Let $(X, d)$ be a metric space. A set $F\subset X$ is closed if and only if for every sequence $\left\{x_n\right\}\subset F$, if $x\in X$ and $x_n\rightarrow x$ then $x\in F$.

Definition of closed set: Set is closed if and only if its complement is open. A Set $U$ is open if and only if $\forall_{x\in U}\exists_{r>0}B(x,r)\subset U$, where $B(x,r)$ is a ball with middle in $x$ and with radius $r$.

It's rather a well-known fact that I used many times while solving problems, but just now I realized that I don't know how to prove it.

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    $\begingroup$ This seems to be more or less the same as this question. The property you wrote is not true for all topological spaces, spaces with this property are called Fréchet–Urysohn spaces. A related property: sequential spaces. See also this blog and perhaps this question. $\endgroup$ – Martin Sleziak Jun 3 '12 at 17:18
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    $\begingroup$ @Martin: I don't think it's the same question. xan: Will you please provide context in your question? You seem to be assuming that $X$ is a metric space without having mentioned it. $\endgroup$ – Jonas Meyer Jun 3 '12 at 17:22
  • $\begingroup$ @MartinSleziak: The information and links in your comment are much appreciated. $\endgroup$ – Jonas Meyer Jun 3 '12 at 17:27
  • $\begingroup$ See also: Closure of Subset of Metric Space by Convergent Sequence at ProofWiki. $\endgroup$ – Martin Sleziak Jun 3 '12 at 17:38

$\Rightarrow$: 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.

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    $\begingroup$ now I see it, thank you very much! $\endgroup$ – xan Jun 3 '12 at 17:39
  • $\begingroup$ @xan: Sure; you're welcome. I'm glad I was able to help you figure this out. $\endgroup$ – T. Eskin Jun 3 '12 at 17:42

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