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

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9  
How about 1/n?, –  Ross Millikan Apr 23 '12 at 0:09

6 Answers 6

up vote 9 down vote accepted

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
2  
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? –  Matt N. 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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