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this is a question that is not addressed in my book directly but I was curious. We just proved that the union of a finite collection of closed sets is also closed, but I was curious about if the union of infinitely many closed sets can be open. This question may not be at the level of the book so perhaps that's why it wasn't addressed.

Just to make things easier, lets imagine sets that are disks in the x-y plane. I can imagine that if there are nested disks inside each other, that in this case the union would clearly be closed.

But what if you could construct an infinite set of disks that together cover the entire real plane. Then in this case, it seems that every point in their union would have an open ball centered around the point that is also contained in the real plane, so that this union of an infinite collection of disks would create an open set.

Is this a correct way of thinking? Or at least on the right track?

I get the feeling that as long as you have no largest individual set that contains all the others, then you won't get a closed set. But I have a feeling there is more subtlety to it.

Thanks everyone

user1236
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  • "I get the feeling..." That conjecture is false. You can get a closed so. For example, if $X_n=[-n,n]$ then $\bigcup X_n = \mathbb R$ is closed. – Thomas Andrews Jan 23 '15 at 05:23
  • BTW: In your example with growing disks you state that the union of growing disc cover the whole plane which is open. But it is also closed (clopen), just like the empty set. – Drunix Jan 23 '15 at 06:50
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    @ThomasAndrews: and that example conveniently deals with the first part of the question as well, "can the infinite union of closed sets be open?". It's so tempting for novices to think of "closed" and "open" as if they're a logical dichotomy, when of course they aren't. – Steve Jessop Jan 23 '15 at 12:53
  • You: I can imagine that if there are nested disks inside each other, that in this case the union would clearly be closed. No, not with an infinite number of disks. For example let $A_n$ be the closed disk in the $x$-$y$ plane with radius $( 7-\frac1n )$. Then the union $\bigcup_{n \in\mathbb{N}} A_n$ is not closed. Actually it is an open disk. – Jeppe Stig Nielsen Jan 23 '15 at 13:37

4 Answers4

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Let $U$ be any open set whatsoever. Each point $x \in U$, taken as a one-point set $\{x\}$, is closed, and $U$ is the union of all these sets.

Daniel McLaury
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We just proved that the union of a finite collection of closed sets is also closed, but I was curious about if the union of infinitely many closed sets can be open.

Sure, one of the common examples being $\mathbb{R}$ with the usual Euclidean topology, $$(0,1) = \bigcup_{n=2}^\infty\left[\frac{1}{n},1-\frac{1}{n}\right]\text{.}$$

Stan Liou
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    Better make those limits from $n = 2$ to infinity, because I don't know what $[1, 0]$ means. – Daniel McLaury Jan 23 '15 at 05:20
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    @DanielMcLaury ok thanks, though the straightforward interpretation would be the empty set. ;) – Stan Liou Jan 23 '15 at 05:22
  • That's a great example too. It's interesting to imagine the intervals expanding like that. Thanks so much for the help! – user1236 Jan 23 '15 at 05:42
  • I guess this also shows how you can get half-open intervals, by fixing one of the two sides. So I guess it follows that closed intervals are a basis for the usual topology on the Real line. – user192680 Jan 23 '15 at 05:51
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    @user192680 the closed intervals don't form a topological basis because they're not open, but you may be interested in a related notion of $F_\sigma$ sets. – Stan Liou Jan 23 '15 at 06:03
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In $\Bbb R$ take $\aleph_0$ closed intervals from $-n$ to $n$ for natural $n$: $$\bigcup_{n\in\Bbb N} [-n,n] = \Bbb R$$

Similary $$\bigcup_{n\in\Bbb N^+} \left([-n,-n+1] \cup [n-1,n]\right) = \Bbb R$$

CiaPan
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Just to address your "feeling" -- using $\mathbb{R}$ as the counter-example might feel like a bit of a trick or a special case, since it's both closed and open. But remember that in $\mathbb{R}$ (and in general in $\mathbb{R}^n$) there are lots of infinite closed sets. Any half-closed interval whose other end is infinity, $[x,\infty)$ or $(-\infty, x]$, is a closed set, and you can easily see how to express those as an infinite union of closed finite sets. So there's no need for any of the things in the union to bound the rest.

Moving on to smaller closed sets, you can express $[0, 1]$ as an infinite union of closed sets, none of which contains all the others. Consider $[1/n, 1]$ for $n > 1$ together with $[0, 0]$.

Your feeling was (not quite) "every closed cover of a closed set has a sub-cover of size 1". That's false, and putting it that way might not mean much to you right now. But remember it for when you hear the definition of a compact set, because you were almost on to something important.

Steve Jessop
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