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A Set is Closed in a First Countable Space iff it contains all its Sequential Limits

Last updated Nov 1, 2022

# Statement

Suppose $(X, \tau)$ is a First Countable Space. Then $K \subset X$ is Closed If and Only If $$K \supset {x : \exists (x_n) \subset K \text{ s.t. } x_{n} \to x}$$ That is, $K$ contains all its sequential limits.

# Proof

TODO