# Sequential characterization of closedness of the set

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 $x\in F$.

[EDIT]: $X=\mathbb{R}^n$ with the usual topology.

Definition of closed set: Set is closed if and only if its complement is open. 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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So what is yours definition of closed set? – Norbert Jun 3 '12 at 17:12
Set is closed if and only if its complement is open. 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 $r$. – xan Jun 3 '12 at 17:14
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. – Martin Sleziak Jun 3 '12 at 17:18
@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. – Jonas Meyer Jun 3 '12 at 17:22
@xan: You should edit that into the question, and remove the inapplicable "general-topology" tag. – Chris Eagle Jun 3 '12 at 17:24

As in comments, I suppose $X=\mathbb{R}^{n}$ (can be any metric space in fact). Let $F\subset X$.
$\Rightarrow$: 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.