Closure of a Set in a First Countable Space is all its Sequential Limits

Last updated Nov 1, 2022

Suppose $(X, \tau)$ is a First Countable Space. Suppose $S \subset X$ Then $$\text{cl} S = {x : \exists (x_n) \subset S \text{ s.t. } x_{n} \to x}$$

Denote $T = {x : \exists (x_n) \subset S \text{ s.t. } x_{n} \to x}$. Since the Closure of a Set is all its Limit Points and A point in a First Countable Space is a Limit Point iff it is a Sequential Limit, we see $$\text{cl} S = \bar{S} = T$$ $\blacksquare$