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Sequentially compact space

Topological space where every sequence has a convergent subsequence

Wikipedia contributorsAdapted for Fathomly · View authors
2 min read
Source updated August 6, 2026

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

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

01Examples and properties

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

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

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

is nonempty. Indeed, a limit of a convergent subsequence is necessarily in the above intersection (this direction holds for any topological space). Conversely, if x is in the above intersection, then let x\in \cdots \subset U_{2}\subset U_{1} be a countable neighborhood base at x. Then, inductively, choose integers n_{i}>0 such that n_{i} is a least integer with the property (1) n_{i}>n_{i-1} and (2) x_{n_{i}}\in U_{i}, which is possible since \mathbb {N} is a well-ordered set. Then 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). Here is how to see this, using only the countable Choice. We have to show "sequentially compact" implies "compact". First, we note X is totally bounded, meaning for each \epsilon >0, there is a finite cover of X consisting of open balls B(x,\epsilon ) of radius \epsilon. Indeed, if it fails for some \epsilon, by countable Choice, choose a sequence x_{n} such that

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

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

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

for some sequence x_{n} with x_{n}\in E_{n}. \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 2^{\aleph _{0}}={\mathfrak {c}} copies of the closed unit interval is an example of a compact space that is not sequentially compact.

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Sources and credits

This article is adapted from the Wikipedia article “Sequentially compact space”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

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