text:
\section{Introduction}

Hajnal and Juh\'asz, in 1969 (see \cite{HajJuh67}) proved that if $X$ is a $T_1$-space, then $|X|\le 2^{s(X)\psi(X)}$, and
if $X$ is a Hausdorff space, then $|X|\le 2^{c(X)\chi(X)}$ and $|X|\le 2^{2^{s(X)}}$. Later Schr\"oder in \cite{Sch93}
proved that if $X$ is a Hausdorff space, then the first inequality could be sharpened to $|X|\le 2^{Us(X)\psi_c(X)}$, and
if $X$ is a Urysohn space, then the second estimation could be improved to $|X|\le 2^{Uc(X)\chi(X)}$. In
\cite[Theorem 12]{Sch93} the author also claimed that if $X$ is a Urysohn space, then the third inequality could be
improved to $|X|\le 2^{2^{Us(X)}}$. The proof of his Theorem 12 is based on \cite[Lemma 11]{Sch93} which states
that if $X$ is a Urysohn space, then $\psi_c(X)\le 2^{Us(X)}$. Unfortunately, there is a gap in the proof of that Lemma 11
and therefore the validity of the last two claims are unknown. We were able to prove both claims only when $X$ is an
$S(3)$-space (see Corollaries \ref{CIG1} and \ref{CIG2}).

Let $n$ be a positive integer. In this paper, for some spaces $X$, we define the cardinal functions
$\chi_n(X)$, $\psi_n(X)$, $s_n(X)$, and $c_n(X)$ called respectively $S(n)$-character, $S(n)$-pseudocharacter,
$S(n)$-spread, and $S(n)$-cellularity and using them we extend the above-mentioned inequalities to the class of
$S(n)$-spaces (see Definition \ref{D1}). In particular, we show that if $X$ is an $S(n)$-space, then
$\left| X\right| \leq 2^{c_{n}(X)\chi_{n}(X)}$ (Theorems \ref{ThGHJ2} and \ref{ThGSc2}).
When $n=1$ we obtain Hajnal and Juh\'asz' inequality $|X|\le 2^{c(X)\chi(X)}$ for Hausdorff
spaces $X$ and the case $n=2$ is the second part of Alas and Ko\v{c}inac' Theorem 2 in \cite{AlaKoc00}, which sharpens
Schr\"oder's inequality $|X|\le 2^{Uc(X)\chi(X)}$ for Urysohn spaces $X$
(see Corollary \ref{CIGSch}).

Our Theorems \ref{THJIG} and \ref{TScIG} contain the generalizations to the class of $S(n)$-spaces of the corresponding
inequalities of Hajnal and Juh\'asz and Schr\"oder mentioned above: Let $k$ be a positive integer. If $X$ is an
$S(k-1)$-space, then $\left| X\right|\leq 2^{s_{2k-1}(X)\psi_{2k-1}(X)}$ and if $X$ is an $S(k)$-space, then
$\left| X\right|\leq 2^{s_{2k}(X)\psi_{2k}(X)}$.

At the end of the paper we generalize Hajnal and Juh\'asz' inequality that if $X$ is a Hausdorff space, then
$|X|\le 2^{2^{s(X)}}$. For that end we find upper bounds for the cardinality of the $S(n)$-pseudocharacter of a space,
where $n$ is any positive integer. In Lemmas \ref{LHJIG} and \ref{LIG} we show that if $n$ is a positive integer and $X$
is an $S(n)$-space, then $\psi(X)\le 2^{s_n(X)}$ and, as a consequence, in Theorem \ref{TISG} we show that
$\left| X\right|\leq 2^{s(X)\cdot 2^{s_{n}(X)}}$. In addition, we show that for every positive integer $k$, the following
are true:

(a) If $X$ is an $S(3k)$-space, then $\psi_{2k}\le 2^{s_{2k}(X)}$ (Lemma \ref{LIG2G1}) and
$\left| X\right|\leq 2^{2^{s_{2k}(X)}}$ (Theorem \ref{TIG2G1});

(b) If $X$ is an $S(3k-2)$-space, then $\psi_{2k-1}(X)\leq 2^{s_{2k-1}(X)}$ (Lemma \ref{LIG2G2}) and
$\left| X\right|\leq 2^{2^{s_{2k-1}(X)}}$ (Theorem \ref{TIG2G2});

(c) If $X$ is an $S(3k-1)$-space, then $\psi_{2k-1}(X)\leq 2^{s_{2k}(X)}$ and $\psi_{2k}(X)\leq 2^{s_{2k-1}(X)}$
(Lemma \ref{LIG2G3} and Lemma \ref{LIG2G4}) and $\left| X\right|\leq 2^{s_{2k-1}(X)\cdot2^{s_{2k}(X)}}$, hence
$\left| X\right|\leq 2^{2^{s_{2k-1}(X)}}$ (Theorem \ref{TIG2G3}).

More results about the cardinality of $S(n)$-spaces involving the cardinal functions $d_n(X)$, $t_n(X)$ and $bt_n(X)$,
called respectively $S(n)$-density, $S(n)$-tightness and $S(n)$-bitightness, are contained in our paper \cite{GotKoc18}.

\section{Preliminaries}

Notations and terminology in this paper are standard as in \cite{Eng89}, \cite{Juh80}, and \cite{Hod84}. Unless otherwise
indicated, all spaces are assumed to be at least $T_1$ and infinite. $\alpha$, $\beta$, $\gamma$, and $\delta$ are ordinal
numbers, while $\kappa$ denotes infinite cardinal; $\kappa^{+}$ is the successor cardinal of $\kappa$. As
usual, cardinals are assumed to be initial ordinals, $\mathbb{N}^{+}$ denotes the set of all positive integers and
$\mathbb{N}=\{0\}\cup \mathbb{N}^{+}$. If $X$ is a set, then $\mathfrak{P}(X)$ and $\left[ X\right] ^{\leq \kappa}$
denote the power set of $X$ and the collection of all subsets of $X$ having cardinality $\leq \kappa$, respectively.

\begin{definition}\label{D1}
Let $X$ be a topological space, $A\subset X$ and $n\in \mathbb{N}^{+}.$ A point $x\in X$ is \emph{$S(n)$-separated
from} $A$ if there exist open sets $U_{i}$, $i=1,2,...,n$ such that $x\in U_{1}$, $\overline{U}_{i}\subset U_{i+1}$ for
$i=1,2,...,n-1$ and $\overline{U}_{n}\cap A=\varnothing $; $x$ is $S(0)$-separated from $A$ if $x\notin \overline{A}.$
$X$ is an \emph{$S(n)$-space} \cite{Vig69} if every two distinct points in $X$ are $S(n)$-separated.

Now, let $n\in \mathbb{N}$. The set $\cl_{\theta ^{n}}A=\{x\in X$ : $x$ is not $S(n)$-separated from $A\}$ is called
$\theta ^{n}$\emph{-closure} of $A$ \cite{DikGiu88}. $A$ is $\theta ^{n}$\emph{-closed} \cite{PorVot73a} if
$\cl_{\theta ^{n}}(A)=A$; $U\subset X$ is $\theta ^{n}$-\emph{open} if $X\backslash U$ is $\theta ^{n}$-closed; and
$A$ is \emph{$\theta ^{n}$-dense} in $X$ if $\cl_{\theta ^{n}}(A)=X$.
\end{definition}

It follows directly from Definition \ref{D1} that $S(1)$ is the class of Hausdorff spaces and $S(2)$ is the class of Urysohn
spaces. Since we are going to consider here only $T_1$-spaces, $S(0)$-spaces will be exactly the $T_1$-spaces. Also,
$\cl_{\theta ^{0}}(A)=\overline{A}$ and $\cl_{\theta ^{1}}(A)=\cl_{\theta }(A)$ - the so called
\emph{$\theta $-closure of $A$} \cite{Vel66}. We want to emphasize here that in general, when
$n \in \bN^+$, the $\theta^n$-closure operator is not idempotent, hence it is not a Kuratowski closure operator. In
particular $\cl_\theta(\cl_\theta(A))\ne\cl_{\theta^2}(A)$. Finally, we
note that if $X$ is a space and $n \in \bN^+$, then $\cl_{\theta^n}(U)=\cl_{\theta^{n-1}}(\overline{U})$ whenever $U$ is an open subset of $X$ (see
\cite[Lemma 1.4(c)]{DikGiu88}).

In this paper it will be more convenient for us to think about $S(n)$-spaces in more 'symmetric' way similar to the way how
$S(n)$-spaces are defined in \cite{DikTho95}, \cite{DikWat95} or \cite{PorVot73a} but here we are going to use different
terminology and notation.

\begin{definition}\label{SnIG}
Let $X$ be a topological space, $U\subseteq X$, $x\in U$ and $k\in\bN^+$.
We will say that $U$ is an \emph{$S(2k-1)$-neighborhood of $x$} if there exist open sets $U_{i}$, $i=1,2,...,k$,
such that $x\in U_{1}$, $\overline{U}_{i}\subset U_{i+1}$, for $i=1,2,...,k-1$, and $U_{k}\subseteq U$.
We will say that $U$ is an \emph{$S(2k)$-neighborhood of $x$} if there exist open sets $U_{i}$, $i=1,2,...,k$,
such that $x\in U_{1}$, $\overline{U}_{i}\subset U_{i+1}$, for $i=1,2,...,k-1$, and $\overline{U}_{k}\subseteq U$.

Let $n\in\bN^+$. When a set $U$ is an $S(n)$-neighborhood of a point $x$ and it is an open (closed) set in $X$,
we will refer to it as open (closed) \emph{$S(n)$-neighborhood of $x$}. A set $U$ will be called $S(n)$-open
($S(n)$-closed) if $U$ is open (closed) and there exists at least one point $x$ such that $U$ is an open (closed)
$S(n)$-neighborhood of $x$.
\end{definition}

\begin{remark}
We note that in what follows every $S(2k-1)$-open set $U$ in a space $X$, where $k\in\bN^+$, will be considered as a
fixed chain of $k$ nonempty sets $U_{i}$, $i=1,2,...,k$, such that $\overline{U}_{i}\subset U_{i+1}$, for $i=1,2,...,k-1$,
and $U_{k}\subseteq U$. (In fact, most of the time we will assume that $U_k=U$). Sometimes, when we need to refer to
the first set $U_1$ in that chain we will use the notation $U(k)$, i.e. by definition $U(k)=U_1$.
\end{remark}

Now, using the terminology and notation introduced in Definition \ref{SnIG} it is easy to see that the following propositions
are true.

\begin{proposition}\label{P1}
Let $X$ be a topological space, $x\in X$ and $k\in\bN^+$.

{\rm (a)} Every closed $S(2k-1)$-neighborhood of $x$ is a closed $S(2k)$-neighborhood of $x$.

{\rm (b)} Every $S(2k)$-neighborhood of $x$ contains a closed $S(2k)$-neighborhood of $x$; hence it contains a closed
(and therefore an open) $S(2k-1)$-neighborhood of $x$. Thus, every $S(2k)$-neighborhood of $x$ is an
$S(2k-1)$-neighborhood of $x$.

{\rm (c)} Every $S(2k+1)$-neighborhood of $x$ contains an open $S(2k+1)$-neighborhood of $x$; hence it contains an
open (and therefore a closed) $S(2k)$-neighborhood of $x$.
Thus, every $S(2k+1)$-neighborhood of $x$ is an $S(2k)$-neighborhood of $x$.
\end{proposition}

\begin{proposition}\label{P2}
Let $X$ be a topological space and $k\in\bN^+$.

{\rm (a)} $X$ is an $S(2k-1)$-space if and only if every two distinct points of $X$ can be separated by disjoint (open)
$S(2k-1)$-neighborhoods.

{\rm (b)} $X$ is an $S(2k)$-space if and only if every two distinct points of $X$ can be separated by disjoint closed
$S(2k-1)$-neighborhoods.

{\rm (c)} $X$ is an $S(2k)$-space if and only if every two distinct points of $X$ can be separated by disjoint (closed)
$S(2k)$-neighborhoods.

{\rm (d)} $X$ is an $S(2k+1)$-space if and only if every two distinct points of $X$ can be separated by disjoint open
$S(2k)$-neighborhoods.
\end{proposition}

\begin{definition}\label{SnClIG}
Let $X$ be a topological space, $A\subseteq X$ and $k\in\bN^+$. We will say that a point $x$ is in the $S(2k-1)$-closure
of $A$ if and only if every (open) $S(2k-1)$-neighborhood of $x$ intersects $A$ and we will say that a point $x$ is in the
$S(2k)$-closure of $A$ if and only if every (closed) $S(2k)$-neighborhood (or equivalently, every closed
$S(2k-1)$-neighborhood) of $x$ intersects $A$. For $n\in\bN^+$, the
$S(n)$-closure of $A$ will be denoted by $\theta_n(A)$. $A$ is $\theta_{n}$\emph{-closed} if
$\theta_{n}(A)=A$ and $U\subset X$ is $\theta_{n}$-\emph{open} if $X\setminus U$ is $\theta_{n}$-closed, or
equivalently, $U\subset X$ is $\theta_{n}$-\emph{open} if $U$ is an $S(n)$-neighborhood of every $x\in U$.
Finally, $A$ is $\theta_n$-dense in $X$ if $\theta_n(A)=X$.
\end{definition}

Clearly, for every  $n\in\bN^+$, every $\theta_n$-open set is open and every set of the form $\theta_n(A)$, where
$A\subseteq X$, is a closed set. Also, it follows directly from Definition \ref{SnClIG} that
$\theta_1(A)=\cl(A)=\overline{A}$ is the usual closure operator in $X$
and $\theta_2(A)=\cl_\theta(A)$ is the $\theta$-closure operator introduced by Veli\v{c}ko \cite{Vel66}. We also note
that, except for the case $n=1$, for many $A\subset X$ we may have $\theta_n(\theta_n(A))\ne\theta_n(A)$, or in other
words, the $\theta_n$-closure operator is not idempotent. More information about the
closure operator $\theta_n$ is contained in \cite{DikGiu88}, \cite{DikTho95} and \cite{DikWat95}.

\begin{definition}\label{DPsi} Let $k\in \bN^+$ and $X$ be a topological space.

\emph{(a)} A family $\{U_\alpha :\alpha < \kappa\}$ of open $S(2k-1)$-neighborhoods of a point $x\in X$ will be called
\emph{an open $S(2k-1)$-neighborhood base at the point $x$} if for every open $S(2k-1)$-neighborhood $U$ of $x$
there is $\alpha<\kappa$ such that $U_\alpha\subseteq U$.

\emph{(b)} An $S(2k-1)$-space $X$ is of \emph{$S(2k-1)$-character} $\kappa$, denoted by
$\chi_{2k-1}(X)$, if $\kappa$ is the smallest infinite cardinal such that for each point $x\in X$ there exists an open
$S(2k-1)$-neighborhood base at $x$ with cardinality at most $\kappa$. In the case $k=1$ the $S(1)$-character
$\chi_1(X)$ coincides with the usual character $\chi(X)$.

\emph{(c)} An $S(2k)$-space $X$ is of \emph{$S(2k)$-character} $\kappa$, denoted by
$\chi_{2k}(X)$, if $\kappa$ is the smallest infinite cardinal such that for each point $x\in X$ there exists a family $\sV_x$
of closed $S(2k-1)$-neighborhoods of $x$ such that $|\sV_x|\le\kappa$ and if $W$ is an open $S(2k-1)$-neighborhood of
$x$, then $\overline{W}$ contains a member of $\mathcal{V}_x$. In the case $k=1$ the $S(2)$-character
$\chi_2(X)$ coincides with the cardinal function $k(X)$ defined in \cite{AlaKoc00}.

\emph{(d)} An $S(k-1)$-space $X$ is of \emph{$S(2k-1)$-pseudocharacter} $\kappa$, denoted by
$\psi_{2k-1}(X)$, if $\kappa$ is the smallest infinite cardinal such that for each point $x\in X$ there exists a family
$\{U_\alpha :\alpha < \kappa\}$ of $S(2k-1)$-open neighborhoods of $x$ such that
$\{x\}=\bigcap \{U_\alpha :\alpha < \kappa \}.$ In the case $k=1$ the pseudocharacter $\psi_1(X)$ coincides with
the usual pseudocharacter $\psi(X)$.

\emph{(e)} An $S(k)$-space $X$ is of \emph{$S(2k)$-pseudocharacter} $\kappa$,
denoted by $\psi_{2k}(X)$, if $\kappa$ is the smallest infinite cardinal such that for each point $x\in X$ there exists a
family $\{U_\alpha :\alpha < \kappa \}$ of $S(2k-1)$-open neighborhoods of $x$ such that
$\{x\}=\bigcap \{\overline{U}_\alpha :\alpha < \kappa\}.$ In the case $k=1$ the pseudocharacter $\psi_2(X)$
coincides with the closed pseudocharacter $\psi_{c}(X)$.
\end{definition}

In relation to Definition \ref{DPsi}(c) we recall that for a topological space $X$, $k(X)$ is the smallest infinite cardinal
$\kappa$ such that for each point $x\in X$, there is a
collection $\mathcal{V}_x$ of closed neighborhoods of $x$ such that
$|\mathcal{V}_x|\le \kappa$ and if $W$ is a neighborhood of $x$,
then $\overline{W}$ contains a member of $\mathcal{V}_x$ \cite{AlaKoc00}.
Clearly, $k(X)\le\chi(X)$.
As it was noted in \cite{AlaKoc00}, $k(X)$ is equal to the character of the semiregularization of $X$. We also note
that if $k\in\bN^+$, then $\psi_{2k-1}(X)\le\psi_{2k}(X)\le\psi_{2k+1}(X)\le\psi_{2k+2}(X)$ and
$\chi_{2k}(X)\le\chi_{2k-1}(X)$, whenever they are defined.

\begin{remark}\label{RemPsi}
Since the $\theta$-closure operator is not idempotent, for each positive integer $n$, a different type of pseudocharacter
could be defined by requiring each point $x\in X$ to have a family of open neighborhoods
$\{U_\alpha :\alpha < \kappa\}$ such that
$\{x\}=\bigcap\{\cl_\theta(\cl_\theta(\ldots\cl_\theta(U_\alpha)\ldots)):\alpha < \kappa\},$ where the
$\theta$-closure operator is repeated $n$ times. We used the notation
$\psi_{\theta^n}(X)$ in \cite{GotUnp3} to denote that pseudocharacter and it is not difficult to see that
$\psi_{\theta^n}(X)$ and the pseudocharacter $\psi_n(X)$ defined in Definition \ref{DPsi} are different.
\end{remark}

\begin{definition}\label{D6}
Let $k\in \bN^+$ and $X$ be a topological space.

\emph{(a)} We shall call a subset $D$ of $X$ \emph{$S(2k-1)$-discrete} if for every $x\in D$, there is an
open $S(2k-1)$-neighborhood $U$ of $x$ such that $U\cap D=\{x\}$, and we define the \emph{$S(2k-1)$-spread} of
$X$, denoted by $s_{2k-1}(X)$, to be $\sup \{\left| D\right| :D$ is $S(2k-1)$-discrete subset of $X\}+\aleph_0$.

\emph{(b)} We shall call a subset $D$ of $X$ \emph{$S(2k)$-discrete} if for every $x\in D$, there is an
open $S(2k-1)$-neighborhood $U$ of $x$ such that $\overline{U}\cap D=\{x\}$, and we define the
\emph{$S(2k)$-spread} of $X$, denoted by $s_{2k}(X)$, to be $\sup \{\left| D\right| :D$ is $S(2k)$-discrete subset of
$X\}+\aleph_0$.
\end{definition}

It follows immediately from Definition \ref{D6} that a set $D$ in a topological space $X$ is discrete if and only if $D$ is
$S(1)$-discrete and a set $D$ is Urysohn-discrete if and only if $D$ is $S(2)$-discrete. Therefore $s_1(X)$ is the usual
spread $s(X)$ and $s_2(X)$ is the Urysohn spread $Us(X)$ defined in \cite{Sch93}.

\begin{definition}\label{DCelIG}
Let $X$ be a topological space and $k\in \bN^+$.

\emph{(a)} We shall call a family $\sU$ of pairwise disjoint non-empty
$S(2k-1)$-open subsets of $X$ \emph{$S(2k-1)$-cellular} and we define the \emph{$S(2k-1)$-cellularity} of
$X$, denoted by $c_{2k-1}(X)$, to be $\sup \{\left| \mathcal{U}\right| :\mathcal{U}$ is an
$S(2k-1)$-cellular family in $X\}+\aleph_0$.

\emph{(b)} We shall call a family $\sU$ of non-empty $S(2k-1)$-open subsets
of $X$ \emph{$S(2k)$-cellular} if for every distinct $U_1,U_2\in\sU$ we have
$\overline{U}_1\cap\overline{U}_2=\emptyset$ and we define the \emph{$S(2k)$-cellularity} of a space
$X$, denoted by $c_{2k}(X)$, to be $\sup \{\left| \mathcal{U}\right| :\mathcal{U}$ is an
$S(2k)$-cellular family in $X\}+\aleph_0$.
\end{definition}

\begin{remark} Let $X$ be a topological space.

\emph{(a)} $c_1(X)$ coincides with the usual cellularity $c(X)$ of the space $X$;

\emph{(b)} $c_2(X)$ coincides with the Urysohn cellularity $Uc(X)$ of $X$ introduced in \cite{Sch92};

\emph{(c)} If $n, m\in\bN^+$ and $n<m$, then $c_n(X)\ge c_m(X)$. For example of a space $X$ such that
$c_1(X)>c_2(X)$ see \cite[Example 4]{Sch92}.

\emph{(d)} If in Definition \ref{DCelIG}(b) we require $\sU$ to be a family of pairwise disjoint non-empty
$S(2k)$-open subsets of $X$ (similar to \ref{DCelIG}(a)), then, according to Proposition \ref{P1}, we would get the
definitions of $S(2k+1)$-cellular family and $S(2k+1)$-cellularity defined in \ref{DCelIG}(a).
\end{remark}

\section{Cardinal inequalities for $S(n)$-spaces}

Hajnal and Juh\'asz proved that if $X$ is a $T_1$-space, then $|X|\le 2^{s(X)\psi(X)}$, and if $X$ is a Hausdorff space,
then $|X|\le 2^{c(X)\chi(X)}$ and $|X|\le 2^{2^{s(X)}}$ (see \cite{HajJuh67}, \cite{Juh80} or \cite{Hod84}).
Schr\"oder in \cite{Sch93} sharpened the first two estimations by
showing that if $X$ is a Hausdorff space, then $|X|\le 2^{Us(X)\psi_c(X)}$, and if $X$ is a Urysohn space, then
$|X|\le 2^{Uc(X)\chi(X)}$, or, in our notation, $\left|X\right| \leq 2^{s_2(X)\psi_2(X)}$ and
$\left| X\right| \leq 2^{c_2(X)\chi (X)}$.

Below we formulate and prove the counterpart of the above inequalities for $S(n)$-spaces, where $n$ is
any positive integer.
We begin with extending the two inequalities $\left| X\right| \leq 2^{c(X)\chi (X)}$  and
$\left| X\right| \leq 2^{c_2(X)\chi (X)}$ to $S(n)$-spaces.

The following lemma for $k=1$ was proved by Charlesworth (see \cite[Theorem 3.2]{Cha77} or
\cite[Proposition 3.4]{Hod84}).

\begin{lemma}\label{LChIG}
Let $k\in \bN^+$, $X$ be a topological space and $\kappa=c_{2k-1}(X)$. If $\sU$ is a family of
$S(2k-1)$-open subsets of $X$, then there exists a subfamily $\sV$ of $\sU$ such that $|\sV|\le\kappa$ and
$\bigcup\{U(k):U\in \sU\}\subseteq \theta_{2k-1}(\bigcup\{V:V\in \sV\})$.
\end{lemma}

\begin{proof}
Let $\sW=\{W:W\ne\emptyset\text{ is $S(2k-1)$-open}, W\subseteq U, U\in \sU\}$. Using Zorn's lemma we can find
a maximal $S(2k-1)$-cellular family $\sW'\subseteq\sW$. Then $|\sW'|\le\kappa$. For each $W\in\sW'$ we fix
$U_W\in \sU$ such that $W\subseteq U_W$ and let $\sV=\{U_W:W\in\sW'\}$. Then $|\sV|\le\kappa$. Now, suppose that
$\bigcup\{U(k):U\in \sU\}\subsetneq \theta_{2k-1}(\bigcup\{V:V\in \sV\})$. Then it is not difficult to see that there exist
$U\in \sU$, $x\in U(k)\setminus \theta_{2k-1}(\bigcup\{V:V\in \sV\})$ and an
$S(2k-1)$-open set $U_x$ such that $x\in U_x(k)\subset U(k)$, $U_x\subseteq U$ and
$U_x\cap \bigcup\{V:V\in \sV\}=\emptyset$. Clearly, $U_x\in\sW$. Hence, $\{U_x\}\cup\sW'$ is an $S(2k-1)$-cellular
family that properly contains $\sW'$ and therefore $\sW'$ is not maximal -- contradiction.
\end{proof}

The next lemma for $k=1$ was proved by Schr\"oder (see \cite[Lemma 7]{Sch92}).

\begin{lemma}\label{LScIG}
Let $k\in \bN^+$, $X$ be a topological space and $\kappa=c_{2k}(X)$. If $\sU$ is a family of
$S(2k-1)$-open subsets of $X$, then there exists a subfamily $\sV$ of $\sU$ such that $|\sV|\le\kappa$ and
$\bigcup\{U(k):U\in\sU\}\subseteq\theta_{2k}(\bigcup\{\overline{V}:V\in \sV\})$.
\end{lemma}

\begin{proof}
Let $\sW=\{W:W\ne\emptyset\text{ is $S(2k-1)$-open}, W\subseteq U, U\in \sU\}$. Using Zorn's lemma we can find
a maximal $S(2k)$-cellular family $\sW'\subseteq\sW$. Then $|\sW'|\le\kappa$. For each $W\in\sW'$ we fix
$U_W\in \sU$ such that $W\subseteq U_W$ and let $\sV=\{U_W:W\in\sW'\}$. Then $|\sV|\le\kappa$. Now, suppose that
$\bigcup\{U(k):U\in\sU\}\subsetneq \theta_{2k}(\bigcup\{\overline{V}:V\in \sV\})$. Then it is not difficult to see that
there exist $U\in \sU$, $x\in U(k)\setminus \theta_{2k}(\bigcup\{\overline{V}:V\in \sV\})$ and an
$S(2k-1)$-open set $U_x$ such that $x\in U_x(k)\subseteq U(k)$, $U_x\subseteq U$, and
$\overline{U}_x\cap \bigcup\{\overline{V}:V\in \sV\}=\emptyset$. Clearly, $U_x\in\sW$. Hence,
$\{U_x\}\cup\sW'$ is an $S(2k)$-cellular
family that properly contains $\sW'$ and therefore $\sW'$ is not maximal -- contradiction.
\end{proof}

The following two theorems, which proof is based on Lemma \ref{LChIG} and Lemma \ref{LScIG}, give an upper bound
for the cardinality of an $S(n)$-space, where $n\in\bN^+$, as a function of the $S(n)$-cellularity and the character of the
space. The inequality for Hausdorff spaces (the case $k=1$ in Theorem \ref{ThGHJ2}) was proved by Hajnal and Juh\'asz
(see \cite{HajJuh67}, \cite[Theorem 2.15(b)]{Juh80} or \cite[Theorem 2]{Pol74}) and for Urysohn spaces (the case $k=1$
in Theorem \ref{ThGSc2}) was proved by Schr\"oder (see \cite[Theorem 9]{Sch93}).

\begin{theorem}\label{ThGHJ2}
Let $k\in \bN^+$. If $X$ is an $S(2k-1)$-space, then $\left| X\right| \leq 2^{c_{2k-1}(X)\chi_{2k-1}(X)}$.
\end{theorem}
\begin{proof}
Let $k\in \bN^+$, $\kappa=c_{2k-1}(X)\chi_{2k-1}(X)$ and for every $x\in X$ let $\sB(x)$ be an open
$S(2k-1)$-neighborhood base at $x$ such that $|\sB(x)|\le\kappa$.
By transfinite recursion we construct an increasing sequence
$(A_\alpha)_{\alpha<\kappa^+}$ of subsets of $X$ and a sequence of families of open sets
$(\sU_\alpha)_{\alpha<\kappa^+}$ as follows:
\begin{itemize}
\item[(a)] $\sU_0=\emptyset$ and $A_0=\{x_0\}$, where $x_0$ is an arbitrary point of $X$;
\item[(b)] $|A_\alpha|\le\ 2^\kappa$ for every $\alpha<\kappa^+$;
\item[(c)] $\sU_\alpha=\bigcup\{\sB(x):x\in\bigcup_{\beta < \alpha}A_\beta\}$, for every $0<\alpha<\kappa^+$;
\item[(e)] For $0<\alpha<\kappa^+$, if $\sW=\{\sV_\gamma:\gamma<\kappa\}$ is a family of subsets of $X$ such that
each $\sV_\gamma$ is the union
of at most $\kappa$ many elements of $\sU_\alpha$ and
$\bigcup_{\gamma<\kappa}\theta_{2k-1}(\sV_\gamma)\ne X$, then we pick a point
$x_\sW\in X\setminus\bigcup_{\gamma<\kappa}\theta_{2k-1}(\sV_\gamma)$. Let $E$ be the set of all such points
$x_\sW$. Then we set $A_{\alpha}=E\cup\bigcup_{\beta<\alpha}A_\beta$. Since
$|\bigcup_{\beta<\alpha}A_\beta|\le 2^\kappa$, we have $|\sU_\alpha|\le 2^\kappa$ and therefore
$|E|\le ((2^\kappa)^\kappa)^\kappa=2^\kappa$. Thus $|A_\alpha|\le 2^\kappa$ and therefore (b) is satisfied.
\end{itemize}

Now, let $A=\bigcup_{\alpha<\kappa^+}A_\alpha$. Since $|A|\le\kappa^+\cdot 2^\kappa=2^\kappa$, to
finish the proof it is sufficient to show that $X=A$. Suppose that
there exists a point $y\in X\setminus A$ and let $\sB(y)=\{B_\delta:\delta<\kappa\}$. For each $\delta<\kappa$ let
$\sV_\delta=\{V:V\in \sB(x), x\in A, B_\delta\cap V=\emptyset\}$. Since $X$ is an
$S(2k-1)$-space and $\sB(x)$, for each $x\in X$, is an open $S(2k-1)$-neighborhood base at $x$,
$A\subseteq\bigcup_{\delta<\kappa}\bigcup\{V(k):V\in\sV_\delta\}$. It follows from Lemma \ref{LChIG} that for every
$\delta<\kappa$ we can find a subfamily $\sW_\delta$ of $\sV_\delta$ such that $|\sW_\delta|\le\kappa$ and
$\bigcup\{V(k):V\in\sV_\delta\}\subseteq\theta_{2k-1}(\bigcup\sW_\delta)$. Using the fact that
$|\bigcup_{\delta<\kappa}\sW_\delta|\le\kappa$ and that $\kappa^+$ is a regular cardinal we can find
$\alpha<\kappa^+$ such that $\bigcup_{\delta<\kappa}\sW_\delta\subseteq \sU_\alpha$. Since for each $\delta<\kappa$
we have $B_\delta\cap\bigcup\{W:W\in\sW_\delta\}=\emptyset$,
$y\notin\bigcup_{\delta<\kappa}\theta_{2k-1}(\bigcup\sW_\delta)$. Then it follows from the
construction of the set $A_\alpha$ that there exists
$x\in A_\alpha\cap \left(X\setminus \bigcup_{\delta<\kappa}\theta_{2k-1}(\bigcup\sW_\delta)\right)$.
Thus $x\in A_\alpha\subseteq A\subseteq\bigcup_{\delta<\kappa}\bigcup\{V(k):V\in\sV_\delta\}\subseteq\bigcup_{\delta<\kappa}\theta_{2k-1}(\bigcup\sW_\delta)$ -- contradiction.
\end{proof}

\begin{theorem}\label{ThGSc2}
Let $k\in \bN^+$. If $X$ is an $S(2k)$-space, then $\left| X\right| \leq 2^{c_{2k}(X)\chi_{2k}(X)}$.
\end{theorem}
\begin{proof}
Let $n\in \mathbb{N}$, $\kappa=c_{\theta ^n}(X)\chi_{2k}(X)$ and for every $x\in X$ let $\sB(x)$ be a family of open
$S(2k-1)$-neighborhoods of $x$ witnessing the fact that $\chi_{2k}(X)\le\kappa$ (hence, $|\sB(x)|\le\kappa$).
By transfinite recursion we construct an increasing sequence
$(A_\alpha)_{\alpha<\kappa^+}$ of subsets of $X$ and a sequence of families of open sets
$(\sU_\alpha)_{\alpha<\kappa^+}$ as follows:
\begin{itemize}
\item[(a)] $\sU_0=\emptyset$ and $A_0=\{x_0\}$, where $x_0$ is an arbitrary point of $X$;
\item[(b)] $|A_\alpha|\le\ 2^\kappa$ for every $\alpha<\kappa^+$;
\item[(c)] $\sU_\alpha=\bigcup\{\sB(x):x\in\bigcup_{\beta < \alpha}A_\beta\}$, for every $0<\alpha<\kappa^+$;
\item[(e)] For $0<\alpha<\kappa^+$, if $\sW=\{\sV_\gamma:\gamma<\kappa\}$ is a family of subsets of $X$ such that
each $\sV_\gamma$ is the union of the closures of at most $\kappa$ many elements of $\sU_\alpha$ and
$\bigcup_{\gamma<\kappa}\theta_{2k}(\sV_\gamma)\ne X$, then we pick a point
$x_\sW\in X\setminus\bigcup_{\gamma<\kappa}\theta_{2k}(\sV_\gamma)$. Let $E$ be the set of all such points
$x_\sW$. Then we set $A_{\alpha}=E\cup\bigcup_{\beta<\alpha}A_\beta$. Since
$|\bigcup_{\beta<\alpha}A_\beta|\le 2^\kappa$,
we have $|\sU_\alpha|\le 2^\kappa$ and therefore $|E|\le ((2^\kappa)^\kappa)^\kappa=2^\kappa$.
Thus $|A_\alpha|\le 2^\kappa$ and therefore (b) is satisfied.
\end{itemize}

Now, let $A=\bigcup_{\alpha<\kappa^+}A_\alpha$. Since $|A|\le\kappa^+\cdot 2^\kappa=2^\kappa$, to
finish the proof it is sufficient to show that $X=A$. Suppose that
there exists a point $y\in X\setminus A$ and let $\sB(y)=\{B_\delta:\delta<\kappa\}$. For each $\delta<\kappa$ let
$\sV_\delta=\{V:V\in \sB(x), x\in A, \overline{B}_\delta\cap\overline{V}=\emptyset\}$. Since $X$ is an
$S(2k)$-space and $\sB(x)$, for each $x\in X$, is a closed $S(2k-1)$-neighborhood base at $x$,
$A\subseteq\bigcup_{\delta<\kappa}\bigcup\{V(k):V\in\sV_\delta\}$. It follows from Lemma \ref{LScIG} that for every
$\delta<\kappa$ we can find a subfamily $\sW_\delta$ of $\sV_\delta$ such that $|\sW_\delta|\le\kappa$ and
$\bigcup\{V(k):V\in\sV_\delta\}\subseteq\theta_{2k}\left(\bigcup\{\overline{W}:W\in\sW_\delta\}\right)$. Using the fact that
$|\bigcup_{\delta<\kappa}\sW_\delta|\le\kappa$ and that $\kappa^+$ is a regular cardinal we can find
$\alpha<\kappa^+$ such that $\bigcup_{\delta<\kappa}\sW_\delta\subseteq \sU_\alpha$. Since for each $\delta<\kappa$
we have $\overline{B}_\delta\cap\bigcup\{\overline{W}:W\in\sW_\delta\}=\emptyset$,
$y\notin\bigcup_{\delta<\kappa}\theta_{2k}\left(\bigcup\{\overline{W}:W\in\sW_\delta\}\right)$. Then it follows from the
construction of the set $A_\alpha$ that there exists
$x\in A_\alpha\cap \left(X\setminus \bigcup_{\delta<\kappa}\theta_{2k}\left(\bigcup\{\overline{W}:W\in\sW_\delta\}\right)\right)$.
Thus $x\in A_\alpha\subseteq A\subseteq\bigcup_{\delta<\kappa}\bigcup\{V(k):V\in\sV_\delta\}\subseteq\bigcup_{\delta<\kappa}\theta_{2k}\left(\bigcup\{\overline{W}:W\in\sW_\delta\}\right)$ -- contradiction.
\end{proof}

The case $k=1$ in the above theorem improves Schr\"oder's inequality \cite[Theorem 9]{Sch92} and is the second part
of Theorem 2 given in \cite{AlaKoc00} without a proof. We write it below with more familiar notation.

\begin{corollary}\label{CIGSch}
If $X$ is a Urysohn space, then $\left| X\right| \leq 2^{Uc(X)k(X)}$.
\end{corollary}

Now, we are going to formulate and prove the counterpart of the two inequalities $\left| X\right| \leq 2^{s(X)\psi (X)}$
and $\left| X\right| \leq 2^{s_2(X)\psi_2 (X)}$ for $S(n)$-spaces.

The following lemma for the case $k=1$ was proved by \v{S}apirovski\v{i} (see \cite[Proposition 4.8]{Hod84}).

\begin{lemma}\label{LSapIG}
Let $k\in \mathbb{N}^{+}$, $X$ be a topological space, $\kappa=s_{2k-1}(X)$ and $C\subseteq X$. For each
$x\in C$ let $U^x$ be an open $S(2k-1)$-neighborhood of $x$ and let $\mathcal{U}=\{U^x:x\in C\}$.
Then there exist an $S(2k-1)$-discrete subset $A$ of $C$ such that $\left| A\right| \leq \kappa$ and $C\subseteq\theta_{2k-1}(A)\cup \bigcup \left\{U^x:x\in A\right\}$.
\end{lemma}

\begin{proof}
Let $\{U_\alpha:\alpha<\beta\}$ be a well-ordering of the elements of $\sU$ and let $x_0\in C\cap U_0(k)$.
Suppose that the points $\{x_{\alpha_\gamma}:\gamma<\delta\}$ have already been chosen. Let
$\alpha_\delta$ be the first ordinal greater than $\alpha_\gamma$, for each $\gamma<\delta$, such that
there exists $x_{\alpha_\delta}\in (C\cap U_{\alpha_\delta}(k))\setminus\left[\left(\bigcup_{\gamma<\delta}U_{\alpha_\gamma}\right)\cup\theta_{2k-1}(\{x_{\alpha_\gamma}:\gamma<\delta\})\right]$. We finish this selection when such a
point $x_{\alpha_\delta}$ does not exist, i.e. when $C\subseteq \left(\bigcup_{\gamma<\delta}U_{\alpha_\gamma}\right)\cup\theta_{2k-1}(\{x_{\alpha_\gamma}:\gamma<\delta\})$.
Therefore if $A$ is the set of selected points, then $C\subseteq\theta_{2k-1}(A)\cup \bigcup \left\{U^x:x\in A\right\}$.

To finish the proof we need to show that $A$ is an $S(2k-1)$-discrete set in $X$ and therefore $|A|\le \kappa$.
Let $x_{\alpha_{\gamma_0}}\in A$. Then
$x_{\alpha_{\gamma_0}}\notin \theta_{2k-1}(\{x_{\alpha_\gamma}:\gamma<\gamma_0\})$.
Hence, there exists an open $S(2k-1)$-neighborhood $V_{\alpha_{\gamma_0}}$ of $x_{\alpha_{\gamma_0}}$ such that
$V_{\alpha_{\gamma_0}}\cap\{x_{\alpha_\gamma}:\gamma<\gamma_0\}=\emptyset$. Now, suppose that there exists
$\gamma_1$ such that $x_{\alpha_{\gamma_1}}\in A\cap V_{\alpha_{\gamma_0}}\cap U_{\alpha_{\gamma_0}}$.
Since $x_{\alpha_{\gamma_1}}\in V_{\alpha_{\gamma_0}}$, we have $\gamma_1\ge \gamma_0$. From
$x_{\alpha_{\gamma_1}}\in U_{\alpha_{\gamma_0}}$ and
$x_{\alpha_{\gamma_1}}\in U_{\alpha_{\gamma_1}}(k)\setminus\left[\left(\bigcup_{\gamma<\gamma_1}U_{\alpha_\gamma}\right)\cup\theta_{2k-1}(\{x_{\alpha_\gamma}:\gamma<\gamma_1\})\right]$
we conclude that $U_{\alpha_{\gamma_0}}\subsetneq \bigcup_{\gamma<\gamma_1}U_{\alpha_\gamma}$, thus
$\gamma_0\ge\gamma_1$. Hence $\gamma_1=\gamma_0$ and therefore
$\{x_{\alpha_{\gamma_0}}\}= A\cap V_{\alpha_{\gamma_0}}\cap U_{\alpha_{\gamma_0}}$ i.e. $A$ is an
$S(2k-1)$-discrete set in $X$.
\end{proof}

The next lemma for the case $n=1$ was proved by Schr\"oder (see \cite[Lemma 6]{Sch93}).

\begin{lemma}\label{LSchIG}
Let $k\in \mathbb{N}^{+}$, $X$ be a topological space, $\kappa=s_{2k}(X)$ and $C\subseteq X$. For each
$x\in C$ let $U^x$ be an open $S(2k-1)$-neighborhood of $x$ and let $\mathcal{U}=\{U^x:x\in C\}$.
Then there exist an $S(2k)$-discrete subset $A$ of $C$ such that $\left| A\right| \leq \kappa $
and $C\subseteq\theta_{2k}(A)\cup \bigcup \left\{\overline{U}^x:x\in A\right\}$.
\end{lemma}
\begin{proof}
Let $\{U_\alpha:\alpha<\beta\}$ be a well-ordering of the elements of $\sU$ and let $x_0\in C\cap U_0(k)$.
Suppose that the points $\{x_{\alpha_\gamma}:\gamma<\delta\}$ have already been chosen. Let
$\alpha_\delta$ be the first ordinal greater than $\alpha_\gamma$, for each $\gamma<\delta$, such that
there exists $x_{\alpha_\delta}\in (C\cap U_{\alpha_\delta}(k))\setminus\left[\left(\bigcup_{\gamma<\delta}\overline{U}_{\alpha_\gamma}\right)\cup\theta_{2k}(\{x_{\alpha_\gamma}:\gamma<\delta\})\right]$. We finish this selection when such a
point $x_{\alpha_\delta}$ does not exist, i.e. when $C\subseteq \left(\bigcup_{\gamma<\delta}\overline{U}_{\alpha_\gamma}\right)\cup\theta_{2k}(\{x_{\alpha_\gamma}:\gamma<\delta\})$.
Therefore if $A$ is the set of selected points, then
$C\subseteq\theta_{2k}(A)\cup \bigcup \left\{\overline{U}^x:x\in A\right\}$.

To finish the proof we need to show that $A$ is an $S(2k)$-discrete set in $X$ and therefore $|A|\le \kappa$.
Let $x_{\alpha_{\gamma_0}}\in A$. Then $x_{\alpha_{\gamma_0}}\notin \theta_{2k}(\{x_{\alpha_\gamma}:\gamma<\gamma_0\})$.
Hence, there exists an open $S(2k-1)$-neighborhood $V_{\alpha_{\gamma_0}}$ of $x_{\alpha_{\gamma_0}}$ such that
$\overline{V}_{\alpha_{\gamma_0}}\cap\{x_{\alpha_\gamma}:\gamma<\gamma_0\}=\emptyset$.
Now, suppose that there exists $\gamma_1$ such that
$x_{\alpha_{\gamma_1}}\in A\cap \overline{V}_{\alpha_{\gamma_0}}\cap \overline{U}_{\alpha_{\gamma_0}}$.
Since $x_{\alpha_{\gamma_1}}\in \overline{V}_{\alpha_{\gamma_0}}$, we have $\gamma_1\ge \gamma_0$. From
$x_{\alpha_{\gamma_1}}\in \overline{U}_{\alpha_{\gamma_0}}$ and
$x_{\alpha_{\gamma_1}}\in U_{\alpha_{\gamma_1}}(k)\setminus\left[\left(\bigcup_{\gamma<\gamma_1}\overline{U}_{\alpha_\gamma}\right)\cup\theta_{2k}(\{x_{\alpha_\gamma}:\gamma<\gamma_1\})\right]$
we conclude that
$\overline{U}_{\alpha_{\gamma_0}}\subsetneq \bigcup_{\gamma<\gamma_1}\overline{U}_{\alpha_\gamma}$,
thus $\gamma_0\ge\gamma_1$. Hence $\gamma_1=\gamma_0$ and therefore
$\{x_{\alpha_{\gamma_0}}\}= A\cap \overline{V}_{\alpha_{\gamma_0}}\cap \overline{U}_{\alpha_{\gamma_0}}$ i.e.
$A$ is an $S(2k)$-discrete set in $X$.
\end{proof}

The following theorem for the case $k=1$ was proved by Hajnal and Juh\'asz \cite{HajJuh67}.

\begin{theorem}\label{THJIG}
Let $k\in \mathbb{N}^{+}$. If $X$ is an $S(k-1)$-space, then
$\left| X\right|\leq 2^{s_{2k-1}(X)\psi_{2k-1}(X)}$.
\end{theorem}
\begin{proof}
Let $k\in \mathbb{N}^+$, $\kappa=s_{2k-1}(X)\psi_{2k-1}(X)$ and for every $x\in X$ let
$\sB(x)=\{B_\alpha^x:\alpha<\kappa\}$ be a family of open $S(2k-1)$-neighborhoods of $x$ such that
$\{x\}=\bigcap\{B_\alpha^x:\alpha<\kappa\}$.
By transfinite recursion we construct an increasing sequence
$(A_\alpha)_{\alpha<\kappa^+}$ of subsets of $X$ and a sequence of families of $S(2k-1)$-open sets
$(\sU_\alpha)_{\alpha<\kappa^+}$ as follows:
\begin{itemize}
\item[(a)] $\sU_0=\emptyset$ and $A_0=\{x_0\}$, where $x_0$ is an arbitrary point of $X$;
\item[(b)] $|A_\alpha|\le\ 2^\kappa$ for every $\alpha<\kappa^+$;
\item[(c)] $\sU_\alpha=\bigcup\{\sB(x):x\in\bigcup_{\beta < \alpha}A_\beta\}$, for every $0<\alpha<\kappa^+$;
\item[(e)] For $0<\alpha<\kappa^+$, if $W$ is the union
of at most $\kappa$ many elements of $\sU_\alpha$ and $\{F_\gamma:\gamma<\kappa\}$ is
a family of subsets of $\bigcup_{\beta < \alpha}A_\beta$ such that $|F_\gamma|\le\kappa$ for
every $\gamma<\kappa$, and $W\cup\bigcup_{\gamma<\kappa}\theta_{2k-1}(F_\gamma)\ne X$, then we pick a point
$x_W\in X\setminus\left(W\cup\bigcup_{\gamma<\kappa}\theta_{2k-1}(F_\gamma)\right)$. Let $E$ be the set of all such
points $x_W$. Then we set $A_{\alpha}=E\cup\bigcup_{\beta<\alpha}A_\beta$. Since
$|\bigcup_{\beta<\alpha}A_\beta|\le 2^\kappa$,
we have $|\sU_\alpha|\le 2^\kappa$ and therefore $|E|\le (2^\kappa)^\kappa=2^\kappa$.
Thus $|A_\alpha|\le 2^\kappa$ and therefore (b) is satisfied.
\end{itemize}

Now, let $A=\bigcup_{\alpha<\kappa^+}A_\alpha$. Since $|A|\le\kappa^+\cdot 2^\kappa=2^\kappa$, to
finish the proof it is sufficient to show that $X=A$. Suppose that
there exists a point $y\in X\setminus A$ and let $\sB(y)=\{B_\delta^y:\delta<\kappa\}$. Then
$X\setminus\{y\}=\bigcup\{C_\delta:\delta<\kappa\}$, where $C_\delta=X\setminus B_\delta^y$
whenever $\delta<\kappa$, and $y\notin\theta_{2k-1}(C_\delta)$ for each $\delta<\kappa$.
For each $\delta<\kappa$ let $D_\delta=A\cap C_\delta$ and for each $x\in D_\delta$ let $B^x\in\sB(x)$ be such that
$y\notin B^x$. Now, we can apply
Lemma \ref{LSapIG} to each $D_\delta$ and the family $\{B^x:x\in D_\delta\}$ to obtain a subset $G_\delta$
of $D_\delta$ such that $|G_\delta|\le\kappa$ and
$D_\delta\subseteq \theta_{2k-1}(G_\delta)\cup\bigcup \left\{B^x:x\in G_\delta\right\}$.
Let $W=\bigcup_{\delta<\kappa}\bigcup\left\{B^x:x\in G_\delta\right\}$. Then
$A\subseteq W\cup(\bigcup_{\delta<\kappa}\theta_{2k-1}(G_\delta))$ and
$y\notin W\cup(\bigcup_{\delta<\kappa}\theta_{2k-1}(G_\delta))$.
Using the fact that $|\bigcup_{\delta<\kappa}G_\delta|\le\kappa$ and that $\kappa^+$ is a regular cardinal
we can find $\alpha<\kappa^+$ such that $\bigcup_{\delta<\kappa}G_\delta\subseteq \sU_\alpha$.
Then it follows from the construction of the set $A_\alpha$ that there exists
$x\in A_\alpha\cap \left(X\setminus \left(W\cup\left(\bigcup_{\delta<\kappa}\theta_{2k-1}(G_\delta)\right)\right)\right)$.
Thus $x\in A_\alpha\subseteq A\subseteq W\cup(\bigcup_{\delta<\kappa}\theta_{2k-1}(G_\delta))$ -- contradiction.
\end{proof}

The next result for the case $k=1$ was proved by Schr\"oder (see \cite[Theorem 8]{Sch93}).

\begin{theorem}\label{TScIG}
Let $k\in \mathbb{N}^{+}$. If $X$ is an $S(k)$-space, then
$\left| X\right|\leq 2^{s_{2k}(X)\psi_{2k}(X)}$.
\end{theorem}
\begin{proof}
Let $k\in \mathbb{N}^+$, $\kappa=s_{2k}(X)\psi_{2k}(X)$ and for every $x\in X$ let
$\sB(x)=\{B_\alpha^x:\alpha<\kappa\}$ be a family of open $S(2k-1)$-neighborhoods of $x$ such that
$\{x\}=\bigcap\{\overline{B}_\alpha^x:\alpha<\kappa\}$.
By transfinite recursion we construct an increasing sequence
$(A_\alpha)_{\alpha<\kappa^+}$ of subsets of $X$ and a sequence of families of $S(2k-1)$-open sets
$(\sU_\alpha)_{\alpha<\kappa^+}$ as follows:
\begin{itemize}
\item[(a)] $\sU_0=\emptyset$ and $A_0=\{x_0\}$, where $x_0$ is an arbitrary point of $X$;
\item[(b)] $|A_\alpha|\le\ 2^\kappa$ for every $\alpha<\kappa^+$;
\item[(c)] $\sU_\alpha=\bigcup\{\sB(x):x\in\bigcup_{\beta < \alpha}A_\beta\}$, for every $0<\alpha<\kappa^+$;
\item[(e)] For $0<\alpha<\kappa^+$, if $W$ is the union
of the closures of at most $\kappa$ many elements of $\sU_\alpha$ and $\{F_\gamma:\gamma<\kappa\}$ is
a family of subsets of $\bigcup_{\beta < \alpha}A_\beta$ such that $|F_\gamma|\le\kappa$ for
every $\gamma<\kappa$, and $W\cup\bigcup_{\gamma<\kappa}\theta_{2k}(F_\gamma)\ne X$, then we pick a point
$x_W\in X\setminus\left(W\cup\bigcup_{\gamma<\kappa}\theta_{2k}(F_\gamma)\right)$. Let $E$ be the set of all such
points $x_W$. Then we set $A_{\alpha}=E\cup\bigcup_{\beta<\alpha}A_\beta$. Since
$|\bigcup_{\beta<\alpha}A_\beta|\le 2^\kappa$,
we have $|\sU_\alpha|\le 2^\kappa$ and therefore $|E|\le (2^\kappa)^\kappa=2^\kappa$.
Thus $|A_\alpha|\le 2^\kappa$ and therefore (b) is satisfied.
\end{itemize}

Now, let $A=\bigcup_{\alpha<\kappa^+}A_\alpha$. Since $|A|\le\kappa^+\cdot 2^\kappa=2^\kappa$, to
finish the proof it is sufficient to show that $X=A$. Suppose that
there exists a point $y\in X\setminus A$ and let $\sB(y)=\{B_\delta^y:\delta<\kappa\}$. Then
$X\setminus\{y\}=\bigcup\{C_\delta:\delta<\kappa\}$, where $C_\delta=X\setminus \overline{B}_\delta^y$
whenever $\delta<\kappa$, and $y\notin\theta_{2k}(C_\delta)$ for each $\delta<\kappa$.
For each $\delta<\kappa$ let $D_\delta=A\cap C_\delta$ and for each $x\in D_\delta$ let $B^x\in\sB(x)$ be such that
$y\notin \overline{B}^x$. Now, we can apply
Lemma \ref{LSchIG} to each $D_\delta$ and the family $\{B^x:x\in D_\delta\}$ to obtain a subset $G_\delta$
of $D_\delta$ such that $|G_\delta|\le\kappa$ and
$D_\delta\subseteq \theta_{2k}(G_\delta)\cup\bigcup \left\{\overline{B^x}:x\in G_\delta\right\}$.
Let $W=\bigcup_{\delta<\kappa}\bigcup\left\{\overline{B}^x:x\in G_\delta\right\}$. Then
$A\subseteq W\cup(\bigcup_{\delta<\kappa}\theta_{2k}(G_\delta))$ and
$y\notin W\cup(\bigcup_{\delta<\kappa}\theta_{2k}(G_\delta))$.
Using the fact that $|\bigcup_{\delta<\kappa}G_\delta|\le\kappa$ and that $\kappa^+$ is a regular cardinal
we can find $\alpha<\kappa^+$ such that $\bigcup_{\delta<\kappa}G_\delta\subseteq \sU_\alpha$.
Then it follows from the construction of the set $A_\alpha$ that there exists
$x\in A_\alpha\cap \left(X\setminus \left(W\cup\left(\bigcup_{\delta<\kappa}\theta_{2k}(G_\delta)\right)\right)\right)$.
Thus $x\in A_\alpha\subseteq A\subseteq W\cup(\bigcup_{\delta<\kappa}\theta_{2k}(G_\delta))$ -- contradiction.
\end{proof}

Now we are going to formulate and prove the counterpart of the inequality $\left| X\right| \leq 2^{2^{s(X)}}$
for $S(n)$-spaces. For that end we need to formulate and prove for $S(n)$-spaces the counterpart of the
inequality $\psi(X)\leq 2^{s(X)}$, which is valid for every Hausdorff space $X$ \cite{Hod84}.

For the case $k=1$ of the following lemma see \cite[Proposition 4.11]{Hod84}.

\begin{lemma}\label{LHJIG}
Let $k\in \bN^{+}$. For every $S(2k-1)$-space $X$, $\psi(X)\leq 2^{s_{2k-1}(X)}$.
\end{lemma}
\begin{proof}
Let $\kappa=s_{2k-1}(X)$ and $x\in X$. For each $y\in X\setminus\{x\}$ let $U_x^y$ and $U_y$ be open
$S(2k-1)$-neighborhoods of $x$ and $y$, respectively, such that $U_x^y\cap U_y=\emptyset$.
Then for the set $C=X\setminus\{x\}$ and the family $\sU=\{U_y:y\in C\}$ we can use Lemma \ref{LSchIG} to find an
$S(2k-1)$-discrete subset $A$ of $C$ such that $|A|\le\kappa$ and
$C\subseteq \theta_{2k-1}(A)\cup\bigcup \left\{U_y:y\in A\right\}$.

For each $z\in \theta_{2k-1}(A)$ and for each open $S(2k-1)$-neighborhood $U$ of $z$ we can choose a point
$z_U\in U\cap U_z\cap A$ and let us
denote by $F_z$ the resulting set. Then $F_z\subseteq A$,
$z\in\theta_{2k-1}(F_z)\subseteq \theta_{2k-1}(\overline{U}_z)$
and since $U_x^z\cap \overline{U}_z=\emptyset$ we have $x\notin \theta_{2k-1}(\overline{U}_z)$.
Thus $x\notin \theta_{2k-1}(F_z)$.

Let $\sF=\{F:F\subseteq A, x\notin\theta_{2k-1}(F)\}$ and for each $F\in\sF$ let $U_F^x=X\setminus\theta_{2k-1}(F)$.
Then $|\sF|\le 2^\kappa$ and $\theta_{2k-1}(A)\subseteq\bigcup\{\theta_{2k-1}(F):F\in\sF\}$. Therefore
$\{X\setminus\overline{U}_y:y\in A\}\cup\{U_F^x:F\in\sF\}$ is an open pseudobase of $x$ with cardinality
$\le 2^\kappa$.
\end{proof}

\begin{lemma}\label{LIG}
Let $k\in \bN^{+}$. For every $S(2k)$-space $X$, $\psi(X)\leq 2^{s_{2k}(X)}$.
\end{lemma}
\begin{proof}
Let $\kappa=s_{2k}(X)$ and $x\in X$. For each $y\in X\setminus\{x\}$ let $U_x^y$ and $U_y$ be open
$S(2k-1)$-neighborhoods of $x$ and $y$, respectively, such that $\overline{U}_x^y\cap\overline{U}_y=\emptyset$.
Then for the set $C=X\setminus\{x\}$ and the family $\sU=\{U_y:y\in C\}$ we can use Lemma \ref{LSchIG} to find an
$S(2k)$-discrete subset $A$ of $C$ such that $|A|\le\kappa$ and
$C\subseteq \theta_{2k}(A)\cup\bigcup \left\{\overline{U}_y:y\in A\right\}$.

For each $z\in \theta_{2k}(A)$ and for each open $S(2k-1)$-neighborhood $U$ of $z$ we can choose a point
$z_U\in \overline{U}\cap \overline{U}_z\cap A$ and let us
denote by $F_z$ the resulting set. Then $F_z\subseteq A$, $z\in\theta_{2k}(F_z)\subseteq \theta_{2k}(\overline{U}_z)$
and since $\overline{U}_x^z\cap \overline{U}_z=\emptyset$ we have $x\notin \theta_{2k}(\overline{U}_z)$.
Thus $x\notin \theta_{2k}(F_z)$.

Let $\sF=\{F:F\subseteq A, x\notin\theta_{2k}(F)\}$ and for each $F\in\sF$ let $U_F^x=X\setminus\theta_{2k}(F)$.
Then $|\sF|\le 2^\kappa$ and $\theta_{2k}(A)\subseteq\bigcup\{\theta_{2k}(F):F\in\sF\}$. Therefore
$\{X\setminus\overline{U}_y:y\in A\}\cup\{U_F^x:F\in\sF\}$ is an open pseudobase of $x$ with cardinality
$\le 2^\kappa$.
\end{proof}

As a corollary of the previous two lemmas we obtain the following theorem:

\begin{theorem}\label{TISG}
Let $n\in \mathbb{N}^{+}$. If $X$ is an $S(n)$-space, then $\left| X\right|\leq 2^{s(X)\cdot 2^{s_{n}(X)}}$.
\end{theorem}
\begin{proof}
The claim follows directly from the fact that every $S(n)$-space is a $T_1$-space, Hajnal and Juh\'asz theorem
(Theorem \ref{THJIG} for $k=1$), Lemma \ref{LHJIG} for $n$-odd and Lemma \ref{LIG} for $n$-even.
\end{proof}

When $n=1$ in Theorem \ref{TISG} we obtain Hajnal and Juh\'asz theorem that if $X$ is a Hausdorff
space, then $|X|\le 2^{2^{s(X)}}$ and when $n=2$, Theorem \ref{TISG} states that if $X$ is a Urysohn space, then
$|X|\le 2^{s(X)\cdot 2^{Us(X)}}$.

We note that in \cite[Lemma 11]{Sch93} the author claims that if $X$ is a Urysohn space, then $\psi_c(X)\le 2^{Us(X)}$,
or in our notation, $\psi_2(X)\le 2^{s_2(X)}$. Based on that lemma the author concluded that for every Urysohn space $X$
we have $|X|\le 2^{2^{Us(X)}}$. Unfortunately, there is a gap in the proof of Lemma 11 in \cite{Sch93} and therefore
the question whether or not either one of both of these claims is true is open. What we can prove in that relation, in
addition to the case $n=2$ of Theorem \ref{TISG}, is the case $k=1$ of Lemma \ref{LIG2G1} and Theorem \ref{TIG2G1}
(see Corollaries \ref{CIG1} and \ref{CIG2}).

\begin{lemma}\label{LIG2G1}
Let $k\in \bN^{+}$. For every $S(3k)$-space $X$, $\psi_{2k}(X)\leq 2^{s_{2k}(X)}$.
\end{lemma}
\begin{proof}
Let $\kappa=s_{2k}(X)$ and $x\in X$. Since $X$ is an $S(3k)$-space, for each $y\in X\setminus\{x\}$ let
$U_1^y(x),\ldots,U_{k}^y(x)$, $U_1(y),\ldots,U_{2k}(y)$ be open sets in $X$ such that
$x\in U_1^y(x)\subseteq\overline{U}_1^y(x)\subseteq\ldots\subseteq U_{k}^y(x)$,
$y\in U_1(y)\subseteq\overline{U}_1(y)\subseteq\ldots\subseteq U_{2k}(y)$ and
$\overline{U}_{k}^y(x)\cap \overline{U}_{2k}(y)=\emptyset$. Then for each $y\in C$,
${U}_{k}(y)$ is an open $S(2k-1)$-neighborhood of $y$.
Therefore for the set $C=X\setminus\{x\}$ and the family $\sU=\{U_{k}(y):y\in C\}$ we can use Lemma \ref{LSchIG} to
find an $S(2k)$-discrete subset $A$ of $C$ such that $|A|\le\kappa$ and
$C\subseteq \theta_{2k}(A)\cup\bigcup \left\{\overline{U}_{k}(y):y\in A\right\}$.

For each $z\in \theta_{2k}(A)$ and for each open $S(2k-1)$-neighborhood $U$ of $z$ we can choose a point
$z_U\in \overline{U}\cap \overline{U}_k(z)\cap A$ and let us
denote by $F_z$ the resulting set. Then $F_z\subseteq A$,
$z\in\theta_{2k}(F_z)\subseteq \theta_{2k}(\overline{U}_k(z))\subseteq\overline{U}_{2k}(z)$
and since $\overline{U}_{k}^z(x)\cap \overline{U}_{2k}(z)=\emptyset$ we have
$x\notin \theta_{2k}(\overline{U}_{2k}(z))$, hence $x\notin \theta_{2k}(F_z)$ and clearly,
$\overline{U}_{k}^z(x)\cap\theta_{2k}(F_z)=\emptyset$.

Let $\sF=\{F:F\subseteq A, x\notin\theta_{2k}(F)\}$ and for each $F\in\sF$ let $U_F^x$ be an open
$S(2k-1)$-neighborhood of $x$
such that $\overline{U}_F^x\cap\theta_{2k}(F)=\emptyset$.
Then $|\sF|\le 2^\kappa$ and $\theta_{2k}(A)\subseteq\bigcup\{\theta_{2k}(F):F\in\sF\}$. Therefore the existence of the
family $\{\overline{U}_k^y(x):y\in A\}\cup\{\overline{U}_F^x:F\in\sF\}$ shows that the $S(2k)$-pseudocharacter at $x$
has cardinality $\le 2^\kappa$.
\end{proof}

\begin{corollary}\label{CIG1}
For every $S(3)$-space $X$, $\psi_2(X)\leq 2^{s_{2}(X)}$.
\end{corollary}

For the case $k=1$ of the following lemma see \cite[Proposition 4.11]{Hod84}.

\begin{lemma}\label{LIG2G2}
Let $k\in \bN^{+}$. For every $S(3k-2)$-space $X$, $\psi_{2k-1}(X)\leq 2^{s_{2k-1}(X)}$.
\end{lemma}
\begin{proof}
Let $\kappa=s_{2k-1}(X)$ and $x\in X$. Since $X$ is an $S(3k-2)$-space, for each $y\in X\setminus\{x\}$ let
$U_1^y(x),\ldots,U_{k}^y(x)$, $U_1(y),\ldots,U_{2k-1}(y)$ be open sets in $X$ such that
$x\in U_1^y(x)\subseteq\overline{U}_1^y(x)\subseteq\ldots\subseteq U_{k}^y(x)$,
$y\in U_1(y)\subseteq\overline{U}_1(y)\subseteq\ldots\subseteq U_{2k-1}(y)$ and
${U}_{k}^y(x)\cap \overline{U}_{2k-1}(y)=\emptyset$. Then for each $y\in C$,
${U}_{k}(y)$ is an open $S(2k-1)$-neighborhood of $y$.
Therefore for the set $C=X\setminus\{x\}$ and the family $\sU=\{U_{k}(y):y\in C\}$ we can use Lemma \ref{LSapIG} to
find an $S(2k-1)$-discrete subset $A$ of $C$ such that $|A|\le\kappa$ and
$C\subseteq \theta_{2k-1}(A)\cup\bigcup \left\{{U}_{k}(y):y\in A\right\}$.

For each $z\in \theta_{2k-1}(A)$ and for each open $S(2k-1)$-neighborhood $U$ of $z$ we can choose a point
$z_U\in {U}\cap {U}_k(z)\cap A$ and let us
denote by $F_z$ the resulting set. Then $F_z\subseteq A$,
$z\in\theta_{2k-1}(F_z)\subseteq \theta_{2k-1}(\overline{U}_k(z))\subseteq\overline{U}_{2k-1}(z)$
and since ${U}_{k}^z(x)\cap \overline{U}_{2k-1}(z)=\emptyset$ we have
$x\notin \theta_{2k-1}(\overline{U}_{2k-1}(z))$, hence $x\notin \theta_{2k-1}(F_z)$ and clearly,
${U}_{k}^z(x)\cap\theta_{2k-1}(F_z)=\emptyset$.

Let $\sF=\{F:F\subseteq A, x\notin\theta_{2k-1}(F)\}$ and for each $F\in\sF$ let $U_F^x$ be an open
$S(2k-1)$-neighborhood of $x$
such that ${U}_F^x\cap\theta_{2k-1}(F)=\emptyset$.
Then $|\sF|\le 2^\kappa$ and $\theta_{2k-1}(A)\subseteq\bigcup\{\theta_{2k-1}(F):F\in\sF\}$. Therefore the existence of
the family $\{{U}_k^y(x):y\in A\}\cup\{{U}_F^x:F\in\sF\}$ shows that the $S(2k-1)$-pseudocharacter at
$x$ has cardinality $\le 2^\kappa$.
\end{proof}

Similarly, for the case $3k-1$, $k\in \bN^{+}$, one can prove the following:

\begin{lemma}\label{LIG2G3}
Let $k\in \bN^{+}$. For every $S(3k-1)$-space $X$, $\psi_{2k-1}(X)\leq 2^{s_{2k}(X)}$.
\end{lemma}
\begin{proof}
Let $\kappa=s_{2k}(X)$ and $x\in X$. Since $X$ is an $S(3k-1)$-space, for each $y\in X\setminus\{x\}$ let
$U_1^y(x),\ldots,U_{k}^y(x)$, $U_1(y),\ldots,U_{2k}(y)$ be open sets in $X$ such that
$x\in U_1^y(x)\subseteq\overline{U}_1^y(x)\subseteq\ldots\subseteq U_{k}^y(x)$,
$y\in U_1(y)\subseteq\overline{U}_1(y)\subseteq\ldots\subseteq U_{2k}(y)$ and
${U}_{k}^y(x)\cap \overline{U}_{2k}(y)=\emptyset$. Then for each $y\in C$,
${U}_{k}(y)$ is an open $S(2k-1)$-neighborhood of $y$.
Therefore for the set $C=X\setminus\{x\}$ and the family $\sU=\{U_{k}(y):y\in C\}$ we can use Lemma \ref{LSchIG} to
find an $S(2k)$-discrete subset $A$ of $C$ such that $|A|\le\kappa$ and
$C\subseteq \theta_{2k}(A)\cup\bigcup \left\{\overline{U}_{k}(y):y\in A\right\}$.

For each $z\in \theta_{2k}(A)$ and for each open $S(2k-1)$-neighborhood $U$ of $z$ we can choose a point
$z_U\in \overline{U}\cap \overline{U}_k(z)\cap A$ and let us
denote by $F_z$ the resulting set. Then $F_z\subseteq A$,
$z\in\theta_{2k}(F_z)\subseteq \theta_{2k}(\overline{U}_k(z))\subseteq\overline{U}_{2k}(z)$
and since ${U}_{k}^z(x)\cap \overline{U}_{2k}(z)=\emptyset$ we have
$x\notin \theta_{2k-1}(\overline{U}_{2k}(z))$, hence $x\notin \theta_{2k-1}(F_z)$ and clearly,
${U}_{k}^z(x)\cap\theta_{2k}(F_z)=\emptyset$.

Let $\sF=\{F:F\subseteq A, x\notin\theta_{2k-1}(F)\}$ and for each $F\in\sF$ let $U_F^x$ be an open
$S(2k-1)$-neighborhood of $x$
such that ${U}_F^x\cap\theta_{2k}(F)=\emptyset$.
Then $|\sF|\le 2^\kappa$ and $\theta_{2k}(A)\subseteq\bigcup\{\theta_{2k}(F):F\in\sF\}$. Therefore the existence of
the family $\{{U}_k^y(x):y\in A\}\cup\{{U}_F^x:F\in\sF\}$ shows that the $S(2k-1)$-pseudocharacter at
$x$ has cardinality $\le 2^\kappa$.
\end{proof}

\begin{lemma}\label{LIG2G4}
Let $k\in \bN^{+}$. For every $S(3k-1)$-space $X$, $\psi_{2k}(X)\leq 2^{s_{2k-1}(X)}$.
\end{lemma}

\begin{proof}
Let $\kappa=s_{2k-1}(X)$ and $x\in X$. Since $X$ is an $S(3k-1)$-space, for each $y\in X\setminus\{x\}$ let
$U_1^y(x),\ldots,U_{k}^y(x)$, $U_1(y),\ldots,U_{2k}(y)$ be open sets in $X$ such that
$x\in U_1^y(x)\subseteq\overline{U}_1^y(x)\subseteq\ldots\subseteq U_{k}^y(x)$,
$y\in U_1(y)\subseteq\overline{U}_1(y)\subseteq\ldots\subseteq U_{2k}(y)$ and
$\overline{U}_{k}^y(x)\cap {U}_{2k}(y)=\emptyset$. Then for each $y\in C$,
${U}_{k}(y)$ is an open $S(2k-1)$-neighborhood of $y$.
Therefore for the set $C=X\setminus\{x\}$ and the family $\sU=\{U_{k}(y):y\in C\}$ we can use Lemma \ref{LSapIG} to
find an $S(2k-1)$-discrete subset $A$ of $C$ such that $|A|\le\kappa$ and
$C\subseteq \theta_{2k-1}(A)\cup\bigcup \left\{{U}_{k}(y):y\in A\right\}$.

For each $z\in \theta_{2k-1}(A)$ and for each open $S(2k-1)$-neighborhood $U$ of $z$ we can choose a point
$z_U\in {U}\cap {U}_k(z)\cap A$ and let us
denote by $F_z$ the resulting set. Then $F_z\subseteq A$,
$z\in\theta_{2k-1}(F_z)\subseteq \theta_{2k-1}({U}_k(z))\subseteq \overline{U}_{2k-1}(z)$
and since $\overline{U}_{k}^z(x)\cap {U}_{2k}(z)=\emptyset$ we have
$x\notin \theta_{2k}({U}_{2k}(z))$, hence $x\notin \theta_{2k-1}(F_z)$ and clearly,
$\overline{U}_{k}^z(x)\cap\theta_{2k-1}(F_z)=\emptyset$.

Let $\sF=\{F:F\subseteq A, x\notin\theta_{2k-1}(F)\}$ and for each $F\in\sF$ let $U_F^x$ be an open
$S(2k-1)$-neighborhood of $x$
such that $\overline{U}_F^x\cap\theta_{2k-1}(F)=\emptyset$.
Then $|\sF|\le 2^\kappa$ and $\theta_{2k-1}(A)\subseteq\bigcup\{\theta_{2k-1}(F):F\in\sF\}$. Therefore the existence of
the family $\{\overline{U}_k^y(x):y\in A\}\cup\{\overline{U}_F^x:F\in\sF\}$ shows that the $S(2k)$-pseudocharacter at
$x$ has cardinality $\le 2^\kappa$.
\end{proof}

Now, as corollaries of the previous four lemmas we obtain the following results:

\begin{theorem}\label{TIG2G1}
Let $k\in \bN^{+}$. If $X$ is an $S(3k)$-space, then $\left| X\right|\leq 2^{2^{s_{2k}(X)}}$.
\end{theorem}
\begin{proof}
The claim follows directly from the fact that every $S(3k)$-space is an $S(k)$-space, Theorem \ref{TScIG},
and Lemma \ref{LIG2G1}.
\end{proof}

\begin{corollary}\label{CIG2}
If $X$ is an $S(3)$-space, then $\left| X\right|\leq 2^{2^{s_{2}(X)}}$.
\end{corollary}

\begin{theorem}\label{TIG2G2}
Let $k\in \bN^{+}$. If $X$ is an $S(3k-2)$-space, then $\left| X\right|\leq 2^{2^{s_{2k-1}(X)}}$.
\end{theorem}
\begin{proof}
The claim follows directly from the fact that every $S(3k-2)$-space is an $S(k-1)$-space, Theorem \ref{THJIG},
and Lemma \ref{LIG2G2}.
\end{proof}

When $n=1$ in Theorem \ref{TIG2G2} we obtain again Hajnal and Juh\'asz theorem that if $X$ is a
Hausdorff space, then $|X|\le 2^{2^{s(X)}}$.

\begin{theorem}\label{TIG2G3}
Let $k\in \bN^{+}$. If $X$ is an $S(3k-1)$-space, then $\left| X\right|\leq 2^{s_{2k-1}(X)\cdot2^{s_{2k}(X)}}$ and
$\left| X\right|\leq 2^{2^{s_{2k-1}(X)}}$.
\end{theorem}
\begin{proof}
The claim follows directly from the fact that every $S(3k-1)$-space is an $S(k-1)$ and $S(k)$-space, Theorem \ref{THJIG},
Theorem \ref{TScIG}, Lemma \ref{LIG2G3}, Lemma \ref{LIG2G4} and the fact that $s_{2k}(X)\le s_{2k-1}(X)$ for
every $k\in \bN^{+}$.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Au Canada, un territoire non organisé (TNO) est une région géographique qui n'est pas une municipalité ou une communauté autochtone.

Alberta

En Alberta, les territoires non organisés sont divisés en districts d'amélioration. Les districts d'améliorations sont des territoires où une faible densité de population et une faible assiette fiscale ne permettent pas de viabiliser une administration locale. Les districts sont administrés par le gouvernement de l'Alberta qui s'occupe de fournir des services à la population, sauf si un conseil est élu.

Nouveau-Brunswick

Au Nouveau-Brunswick, les territoires non organisés sont divisés en districts de services locaux. Les districts sont administrés par le ministère de l'Environnement et des Gouvernements locaux du Nouveau-Brunswick qui s'occupe de fournir des services à la population.

Ontario

En Ontario, les territoires non organisés n'existent que dans le Nord de l'Ontario où il n'y a pas d'administration de comté ou de municipalité. Cependant, quelques communautés situées dans les territoires non organisés ont formé des régies de services locaux. À des fins de recensement, les territoires non organisés ontariens sont nommés selon le nom du district dont ils font partie avec un qualitatif géographique s'il y a plus d'un tel territoire à l'intérieur du même district. Trois territoires non organisés n'avaient aucune population en 2006.

Liste des territoires non organisés de l'Ontario

 Algoma, UNO, North Part
 Algoma, UNO, South East Part
 Cochrane, UNO, North Part
 Cochrane, UNO, South East Part
 Cochrane, UNO, South West Part
 Kenora, UNO
 Manitoulin, UNO, Mainland
 Manitoulin, UNO, West Part
 Nipissing, UNO, North Part
 Nipissing, UNO, South Part
 Parry Sound, UNO, Centre Part
 Parry Sound, UNO, North East Part
 Rainy River, UNO
 Sudbury, UNO, North Part
 Thunder Bay, UNO
 Timiskaming, UNO, East Part
 Timiskaming, UNO, West Part

Québec

Au Québec, un territoire non organisé est une région qui ne fait partie d'aucune municipalité locale et dont la gestion est assurée par la municipalité régionale de comté (MRC) dont il fait partie.

La MRC est tenue d'adopter un règlement de zonage, de lotissement et de construction s'appliquant à ces territoires.

La tenure des terres d'un territoire non organisé est généralement publique. De façon générale, il est situé en milieu forestier inhabité : on compte moins de  dans l'un ou l'autre des 103 territoires. Les établissements permanents comprennent des sites de villégiature, tels que Sagard ou Saint-Guillaume-Nord, ou d'anciens cantons de colonisation n'ayant jamais été municipalisés, tels que Mont-Albert ou Lac-Chicobi. Une MRC peut constituer pour les localités habitées un comité local élu ayant un pouvoir d'étude et de recommandation concernant les sujets d'intérêt pour les résidents.

Notes et références

Liens externes
Organisation municipale au Québec - Voir à la page 10 de ce document une carte montrant l'étendue des territoires non organisés au Québec.[Lien ne fonctionne pas]

Administration territoriale au Québec
Géographie de l'Ontario

redpajama_set_name: RedPajamaWikipedia


text:
Home»Posts tagged with»Willie Nelson
Beachlife Festival Announces Inaugural Lineup: Brian Wilson, Willie Nelson, Bob Weir Headlining CA Fest in May 2019
By Rock Cellar Magazine Staff on January 14, 2019 Latest News
There's announcing your first lineup, and then there's announcing your first lineup in the way the Beachlife Festival did on Monday. Set for picturesque Redondo Beach, California on May 3-5, the fest is just teeming with talent spanning genres and decades, all in the name of what's sure to be a great weekend on the […]
Willie's Reserve: Willie Nelson's Cannabis is at the Forefront of a Growing Industry
By Jeff Slate on October 11, 2018 Featured Articles
"When we get to New York you'll have to come visit me on the tour bus," Willie Nelson said, when I told him how disappointed I was to be interviewing him by phone rather than in person. "We have fun on the bus. I think you'd like it." He then let out a long, hearty […]
Willie Nelson Premieres Studio Version of the Charged 'Vote 'Em Out,' Ahead of Election Day on 11/6 (Listen)
By Rock Cellar Magazine Staff on October 11, 2018 Latest News
Voting is important — and Willie Nelson knows that. Fresh off the release of his Frank Sinatra tribute album My Way and his appearance at a rally for Beto O'Rourke, the challenger to incumbent Senator Ted Cruz (which ruffled feathers among Nelson's fans who … somehow … weren't aware of his political leanings, which have […]
Stream Willie Nelson's New Frank Sinatra Tribute Album 'My Way' (New Music Friday)
By Rock Cellar Magazine Staff on September 13, 2018 Latest News
At 85, Willie Nelson remains one of the hardest-working musicians around, and today he adds to his dense catalog of music with My Way, a musical tribute to the late Frank Sinatra — whom Nelson considered an inspiration and a friend. "I learned a lot about phrasing listening to Frank," Willie said recently in an […]
Willie Nelson Premieres Video for 'One for My Baby (And One More for the Road),' from New 'My Way' LP
This Friday, Sept. 14, Willie Nelson will release My Way, his tribute album to the late Frank Sinatra. The 85-year-old Nelson has previewed his reverent homage to the Rat Pack icon with a few tracks already, and on Monday he premiered the third (and last) pre-release song, his take on "One for My Baby (And […]
Preview Willie Nelson's Sinatra Tribute Album 'My Way' with His New Video for 'I'll Be Around'
By Rock Cellar Magazine Staff on August 27, 2018 Latest News
On September 14, Willie Nelson will add to his ever-growing collection of late-period releases with My Way, a musical tribute to the late Frank Sinatra. Recently, Willie premiered a new clip for his elegant, refined take on the Sinatra song "I'll Be Around," featuring excellent piano work from Buddy Cannon: "I learned a lot about […]
Willie Nelson Premieres Video for 'Summer Wind,' from Frank Sinatra Tribute LP 'My Way' Out Soon
By Rock Cellar Magazine Staff on July 19, 2018 Latest News
At 85, Willie Nelson is busier than most artists decades younger than him. Having just released a powerful new record titled Last Man Standing in time for his birthday in April, he already has another new one in the works. This one is titled My Way, and it's a tribute by Nelson to one of […]
The Wild Feathers Q&A: On 'Greetings from the Neon Frontier' and 'Finding That Magic Spot' with Three Singer/Songwriters
By George A. Paul on July 12, 2018 Featured Articles
Any band with three singer/songwriters could easily get bogged down by disagreements about who's going to handle lead and secondary vocal duties — but not The Wild Feathers. "Most of the time it's pretty straightforward," said bassist Joel King. "You get in where you can fit in. A lot of times, I'll be singing the […]
Happy Early 85th Birthday to Willie Nelson — Stream His Powerful New Album 'Last Man Standing'
By Rock Cellar Magazine Staff on April 27, 2018 Latest News
Over the past few weeks, country legend Willie Nelson has premiered a few songs from his new album, Last Man Standing — and each has shown the living icon at his most vulnerable, raw, and self-reflective. Its release coincides with Nelson's 85th birthday on Sunday, April 29. Today, that album was released, and it's just […]
4/20 Greetings from Willie Nelson: New 'Ready to Roar' Video and a Preview of New Willie's Reserve Cannabis Strand
Next week, Willie Nelson will celebrate his 85th birthday with a new album, Last Man Standing. The country legend has been previewing it for a few weeks now with various videos of songs from the recording sessions, and he's now offered up a fourth selection. Titled "Ready to Roar," the song again features a video […]

redpajama_set_name: RedPajamaCommonCrawl


Q. I need your help and expertise. Now I'm sure all of you own suits and a majority of you are of the MTM variety (which is what I need). I recently got engaged and planning the wedding for next November (2014) and decided to buy suits rather than rent because I think it is pointless to spend that type of money for one day and nothing to show for it in the end. Half of them don't own a suit either, so it would help them a great deal. I have 9 guys to dress, plus the fathers. We are all different sizes which leads me to the MTM course.
To all of you out there..Where can I go (online preferably) for the most affordable yet quality MTM suits?
Pricing is a range right now which is why I'm planning this far out. I don't want to break my groomsmen's pockets but I don't want them looking crummy either.
Also, any ideas for shoes, Dress shirts, ANYTHING!!!
Suitsupply will help you out.
Suitsupply and Indochino are very good choices. But since you have some time, I would watch Jos A Bank for one of their sales (which are usually buy one, get one free). It will probably be easier to deal with live, in-person fittings and most of the men in your party should have one nearby. And don't get scared by snobbery, Jos A Bank is a good store (I did not say GREAT store) and it is well-beyond what you would get at Men's Scarehouse.
Here is the stock SF answer, which is basically what I'm doing. Just let them get a charcoal gray solid suit (or navy, or allow them to choose). This would be the most versatile suit they could own and is formal enough for a wedding. Then, assuming your wedding is at least somewhat formal, have them wear black dress shoes and a white shirt and provide coordinating (not matching) ties. This could be a traditional "wedding tie" (a black and white or silver tie in a small-scale pattern such as houndstooth, glen plaid, or shepherd's check) or a tie that coordinates with the wedding colors.
Now, if you insist on getting the exact same suit, from all I've read, online MTM such as Indochino is very hit-and-miss. If it fits your budget, Kent Wang has very positive reviews, but that's probably too much for your groomsmen. I would bet you'd have better success at a lower price point having them buy OTR suits and take them to the tailor. SuitSupply would also be an option since you're in NYC. That way those who want to pay for MTM have the option, those who don't can get probably the best suit they will get for under $500.
BTW, is it common now to dress the fathers with the wedding party?
Try Proper Suit MTM in NYC. They have a solid charcoal or navy starting at $750. I saw a wedding party picking theirs up and they all looked great.
From what I have seen of Suitsupply they do a good job but their style tends to be a bit more fashion driven in style (peal lapels, higher button stance, much more fitted, etc). From what I've seen Proper Suit has more of a classic modern look that is appropriate for more formal and business settings. Check them both out and see what suits you..
I think you will increase your chances of success with any MTM if you write down a clear list of what you want in fit and details. Take in a photo or an example of a great fitting suit or jacket. Too often there is a communication breakdown between the Sales Associate and customer. Words like "Trim" or "classic" , "narrow", "wide" take on very different meanings with different people.
The good news is that there are some good options all under $1000 a suit. Good luck!
If you still have time, depending on when the wedding is, you can wait until July sales - get everyones sizes, and grab up fitting suits from the end of the MrP sale or TheCorner sale for a very nice fraction of the normal $. I did this and got 8 tuxes for the price of two or so on MrP and had them reimburse me. If your wedding is sooner than that then perhaps someone else reading this may take advantage. Good luck.


text:
Q: Could not build wheels for frozenlist, multidict, yarl I am trying to make a discord bot with nextcord, but when I try to install it using command prompt, it gives me this error:
 × Building wheel for yarl (pyproject.toml) did not run successfully.
  │ exit code: 1
  ╰─> [37 lines of output]
      C:\Users\klama\AppData\Local\Temp\pip-build-env-pknvci1v\overlay\Lib\site-packages\setuptools\config\setupcfg.py:508: SetuptoolsDeprecationWarning: The license_file parameter is deprecated, use license_files instead.
        warnings.warn(msg, warning_class)
      **********************
      * Accelerated build *
      **********************
      running bdist_wheel
      running build
      running build_py
      creating build
      creating build\lib.win-amd64-cpython-311
      creating build\lib.win-amd64-cpython-311\yarl
      copying yarl\_quoting.py -> build\lib.win-amd64-cpython-311\yarl
      copying yarl\_quoting_py.py -> build\lib.win-amd64-cpython-311\yarl
      copying yarl\_url.py -> build\lib.win-amd64-cpython-311\yarl
      copying yarl\__init__.py -> build\lib.win-amd64-cpython-311\yarl
      running egg_info
      writing yarl.egg-info\PKG-INFO
      writing dependency_links to yarl.egg-info\dependency_links.txt
      writing requirements to yarl.egg-info\requires.txt
      writing top-level names to yarl.egg-info\top_level.txt
      reading manifest file 'yarl.egg-info\SOURCES.txt'
      reading manifest template 'MANIFEST.in'
      warning: no previously-included files matching '*.pyc' found anywhere in distribution
      warning: no previously-included files matching '*.cache' found anywhere in distribution
      warning: no previously-included files found matching 'yarl\*.html'
      warning: no previously-included files found matching 'yarl\*.so'
      warning: no previously-included files found matching 'yarl\*.pyd'
      no previously-included directories found matching 'docs\_build'
      adding license file 'LICENSE'
      writing manifest file 'yarl.egg-info\SOURCES.txt'
      copying yarl\__init__.pyi -> build\lib.win-amd64-cpython-311\yarl
      copying yarl\_quoting_c.pyi -> build\lib.win-amd64-cpython-311\yarl
      copying yarl\_quoting_c.pyx -> build\lib.win-amd64-cpython-311\yarl
      copying yarl\py.typed -> build\lib.win-amd64-cpython-311\yarl
      running build_ext
      building 'yarl._quoting_c' extension
      error: Microsoft Visual C++ 14.0 or greater is required. Get it with "Microsoft C++ Build Tools": https://visualstudio.microsoft.com/visual-cpp-build-tools/
      [end of output]

  note: This error originates from a subprocess, and is likely not a problem with pip.
  ERROR: Failed building wheel for yarl
Failed to build frozenlist multidict yarl
ERROR: Could not build wheels for frozenlist, multidict, yarl, which is required to install pyproject.toml-based projects

I've tried updating my pip, but it says "Requirement already satisfied". I've been looking for about 2 hours an a half and still can't find it.

A: Just instal python 3.10. I tried instal Microsoft Visual C++ 14.0... did not work out

A: The error message you posted says it all: you need Microsoft Visual C++ 14.0 or greater.
For those of you here because you got this error after updating your Python Docker image to Python 3.11, it may be because gcc and other C++ libraries aren't included in the image anymore. For example, this can be fixed in 3.11.0-slim-buster by adding the following line to your Dockerfile
RUN apt-get update && \
    apt-get install -y --no-install-recommends build-essential && \
    rm -rf /var/lib/apt/lists/*

redpajama_set_name: RedPajamaStackExchange


How does it feel when your life suddenly changes? It's exhausting yet very exciting. But you might be wondering what I am talking about, right?
Now, don't you think I didn't want to write anything, when in the truth, just almost two months ago I took a step to the unknown and it has been a ride ever since. I changed my job.
In recent years, I was set up in very easy flow working environment that many of you desire, I am sure. I was a freelancer so I worked from home and was in charge of my own day. Pretty much. I woke up on my own schedule mostly, walked the dogs, ate breakfast and started working. All my day was dictated mostly by needs of my clients and my own projects, including this blog. But this sort of workday lasted till 7 or 8 p.m. and I didn't really get to enjoy the day. Plus I was closed at home all the time.
Can you guess how many steps from the recommended 10 000 I did in a day? Hardly 2 000. And with more than 10 hours of work I was closed in my bubble that made me fatter and less active.
I told myself I really need a change so I started to look around for something new. Coincidentaly I got a call from one sweet headhuntress with a proposition to work for one of the biggest Czech companies as a project manager. I was backing off at first, because this position requires me to travel every day. But more I found out about the offer, the more I was inclined to accept because so many things were setting so nicely that I came to conclusion I should accept.
It's a challenge that I have been diving into last two months. Instead of getting up at 8 I have to get up at 5:30 a.m. After I leave, it's an one hour drive and I am just starting to drive so it's sort of a drama for me. Right after it's a day filled with new learnings. After that return to Prague and then household work. That's my new everyday. Maybe a routine for some, and you might consider it a step back. For me it's a step further, a challenge that brings a new regime. Just so you know, I tripled my daily step count, I started eating regularly and lost 3 kilos already, after I get home I have no extra work and can use my time effectively like sewing, exercise and rest. Plus I will learn to drive properly.
I am not going to explain to you about how these changes are bringing a stress and how tired I am. But I am truthfully happy, I can work on things with purpose and I know the things I work on matter. So don't be mad with lack of articles, but when I get the hang of it all, I'll also write more.


text:
films seen
Japan - 60 years old
IMDb and Wikipedia
The undisputed king of Japanese cyberpunk cinema. A real renaissance man who wrote, shot, edited, directed and even acted in many of his own films. If you like your films raw and undiluted, Tsukamoto's oeuvre is a real treasure trove.
Beyond exceptional
Tetsuo: The Iron Man
by Shinya Tsukamoto
Read my full review
Tsukamoto's Tetsuo is an almost perfect cyberpunk endeavor, sporting lush black and white visuals and a superb industrial soundtrack, while providing a nerve-wrecking and overwhelming experience.
Drama, Mystery
Fans of Tsukamoto might be a little surprised by the stilted acting, the strong dramatic undercurrents and the softer, more poetic scenes.
Rare treats
Tetsuo: The Bullet Man
Usually negative feedback lowers people's expectations (which is a good thing), in this case many potential viewers seemed to dismiss this film completely without even giving it a fair chance.
A Snake of June
Rokugatsu no Hebi
Tsukamoto went through his own transformation when the previous millennium ended, A Snake of June is one of the films that illustrates that pretty well.
Drama, Action
Starts off a little restrained, especially for a Tsukamoto film, but once it gets going it becomes an unstoppable force. Well acted, beautifully scored and largely defined by Tsukamoto's trademark camera work, Killing is a film that may be short and light on plot, but leaves a big impression nonetheless. Great stuff.
Tokyo Fist
Tokyo Ken
Tokyo Fist is somewhat of a forgotten Tsukamoto film, which is weird considering its many qualities.
Horror, Mystery
Haze is a landmark in Tsukamoto's oeuvre. The possibilities of DV allowed him to go back to his roots.
Kotoko is another work of love.
Soseiji
Gemini could be considered a good entry-level Tsukamoto since his typical style isn't as demanding as in his other films.
Fires on the Plain
This is not a very happy film, nor is it very subtle.
Nightmare Detective
Akumu Tantei
Japan [2006] - 106 mins
It's still a great film though. Even a slightly watered down Tsukamoto film is way better than most of the J-Horror films released in the past 15 years.
Bullet Ballet
The second time around I was little let down by the acting performances of the secondary cast. For a film that puts more focus on drama than usually the case, that's somewhat of a problem.
Nightmare Detective 2
Akumu Tantei 2
Nightmare Detective 2 might be his most commercial work to date, there is still plenty here to scare away the regular movie fan.
Solid pieces
Tetsuo II: Body Hammer
Adventures of Electric Rod Boy
Denchu Kozo no Boken
Comedy, Horror, Sci-fi
The inoffensive
Kaidan Horror Classics
by Masayuki Ochiai, Shinya Tsukamoto, Sang-il Lee, Hirokazu Koreeda
Ayashiki Bungo Kaidan
Decent anthology, sadly without stand-out entries. Kaidan stories aren't very scary or gory, so don't expect any modern horror action. Tsukamoto and Koreeda deliver the best entries, Lee's one is decent, the only subpar short is coming from Ochiai. Ironically, the only true horror director present.
by Ryuichi Hiroki, Suzuki Matsuo, Miwa Nishikawa, Tetsuo Shinohara, Shinya Tsukamoto
Fîmeiru
Dubious filler
Hiruko the Goblin
Yôkai Hantâ: Hiruko
Comedy, Fantasy, Horror
Big nopes
Venice 70: Future Reloaded
by Hala Abdallah, John Akomfrah, Catherine Breillat, Júlio Bressane, Rama Burshtein, Antonio Capuano, Peter Ho-Sun Chan, Isabel Coixet, Amiel Courtin-Wilson, Claire Denis, Lav Diaz, Amit Dutta, Atom Egoyan, Aleksey Fedorchenko, Frédéric Fonteyne, James Franco, Lluís Galter, Haile Gerima, Aleksey German, Amos Gitai, Sang-soo Hong, Benoît Jacquot, Zhangke Jia, Semih Kaplanoglu, Abbas Kiarostami, Yorgos Lanthimos, Pablo Larraín, Tobias Lindholm, Guido Lombardi, Jazmín López, Samuel Maoz, Pietro Marcello, Brillante Mendoza, Celina Murga, Amir Naderi, Shirin Neshat, Nicolás Pereda, Franco Piavoli, Giuseppe Piccioni, Edgar Reitz, João Pedro Rodrigues, Walter Salles, Paul Schrader, Ulrich Seidl, Luca Severi, Sion Sono, Jean-Marie Straub, Tusi Tamasese, Tariq Teguia, Pablo Trapero, Athina Rachel Tsangari, Teresa Villaverde, Bing Wang, Apichatpong Weerasethakul, Andrew Wonder, Yonfan, Krzysztof Zanussi, Karim Aïnouz, Bernardo Bertolucci, Jan Cvitkovic, Davide Ferrario, Monte Hellman, Marlen Khutsiev, Milcho Manchevski, Franco Maresco, Salvatore Mereu, Ermanno Olmi, Michele Placido, Todd Solondz, Shekhar Kapur, Heeraz Marfatia, Kim Ki-duk, Shinya Tsukamoto
USA [2013] - 120 mins
Browse directors
16 films seen
Lik-Chi Lee
Marco Mak
Ryuhei Kitamura
Takashi Shimizu
Hideo Nakata
more directors


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PLANNING FOR ONE OF THE BIGGEST SOCIAL CHALLENGES IN THE COMING YEARS. Have you thought about the cost of care in later life? One of the biggest challenges of the 21st century is Britain's ageing population. There is continually much talk about later life care in the media, and it's a subject that is going to be one of the biggest social challenges in the coming years.
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Mary Louise (Gullion) Ashman, age 93, passed away on Saturday afternoon, June 10, 2017 at the Indiana Masonic Home. She was a resident of Franklin and a former resident of Morgantown.
Mary Louise was born on September 15, 1923 in Hendricks County, to her parents, the late Burton G. and Vernie E. (Dautrich) Gullion. She was a 1941 graduate of Arsenal Tech High School in Indianapolis. Mary Louise married Harry T. Ashman on September 19, 1942 in Indianapolis. He preceded her in death on April 2, 2013 after 70 devoted years of marriage.
Mary Louise worked at Eli Lilly & Co. during the 1940's. She was a loving wife, mother and homemaker. She was a 50 year member of the Morgantown First Christian Church. Other memberships include the Morgantown Eastern Star, Morgantown Kappa Kappa Sigma Sorority and the Nineveh Senior Center.
Mary Louise enjoyed spending time with her family, especially her grandchildren and great grandchildren. She was an avid reader who also enjoyed sewing, crocheting and collecting out house miniatures.
She is survived by her children, Rebecca (Richard) Soendlin of Indianapolis, Tina Massey & Libby (Robert) Torrance both of Franklin and Kenneth M. (Barbara) Ashman of Naples, FL; 16 grandchildren, several great grandchildren & great-great grandchildren.
Mary Louise was preceded in death by siblings, Kenneth Gullion, Kathryn Gullion and Doris Rahn; and a son in law, Richard Massey.
The Rev. Burt Brock will conduct a funeral service on Thursday at 1 p.m. at Meredith-Clark Funeral Home Cremation & Personalization Center, 179 E. Mulberry Street in Morgantown. Friends may call from 4 to 8 p.m. on Wednesday at the funeral home. Burial will be at Forest Lawn Memory Gardens in Greenwood.
The family would like to thank the Indiana Masonic Home and Heart to Heart Hospice for their kindness and loving care.
Memorial contributions may be sent in honor of Mary Louise to the Morgantown First Christian Church, "Building Fund", PO Box 306, Morgantown, IN 46160.
Condolences to the Ashman family can be made at www.meredith-clark.com.
We were sad to hear of Mary's passing. I attended Morgantown High school with her son Ken and was privileged to know this fine lady. Our deepest sympathy and prayers are with the family.
Blessings and condolences to your family in this time of sorrow. Prayers to all.
Tori, Sam, Jack and Ashman Family: so very sorry for your loss. Sounds as if Mary Louise was a very Special Lady and you were all very lucky to have had her, as a part of your life.


text:
Finance & Capital Market
Industry & Logistics
Media & Promotion
Analysis, Interviews, Opinions
Design International names Dubai office director
Design International, an innovative architectural practice with vast experience on a global scale, has appointed Mark O'Neill RIBA as the new office director for its Dubai office. Regarded as one of the i
Blockchain tech firm Bitfury opens UAE office
The Bitfury Group, a leading full-service blockchain technology company, has expanded its global operations into the UAE, opening a Dubai office aiming to build blockchain-based solutions for governments and institutions across the region.
Injazat Data Systems opens Dubai office
Abu Dhabi-based Injazat Data Systems, a leader in IT, data center and managed services, has announced its expansion into the Emirate of Dubai. This move brings the Abu-Dhabi-based firm closer to its private and public sector clien
GHD expands in UAE with new Dubai office
GHD, a leading global professional services company, is extending its presence in the UAE with the launch of a Dubai office. GHD has been operating continuously in the UAE since 2000, and employs approximately 130 people in Abu Dh
SAI Global opens Dubai office
SAI Global, a leading provider of integrated risk management solutions, has opened an office in Dubai to address increasing demand and growth the company has experienced since it first entered the region in 2015. SAI Global will b
China's BOE sets up EMEA base in Dubai
China-based BOE Technology Group, a supplier of Internet-of-Things technologies, products and services, has opened its Middle East subsidiary in Dubai, UAE. This is another important move taken by BOE in its expansion of the overs
Russia's InfoWatch sets up base in Dubai
InfoWatch Group, a Russian vendor of end-to-end enterprise cybersecurity solution, has announced the opening of its subsidiary InfoWatch Gulf, registered at the Dubai Multi Commodities Centre (DMCC) in Dubai, UAE. The group has ap
Bracewell adds partner to Dubai office
Bracewell, a leader in assisting clients with all types of infrastructure projects, has announced that Adrian Creed has joined the firm as a partner in its Dubai office. Creed has more than 20 years of experience representing clie
UK's Future Thinking opens new Dubai office
UK-based Future Thinking, a top data-driven business intelligence consultancy, has teamed up with its sister agency GemSeek for a new Dubai office, rolling out their combined offering for the Mena region. The new office, based at
Berry Appleman & Leiden opens Dubai office
Berry Appleman & Leiden (BAL), a leading corporate immigration law firm, said it has officially opened its new regional head office in Dubai. Bal previously announced that Hadi Allawi had been hired as managing director for Mi
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Sectors |
Government & Laws
TradeArabia Network


Should I run while I'm pregnant?
Home » Should I run while I'm pregnant?
Running during pregnancy is a contentious topic. The truth is, when it comes down to all forms of exercise, health professionals make the decision based on RISK vs BENEFIT. So when it comes to running during pregnancy, the question is whether there is any risk to you (eg pelvic floor, ankle injury etc) or risk to the baby (overheating for example) vs the benefit (maintain fitness, mental health). The list below is to help you weigh up the risk and benefits so you can make an educated decision on whether you should run during pregnancy.
National guidelines state that there is a general consensus that walking, jogging, cycling, swimming, strengthening exercise and water based exercise are generally safe during an uncomplicated pregnancy. It is stated that for women who participated in physical activity prior to pregnancy (ie running) and are experiencing an uncomplicated pregnancy, that they should continue that activity (ie running) until such a time that it becomes uncomfortable to do so.
For women who were previously inactive – they should be encouraged to commence activity during pregnancy, starting with low intensity exercise (walking or swimming) and slowly progress the minutes of exercise they do.
Basically what this summarises is that if you have always been a runner, and were running consistently prior to your pregnancy, you can continue running as long as there are no complications in your pregnancy and you are comfortable when you run. If you have never been a runner, during pregnancy is not the time to start.
Did you have any pelvic floor symptoms prior to pregnancy?
If you had urine/wind/faeces leaking during running prior to pregnancy, or feelings of prolapse prior to pregnancy, it would be advisable to consider more low impact exercise than running. If you have those symptoms you should be assessed by your gynaecologist, GP or pelvic floor physiotherapist.
During pregnancy there is increased ligamentous laxity, which includes the ligaments of your pelvic floor. Laxity means your ligaments stretch more easily, so the above mentioned symptoms may worsen with increased pressure. You want to take into consideration your long term goals for running, and perhaps taking a break while you are pregnant to protect your pelvic floor, which will allow you to return back to running in the future.
As you progress through your pregnancy and your baby grows, and your body shape and centre of gravity changes, you may feel that when you run you become uncomfortable or get pain in certain areas. You do not want these pains to increase so listen to your body. If you are able to run easily, pain free and comfortable, these are signs you are safe to continue running.
It is highly recommended that you discuss your exercise with your doctor or obstetrician during your pregnancy to know you are medically cleared for exercise. If there is any risk to your or your baby, your doctor will guide you on what exercise to choose. Some women can run until 36 weeks, others until 12 weeks. The most important thing is that you are comfortable, you and your baby are safe, and your mental health is ok. For some women, running is their emotional outlet, so stopping running during pregnancy can be very difficult. If you are having any struggles with this – as many women do; talk to your doctor about it for support and to make an action plan.
Click here to book an appointment with Emily or view our Clinical Pilates timetable here.


text:
Music & Classical Greece
Choirs and Bands
Flight from home to Athens.
(L, D) Arrival in Athens- probably morning. Panoramic City tour, lunch. Hotel check-in. Explore the beauty of Athens and visit all major sights and attractions, many of which are located within walking distance from the hotel, making it even easier and more pleasurable to discover the historic center of Athens and its hidden treasures. Welcome Dinner with traditional Greek music and Greek dance lessons. Overnight Athens.
(B,L,D, Performance) Acropolis and New Acropolis Museum in the morning and lunch. Outdoor concert in Athens' Syntagma Square, which is the main square of Athens with significant pedestrian traffic and located right in front of the Parliament house. Afternoon tour. Dinner in Plaka, the old section of Athens. Overnight Athens.
(B,L) Full-day cruise from 8:00am- 7:00pm to the islands of Hydra, Aegina and Poros. Lunch included. Let Us Delight You with the Beauty of the Greek Islands! Athens One-Day Cruise to 3 Islands of Aegina – Poros – Hydra We thought this would be a great way for the group to see some of the Greek Islands without the big expense of traveling to and staying in the Greek Islands. It is a beautiful cruise with lunch on board. This evening the group arrives back at the hotel late. We can opt to do dinner on own this one evening. Dinner at hotel. Overnight Athens.
(B,L,D- Performance) Today we travel from Athens to Corinth and Nauplion. We will see the Corinth Canal, which connects the Peloponnese to the rest of Greece. It was conceived by the tyrant Periander in the sixth century BC to link the Ionian Sea and Aegean Sea. Its completion, however, didn't come about until the late 1800s after modern Greece gained independence. We will see ancient Corinth, the Temple of Apollo, and the Corinth Archeological Museum. We continue to Nafplio. It is located on the Saronic Gulf and it is often dubbed the most romantic destination in Greece. Napflio, the original capital of the modern State of Greece (1823–1834), boasts stunning Venetian architecture and Ottoman structures and offers a perfect romantic getaway from Athens. This evening an outdoor concert. Overnight in Nauplion.
(B,L,D, Performance) Visit Mycenae - Mycenae is the largest and most important center of the civilization that was named "Mycenaean" after this very citadel. Mycenaean is the culture that dominated mainland Greece, the Aegean islands, and the shores of Asia Minor during the late Bronze Age era (circa 1600-1100 BCE). The Mycenaean Era occupies the tail end of the Helladic Civilization, which flourished in mainland Greece since 3000 BCE. We will also visit Epidaurus. Epidaurus is one of the most popular archaeological sites in Greece. Located on the northeastern side of Peloponnese, in the region of Argolis, Epidaurus is mostly famous for its Ancient Theatre. This theatre was constructed in the late 4th century BC to host religious ceremonial events in honor of god Asclepius, whose healing center was located few steps away. Famous for its symmetry and the incredible acoustics, this classical theatre is surrounded by lush greenery and gives nice view to the valley below. Outdoor concert. Overnight Nauplion.
(B,L,D) This morning we leave Nauplion for Olympia, the site of the original Olympic games. Olympia is an ancient site on Greece's Peloponnese peninsula that hosted the original Olympic Games, founded in the 8th century B.C. Its extensive ruins include athletic training areas, a stadium and temples dedicated to the gods Hera and Zeus. The Archaeological Museum of Olympia exhibits finds from the site, including a statue of Hermes attributed to the sculptor Praxiteles. Dinner and overnight Olympia.
(B,L,D, Performance) Today we depart Olympia for Delphi. We cross the Rion-Antirion bridge to the northern part of Greece and on to Delphi site of the ancient oracle. The Rion–Antirion Bridge, officially the Charilaos Trikoupis Bridge, is one of the world's longest multi-span cable-stayed bridges and the longest of the fully suspended type. It crosses the Gulf of Corinth near Patras, linking the town of Rio on the Peloponnese peninsula to Antirrio on mainland Greece by road. Visit the actual site of the oracle with the stadium, archeological site and the museum of Delphi. Marching and moving outdoor performance from the hotel into the small town of Delphi. Performance in the town square. Overnight Delphi.
(B,L,D) Delphi to Athens. Lunch on the way. Take funicular up to the highest point of Athens Lycabettus Hill. See the view of Athens. Standing 277 meters above sea level, Lycabettus Hill (sometimes spelt Lykavitos) is the highest point of Athens. Although a beautiful walk up via a circular path, it will be a test of endurance and a challenge in summer. There is also a funicular to take you up to the top. The view from Lycabettus Hill is best enjoyed at sunset whilst waiting for the lights of the Acropolis, Temple of Olympian Zeus, Panathenaic Stadium and Ancient Agora to come on. You will also be reminded that Athens is surrounded by sea with spectacular views across the Aegean. Dinner and overnight Athens.
(B,D, Performance) Visit the Athens Archeological museum. The National Archaeological Museum in Athens houses some of the most important artifacts from a variety of archaeological locations around Greece from prehistory to late antiquity. It is considered one of the greatest museums in the world and contains the richest collection of artifacts from Greek antiquity worldwide. Free afternoon for lunch, shopping and explorations. Tonight is our farewell Greek musical exchange concert, indoors at one of the Athenian concert halls with an audience. Farewell Dinner. Overnight Athens.
Depart for the U.S.
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Copyright © 2023 Expedition and Travel. All Rights Reserved.
The safety of your travelers has been always our top priority
Covid-19 Compliance
At Expedition and Travel the safety of your travelers has been always our top priority. We would like to inform you about the extra measures with regards to the virus SARS-CoV-2 and COVID-19 taken by our team, to protect your travelers.
We enrolled in POSI SHIELD initiative which is a program for infection prevention and control that consists of creation of hygiene and operation protocols, enforcement of personal protective equipment use, personnel training, creation of Prevention Team and introduction of POSI efficiency validation procedure
Hand sanitizers have been added to all our buses, hotels and restaurants we cooperate
Enhanced the frequency of cleaning and disinfection of all areas ie bus, restaurants, hotels
A trip to Greece and Mediterranean is a life time experience! With this in mind our team will ensure that your travelers will have an unforgettable experience with us while following the latest government guidelines and global prevention best practices.
While we are looking forward welcoming you in Greece we remain at your disposal should you need any further information.


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