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Theorem 2.24.

(a) For any collection ${G_\alpha}$ of open sets, $\cup_\alpha G_\alpha%$ is open.

(b) For any coll ection ${F_\alpha}$ of closed sets, $\cap_\alpha F_\alpha%$ is closed.

(c) For any finite collection $G_1,...,G_n$ of open sets, $\cap_{i=1}^n G_i%$ is open.

(d) For any fi nite collection $F_1,...,F_n$ of closed sets, $\cup_{i=1}^n F_i%$ is closed.

Here is the proof of (c) in the book.

Proof.$\quad$ Put $H=\cap_{i=1}^n G$. For any $x\in H$, there exist neighborhoods $N_i$ of $x$, with radii $r_i$ such that $N\subset G_i(i=1,...,n)$. Put $r=\min(r_1,...,r_n)$, and let $N$ be the neighborhood of $x$ of radius $r$. Then $N\subset G_i$ for $i=1,...,n$, so that $N\subset H$, and $B$ is open.

So, why can't we use the similar process to prove $\cap_{i=1}^\infty G_i%$ is open? Can't we get the $r=\min(r_1,...,r_n,...)$ when $n$ is for all the natural numbers?

Community
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Johnny Ji
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  • In fact, we have a "negation" of what you have asked: every closed set is a countable intersection of open sets. (But since $\Bbb{R}$ is connected, a open set cannot be closed unless it is $\Bbb{R}$ or $\emptyset$) see https://math.stackexchange.com/questions/504798/closed-set-as-a-countable-intersection-of-open-sets – Li Chun Min Jul 01 '17 at 14:18
  • Look at example 2.25. If you try to prove $0$ is an interior point of $\cap G_n$, you will have $r_n=1/n$ for each $G_n=(−1/n,1/n)$. But you can't take a minimum from this sequence of $r_n$ since $\inf r_n=0$. – Li Chun Min Jul 01 '17 at 14:20

4 Answers4

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Actually the "similar process" mentioned by you is not similar.

For a collection of arbitrary sets $G$s, you can only know that $\min$ always exists if there are finitely many (because you could check each one of them, and find the min one) - however you cannot do so if there are infinitely many. Thus you do not know if $\min$ exists or not (but we could say about its infimum, which does not help because we want to find a G out of the collection).

Jay Zha
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  • Is this a theorem or we just use it when we prove other theorems? BTW, if $G_1={1}, G_2={2},...,G_n={n},...$when n is for all the natural numbers. The sets $Gs$ are infinite, but I know the min is always $G_1$. – Johnny Ji Jul 01 '17 at 14:06
  • @JohnnyJi Yes, for a specific infinite collection of sets, you might know that. What I meant is for arbitrary sets in general, like what the book tries to prove, then you cannot be certain about that. Could you give it a try for the other theorem, and let me know where you get stuck. – Jay Zha Jul 01 '17 at 14:09
  • I got it. Can I say we can get the min for ascending order sets, but we can't get the min for arbitrary sets. – Johnny Ji Jul 01 '17 at 14:25
  • @JohnnyJi I guess you could say so, as in that case the first is the "smallest" (small by your definition). But it is not necessary for ascending, e.g. 1,2,4,3,5,6,7... also has the minimum. I updated my answer according to your comment to clarify that "min" meaning the element is within the collection, because here you are not just only caring about the value itself. – Jay Zha Jul 01 '17 at 14:31
  • Yeah, Thanks for you patience. – Johnny Ji Jul 01 '17 at 14:35
  • @JohnnyJi You're welcome, glad to help :) – Jay Zha Jul 01 '17 at 14:36
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  1. mind that there is a difference between the $infimum$ and the $minimum$ of a set. For instance, take $$A:=\{\frac{1}n; n \in \mathbb{N} \} \subset \mathbb{R}$$ This set does not have a minimal element, since $0 \notin A$, but $inf(A)=0$. If the $min$ exists, it is the same as the $inf$, but the existance of the latter does not guarantee the existence of the first. So in your set of indices $\{r_1,r_2,\dots\}$ there is not necesserily a minimal element.

  2. Another (counter-) example why your proposition is not true: Consider $$B:=\bigcap_{k=1}^{\infty} B_k(0) \subset \mathbb{R}^n,$$ where $B_k(0):=\{x \in \mathbb{R}^n:|x|< 1+1/k, \space k \in \mathbb{N}\} $. Cleary, $B_k(0) $ is open for every $k \in \mathbb{N}$, but $B= \{x \in \mathbb{R}^n:|x|\leq 1 \}$ is the closed unit ball and therefore not open.

Simonsays
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Well, a trivial but critical requirement is that $r$ exists. However, an arbitrary subset of real numbers doesn't necessarily have a minimum, unless provided that it's finite. That is to say, $\min(r_1,r_2,r_n,...)$ doesn't necessarily exist. So the proof in case of $G$'s finite couldn't be extended to the cases when $G$'s are infinite.

Actually, extending a theorem is rather risky, unless you clearly understand the theorems used in the proof.

B. Goddard
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Bill Tao
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You can be looking for a min forever, so it's very different to have finitely many subsets or infinite subsets. You need to prove that this minium exists before apply this way to prove it.

This will help you to understand why we cannot assume the existence of such min:

Maximum Number in an Infinite set

Relure
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