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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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    This seems to be more or less the same as [this question](http://math.stackexchange.com/questions/125287/what-are-the-minimal-conditions-on-a-topological-space-for-its-closure-to-equal). 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](http://en.wikipedia.org/wiki/Sequential_space). See also [this blog](http://dantopology.wordpress.com/tag/sequential-space/) and perhaps [this question](http://math.stackexchange.com/questions/132352/).2012-06-03
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    @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.2012-06-03
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    @MartinSleziak: The information and links in your comment are much appreciated.2012-06-03
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    See also: [Closure of Subset of Metric Space by Convergent Sequence](http://www.proofwiki.org/wiki/Closure_of_Subset_of_Metric_Space_by_Convergent_Sequence) at ProofWiki.2012-06-03

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$\Rightarrow$: 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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    now I see it, thank you very much!2012-06-03
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    @xan: Sure; you're welcome. I'm glad I was able to help you figure this out.2012-06-03