# give an example of an infinite class of closed sets whose union is not closed.

give an example of an infinite class of closed sets whose union is not closed. Thanks for your help

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How about 1/n?, –  Ross Millikan Apr 23 '12 at 0:09

I think probably the most instructive example is considering $\displaystyle A_n=\left[\frac{1}{n},\infty\right)$.

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thank you very much for your help –  Chalie Her Apr 23 '12 at 0:28
Why is it the most instructive example? –  lhf Apr 23 '12 at 1:32

Every subset $S\subset X$ of a Hausdorff space is the union of its singleton subsets, which are closed : $$S=\bigcup_{s\in S} \lbrace s\rbrace$$

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I like this one the most. It's more generic! –  Don Larynx Oct 16 '13 at 22:15

Can you express $(0,1)$ as an increasing union of closed sets? Maybe find a pair of sequences $a_n$ and $b_n$ with $a_n$ decreasing to $0$ and $b_n$ increasing to $1$? Then you can try taking $[a_n,b_n]$ and see if that works.

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thank you very much for your help –  Chalie Her Apr 23 '12 at 0:28

the union of intervals of the form $(1/n ,1- 1/n ) = (0,1)$ will be a example. the behaviour of the interval already stated above.

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How are these intervals closed? –  Rudy the Reindeer Oct 5 '12 at 16:26

As another example, let $X$ be any infinite set, and consider the cofinite topology on $X$ (ie all open sets are either the empty set or sets whose complement is finite). Every proper closed subset of $X$ is finite. So, fixing an element $x_0\in X$, we have the union closed sets equaling an open set: $$X\setminus\{x_0\}=\bigcup\limits_{x\not=x_0} \{x\}$$

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Here is a good example which clearly shows that the infinite union of closed sets may not be closed. consider the usual topology on $\mathbb{R}$, and let $\mathcal{C}$ be the collection of all closed sets of the form $( -\infty , \frac n{n+1} ]$ where $n \geq 1$. Then $\bigcup \mathcal{C} = ( - \infty , 1 )$, which is open. So this union of infinitely many closed sets is open.

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