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

Caleb Stanford
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Chalie Her
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10 Answers10

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

Alex Youcis
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  • thank you very much for your help – Chalie Her Apr 23 '12 at 00:28
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    Why is it the most instructive example? – lhf Apr 23 '12 at 01:32
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    Because $\bigcup_{n \in \mathbb N} A_n$ gives you the open interval $(0,\infty)$,which we know is not closed. –  Jul 20 '19 at 14:43
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    If you want to get an argument for this, consider the complement of this set, which is $(-\infty,0]$. We cannot find a ball around the point $0$ of radius $\epsilon$ that lies purely in the set $\forall \epsilon>0 \quad (0-\epsilon, 0 + \epsilon) \not \subseteq (-\infty, 0]$. therefore the complement set $(-\infty,0]$ is not open and the original set $[0, \infty)$ cannot be closed.

    Another argument is that the only clopen sets in $\mathbb R$ are the trivials subsets $\emptyset, \mathbb R$ so if a proper nonempty subset is open, it is not closed.

     –  Jul 20 '19 at 14:43
30

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

Georges Elencwajg
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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.

IAmNoOne
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poorvi
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  • How are these intervals closed? – Rudy the Reindeer Oct 05 '12 at 16:26
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    This 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. – ShrewdCleric Oct 04 '17 at 01:33
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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.

Arturo Magidin
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5

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.

user642796
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shah faisal
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3

Rationals. $\mathbb{Q} = \cup_r \{r\}$ is a countable union of closed sets.

2

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: $$X\setminus\{x_0\}=\bigcup\limits_{x\not=x_0} \{x\}$$

0

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.

ryaron
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Consider the sets $[\frac{1}{n}, 2]$. Each of these sets is clearly closed. However, their union over all $\mathbb{N}$ is $(0, 2]$ which is neither open nor closed.

P.S: Note that the union cannot contain 0 because there is no $n \in \mathbb{N}$ such that $\frac{1}{n} = 0$.

Kartik Sreenivasan
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Since singletons in R are closed in usual topology. We can think about infinite class of singletons {x} where x belongs to (0,1] then there union will be (0,1] which is not closed in R.

Navid Gujjar
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