# Sequence of closed set in topological space

How do I find an example for a sequence of closed sets in a topological space $X$ whose union is not closed? Would $\{{1\over n}\} n\to \infty$ work?

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Yes. This would work. Other example may include $\{q\}$ for $q\in\Bbb Q$. It is a countable collection so we can make it into a sequence.
As well the classical example: for $n\in\Bbb N$ let $F_n=[\frac1n,\frac n{n+1}]$, then $\bigcup_{n=1}^\infty F_n=(0,1)$ which is open and not closed.
To see that this union equals $(0,1)$ note that given $x\in(0,1)$ there is some $n$ such that $\frac1n<x<\frac n{n+1}$, therefore $x\in F_n$ and so $x\in\bigcup F_n$; and on the other hand if $x\leq 0$ or $1\leq x$ then $x\notin F_n$ for all $n$ so it cannot be in the union.
also $[{1\over n}, {n\over {n+1}}]$ is a closed set, how do I construct a sequence of this set, would it just be from $n=1$ to $\infty$? –  Akaichan Feb 18 '13 at 2:22
@IvordesGreenleaf: Think about the sequence $1/n$. If the union of these sets were closed, then the limit point of this sequence must be in the set. –  Clayton Feb 18 '13 at 2:32
@IvordesGreenleaf Asaf gives a closed set $[\frac{1}{n}, \frac{n}{n+1}]$, whose definition depends on $n$, so indeed he means: let $F_n$ be this closed set, for every $n \in \mathbb{N}$. Then their union $\cup_n F_n$ equals $(0,1)$ and this set is not closed. Do not write $n \rightarrow \infty$ when defining a sequence, but this is used for limits, not the definition of a sequence. –  Henno Brandsma Feb 18 '13 at 7:11