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Question: {$B_n$}$ \in \Bbb R$ is a family of closed sets. Prove that $\cup _{n=1}^\infty B_n$ is not necessarily a closed set.

What I thought: Using a counterexample: If I say that each $B_i$ is a set of all numbers in range $[i,i+1]$ then I can pick a sequence $a_n \in \cup _{n=1}^\infty B_n$ s.t. $a_n \to \infty$ (because eventually the set includes all positive reals) and since $\infty \notin \Bbb R$ then $\cup _{n=1}^\infty B_n$ is not a closed set.

Is this proof correct? Thanks

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  • $\begingroup$ The notation $\{B_n\} \in \Bbb R$ is not correct. You probably meant that $\{B_n\}$ is a family/sequence of closed sets and each $B_n\subseteq\Bbb R$, i.e., each of the sets is a subset of real line. $\endgroup$ Apr 24, 2013 at 13:04
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    $\begingroup$ What might be interesting to notice (although it might be a little too advanced, if this is your first course in topology) that the family of sets in your example is locally finite. Union of a locally finite system of closed sets is again a closed set. $\endgroup$ Apr 24, 2013 at 13:06
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    $\begingroup$ Let $B_n=\{\frac 1 n\}$ is closed in $\mathbb R$, then $\bigcup B_n=\{\frac 1 n: n\in N^+\}$ is not closed in $\mathbb R$. $\endgroup$
    – Paul
    May 2, 2013 at 10:32
  • $\begingroup$ The set containing all the reals is a closed set. infinity is not a point in in the universe so it is not a limit point of a_n. a_n -> c has c being a limit point only if c is finite. Hint: Let a_n -> 0 but not have 0 in any of the sets. Let B_i = [a_n, something] and a_n -> 0. then what happens? $\endgroup$
    – fleablood
    Aug 21, 2016 at 2:53
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    $\begingroup$ For future visitors: this can be true under some extra conditions, like the family of closed sets that one is taking the union over being locally finite. See math.stackexchange.com/a/2269119/405572, or planetmath.org/…, or en.wikipedia.org/wiki/Locally_finite_collection $\endgroup$
    – D.R.
    Jul 29, 2023 at 7:08

6 Answers 6

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Every subset of $\mathbb R^ n$ is a union of closed sets, namely, the one-point sets consisting of each one of its points.

Yet not all subsets of $\mathbb R^n$ are closed!

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  • $\begingroup$ I had a typo.... $\endgroup$
    – jreing
    Apr 24, 2013 at 8:27
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    $\begingroup$ That's a nice proof. +1 $\endgroup$ Apr 24, 2013 at 9:00
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    $\begingroup$ Does not $\cup^{\infty}_{n=1}B_n$ not imply countable union? In that sense $\mathbb{R}^n$ would not qualify. $\endgroup$
    – user45099
    Apr 24, 2013 at 10:39
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    $\begingroup$ @user57 thankfully there are countable subsets of $\mathbb{R}$ which are not closed. $\endgroup$
    – Dan Rust
    Apr 24, 2013 at 12:08
  • $\begingroup$ Ah, yes sure. Sorry. $\endgroup$
    – user45099
    Apr 24, 2013 at 12:53
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Think of the union of the $B_{n} = [1/n, 1]$.

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  • $\begingroup$ is there a major difference between [i,i+1] and [\frac 1n,1]? $\endgroup$
    – jreing
    Apr 24, 2013 at 8:29
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    $\begingroup$ I'm sorry but could you elaborate on why $[i, i+1]$ is not closed in $\mathbb R$ with the standard topology? $\endgroup$ Apr 24, 2013 at 8:54
  • $\begingroup$ Sorry all, I thought I had seen parentheses instead of brackets. Better check my spectacles. Apologies. $\endgroup$ Apr 24, 2013 at 11:42
  • $\begingroup$ This is a good example. If we let $n$ begin from $2$, the result will be $(0,1]$, which is not closed. $\endgroup$
    – Boar
    Dec 22, 2021 at 8:00
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Consider the topological space (subspace of$\mathbb{R}$ (real numbers) with the usual topology given by $\epsilon$-neighborhood) given by $$\{0\}\cup\{\frac{1}{n}:n\in\mathbb{N}_{>0}\}$$ Then, for every $n\in\mathbb{N}_{>0}$ the subset $$\{\frac{1}{n}\}$$is both open and closed, but the countable union $$\displaystyle\bigcup_{n>0}\{\frac{1}{n}\}$$ is precisely the set $$\{\frac{1}{n}:n>0\}$$ which is not closed, for example because $0$ lies in its closure, but not in the set itself.

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But since $\infty\notin\Bbb R$ we cannot use it as a counterexample. To see that is indeed the case note that your union is the set $[1,\infty)$, which is closed. Why is it closed? Recall that $A\subseteq\Bbb R$ is closed if and only if every convergent sequence $a_n$ whose elements are from $A$, has a limit inside $A$. So we need sequences whose limits are real numbers to begin with, and so sequences converging to $\infty$ are of no use to us. On the other hand, if $a_n\geq 1$ then their limit is $\geq 1$, so $[1,\infty)$ is closed.

You need to try this with bounded intervals whose union is bounded. Show that in such case the result is an open interval.

Another option is to note that $\{a\}$ is closed, but $\Bbb Q$ is the union of countably many closed sets. Is $\Bbb Q$ closed?

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  • $\begingroup$ Got it now. Thanks for all. $\endgroup$
    – jreing
    Apr 24, 2013 at 9:05
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In your example, $\bigcup B_i = [1,\infty)$. But this is a closed set so it does not provide a counter example.

Consider $B_n = [0, 1 - \frac{1}{n}]$. Then every $x \in [0,1)$ is in $\bigcup_{n \in \mathbb N_{>0}} B_n$. But what about the end point $1$?

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The way I do this is by first proving that the countable intersection of open sets need not be open by this counterexample: $$\bigcap_{\infty}\left (-\frac{1}{n},\frac{1}{n}\right) = \left\{ {0} \right\} $$ Then because a closed set is the complement of an open set we get that the countable union of closed sets need not be closed because $$\left(\bigcap_{i=1}^nA_i\right)^c=\bigcup_{i=1}^nA_i^c\;.$$

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