Give an example of an infinite class of closed sets whose union is not closed.
Infinite class of closed sets whose union is not closed
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9How about 1/n?, – 2012-04-23
9 Answers
Example in R, take I= [0,1] with the usual topology inherited from R. Then the union of all rationals in I Is a union of singletons, yet it's complement is not open since every ball around an irrational point contains rational points.
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 $
I think probably the most instructive example is considering $\displaystyle A_n=\left[\frac{1}{n},\infty\right)$.
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11Why is it the most instructive example? – 2012-04-23
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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0thank you very much for your help – 2012-04-23
The union of intervals of the form $\left[\frac{1}{n} ,1- \frac{1}{n}\right]$ will be one example: $ \bigcup_{n=2}^\infty \left[\frac{1}{n} ,1- \frac{1}{n}\right] = (0,1) $ The behaviour of the interval is already stated above.
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0This is superbly old, but in the interest of sharing, they're closed because each interval contains its own bounds, and by the definition of a closed set - a set which contains its boundary points - we know the intervals are closed. – 2017-10-04
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: $X\setminus\{x_0\}=\bigcup\limits_{x\not=x_0} \{x\}$
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.