Closed sets contain their limits

Statement

Let $(X, d)$ be a metric space and let $F$ be a closed subset of $X$. Let $(x_n)$ be a sequence of points of $F$ that converges to a point $x \in X$. Then $x \in F$.

Remarks

Proof: if $x \notin F$, then $X \setminus F$ is open and contains $x$. Thus some ball $B(x, \varepsilon)$ misses $F$. But $x_n \in B(x, \varepsilon)$ for large $n$, a contradiction. In a metric space the converse is also true: a set that contains the limits of all its convergent sequences is closed. This follows from closure is the set of limits of sequences. The statement holds in every topological space, with the same proof.

Explanations

L1 human

You cannot leave a closed set by a limit of a sequence. For example, $1/n \to 0$ in $\mathbb{R}$, and $0$ is not in $(0, 1]$, which shows that $(0, 1]$ is not closed.