Sequential characterization of closedness of the set

Solution 1:

$\Rightarrow$: Suppose that $F$ is closed and let $\{x_{n}\}\subset F$ so that $x_{n}\to x$ for some $x\in X$. We show that $x\in F$. Let $U$ be any nhood of $x$. Since $x_{n}\to x$ there exists $k\in\mathbb{N}$ so that $x_{n}\in U$ for all $n\geq k$. In particular, $U\cap F\neq \emptyset$ (since e.g. $x_{k}\in U\cap F$). Since $U$ was an arbitrary nhood of $x$, this shows that $x$ is in the closure of $F$, which is equal to $F$ since $F$ is a closed set. Hence $x\in F$.

$\Leftarrow$: We show that $F$ is closed provided the latter property. Let $x$ be any element in the closure of $F$: we show that $x\in F$. Choose $x_{n}\in B(x,\frac{1}{n})\cap F$ for all $n\in\mathbb{N}$ (such $x_{n}$ exists since $x$ is in the closure of $F$, whence every nhood of $x$ intersects $F$). Now $x_{n}\to x$ and by assumption of this direction we have $x\in F$. Hence the closure of $F$ is a subset of $F$, whence they are in fact equal since a set is always subset of its closure. But this means that $F$ is a closed set.