Complete metric space

Sequentially compact space

A topological space is said to be sequentially compact iff every sequence in has a convergent subsequence with a limit in . #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 be a second-countable topological space, e.g. a Metric space. Then is compact iff it is sequentially compact. #m/thm/topology

Proof

Let be a first-countable Compact space, and be a sequence with end pieces . The intersection of finite end pieces will always be inhabited, since , so by Complement characterisation and thus the intersection of closures . Since Limit points are points contained in the closure of every end piece, has at least one limit point, and since Limit points are limits of convergent subsequences in a first-countable space, has a convergent subsequence. Therefore is sequentially compact.

For the converse, let be a second-countable sequentially compact space. Second countable implies Lindelöf, so without loss of generality we can take a countable open cover . Let be partial unions of the open cover. Assume has no finite subcover, so for all , so we can construct a sequence with . Since is sequentially compact, has a convergent subsequence, and the limit thereof is a limit point of because Limit points are limits of convergent subsequences in a first-countable space. Therefore let be a Limit point of . Since covers , for some , so both and are open neighbourhoods of . Since is a limit point of , its neighbourhood contains infinite , which contradicts our construction. Therefore 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

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

Footnotes

  1. Since at least one element must be repeated infinitely many times in a sequence by the Pigeonhole principle, yielding an eventually constant subsequence. ↩

  2. 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. ↩