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\begin{document}

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\section*{sequentially compact topological space}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{relationship_to_compactness}{Relationship to Compactness}\dotfill \pageref*{relationship_to_compactness} \linebreak
\noindent\hyperlink{Examples}{Examples and counter-examples}\dotfill \pageref*{Examples} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A [[topological space]] is called \emph{sequentially compact} if every [[sequence]] of points in that space has a sub-sequence which [[convergence|converges]]. In general this concept neither implies nor is implied by that of actual [[compact space|compactness]], but for some types of [[topological spaces]], such as [[metric spaces]], it is equivalent.

[[compact space|Compactness]] is an extremely useful concept in [[topology]]. The basic idea is that a [[topological space]] is [[compact space|compact]] if it isn't ``fuzzy around the edges''.

Whilst one can study a topological space by itself, it is often useful to probe it with known spaces. A common choice for topological spaces, and in particular metric spaces, is to use the natural numbers $\mathbb{N}$, and the [[one-point compactification]] $\mathbb{N} \cup \{*\}$ of the natural numbers. This is more traditionally known as studying the topology using sequences and convergent sequences.

Thus one can ask, ``Can I detect compactness using probes from $\mathbb{N}$, and $\mathbb{N} \cup \{*\}$?''. The short answer to this is ``No'', but that just reveals that the question was too restrictive. Rather, one should ask ``What does compactness look like if all I'm allowed to use are probes from $\mathbb{N}$ and $\mathbb{N} \cup \{*\}$?''. The answer to that question is ``sequential compactness''.

Thus \textbf{sequential compactness} is what [[compact space|compactness]] looks like if all one has to test it are sequences.

\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

\begin{defn}
\label{seqcpt}\hypertarget{seqcpt}{}
A topological space is \textbf{sequentially compact} if every [[sequence]] in it has a [[convergence|convergent]] [[subsequence]].

\end{defn}
\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

The following is a list of properties of and pertaining to sequentially compact spaces.

\begin{enumerate}%
\item For a [[metric space]], the notions of sequential compactness and compactness coincide. See at \emph{[[sequentially compact metric spaces are equivalently compact metric spaces]]}.


\item The [[Eberlein-Smulian theorem|Eberlein–Šmulian theorem]] states that in a [[Banach space]], for a subset with regard to the [[weak topology]], compactness and sequentially compactness are both equivalent to the weaker notion of [[countably compact space|countable compactness]].


\item A [[countable set|countable]] product of sequentially compact spaces is again sequentially compact.

Let $\{X_k\}$ be a countable family of sequentially compact spaces. Let $(a_l)$ be a sequence in $\prod X_k$. For each $m$ we recursively define an infinite subset $A_m \subseteq A_{m-1} \subseteq \mathbb{N}$ with the property that the sequence $(a_l)_{l \in A_m}$ converges when projected down to $\prod_{k=1}^m X_k$. Let $l_m = \min\{A_l\}$. Consider the sequence $(a_{l_m})$. For each $k$, we choose a limit $x_k$ of the projection of $(a_l)_{l \in A_k}$ to $X_k$. Let $x = (x_k) \in \prod X_k$. Let $U$ be a neighbourhood of $x$. Then there is some $n \in \mathbb{N}$ and neighbourhood $U_n \subseteq \prod_{k=1}^n X_k$ of $(x_k)_{k=1}^n$ such that $U$ contains the preimage of $U_n$. For $m \ge n$, the sequence $(l_m)$ is contained in $A_n$ and so the image of $(a_{l_m})$ converges to $(x_k)_{k=1}^n$. Hence there is some $r$ such that for $m \ge r$, the projection of $a_{l_m}$ lies in $U_n$. Hence for $m \ge r$, $a_{l_m} \in U$. Thus $(a_{l_m})$ converges to $(x_k)$ and so $\prod X_k$ is sequentially compact.

This shows that the example of a compact space that is not sequentially compact is about as simple as can be.


\item The theorem that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism has a counterpart for sequentially compact spaces.

\begin{theorem}
\label{SeqCptReg}\hypertarget{SeqCptReg}{}
Let $\mathcal{T}_1$ and $\mathcal{T}_2$ be two topologies on a set $X$ such that:

\begin{enumerate}%
\item $\mathcal{T}_1 \supseteq \mathcal{T}_2$ (equivalently, the identity map on $X$ is continuous as a map $(X,\mathcal{T}_1) \to (X, \mathcal{T}_2)$)
\item $\mathcal{T}_1$ is sequentially compact
\item $\mathcal{T}_2$ is [[completely regular space|completely regular]] and singleton sets are $G_\delta$-[[G-delta set|sets]],

\end{enumerate}
then $\mathcal{T}_1 = \mathcal{T}_2$.

\end{theorem}
\begin{proof}
Let $V \subseteq X$ be such that $V \notin \mathcal{T}_2$. Then it must be non-empty and there must be a point $v \in V$ such that $V$ is not a neighbourhood of $v$. As $\mathcal{T}_2$ is completely regular and singleton sets are $G_\delta$ sets, there is a continuous function $g \colon (X, \mathcal{T}_2) \to \mathcal{R}$ such that $g^{-1}(0) = \{v\}$. Since $V$ is not a neighbourhood of $v$, for each $n \in \mathbb{N}$, the set $g^{-1}(-\frac1n, \frac1n)$ is not wholly contained in $V$. Thus for each $n$ there is a point $x_n \in X$ such that $x_n \notin V$ and $|g(x_n)| \lt \frac1n$. As $\mathcal{T}_1$ is sequentially compact, this sequence has a $\mathcal{T}_1$-convergent subsequence, say $(x_{n_k})$ converging to $y$. Since $g(x_n) \to 0$, $g(x_{n_k}) \to 0$ and thus $g(y) = 0$. Thus $y = v$ and so $(x_{n_k}) \to v$ in $\mathcal{T}_1$. As $x_{n_k}
\notin V$ for all $n_k$, and $v \in V$, it must be the case that $V$ is not a $\mathcal{T}_1$-neighbourhood of $v$. Hence $V \notin \mathcal{T}_1$. Thus $\mathcal{T}_1 \subseteq \mathcal{T}_2$, whence they are equal.

\end{proof}

\item The [[image]] of a sequentially compact space $X$ under a continuous map $f: X \to Y$ is also sequentially compact. For suppose $y_n$ is a sequence in $f(X)$, say $y_n = f(x_n)$. Then $x_n$ has a convergent subsequence $x_{n_j}$, converging to $x$ say, and by continuity $y_{n_j} = f(x_{n_j})$ converges to $f(x)$.



\end{enumerate}
\hypertarget{relationship_to_compactness}{}\subsubsection*{{Relationship to Compactness}}\label{relationship_to_compactness}

[[compact space|Compactness]] does not imply sequentially compactness, nor does sequentially compactness imply compactness, without further assumptions, see at \emph{\hyperlink{Examples}{Examples and counter-examples}} below.

In [[metric spaces]] for example both notions coincide, see at [[sequentially compact metric spaces are equivalently compact metric spaces]]. (This is a consequence of the [[Lebesgue number lemma]] and the fact that [[sequentially compact metric spaces are totally bounded]].)

This is \emph{not} a contradiction to the statement that compact is equivalent to every [[net]] having a convergent subnet: Given a sequence in a compact space, its convergent \emph{subnet} need not be a \emph{subsequence} (see [[net]] for a definition of subnet).

\hypertarget{Examples}{}\subsection*{{Examples and counter-examples}}\label{Examples}

A [[metric space]] is sequentially compact precisely if it is [[compact topological space|compact]]. See at \emph{[[sequentially compact metric spaces are equivalently compact metric spaces]]}.

In general neither of these two properties implies the other:

\begin{itemize}%
\item Examples of a sequentially compact spaces which are not compact are given in in (\hyperlink{BuskesRooij97}{Buskes-Rooij 97, example 13.5}, \hyperlink{Patty08}{Patty 08, chapter 4, example 13}, \hyperlink{Vermeeren10}{Vermeeren 10, prop. 18}). The [[long line]] is such an example.


\item An example of a [[compact topological space]] which is not sequentially compact is given in (\hyperlink{SteenSeebach70}{Steen-Seebach 70, item 105}), recalled at \hyperlink{Vermeeren10}{Vermeeren 10, prop. 17}. See at \href{compact+space#CompactSpacesWhichAreNotSequentiallyCompact}{compact space -- Compact spaces which are not sequentially compact}.



\end{itemize}
\begin{example}
\label{ACompactSpaceNotSequentiallyCompact}\hypertarget{ACompactSpaceNotSequentiallyCompact}{}
\textbf{(a compact space which is not sequentially compact)}

Consider the following uncountable power of the [[discrete space]] $2$, namely the [[product topological space]] (with its [[Tychonoff topology]])

\begin{displaymath}
X \coloneqq \underset{f: \mathbb{N} \to 2}{\prod} Disc(\{0,1\})
\end{displaymath}
of copies of the discrete space on two elements, indexed by functions $f: \mathbb{N} \to \{0, 1\}$. Since $Disc(\{0,1\})$ is a [[finite topological space|finite]] [[discrete topological space]] it is clearly compact. Therefore the [[Tychonoff theorem]] says that also $X$ is compact.

Let $(x_n)_{n: \mathbb{N}}$ be the sequence in $X$ given by the [[duality|double-dual embedding]]

\begin{displaymath}
\mathbb{N} \to 2^{2^\mathbb{N}},
\end{displaymath}
i.e., define $x_n$ to have coordinate at $f: \mathbb{N} \to 2$ given by $(x_n)_f = f(n)$. We claim this sequence has no subsequence that converges in $X$, so that $X$ is not sequentially compact. We will argue by contradiction. (This [[refutation by contradiction]], refuting sequential compactness, is [[constructive logic|constructively valid]]. The argument above that affirms compactness, via the Tychonoff theorem, is not constructive.)

Suppose instead some subsequence $(x_{(n_k)})_{k \in \mathbb{N}}$ converges to some $x \in X$. Choose any $f: \mathbb{N} \to 2$ that is not eventually constant on the subsequence $(n_k)_{k: \mathbb{N}}$; for example, define $f: \mathbb{N} \to 2$ by $f(n_k) = k\; mod\; 2$, else $f(n) = 0$ if $n$ does not appear in the subsequence. Consider the open set $U_f = \{x_f\} \times \prod_{g: g \neq f} \{0, 1\}$, which is an open neighborhood of $x$. In order to have $x_{n_k} \in U_f$ for all $k \geq k_0$, we'd have to have $f(n_k) = x_f$ for all $k \geq k_0$, in other words $f$ would be eventually constant on the subsequence $n_k$. Contradiction.

\end{example}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[compact topological space]]


\item [[countably compact topological space]]


\item [[paracompact topological space]]


\item [[locally compact topological space]]


\item [[compactly generated topological space]]



\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

\begin{itemize}%
\item Buskes, van Rooij, \emph{Topological Spaces: From Distance to Neighbourhood}, Springer 1997


\item Lynn Steen, J. Arthur Seebach, \emph{Counterexamples in Topology}, Springer-Verlag, New York (1970) 2nd edition, (1978), Reprinted by Dover Publications, New York, 1995


\item Wayne Patty, \emph{Foundations of Topology}, Jones and Bartlett Publishers (2008)


\item [[Stijn Vermeeren]], \emph{Sequences and nets in topology}, 2010 (\href{http://stijnvermeeren.be/download/mathematics/nets.pdf}{pdf})



\end{itemize}
See also

\begin{itemize}%
\item Wikipedia, \emph{\href{https://en.wikipedia.org/wiki/Sequentially_compact_space}{Sequentially compact space}}

\end{itemize}
[[!redirects sequentially compact space]] [[!redirects sequentially compact spaces]]

[[!redirects sequentially compact]] [[!redirects sequential compactness]]

[[!redirects sequentially compact metric space]] [[!redirects sequentially compact metric spaces]]



\end{document}