Sequentially compact space

A topological space $𝑋$ is said to be sequentially compact iff every sequence $(𝑥 𝑛) \infty 𝑛 = 1$ in $𝑋$ has a convergent subsequence with a limit in $𝑋$ . #m/def/topology

In general, sequential compactness is neither weaker nor stronger than compactness. However, the Main theorem describes when these conditions are equivalent.

Main theorem

Let $𝑋$ be a second-countable topological space, e.g. a Metric space. Then $𝑋$ is compact iff it is sequentially compact. #m/thm/topology

Proof

Let $𝑋$ be a first-countable Compact space, and $(𝑥 𝑛) \infty 𝑛 = 1 \in 𝑋$ be a sequence with end pieces $𝑀 𝑚 = {𝑚 𝑛} \infty 𝑛 = 𝑚$ . The intersection of finite end pieces will always be inhabited, since $𝑚 \geq 𝑛 ⟹ 𝑀 𝑚 \subseteq 𝑀 𝑛$ , so by Complement characterisation $⋂ \infty 𝑛 = 1 𝑀 𝑛 \neq \emptyset$ and thus the intersection of closures $⋂ \infty 𝑛 = 1 𝑀 𝑛 \neq \emptyset$ . Since Limit points are points contained in the closure of every end piece, $𝑥 𝑛$ has at least one limit point, and since Limit points are limits of convergent subsequences in a first-countable space, $𝑥 𝑛$ has a convergent subsequence. Therefore $𝑋$ is sequentially compact.

For the converse, let $𝑋$ be a second-countable sequentially compact space. Second countable implies Lindelöf, so without loss of generality we can take a countable open cover $(𝑈 𝑛) \infty 𝑛 = 1 \subseteq 𝑋$ . Let $𝑉 𝑛 = ⋃ 𝑛 𝑖 = 1 𝑈 𝑖$ be partial unions of the open cover. Assume $𝑈 𝑛$ has no finite subcover, so $𝑋 ∖ 𝑉 𝑛 \neq \emptyset$ for all $𝑛 \in ℕ$ , so we can construct a sequence $(𝑥 𝑛) \infty 𝑛 = 1 \in 𝑋$ with $𝑥 𝑛 \notin 𝑉 𝑛$ . Since $𝑋$ is sequentially compact, $𝑥 𝑛$ has a convergent subsequence, and the limit thereof is a limit point of $𝑥 𝑛$ because Limit points are limits of convergent subsequences in a first-countable space. Therefore let $𝑎 \in 𝑋$ be a Limit point of $𝑥 𝑛$ . Since $𝑈 𝑛$ covers $𝑋$ , $𝑎 \in 𝑈 𝑚$ for some $𝑚 \in ℕ$ , so both $𝑈 𝑚$ and $𝑉 𝑚$ are open neighbourhoods of $𝑎$ . Since $𝑎$ is a limit point of $𝑥 𝑛$ , its neighbourhood $𝑉 𝑚$ contains infinite $𝑥 𝑛$ , which contradicts our construction. Therefore $𝑈 𝑛$ must have a finite subcover.

Note the forward statement only requires the First countability axiom, whereas the converse requires both first-countability and Lindelöf.

Properties

Any finite subspace is compact¹
Any compact subspace is closed and bounded²
Closed subspaces of a compact space are compact
Heine-Borel theorem: For Euclidean space, a subset is compact iff. it is closed andbounded

#state/tidy | #lang/en | #SemBr

Since at least one element must be repeated infinitely many times in a sequence by the Pigeonhole principle, yielding an eventually constant subsequence. ↩
Closedness follows from the fact that it must be sequentially closed (since subsequences of a convergent sequence converge to the same limit). Boundedness is trivial, since an unbounded set contains divergent sequences. ↩

Sequentially compact space

Main theorem

Properties

Footnotes