Sequentially compact space

In mathematics, a topological space ${\displaystyle X}$ is sequentially compact if every sequence of points in ${\displaystyle X}$ has a convergent subsequence converging to a point in ${\displaystyle X}$.

Every metric space is naturally a topological space, and for metric spaces, the notions of compactness and sequential compactness are equivalent (countable choice suffices; see below). However, there exist sequentially compact topological spaces that are not compact, and compact topological spaces that are not sequentially compact.

The space of all real numbers with the standard topology is not sequentially compact; the sequence ${\displaystyle (s_{n})}$ given by ${\displaystyle s_{n}=n}$ for all natural numbers ${\displaystyle n}$ is a sequence that has no convergent subsequence.

On a first countable space, a sequence ${\displaystyle x_{n}}$ has a convergent subsequence if and only if

${\displaystyle \bigcap _{n}{\overline {\{x_{m}\mid m\geq n\}}}}$

is nomempty. Indeed, a limit of a convergent subsequence is necessarily in the above intersection (this direction holds for any topological space). Conversely, if ${\displaystyle x}$ is in the above intersection, then let ${\displaystyle x\in \cdots \subset U_{2}\subset U_{1}}$ be a countable neighborhood base at ${\displaystyle x}$. Then, inductively, choose integers ${\displaystyle n_{i}>0}$ such that ${\displaystyle n_{i}}$ is a least integer with the property (1) ${\displaystyle n_{i}>n_{i-1}}$ and (2) ${\displaystyle x_{n_{i}}\in U_{i}}$, which is possible since ${\displaystyle \mathbb {N} }$ is a well-ordered set. Then ${\displaystyle x_{n_{j}}\to x}$.

A point in the above intersection is called a cluster point. Thus, for first countable spaces, the definition of a sequentially compact space is the same as saying that each sequence in the space has a cluster point.

If a space is a metric space, then it is sequentially compact if and only if it is compact (cf. Heine–Borel theorem § Generalization).¹ Here is how to see this, using only the countable Choice. We have to show "sequentially compact" implies "compact". First, we note ${\displaystyle X}$ is totally bounded, meaning for each ${\displaystyle \epsilon >0}$, there is a finite cover of ${\displaystyle X}$ consisting of open balls of radius ${\displaystyle \epsilon }$. Indeed, if it fails for some ${\displaystyle \epsilon }$, by countable Choice, choose a sequence ${\displaystyle x_{n}}$ such that

${\displaystyle x_{n}\not \in B(x_{1},\epsilon )\cup \cdots \cup B(x_{n-1},\epsilon ).}$

This sequence ${\displaystyle x_{n}}$ has no convergent subsequence, a contradiction. It follows that ${\displaystyle X}$ has a countable base. Hence, it is enough to show ${\displaystyle X}$ is countably compact; i.e., each descending sequence ${\displaystyle E_{1}\supset E_{2}\supset \cdots }$ of nonempty closed subsets has nonempty intersection. But this is clear since

${\displaystyle \emptyset \neq \cap _{n}{\overline {\{x_{m}\mid m\geq n\}}}\subset \cap _{n}E_{n}}$

for some sequence ${\displaystyle x_{n}}$ with ${\displaystyle x_{n}\in E_{n}}$. ${\displaystyle \square }$

The first uncountable ordinal with the order topology is an example of a sequentially compact topological space that is not compact. The topological product of ${\displaystyle 2^{\aleph _{0}}={\mathfrak {c}}}$ copies of the closed unit interval is an example of a compact space that is not sequentially compact.²

A topological space ${\displaystyle X}$ is said to be limit point compact if every infinite subset of ${\displaystyle X}$ has a limit point in ${\displaystyle X}$, and countably compact if every countable open cover has a finite subcover. In a metric space, the notions of sequential compactness, limit point compactness, countable compactness and compactness are all equivalent (if one assumes the axiom of choice).

In a sequential (Hausdorff) space sequential compactness is equivalent to countable compactness.³

There is also a notion of a one-point sequential compactification—the idea is that the non convergent sequences should all converge to the extra point.⁴