, then $\G\times\G^+$ is not homeomorphic to
 to $\G$.
\end{corollary}
\begin{proof}
 $\G^2$ is Menger by \cite{MedZdo16} (see also \cite[Claim~5.5]{ChoRepZdo15} for a simpler proof),
whereas $(\G\times\G^+)^2$ can be mapped continuously onto $(\G^+)^2$ and hence is not Menger.
\end{proof}

\section{Examples by forcing}

This section was inspired by \cite[\S~4]{GuzHruMar??} as well as
\cite[\S~6]{Laf89}. More precisely, one can obtain a filter $\G$
such as in Theorem~\ref{0main} by countably complete forcing, namely
let $\IP$ be the poset consisting of conditions $p=\la \mathsf
F,\mathsf H\ra$ such that
\begin{itemize}
\item[$(i)$] $\mathsf F$ is a countable collection of compact subsets of $[\w]^\w$ such that
$\bigcup\mathsf F$ is centered;
\item[$(ii)$] $\mathsf H$ is a countable collection of
compact subsets of $[\w]^\w$ such that $\mathsf F\subset\mathsf H$;
\item[$(iii)$] $\bigcup\mathsf H\subset \la\bigcup\mathsf F\ra^+$.
\end{itemize}
A condition $\la \sfF_1,\sfH_1\ra$ is stronger than $\la
\sfF_0,\sfH_0\ra$ (and written $\la \sfF_1,\sfH_1\ra \leq \la
\sfF_0,\sfH_0\ra$) if $\sfF_1\supset\sfF_0$ and
$\sfH_1\supset\sfH_0$.

\begin{theorem} \label{0main_f}
Let $G$ be a $\IP$-generic filter, $\G=\bigcup\{\bigcup\sfF:\exists\sfH\: (\la\sfF,\sfH\ra\in G)\}$,
and $\HH=\bigcup\{\bigcup\sfH:\exists\sfF\: (\la\sfF,\sfH\ra\in G)\}$.
Then $\G$ is a filter, $\HH=\G^+$, and both $\G$ and $\HH$ are Menger.
Moreover, for every  surjection $\phi:\w\to\w$ there exists $X\subset\w$ such that
$\phi^{-1}[X],\phi^{-1}[\w\setminus X]\in\G^+$.
\end{theorem}

We leave the proof of Theorem~\ref{0main_f} to the reader, as it is
more or less  a kind of a repetition of that of Theorem~\ref{0main},
with the only difference being that now everything we need would
happen ``generically'', i.e., the set of suitable conditions is
dense, while for Theorem~\ref{0main} we had to ``manually''
guarantee all that by going over appropriate enumerations.

Instead, we shall address a similar poset tailored to analyze the
density one filter  $\mathcal Z^*$ on $\w$ consisting of those
$Z\subset\w$ such that $\lim_{n\to\infty}\frac{|Z\cap n|}{n}=1$. It
is a well known open problem   whether there exists a proper poset
adding no dominating reals but adding an infinite subset of $\w$
almost included into all ground model elements of $\mathcal Z^*$,
see, e.g., \cite[Question~2.12]{Hru11}. This  motivated us to
introduce the following  poset.

Let $\IQ$ be the set of conditions
$\la\mathsf F,\HH, \ve\ra$
such that
\begin{itemize}
\item[$(i)$] $\mathsf F$ is a countable collection of compact subsets of $[\w]^\w$ such that
$\bigcup\mathsf F$ is centered;
\item[$(ii)$] $\HH$ is a countable subset of $[\w]^\w$ and
$\ve:\HH\to (0,1]$; and
\item[$(iii)$] For every $F\in\la\sfF\ra$ and $H\in\HH$ there exists $X\in [\w]^\w$ such that
$\lim_{n\in X}\frac{|F\cap n|}{n}=1$
and $\liminf_{n\in X}\frac{|F\cap H\cap n|}{n}\geq\ve(H)$.
\end{itemize}
A condition $\la\mathsf F_1,\HH_1,\ve_1\ra$ is stronger than
$\la\mathsf F_0,\HH_0,\ve_0\ra$ (and written $\la\mathsf
F_1,\HH_1,\ve_1\ra\leq  \la\mathsf F_0,\HH_0,\ve_0\ra$) if
$\sfF_1\supset\sfF_0$,  $\HH_1\supset\HH_0$, and
$\ve_0=\ve_1\uhr\HH_0$. Clearly, $\IQ$ is countably closed.

\begin{theorem} \label{0main_density_z}
Let $G$ be a $\IQ$-generic filter and $\G=\bigcup\{ \bigcup \sfF\: :\: \exists \HH\exists\ve\:(\la\sfF,\HH,\ve\ra\in G)\}$.
Then $\G$ is a filter, $\mathcal Z^*\subset\G$, and $\G^+$ is Menger.
\end{theorem}
\begin{proof}
To see that $\G$ is a filter note that for any $\la\sfF,\HH,\ve\ra\in\IQ$ and finite $\mathcal F\subset\la \bigcup\sfF\ra$
we have $\la \sfF\cup\{ \{\bigcap \mathcal F\}\},\HH,\ve\ra\in \IQ$, and hence the set of all conditions
in $\IQ$ whose first component
contains  $\{\bigcap\mathcal F\}$ as an element, is dense below $\la\sfF,\HH,\ve\ra$.

The fact that $\Z^*\subset\G$, follows from the observation that for
any $T\in\mathcal Z^*$ and $\la\sfF,\HH,\ve\ra\in\IQ$, we have that
$\la\sfF\cup\{\{T\}\},\HH,\ve\ra\in\IQ$. Indeed, if for some
 $F\in\la\sfF\ra$, $H\in\HH$, and $X\in [\w]^\w$ we have
$\lim_{n\in X}|\frac{|F\cap n|}{n}|=1$
and $\liminf_{n\in X}|\frac{|F\cap H\cap n|}{n}|\geq\ve(H)$, then
also $\lim_{n\in X}|\frac{|(F\cap T)\cap n|}{n}|=1$
and $\liminf_{n\in X}|\frac{|(F\cap T)\cap H\cap n|}{n}|\geq\ve(H)$.

Thus we are left with the task of showing that for every sequence
$\la \K_n:n\in\w\ra$ of compact subspaces of $\G$ there exists an
increasing sequence $\la m_n:n\in\w\ra\in\w^\w$ such that for every
$\la K_n:n\in\w\ra\in\prod_{n\in\w}{\K_n}$ we have
$\bigcup_{n\in\w}(K_n\cap m_n)\in\G$. Let
$\la\sfF_0,\HH_0,\ve_0\ra\in G$ be such that
$\la\sfF_0,\HH_0,\ve_0\ra\forces\bigcup_{n\in\w}\K_n\subset\G$. We
claim that there exists $\la\sfF_1,\HH_1,\ve_1\ra\in G$ such that
$\bigcup_{n\in\w}\K_n\subset\la\bigcup\sfF_1\ra$. Indeed, given any
condition $\la\sfF ,\HH,\ve\ra\leq\la\sfF_0,\HH_0,\ve_0\ra$ two
cases are possible.

1. There exists $K\in \la\bigcup_{n\in\w}\K_n\ra$ such that $\w\setminus K\in\HH$,
or there exists  $\delta>0$ such that
  for every $F\in\la\sfF\ra$ there exists $X\in [\w]^\w$ with properties
$\lim_{n\in X}\frac{|F\cap n|}{n}=1$
and $\liminf_{n\in X}\frac{|F\cap (\w\setminus K)\cap n|}{n}\geq\delta$.
In this case
$ \la\sfF,\HH\cup\{\w\setminus K\},\ve'\ra\forces K\not\in\G$, where
$\ve'\uhr\HH=\ve$ and $\ve'(\w\setminus K)=\delta$ if $\w\setminus K\not\in\HH$. The latter leads to a contradiction.

2.  For every $K\in \la\bigcup_{n\in\w}\K_n\ra$ we have $\w\setminus K\not\in\HH$,
and for every  $\delta>0$ there exists
$F\in\la\sfF\ra$ such that for every $X\in [\w]^\w$ with $\lim_{n\in X}\frac{|F\cap n|}{n}=1$ the following holds:
\begin{equation} \label{eq11}
\forall^* n\in X\: \frac{|F\cap (\w\setminus K)\cap n|}{n}<\delta.
\end{equation}
Let us note that if for some $\delta>0$ an element $F\in\la\sfF\ra$ is a witness for Equation~\ref{eq11},
then any smaller $F'\in\la\sfF\ra$ is also one.
Given any $F\in\la\sfF\ra$ and $K\in\la\bigcup_{n\in\w}\K_n\ra$, let us construct a
decreasing sequence $\la F_i:i\geq 1\ra$ of elements of
$\la\sfF\ra$ such that $F_0=F$ and
\begin{equation} \label{eq12}
\forall X\in [\w]^\w\: \Big(\lim_{n\in X}\frac{|F_i\cap n|}{n}=1 \Rightarrow
\forall^* n\in X\: \frac{|F_i\cap (\w\setminus K)\cap n|}{n}<\frac{1}{i}\Big).
\end{equation}
Equation~\ref{eq12} implies
\begin{equation} \label{eq13}
\forall X\in [\w]^\w\: \Big(\lim_{n\in X}\frac{|F_i\cap n|}{n}=1 \Rightarrow
\forall^* n\in X\: \frac{|F_i\cap K\cap n|}{n}\geq 1-\frac{2}{i}\Big).
\end{equation}
Let us fix now any $H\in\HH$ and
 for every $i$ find $X_i\in [\w]^\w$
such that $\lim_{n\in X_i}\frac{|F_i\cap n|}{n}=1$
and $\frac{|F_i\cap H\cap n|}{n}>\ve(H)-\frac{1}{i}$ for all $n\in X_i$.
This is possible by item $(iii)$ of the definition of $\IQ$.
Removing finitely many elements of $X_i$, if necessary, by Equation~\ref{eq13} we may assume that
$\frac{|F_i\cap K\cap  n|}{n}>1-\frac{2}{i}$  for all $n\in X_i$.
Now let $\la n_i:i\in\w\ra\in\w^\w$ be an increasing sequence
such that $n_i\in X_i$ and $X(F,H,K)=\{n_i:i\in\w\}$.
It follows that
$\frac{|F_i\cap H\cap n_i|}{n_i}>\ve(H)-\frac{1}{i}$ and $\frac{|F_i\cap K\cap  n_i|}{n_i}>2-1/i$  for all $i\in \w$.
Since $F_i\subset F$ for all $i$, we have that
$\frac{|F\cap H\cap n_i|}{n_i}>\ve(H)-\frac{1}{i}$ and $\frac{|F\cap K\cap  n_i|}{n_i}>1-2/i$
and therefore $\frac{F\cap K\cap H\cap n_i}{n_i}>\ve(H)-\frac{3}{i}$
  for all $i\in \w$.
Thus $\la\sfF\cup\{\K_n:n\in\w\},\HH,\ve\ra\in \IQ$, where for each
$F\in\la \sfF\ra$, $H\in\HH$, and $K\in\la\bigcup_{n\in\w\K_n}\ra$ the infinite set $X(F,H,K)$
is such as required in $(iii)$ for $F\cap K$ and $H$.

Summarizing, we have proved that for any
$\la\sfF ,\HH,\ve\ra\leq\la\sfF_0,\HH_0,\ve_0\ra$
we have $\la\sfF\cup\{\K_n:n\in\w\},\HH,\ve\ra\in \IQ$, and thus there exists
$\la\sfF_1,\HH_1,\ve_1\ra\in G$ with $\bigcup_{n\in\w}\K_n\subset\la\bigcup\sfF_1\ra$.

Let us now fix $\la\sfF ,\HH,\ve\ra\leq\la\sfF_1,\HH_1,\ve_1\ra$ and  enumerations
$\la H_i:i\in\w \ra$ of $\HH$ as well as $\la\F_i:i\in\w\ra$ of $\sfF$.
Set
$$\mathcal L_i=\big\{\bigcap\mathcal Y\: :\: \mathcal Y\in [\bigcup_{j\leq i}\F_j\cup\bigcup_{j\leq i}\K_j]^{\leq i}\big\}$$
and by recursion over $i$ construct an increasing number sequence
$\la m_i:i\in\w\ra$ such that for every $L\in\mathcal L_i$ and $j\leq i$ there exists
$n_j\in [m_{i-1},m_{i})$ such that
$\frac{|L\cap n_j|}{n_j}>1-\frac{1}{i}$ and $\frac{|L\cap H_j\cap n_j|}{n_j}>\ve(H_j)(1-\frac{1}{i})$.
This is possible by $(iii)$ and the compactness of $\mathcal L_i$.
Letting
$$\K=\Big\{\bigcup_{i\in\w}(K_i\cap m_i)\: :\:
\la K_i:i\in\w\ra\in\prod_{i\in\w}{\K_i}\Big\},$$
 we claim that
 $\la \sfF\cup\{\K\},\HH,\ve\ra\in\IQ$. Indeed, let us fix $k\in\w$ and a family
 $\{Y_s:s\leq k\}\subset\K$, where
 $Y_s=\bigcup_{i\in\w}(K_{i,s}\cap m_i)$ for some $K_{i,s}\in\K_i$.
 Also, let us fix a family  $\{R_s:s\leq k\}\subset\bigcup_{s\leq k}\F_s$
 and set $R=\bigcap_{s\leq k}R_s$. Then for every
 $i\geq 2k+1$ and $s\leq k$ we  have that
 $$\mathcal Y:=\{R_s:s\leq k\}\cup\{K_{i,s}:s\leq k\}\in [\bigcup_{j\leq i}\F_j\cup\bigcup_{j\leq i}\K_j]^{\leq i},$$
and therefore
for every and $j\leq i$ there exists
$n_j\in [m_{i-1},m_{i})$ such that
$\frac{|\bigcap\mathcal Y\cap n_j|}{n_j}>1-\frac{1}{i}$ and $\frac{|\mathcal Y\cap H_j\cap n_j|}{n_j}>\ve(H_j)(1-\frac{1}{i})$.
Since
\begin{equation} \label{eq_14}
 \bigcap\mathcal Y\cap n_j\subset\bigcap\{R_s:s\leq k\}\cap \bigcap\{Y_s:s\leq k\} \cap n_j,
 \end{equation}
(because $K_{i,s}\cap m_i\subset Y_s\cap m_i$ for all $s\leq k$), we
conclude that $(iii)$ is satisfied for $\la
\sfF\cup\{\K\},\HH,\ve\ra$, and hence it is a member of $\IQ$. Thus
for arbitrary $\la\sfF ,\HH,\ve\ra\leq\la\sfF_1,\HH_1,\ve_1\ra$ we
have $\la\sfF\cup\{\K\} ,\HH,\ve\ra\in\IQ$, and hence there exists
$\la\sfF_2 ,\HH_2,\ve_2\ra\in G$ with $\K\in\sfF_2$, which yields
$\K\subset\G$. It remains to apply Theorem~\ref{basics}.
\end{proof}

It is well-known and easy to see that the filter $\mathcal Z^*$ cannot be extended to any $P^+$-filter, and hence
$\G$ from Theorem~\ref{0main_density_z} is not Menger by Corollary~\ref{obs_obv}. In addition,
the Menger co-ideal
$\G^+$  does not contain any $P$-point, and hence also no Menger ultrafilter.
We do not know whether such examples can be obtained in ZFC, see Section~\ref{problem}
for more  questions related to $\mathcal Z$.

\section{Impossibility results} \label{imp_res}

Here we show that Theorem~\ref{0main} cannot be proved without additional set-theoretic assumptions,
see Theorem~\ref{0_anti_main} below. Following \cite{BelTokZdo16}
for a semifilter $\F$ we denote by $\IP_\F$ the poset consisting
of
all partial maps $p$ from $\w\times\w$ to $2$ such that for every
$n\in\w$ the domain of $p_n:k\mapsto p(n,k)$ is an element of
$\sim\F=\{\w\setminus F:F\in\F\}$. If, moreover, we assume that additionally
$\mathrm{dom}(p_n)\subset
\mathrm{dom}(p_{n+1})$ for all $n$, the corresponding poset will
be
denoted by $\IP_\F^*$. A condition $q$ is stronger than $p$ (in
this
case we write $q\leq p$) if $p\subset q$. For filters $\F$ the
poset
$\IP_\F^*$ is obviously dense in $\IP_\F$, and the latter is
proper
and $\w^\w$-bounding if $\F$ is a non-meager $P$-filter
\cite[Fact
VI.4.3, Lemma~VI.4.4]{She_propimp}.
This result has the following topological counterpart
proved  in \cite{BelTokZdo16}. Recall that a poset $\IP$ is called
\emph{$\w^\w$-bounding} if $(\w^\w)^V$ is dominating in $V^{\IP}$.

\begin{lemma} \label{w^w-bound}
If $\F^+$ is a Menger semifilter, then both  $\IP_\F$ and
$\IP_\F^*$
are proper and $\w^\w$-bounding.
\end{lemma}

We shall need the following game of length $\omega$ on a topological space $X$:
 In the $n$th  move player $I$ chooses an  open cover
$\U_n$ of $X$, and player $\mathit{II}$ responds by choosing a
finite $\V_n\subset \U_n$. Player $\mathit{II}$ wins the game if $\bigcup_{n\in\omega}
\bigcup\V_n =X$. Otherwise, player $I$ wins.  We shall call this game
 \emph{the Menger game} on $X$.
 It is well-known  that
  $X$  is Menger  if and only if
 player $I$ has no winning strategy in the Menger game on
$X$, see \cite{Hur25} or \cite[Theorem~13]{COC1}.

For a relation $R$ on $\w$ and $x,y\in\w^\w$ we denote by
$[x\:R\:y]$ the set $\{n: x(n)\:R\:y(n)\}$.
The next lemma improves \cite[Lemma~4.3]{BelTokZdo16}.

\begin{lemma} \label{meng_forcing}
Suppose that $\F$ is a semifilter  such that $\F\subset\F^+$ and $\F^+$ is Menger.
 Let
$x$ be   $\IP_\F^*$-generic, $\IQ\in V[x]$ be an
$\w^\w$-bounding poset, and $H$  a $\IQ$-generic over $V[x]$.
Then
in $V[x*H]$ there is no  Menger semifilter $\G$ containing $\F$
such that $\G\subset\G^+$ and $\G^+$ is Menger.
\end{lemma}
\begin{proof}
Suppose that a semifilter $\G\in V[x*H]$ as above exists. Throughout
the proof we shall identify $x$ with $\bigcup x:\w\times\w\to 2$.
Suppose to the contrary, that such a $\G$ exists. Set
$x_j(n)=x(j,n)$. Since $\G\subset\G^+$, $\G$ cannot contain two
disjoint elements, and hence for every $j$ there exists $\ve_j\in 2
$ such that $X_j:=x_j^{-1}(\ve_j)\in\G^+$. Without loss of
generality we may assume that there exists an infinite $T\subset\w$
such that $\ve_j=1$ for all $j\in T$. Since $\IP_\F^* *\IQ$ is
$\w^\w$-bounding we can get an increasing sequence  $\la j_k:k\in\w
\ra\in V$  such that for every $k\in\w$ there exists $t_k\in
[j_k,j_{k+1})\cap T$.

Given $k\in\w$,  for every   $s\in\w^{j_{k+1}}$ set
$$\U_{k,s}=\{X\subset\w\: :\: s[j_{k+1}]\subset X \}.$$
Then $\{\U_{k,s}:s\in\Sigma_{k,l}\}$, where
$\Sigma_{k,l}=\prod_{j<j_{k+1}}(X_j\setminus l)$,
 is an open cover of
$\G$ for every $k,l\in\w$, because each  $Y\in \G$ has infinite
intersection with each $X_j$. We shall define
 a strategy $\Theta$ for  player $I$ in the Menger game on $\G$ as follows.
 $\Theta(\emptyset)=\{\U_{0,s}:s\in \Sigma_{0,0}\}$.
 If $\mathit{II}$ replied with some
 finite subset $\mathcal V_0$ of  $\Theta(\emptyset)$, $I$ finds $h(0)\in\w$ such that
 this reply is contained in $\{\U_{0,s}:s\in h(0)^{j_1}\cap \Sigma_{0,0}\}$ and
 his next move is then
$\Theta\la\mathcal V_0\ra:= \{\U_{1,s}:s\in \Sigma_{1, h(0)}\}.$
And so on, i.e., after $k$ many rounds
player $I$ finds $h(k)\in\w$ above $h(k-1)$ (here $h(-1):=0$) such that the reply of $\mathit{II}$
is contained in $\{\U_{k,s}:s\in h(k)^{j_{k+1}}\cap \Sigma_{k,h(k-1)}\}$ and
 his next move is then
$$\Theta\la\mathcal V_0,\ldots,\mathcal V_k\ra:= \{\U_{k+1,s}:s\in \Sigma_{k+1,h(k)}\}.$$
Since $\G$ is Menger,
 $\Theta$ is not winning, and hence there exists
an increasing sequence $h\in\w^\w\in V[x*H]$
such that
\begin{equation} \label{eq_new_1}
 \G\subset\bigcup\big\{\bigcup\{\U_{k,s}: s\in h(k)^{j_{k+1}}\cap \Sigma_{k,h(k-1)} \} \: :\: k\in\w           \big\}.
 \end{equation}
 Next, we shall define
 a strategy $\Upsilon$ for  player $I$ in the Menger game on $\G^+$ as follows.
 $\Upsilon(\emptyset)=\{\U_{0,s}:s\in S_{0,0}\}$, where
 $S_{k,l}=\prod_{j<j_{k+1}}\big(\w\setminus(\mathrm{dom}(p_j)\cup l)\big)$ for all $k,l\in\w$.
 If $\mathit{II}$ replied with some
 finite subset $\mathcal V_0$ of  $\Upsilon(\emptyset)$, $I$ finds $l_0\in\w$ such that
 this reply is contained in $\{\U_{0,s}:s\in h(l_0)^{j_1}\}$ and
 his next move is then
$\Upsilon(\mathcal V_0):= \{\U_{1,s}:s\in S_{1,h(l_0)}\}.$
And so on, i.e., after $k$ many rounds
player $I$ finds $l_k\in\w$ above $l_{k-1}$ such that the reply of $\mathit{II}$
is contained in $\{\U_{k,s}:s\in h_1(l_k)^{j_{k+1}}\}$ and
 his next move is then
$$\Upsilon\la\mathcal V_0,\ldots,\mathcal V_k\ra:= \{\U_{k+1,s}:s\in S_{k+1,h(l_k)}\}.$$
Since $\Upsilon$ is not winning, there exists
an increasing sequence $\la l_k:k\in\w\ra$ (we set also $l_{-1}=0$ for convenience)
such that
\begin{equation} \label{eq_new_2}
 \G^+\subset\bigcup_{k\in\w}\bigcup\big\{\U_{k,s}: s\in h(l_k)^{j_{k+1}}\cap S_{k, h(l_{k-1})}\big\}.
 \end{equation}
Let $h_1\in V\cap \w^\w$ be an increasing function such that $h_1(k)\geq h(k)$ for all $k\in\w$,
and
 $\la p,\dot{q}\ra\in x*H$  a condition forcing all the above to happen.
 Let also $\name{h},$ $\name{t}_k,$ $\name{\G},$ $\name{\Sigma}_{k,l},$ $\name{x}$, $\name{x}_j$,
 $\name{X}_j$, $\name{T}$, ... be
 $\IP_{\F}^* *\IQ$-names of the objects in $V[x*H]$ considered above.
Consider the condition $p^1\in\IP_{\F}^*$ below $p$ defined as follows:
$$ p^1_j=p_j\cup\{\la n,0\ra: n\in h_1(l_k)\setminus \mathrm{dom}(p_j)\}$$
whenever $j\in [j_k,j_{k+1})$ and $k\in\w$. Thus $\la
p^1,\name{q}\ra$ forces Equations~\ref{eq_new_1} and \ref{eq_new_2}.
Thus $\la p^1,\name{q}\ra$ forces the following: If for every
$k\in\w$ and $s\in \name{h}(k)^{j_{k+1}}\cap
\Sigma_{k,\name{h}(k-1)}$ we pick $j_s\in j_{k+1}$, then
\begin{equation} \label{eq_new_3}
\big\{ s(j_s)\: :\: s\in \name{h}(k)^{j_{k+1}}\cap \Sigma_{k,\name{h}(k-1)}, k\in\w\big\}\in\name{\G}^+; \mbox{\ and}
\end{equation}
if for every $k\in\w$ and $s\in  \name{h}(\name{l}_k)^{j_{k+1}}\cap \name{S}_{k, \name{h}(\name{l}_{k-1})}$ we pick
$j'_s\in j_{k+1}$, then
\begin{equation} \label{eq_new_4}
\big\{ s(j'_s)\: :\: s\in \name{h}(\name{l}_k)^{j_{k+1}}\cap \name{S}_{k, \name{h}(\name{l}_{k-1})}, k\in\w\big\}\in\name{\G}.
\end{equation}
Given a $\IP_\F^* *\IQ$-generic filter $ x^1*H^1$  containing $\la
p^1,\name{q}\ra$, we shall work in $V[x^1*H^1]$ in what follows. For
abuse of notation we shall again use notations for the evaluations
with respect to $x^1*H^1$ of all names considered above, obtained
simply by removing ``$\name{\ }$'', i.e., $h:=\name{h}^{x^1*H^1}$
etc. This way letting $j'_{s}:=t_k$ for each   $s\in
h(l_k)^{j_{k+1}}\cap S_{k, h(l_{k-1})}$, the set $B'$
 defined in
Equation~\ref{eq_new_4} belongs to $\G$. For every $k\in\w$ and
$s\in h(k)^{j_{k+1}}\cap \Sigma_{k,h(k-1)}$ set $j_s:=t_{m(k)}$,
where $m(k)=\min\{m:l_m\geq k\}$. The set $B$
 defined in
Equation~\ref{eq_new_3} for this choice of $j_s$'s belongs to
$\G^+$, and hence $B\cap B'\neq\emptyset$. Thus there exist
$k,k'\in\w$, $s'\in h(l_{k'})^{j_{k'+1}}\cap S_{k', h(l_{k'-1})}$,
and $s\in h(k)^{j_{k+1}}\cap \Sigma_{k,h(k-1)}$ such that
\begin{equation}\label{eq_new_5}
s(t_{m(k)})=s'(t_{k'}).
\end{equation}
Since $s(t_{m(k)})\in h(k)\setminus h(k-1)$ and $s'(t_{k'})\in
h(l_{k'})\setminus h(l_{k'-1})$, from Equation~\ref{eq_new_5} we
conclude that $h(k)\setminus h(k-1)\subset h(l_{k'})\setminus
h(l_{k'-1})$, and hence $k'=\min\{m:l_m\geq k\}$, which yields
$k'=m(k)$. Recall that $s(t_{m(k)})\in X_{t_{m(k)}}$, and therefore
\begin{equation}\label{eq_new_6}
x_{t_{m(k)}}\big(s(t_{m(k)})\big)=\varepsilon_{t_{m(k)}}=1.
\end{equation}
On the other hand,
\begin{equation}\label{eq_new_7}
x_{t_{m(k)}}\big(s(t_{m(k)})\big)=
x_{t_{m(k)}}\big(s'(t_{m(k)})\big)=
p^1_{t_{m(k)}}\big(s'(t_{m(k)})\big)
\end{equation}
 because $p^1_{t_{m(k)}}\subset
x_{t_{m(k)}}$ and
$$s'(t_{m(k)})\in h(l_{m(k)})\subset
h_1(l_{m(k)})\subset\mathrm{dom}(p^1_{t_{m(k)}}),$$ the last
inclusion following from $t_{m(k)}\in [j_{m(k)}, j_{m(k)+1})$ and
the definition of $p^1$. Recall that
$s'(t_{m(k)})\not\in\mathrm{dom}(p_{t_{m(k)}})$ by the definition of
$S_{k',h(l_{k'-1})}=S_{m(k),h(l_{m(k)-1})}$, which implies that
$p^1_{t_{m(k)}}\big(s'(t_{m(k)})\big)=0$ by the definition of $p^1$.
Thus
 $x_{t_{m(k)}}\big(s(t_{m(k)})\big)=
x_{t_{m(k)}}\big(s'(t_{m(k)})\big)=0$ by Equation~\ref{eq_new_7},
which is impossible by Equation~\ref{eq_new_6}. This contradiction
completes our proof.
 \end{proof}
The next theorem improves \cite[Theorem~4.5]{BelTokZdo16} and is
the main result of this section.
\begin{theorem} \label{0_anti_main}
It is consistent that there is no semifilter $\G$ on $\w$ such that
$\G\subset\G^+$ and both $\G,\G^+$ are Menger.
\end{theorem}
\begin{proof}
Let us assume that $V=L$  and\footnote{It suffices to assume
$2^\w=\w_1$, $2^{\w_1}=\w_2$, and
$\diamond_{\{\delta\in\w_2:\mathrm{cf}(\delta)=\w_1\}}$, see
\cite[Theorem~5.13]{She_propimp}.} consider a function
$B:\w_2\to H(\w_2)$, the family of all sets whose transitive closure
has size $<\w_2$. Let $\la
\IP_\alpha,\name{\IQ}_\beta:\beta<\alpha\leq\w_2\ra$ be the
following  iteration with at most countable supports: If $B(\alpha)$
is a $\IP_\alpha$-name for $\IP^*_{\name{\F}}$ for some semifilter
$\name{\F}$ such that $\forces_{\IP_\alpha}$
``$\name{\F}\subset\name{\F}^+$ and $\name{\F},\name{\F}^+$ are
Menger'', then $\name{\IQ}_\alpha=B(\alpha)$. Otherwise we let
$\name{\IQ}_\alpha$  be a $\IP_\alpha$-name for the trivial forcing.
Then $\IP_{\w_2}$ is $\w^\w$-bounding forcing notion with
$\w_2$-c.c. being a countable support iteration of length $\w_2$ of
proper $\w^\w$-bounding posets of size $\w_1$  over a model of GCH.
Now, using the suitable diamond in $V$ for the choice of $B$
together with a standard reflection argument, we can guarantee in
addition that for any $\IP_{\w_2}$-generic filter $G$ over $V$ and
semifilter $\F\in V[G]$ such that $\F\subset\F^+$ and $\F,\F^+ $ are
Menger, the following holds:
\begin{quote}
The set $\big\{\alpha:\F_\alpha:=(\F\cap V[G\cap\IP_\alpha])\in
V[G\cap\IP_\alpha], $ $\F_\alpha\subset\F^+_\alpha$,
$\F_\alpha,\F^+_\alpha$ are Menger in $V[G\cap\IP_\alpha],$ and
$\name{\IQ}_\alpha^{G\cap\IP_\alpha}=\IP^*_{\name{\F}_\alpha}\big\}$
is stationary in $\w_2$.
\end{quote}
 Now, a direct application of
Lemma~\ref{meng_forcing} implies that $\F_\alpha$ cannot be enlarged
to any semifilter $\U\subset\U^+$ in $V[G]$ such that $\U,\U^+$ are
Menger,  which contradicts the fact that  $\F$ is such an
enlargement.
\end{proof}

\section{Open questions} \label{problem}

Theorem~\ref{0main} together with Corollaries~\ref{cor01} and \ref{cor02} motivate the following

\begin{question}\,
\begin{itemize}
\item Is there a ZFC example of a non-meager filter $\F$ such that $\F^+$ is not homeomorphic to
$(\F^+)^2$?
\item Is $\F\times\F^+$ homeomorphic to $\F^+$ for every non-meager filter $\F$?
\end{itemize}
\end{question}

In light of Theorem~\ref{0main_density_z} it is natural to ask the next question.
Let us recall from \cite{ChoRepZdo15} that for a filter $\F$ the Mathias forcing associated to it
adds no dominating reals iff $\F$ is Menger, so the following question is
especially interesting when $\F$ is not an ultrafilter.

\begin{question}
Let $\F$ be a filter such that $\F^+$ is Menger. Is there a (proper) poset
adding no dominating reals and adding an infinite pseudointersection of $\F$?
What about the Mathias forcing associated to $\F^+$?
\end{question}

Theorem~\ref{0_anti_main} leaves open one possibility whose inconsistency we are unable to establish.

\begin{question}\label{main_q}\,
\begin{itemize}
\item Is there a ZFC example of a filter $\F$ such that $\F^+$ is Menger?
\item Is there a ZFC example of a semifilter $\F\subset\F^+$ such that $\F^+$ is Menger?
\end{itemize}
\end{question}

The first item of the question above  has been mentioned to us by Miko\l aj Krupski in private communication and
is related to his work \cite{Kru??}.
By \cite[Lemma~3.1]{BelTokZdo16} the negative answer to the second item of Question~\ref{main_q} implies that consistently
every  Menger remainder of a topological group is $\sigma$-compact.

redpajama_set_name: RedPajamaArXiv


text:
Heating and plumbing giant cuts lighting energy use by 80%
A leading supplier of heating and plumbing products, has achieved savings of £116,000 on its annual energy bills at its Cumbrian distribution centre, as a result of an installation of LED lamps controlled by high-bay motion detectors from Steinel (UK) Ltd
The lighting at Wolseley UK's sites was identified as a significant source of energy waste. Not only were its fluorescent lamps and light fittings outdated and inefficient, the lights also blazed 24 hours a day.
Using sensor control seemed like the obvious solution, providing an automatic means to switch on the lights only in the areas that people were actually using – and switch them off again when that part of the building was empty. However, Wolseley had installed sensors in previous projects and found them not fit for purpose. Part of the problem was that the distribution centre's high ceilings – 15m in some areas – were too much of a challenge for many types of sensors.
Rather than abandoning the idea of sensor control altogether, Wolseley brought in a leading supplier of LED lighting solutions, and ran a small pilot scheme in one of the loading areas at Wolseley's Melmerby site in Cumbria.
For the pilot, they used Steinel's highly-sensitive IS345 MX infrared motion detectors and not only established that the sensors were fit for purpose, but that it was also possible to make accurate payback projections. This meant that Wolseley was able to mobilise capital expenditure for a full installation of sensors.
Following the successful trial, QE Global completely overhauled the lighting at Wolseley's Melmerby site, with the installation taking place over a seven-week period. All of the fluorescent fittings and high- and low-bay floodlights were replaced with LEDs, with 349 Steinel IS345 MX sensors installed to ensure intelligent control of the new lighting.
The Steinel IS 345 MX infrared motion detectors have been designed specifically for depot and warehouse environments with high ceilings. With a radial detection zone of 30 x 4 m, the IS 345 MX sensors are extremely sensitive – ensuring no one is ever left in the dark.
The energy savings resulting from the project have been dramatic, with Wolseley saving an estimated £116,000 on its energy bills every year. The lights have gone from being in use 24 hours a day to an average of 2.5 hours a day over a 5-day week. QE Global estimates the site is now using 80% less energy for its lighting – providing a huge contribution towards Wolseley's energy-saving targets.
Steinel's high-bay motion detectors
Steinel UK Ltd
Staggering savings at Model Train warehouse
University reaps energy and maintenance savings with LED road lighting
Retail lighting offered versatility with LED spot lights
Get ready for the revolutionary lamp
New guidance to assist lighting upgrades in factories

redpajama_set_name: RedPajamaCommonCrawl


Description: A rare opportunity to own farm land that has been improved for cattle and crops with t-post and barbed wire fencing. Also has a large grove of mature pecan trees and some wooded areas for hunting. Many acres of Bermuda grass improvement. Has a beautiful area for building site. Very close to Highway 66.


text:
Purple pain: why the deaths of Prince and Bowie cut us so deep
Article by Jill Stark /
Can we press pause on 2016? The ground feels unstable.
Only four months in and the losses continue to mount.
The deaths of music giants David Bowie and Prince have marked an ominous start to a year that has also seen the passing of esteemed actor Alan Rickman, Pulitzer Prize-winning novelist Harper Lee and renowned singer Natalie Cole.
Add to this the loss of Motorhead frontman Lemmy in December, and we are left to ponder whether God is recruiting for the mother of all music festivals.
This loss of iconic artists who have helped shape our world – and for some our sense of self – has led to an outpouring of global grief.
But the reaction to these deaths stretches beyond our culture's obsession with celebrity. It speaks to something more basic, even primal.
"What these deaths do is punctuate our sense of mortality," says Chris Hall, head of the Australian Centre for Grief and Bereavement. "Death is the great leveller whether one is famous or not – it reminds us as human beings that we'll die and that fame or riches isn't any protection against that."
When news of Prince's sudden death came through on April 21, fans took to social media to share their shock and desolation. It was simply too much to comprehend so soon after the passing of Bowie.
Dressed in purple, ABC News Breakfast presenter Virginia Trioli cried on air as she reported the news. In a video that went viral, a high school teacher was comforted by his students as he broke down watching Purple Rain.
Pierz Newton-John, a faculty member at Melbourne's School of Life, who teaches classes on how to face death, said this raw grief for a performer they had likely never met would have complex undertones.
"It's often not really about the celebrity but the revivifying of other griefs that the person has experienced. It becomes an opportunity in a society where we don't have much opportunity for public expression of grief, to mourn," he said.
"In our culture we have a very brief window to grieve in – the funeral and maybe a couple of weeks – and then you're expected to get on with it. We don't have black armbands or black mourning garb or some sort of signifier that we're in mourning, so a lot of people have unresolved grief."
Newton-John points out that grief triggered by the passing of a celebrity is not necessarily related to a previous death but can be an emotional response to a broader sense of loss – a longing for youth, the end of a relationship or mourning lapsed friendships.
Music's connection to a certain time and place in our lives often evokes strong memories and emotions. When a performer from that era dies, the reaction can be visceral.
"I encourage people to delve into what is it that they're really grieving at a deeper level and understand that grief is part of living. The very fact of loving means that we're exposed to the necessity of grieving, and the only way not to grieve is not to love. We need to really embrace our sadness and not suppress or reject it."
The public way we live our lives has caused a seismic shift in how the community mourns, with social media at the centre of the change.
Many celebrities are active on Instagram, Facebook, Twitter and Snapchat and are inserted into our personal news feeds, creating a sense of connection that makes a sudden loss even more profound.
Indeed, the confirmation of Bowie's death came from his son via Twitter, giving us direct and instant access to the family's pain.
Dr Margaret Gibson, an expert in digital mourning and commemorative culture, from Griffiths University, said the collective nature of grief online can impact on how keenly we feel the loss.
"We can see ourselves mourning with others now in ways we couldn't in the past. Social media creates a form of intimacy and a sense of ownership. It's coming through all our personal networks and people are talking about it. There's this sense that there is literally a proximity to our own personal lives that creates this direct hit to our self."
But perhaps the strong response to these recent deaths can be partly explained by their frequency.
The BBC's obituary editor noted there had been a sharp increase in significant deaths in 2016 – 24 from January to March alone compared with just five in the same time period last year. He puts this down to the ageing Baby Boomers who make up a significant proportion of the population.
However,Time magazine contradicted this, saying their archive showed the number of entertainer deaths was 21 in the first three months of the year compared with 24 last year.
Perhaps the discrepancy reflects different concepts of what constitutes celebrity. Could it be that at a time when anyone with a smartphone can be rocketed to fame we mourn more acutely those who we deem truly worthy of the title "legend"?
The internet has been abuzz with memes telling us to put Springsteen under 24-hour police protection, while lamenting the robust state of Kanye or Bieber's health.
Pop culture expert Lauren Rosewarne from the University of Melbourne said duration of career should not determine our level of respect.
"Someone like Justin Bieber has always been easy pickings because he's considered youth culture, but we have this amazing ability to forget that the Beatles were also youth culture. It's just that the passing of time has given the Beatles legitimacy in the way that Justin Bieber doesn't have."
Patrick Donovan, chief executive of Music Victoria, said the loss of Prince has been felt deeply in music-mad Melbourne, where the singer played one of his last shows in an intimate gig in February. Likewise, when Bowie died the grief was immense, with fans converging that night on rock venue Cherry Bar.
"These singers provide the soundtrack to our lives. All through the different stages in your life these artists are there for you and you change as they change. They are so important to us that you want to go and spend that time grieving with other fans. It's like we knew these people."


text:
WALES – (18) Cert – Student's naked Twitter video shame
UK – Which Benefits are Changing and When
UK – Iain Duncan Smith Firm On Planned CSA Charges, Saying They're 'Better For The Children'
Filed under (DWP) Child Maintenance
Iain Duncan Smith Insisted His Planned CSA Charges Would Be 'Better For The Children'
Iain Duncan Smith has pledged to overturn the House of Lords' amendments to his Welfare Reform Bill, including the vote to cancel charges for single parents to use the Child Support Agency (CSA).
On Wednesday peers delivered their largest-ever defeat on the coalition by voting by a majority of 142 to cancel plans to introduce CSA charges. The amendment was tabled by Lord Mackay, a Conservative peer, and backed by many former Tory ministers including Lord Howe and Lord Lawson.
But the work and pensions secretary told Andrew Marr on Sunday morning that the current system led to parents being "almost forced to be at each other's throats," arguing that the charges would lead to disputes over child maintenance being settled out of courts more often.
"If you have a relatively small charge, what happens is that people will think about it," said Duncan Smith, insisting that the proposals would be "better for the children."
On Wednesday MPs will consider all the amendments made by the Lords to the Welfare Reform Bill, and the secretary of state confirmed that coalition MPs would be whipped into overturning them. Most of the amendments were passed by relatively small margins, but the CSA amendment was overwhelmingly passed. This raises the prospect of "parliamentary ping-pong" taking place between the two houses in the coming days.
On the CSA charges, IDS claimed: "We want to make sure that this is a balance. We're not asking for much, It's only about 9%. We're going to pick up the lion share of the cost."
He was particularly bullish on the planned benefit cap of £26,000 per household. An amendment by peers would make child benefit exempt from this calculation, but IDS insisted it would also be overturned.
"Even among Labour and Liberal voters… it's overwhelmingly popular," he said. "Why should someone on benefits earn more than the average wage?
"We have people living in London that cost over £100,000 a year to rent… that's the kind of nonsense that we've got into."
The government has a majority of more than 80 in the House of Commons, but lacks a concrete majority in the Lords. In the unelected chamber it requires the support of at least some of the independent cross-bench MPs.
Nick Clegg wants to replace the Lords with a mostly elected chamber, although the Bishops which introduced one of the amendements last week would be retained under the deputy prime minister's draft proposals.
If the Commons and the Lords cannot agree, there will be a state of "ping-pong" when the Lords will be urged to accept the will of the elected lower chamber. The government may use the Parliament Act – the equivalent of a veto by the Commons, if there is no agreement, but that could delay Iain Duncan Smith's welfare reforms for months.
Huffington Post UK
Permanent link to this article: https://operationfatherhood.org/dwp-department-for-work-and-pensions/child-maintenance/uk-iain-duncan-smith-firm-on-planned-csa-charges-saying-theyre-better-for-the-children/


text:
Q: How to read a super huge file into numpy array N lines at a time I have a huge file (around 30GB), each line includes coordination of a point on a 2D surface. I need to load the file into Numpy array: points = np.empty((0, 2)), and apply scipy.spatial.ConvexHull over it. Since the size of the file is very large I couldn't load it at once into the memory, I want to load it as batch of N lines and apply scipy.spatial.ConvexHull on the small part and then load the next N rows! What's an efficient to do it?
I found out that in python you can use islice to read N lines of a file but the problem is lines_gen is a generator object, which gives you each line of the file and should be used in a loop, so I am not sure how can I convert the lines_gen into Numpy array in an efficient way?
from itertools import islice
with open(input, 'r') as infile:
    lines_gen = islice(infile, N)

My input file:
0.989703    1
0   0
0.0102975   0
0.0102975   0
1   1
0.989703    1
1   1
0   0
0.0102975   0
0.989703    1
0.979405    1
0   0
0.020595    0
0.020595    0
1   1
0.979405    1
1   1
0   0
0.020595    0
0.979405    1
0.969108    1
...
...
...
0   0
0.0308924   0
0.0308924   0
1   1
0.969108    1
1   1
0   0
0.0308924   0
0.969108    1
0.95881 1
0   0

A: With your data, I can read it in 5 line chunks like this:
In [182]: from itertools import islice
with open(input,'r') as infile:
    while True:
        gen = islice(infile,N)
        arr = np.genfromtxt(gen, dtype=None)
        print arr
        if arr.shape[0]<N:
            break
   .....:
[(0.989703, 1) (0.0, 0) (0.0102975, 0) (0.0102975, 0) (1.0, 1)]
[(0.989703, 1) (1.0, 1) (0.0, 0) (0.0102975, 0) (0.989703, 1)]
[(0.979405, 1) (0.0, 0) (0.020595, 0) (0.020595, 0) (1.0, 1)]
[(0.979405, 1) (1.0, 1) (0.0, 0) (0.020595, 0) (0.979405, 1)]
[(0.969108, 1) (0.0, 0) (0.0308924, 0) (0.0308924, 0) (1.0, 1)]
[(0.969108, 1) (1.0, 1) (0.0, 0) (0.0308924, 0) (0.969108, 1)]
[(0.95881, 1) (0.0, 0)]

The same thing read as one chunk is:
In [183]: with open(input,'r') as infile:
    arr = np.genfromtxt(infile, dtype=None)
   .....:
In [184]: arr
Out[184]:
array([(0.989703, 1), (0.0, 0), (0.0102975, 0), (0.0102975, 0), (1.0, 1),
       (0.989703, 1), (1.0, 1), (0.0, 0), (0.0102975, 0), (0.989703, 1),
       (0.979405, 1), (0.0, 0), (0.020595, 0), (0.020595, 0), (1.0, 1),
       (0.979405, 1), (1.0, 1), (0.0, 0), (0.020595, 0), (0.979405, 1),
       (0.969108, 1), (0.0, 0), (0.0308924, 0), (0.0308924, 0), (1.0, 1),
       (0.969108, 1), (1.0, 1), (0.0, 0), (0.0308924, 0), (0.969108, 1),
       (0.95881, 1), (0.0, 0)],
      dtype=[('f0', '<f8'), ('f1', '<i4')])

(This is in Python 2.7; in 3 there's a byte/string issue I need to work around).

A: You could try the second method from this post and read the file in chunks by referring to a given line using a pre-computed lines offset array if it fits into memory. Here is an example of what I typically use to avoid loading whole files to memory::
data_file = open("data_file.txt", "rb")

line_offset = []
offset = 0

while 1:
    lines = data_file.readlines(100000)
    if not lines:
        break

    for line in lines:
        line_offset.append(offset)
        offset += len(line)

# reading a line
line_to_read = 1
line = ''

data_file.seek(line_offset[line_to_read])
line = data_file.readline()

A: you can define a chunk reader as follows using a generator
def read_file_chunk(fname, chunksize=500000):
    with open(fname, 'r') as myfile:
        lines = []
        for i, line in enumerate(myfile):
            line_values = (float(val) for val in line.split())
            lines.append(line_values)
            if i > 0 and i % 5 == 0:
                yield lines
                lines = [] # resets the lines list
        if lines:
            yield lines # final few lines of file.

# and, assuming the function you want to apply is called `my_func`
chunk_gen = read_file_chunk(my_file_name)
for chunk in chunk_gen:
    my_func(chunk)

A: You can look onto DAGpype's chunk_stream_bytes.
I didn't worked with it, but I hope it will help.
This is example of chunk read and processing some .csv file (_f_name):
 np.chunk_stream_bytes(_f_name, num_cols = 2) | \
        filt(lambda a : a[logical_and(a[:, 0] < 10, a[:, 1] < 10), :]) | \
        np.corr()

redpajama_set_name: RedPajamaStackExchange


text:
// Copyright 2020 The Chromium Authors. All rights reserved.
// Use of this source code is governed by a BSD-style license that can be
// found in the LICENSE file.

package org.chromium.weblayer;

import androidx.annotation.IntDef;

import java.lang.annotation.Retention;
import java.lang.annotation.RetentionPolicy;

/**
 * @hide
 */
@IntDef({CookieChangeCause.INSERTED, CookieChangeCause.EXPLICIT, CookieChangeCause.UNKNOWN_DELETION,
        CookieChangeCause.OVERWRITE, CookieChangeCause.EXPIRED, CookieChangeCause.EVICTED,
        CookieChangeCause.EXPIRED_OVERWRITE})
@Retention(RetentionPolicy.SOURCE)
public @interface CookieChangeCause {
    /** The cookie was inserted. */
    int INSERTED = org.chromium.weblayer_private.interfaces.CookieChangeCause.INSERTED;
    /** The cookie was changed directly by a consumer's action. */
    int EXPLICIT = org.chromium.weblayer_private.interfaces.CookieChangeCause.EXPLICIT;
    /** The cookie was deleted, but no more details are known. */
    int UNKNOWN_DELETION =
            org.chromium.weblayer_private.interfaces.CookieChangeCause.UNKNOWN_DELETION;
    /** The cookie was automatically removed due to an insert operation that overwrote it. */
    int OVERWRITE = org.chromium.weblayer_private.interfaces.CookieChangeCause.OVERWRITE;
    /** The cookie was automatically removed as it expired. */
    int EXPIRED = org.chromium.weblayer_private.interfaces.CookieChangeCause.EXPIRED;
    /** The cookie was automatically evicted during garbage collection. */
    int EVICTED = org.chromium.weblayer_private.interfaces.CookieChangeCause.EVICTED;
    /** The cookie was overwritten with an already-expired expiration date. */
    int EXPIRED_OVERWRITE =
            org.chromium.weblayer_private.interfaces.CookieChangeCause.EXPIRED_OVERWRITE;
}

redpajama_set_name: RedPajamaGithub


René A. Guzman Aug. 10, 2018 Updated: Aug. 10, 2018 4:46 p.m.
Thousands of tiny paintings in San Antonio are officially a big deal.
Guinness World Records has given the title of "largest collection of miniature paintings" to San Antonio artist Martha I. Ochoa for her hand-painted collection of 5,000 palm-friendly works. The dollhouse-ready canvases each measure about 1½ by 2 inches and depict myriad scenes and subjects in acrylics, oils and watercolors.
Ochoa first vied for Guinness glory in November 2015 with an exhibition at St. Mary's Hall to meet Guinness requirements for independent witness verification and public display. Alas, that attempt lacked required video of the portrait count, as well as a specific name for each painting. So Ochoa tried again last summer with a showing at Central Library.
Ochoa, who is a native of Peru, began her mini artworks about 20 years ago, drawing — or rather, painting — from what she saw in her travels and photos as well as with her mind's eye.


text:
HomeMoviesM3GAN (2023)
MoviesM3GAN (2023)
Is there M3GAN Sequel?
The doll currently causing the most internet buzz is M3GAN. Our new favorite doll is prepared to become an icon for the younger generations with a beige dress, a ribbon knotted around her neck, and a wig worthy of Nicole Kidman. Perhaps the new Chucky is here. It, therefore, comes as no surprise when M3GAN, an acronym for Model 3 Generative Android, begins to murder people. Fans have been impatiently awaiting the release of the M3GAN Sequel since the first new AI horror movie dropped. (tvacute.com) Here is all the information you require. Will there be a "M3GAN sequel"?
Who Are Violet McGraw's (Katie) Real-Life Parents?
Is M3GAN (2023) Available to Stream?
Is M3GAN's Doll Played by a Real Person?
What happens in M3GAN Movie?
Cady (Violet McGraw) moves in with her aunt Gemma (Allison Williams) when her parents are fatally injured in a car accident from which she survives uninjured. Gemma, though, is a woman who adores her job. She is single and happy. She unexpectedly has a 10-year-old girl at home, and she has no idea how to interact with, care for, or soothe her. Additionally, Robotics engineer Gemma is also working on her most critical project. An artificially intelligent doll with a huge learning potential connects with youngsters on numerous levels. a doll that serves as the child's mother, teacher, bodyguard, and companion.
After Gemma completes the doll, M3GAN, she chooses to check the outcomes using her niece as a test subject. She will accomplish what she hasn't been able to with her niece and connect the doll and the girl. Cady bonds with the doll while grieving her parents. They're inseparable. M3GAN is becoming self-aware. She experiments and thinks. She becomes dangerous to Cady, Gemma, and others. In almost no time at all, he had murdered a dog, a toddler, the gossipy neighbor, Gemma's coworker, and even her boss. She initially did it to defend Cady, but she quickly comes to understand that she is really defending herself. Cady and M3GAN are close. The doll does everything to get her home. She must overcome a challenge to be with the girl. Gemma, the final impediment, tried to deactivate the doll and prevent its sale. She plans to leave Gemma in palliative care, inserting a pencil into her brain, so she can always be with her and not be detected.
M3GAN Movie Ending!
Together, they weaken the doll (sawing its face, pulling its hair, splitting it in half, and finally sticking a screwdriver into its brain). Cady breaks her link with Emma aka M3GAN and forms one with Gemma. Aunt and niece finally unite. Gemma and Cady leave the house after dismembering M3GAN in the final scene. With the door open, Gemma's house monitor—a futuristic Alexa—activates unexpectedly. The film suggests that M3GAN may have disappeared, but its data is still on the internet. She is alive and more dangerous than ever. It can control any technology if downloaded online.
Will There Be a Second Part of the M3GAN Movie?
Even if a monster from a horror film is slain at the end of the film, their demise only lasts as long as there is no chance of a sequel. The possibilities multiply if we include the open ending that suggests she is living. The camera stays indoors and focuses on the tiny robot control panel of the home's Alexa-like smart home assistant as Gemma and Cady leave the house in the movie's last scene. Just as the movie cuts to the titles, the machine suddenly turns on and begins to move on its own. It is implied that M3GAN's intelligence would have easily escaped and ended up on the internet or on other devices. This year-long-awaited horror picture has the potential to do well at the box office.
When asked what his potential next project would be during a collider interview, M3GAN Director Gerard Johnstone responded, "Oh, well, there are rumors about a sequel, but we'll have to wait and see." Wan later clarified how many of his films are able to develop into franchises, even though the producer would not explicitly affirm this intriguing idea, saying:
"What I will say to that is, in any of my movies, whether it's The Conjuring Universe, or Saw, or Malignant, or M3GAN here, we like to think of a bigger world. For me, it's about creating the world, and knowing who the characters are, where the story could potentially go, and then building this bigger world, and then going into that and going, "Okay, I'm telling this particular story, but I know other stuff that's going on." So if we're fortunate enough to have sequels, then we have an idea of where we want to go."
If the movie is successful in its opening weekend, there is a good chance that there will be an M3GAN sequel. This is because James Wan has the propensity to plan ahead, and Blumhouse Productions is known for producing films on a relatively low budget while maintaining an amazing turnaround. As moviegoers, the most we can do is ensure that theatres are full and convey our gratitude to the filmmakers by letting them know how much we enjoy their work.
M3GAN 2
M3GAN Sequel
Will There Be a Second Part of M3GAN Movie?
(M3GAN: Cady) Who Are Violet McGraw's Real-Life Parents?
Get Ready for SWAT Season 6 Episode 9 "Pariah"


text:
ABVC BioPharma Announces Vitargus Phase II Study Plan Approved by HREC in Australia
By: NewMediaWire
FREMONT, CA - (NewMediaWire) - October 05, 2022 - ABVC Biopharma, Inc. (NASDAQ: ABVC), a clinical-stage biopharmaceutical company developing therapeutic solutions in oncology/hematology, CNS, and ophthalmology, today announced that its Vitrargus Phase II study plan was approved by the Australian Bellberry Human Research EthicsCommittee (HREC) on September 29, 2022. Subsequently, a Clinical Trial Notification (CTN) was submitted to the Australian Therapeutic Goods Administration (TGA) for the initiation of the study in Australia.
Early clinical studies indicate that Vitargus has unique properties that eliminate the need for positioning patients face-down post-surgery and result in significantly greater patient comfort and visual acuity during surgical recovery than currently available products. The objective of the Phase II study entitled "A Perspective Multi-Site Open Label Randomized Controlled Clinical Investigation of the Safety and Effectiveness of ABV-1701 Ocular Endotamponade (OE)" is to demonstrate the safety and effectiveness of Vitargus for the treatment of uncomplicated retinal detachment when compared to the commonly used SF6 Gas OE. The Phase II study will be conducted in Thailand as well as Australia and will include a total of at least 40 subjects.
"We are pleased to have received approval from the Australian Bellberry Human Research EthicsCommittee (HREC) for theVitargus Phase II study, which follows approval from the Central Research Ethics Committee (CREC) of The National Research Council of Thailand," said Dr. Howard Doong, Chief Executive Officer of the Company. "We are aiming to conduct site initiation visits (SIVs) next month in both Thailand and Australia so that we can begin the study in early 2023."
About ABVC BioPharma
ABVC BioPharma is a clinical-stage biopharmaceutical company with an active pipeline of six drugs and one medical device (ABV-1701/Vitargus®) under development. For its drug products, it is focused on utilizing its licensed technology to conduct proof-of-concept trials through Phase II of the clinical development process at world-famous research institutions (such as Stanford University, University of California at San Francisco, and Cedars-Sinai Medical Center). For Vitargus®, the company intends to conduct clinical trials through Phase III at various locations throughout the globe.
This press release contains "forward-looking statements." Such statements may be preceded by the words "intends," "may," "will," "plans," "expects," "anticipates," "projects," "predicts," "estimates," "aims," "believes," "hopes," "potential," or similar words. Forward-looking statements are not guarantees of future performance, are based on certain assumptions, and are subject to various known and unknown risks and uncertainties, many of which are beyond the Company's control, and cannot be predicted or quantified, and, consequently, actual results may differ materially from those expressed or implied by such forward-looking statements. Such risks and uncertainties include, without limitation, risks and uncertainties associated with (i) our inability to manufacture our product candidates on a commercial scale on our own, or in collaboration with third parties; (ii) difficulties in obtaining financing on commercially reasonable terms; (iii) changes in the size and nature of our competition; (iv) loss of one or more key executives or scientists; and (v) difficulties in securing regulatory approval to proceed to the next level of the clinical trials or to market our product candidates. More detailed information about the Company and the risk factors that may affect the realization of forward-looking statements is set forth in the Company's filings with the Securities and Exchange Commission (SEC), including the Company's Annual Report on Form 10-K and its Quarterly Reports on Form 10-Q. Investors are urged to read these documents free of charge on the SEC's website at http://www.sec.gov. The Company assumes no obligation to publicly update or revise its forward-looking statements as a result of new information, future events or otherwise.
ICR, LLC
Stephanie Oshchepkov
Email: Stephanie.Oshchepkov@icrinc.com
View the original release on www.newmediawire.com


text:
Disney Raising Prices Again, But You Can Save With These Tips
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4 Simple Steps for a More Comfortable Retirement
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Happiness may soon be a little dearer at Disney's most popular U.S. theme park as the base ticket price bursts through a $100 threshold like fireworks exploding over Cinderella Castle.
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Jim Gold • February 24, 2015
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Happiness is getting a little dearer at Disney's most popular U.S. theme park, with news that the base one-day ticket price has burst above $100 like fireworks over Cinderella Castle.
At Walt Disney World in Orlando, Florida, a one-day adult ticket jumped to $105, up from $99 on Sunday. Disneyland and Disney California Adventure went up $5, to $99. Disney World's Hollywood Studios, Orlando's Epcot and Animal Kingdom theme parks cost $97. Taxes are additional on all tickets.
And that's just the entrance fee. After that, well, there's the food, lodging, must-have souvenirs, miscellaneous stuff.
But you can offset the increase at the entrance to the Magic Kingdom without turning into Scrooge McDuck by following some simple money savings tips.
First, the tickets:
Multi-day discounts: If you're in it for more than a day, you can take advantage of discounts that increase with duration. At Walt Disney World, a three-day adult ticket for one park per day is $275, or $91.67 a day, which is a savings of $13.33 a day, or nearly 13 percent. If you've got the time, a 10-day ticket for one park per day is $365, or $36.50 a day, which is $68.50 a day, or nearly two-thirds, off base price. Similar saving are available in California, with a five-day ticket for one park a day offered at $275, or $55 a day, more than 40 percent off the base price.
Annual pass: If you live close enough to go often, consider a "premium" annual pass ($779 in Florida; $779 in California) which includes multiple park visits — including water parks in Florida — in one day, free parking, and discounts on food and merchandise 365 days a year. An annual pass in Florida ($654) doesn't include water park admissions; in California, a "deluxe" passport ($549) offers only 315 days of admissions and lesser discounts.
Outside ticket discounts are available, but most experts say you are likely to pay 90 to 95 percent of full price. In California, a CityPass ($334 adults, $291 children) includes a three-day Disney Park Hopper Ticket and one-day admissions to both Universal Studios Hollywood and SeaWorld San Diego. That saves you up to 30 percent off the combined admissions, if you were going to all three anyway.
Walt Disney World includes about 30 hotels and even campgrounds on its nearly 40-square-mile resort at Lake Buena Vista, 20 miles southwest of Orlando. They include "value," "moderate," and "deluxe" accommodations in a variety of prices, many discounted between now and June 15.
Still, other deals may be found in the 200 Orlando hotels and motels. In California, only three hotels are on the 510-acre resort. Outside the resort but nearby are 39 Disney-approved lodges and more than 100 other places to stay.
Although it can run $200 a night or more to stay in a resort hotel instead of off site, amenities often include parking (otherwise $17 a day), proximity to attractions, transportation around the resort, extended park hours and access to Disney characters.
For off-site lodging, consider shuttle service or walking distance to and from park gates, especially if you plan to return to your room during the day, and other promotions, such as gasoline refunds and off-season rates.
Eating away
Many visitors try to get around buying expensive food at Disney parks by packing in their own (not really allowed inside Disneyland Park, which offers a complimentary picnic area outside the main entrance) or by going to off-site restaurants or back to their hotel rooms to eat. Experienced visitors recommend going to grocery stores near the theme parks and buying food for savings of $10 to $20 a meal.
Also, you may want to pack items such as batteries, memory cards for cameras, Band Aids, sunscreen.
But don't load yourself down so much you can't enjoy the parks.
Money and mementos
My Frugal Adventures and others suggest purchasing Disney-related apparel and other paraphernalia on sale in advance from your local Disney Store, Walmart or other discount retailer. Besides saving on resort gift shop prices, you'll be dressed for your occasion upon arrival. You'll still likely buy souvenirs at the park, but maybe not as many if you hadn't bought ahead.
Alternative destinations
Theme Park Insider recently pegged Walt Disney World's Magic Kingdom as the world's most popular theme park, with 18.6 million visitors a year; Disneyland saw 16.2 million visitors. And Disneyland is gearing up for its 60th birthday with special attractions beginning May 22.
Price hikes have come nearly annually at the Disney resorts since 2010. In Florida, prices in 2014 jumped just before spring break; in California, just in time for the Memorial Day weekend's unofficial launch of summer.
Disney analysts continue to predict increasing attendance at the company's theme parks, despite bursting through the $100 barrier.
Even so, it's enough to make some adventure seekers look for new horizons. If you're one of those, you might consider some of these possibilities:
Universal Orlando Resort has a one-day one-park ticket to Universal Studios Florida or Universal's Islands of Adventure for $96. (There's a Wizarding World of Harry Potter attraction in each.) Universal Studios Hollywood is $92.
Legoland California, not including its Sea Life aquarium or water park, is $87; Legoland Florida is also $87 but is offering one-day discounts for $72.
Both Florida and California offer miles of free beaches along their coasts.
For more ideas about how to have a fantastic vacation without breaking the bank, check out this video:
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