First countability axiom

Sequential closedness

Let be a topological space. A subset is sequentially closed iff every convergent sequence in such that has . topology Every closed set is sequentially closed.

Proof

Let be closed, , and be sequence in such that . But since is an open neighbourhood of , there exists some such that for all , so for all , a contradiction.

Main theorem

Let be first-countable space. Then is closed iff it is sequentially closed. topology

Proof

The forward direction is given above. For the converse, let such that for every sequence , implies . Suppose is not open in Then there exists some such that for every neighbourhood of , . Since is first-countable, we may construct a nested neighbourhood basis , for which for all . One may then construct a sequence such that for all . But then , contradicting the requirement that be sequentially closed. Hence must be open, whence is closed.

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