1

A common introductory topology question is to prove or disprove the following claim:

The intersection of an infinite number of open sets is open.

The claim is false, and a counter-example I've seen is this:

Consider the set of open intervals of the real numbers $A = \{A_n | A_n = (-1/n,1/n), n\in \mathbb{N}\}$. The intersection of this set is $\cap_{n=1}^\infty A_n = \{0\}$. This is a singleton, which is not open, and therefore the claim is false.

I am a bit confused by this counter-example. It seems that we've ignored something very important - somehow we took a limit (or something similar) and stated that "as $n\rightarrow \infty$, $A_n \rightarrow (0,0)$". It's that $\rightarrow$ that I have an issue with. Although it's certainly true that the intervals approach the interval $(0,0)$, why is it true that the intersection contains this interval? This seems as though we actually let $n=\infty$, but if that's the case it seems that we can get around this entire counter-example by claiming that infinity is not a natural number.

So, why is this counter-example valid? (assuming it is, but it shows up in Schaum's Outlines, so I'm assuming it's valid...)

Michael Stachowsky
  • 2,943
  • 11
    No, we didn't say $A_n\to(0,0)$ (well you did, but the rest of us didn't). What is the case is that the only real number in all of sets $(-1/n,1/n)$ is zero. – Angina Seng Oct 30 '17 at 20:36
  • Oh! Of course. Thanks – Michael Stachowsky Oct 30 '17 at 20:37
  • 2
    "As $n\to\infty$, $A_n\to (0,0)$." You are correct that this is nonsense and irrelevant to the problem. –  Oct 30 '17 at 20:38
  • Btw, for every example of an infinite intersection of opens not being open, there is a corresponding example of an infinite union of closed sets not being closed; maybe this is easier for some to think about. E.g. $\displaystyle \bigcup_{n=2}^\infty [1/n, 1!-!1/n] = (0,1)$. – Kaj Hansen Oct 30 '17 at 20:58
  • I don't have the time, energy or inclination right now. But this question has been asked uncountably many times before. Please, put some effort into perusing the site before posting something... – Asaf Karagila Oct 30 '17 at 22:31
  • Quick example, https://math.stackexchange.com/questions/1620526/intersection-of-all-open-intervals-of-the-form-frac1n-frac1n-is, which I found as one of the first few results of searching for "intersection of open intervals". – Asaf Karagila Oct 31 '17 at 00:29
  • @AsafKaragila: I did put effort in, as did the other commenters and the person who answered. My question was to address a very specific misunderstanding of mine, which I feel was done very well – Michael Stachowsky Oct 31 '17 at 00:30
  • There is a difference between putting effort into your question, and into searching if your question has been asked before. Yes, you did the former (which is why I didn't downvote); but apparently not the latter. And I can assure you that this question has been asked inside and out from every possible angle. – Asaf Karagila Oct 31 '17 at 00:32
  • Another example, https://math.stackexchange.com/questions/2274229/problem-with-a-limit-and-the-infinite-intersection-of-open-sets, which I found quickly by searching for "intersection open sets limit"... – Asaf Karagila Oct 31 '17 at 00:36
  • While it's certainly true that other similar questions have been asked, I will respectfully admit that I don't feel bad about that. Asking allowed me to present my confusion exactly and I'm happy with the responses. I don't really see a significant downside to asking questions again. They can always be ignored by those who are tired of answering them. – Michael Stachowsky Oct 31 '17 at 01:08
  • An infinite intersection is not a limit. By definition, a point is contained in $\bigcap_{n=1}^{\infty}A_n$ if and only if it is contained in every $A_n$. It's easy to check that $0$ is the only point that is contained in every $A_n = (-1/n, 1/n)$. –  Oct 31 '17 at 01:18
  • Michael, I wish you to start answering questions at some point, and I wish you to do so for many years. When you'll grow up and answer the same answer so many times, you will see the issue. And it's not like I would answer the same question every year to a different set of students. Here the same question was written down, publicly, and received so many wonderful answers, publicly. So this is generally a waste of good energy on the side of those who actually answer the question. This behavior that you describe is sometimes pushed to the limit and is then classified as "help vampirism". – Asaf Karagila Oct 31 '17 at 01:23
  • @bof: edited, thank you. – Michael Stachowsky Oct 31 '17 at 10:27
  • @AsafKaragila: I'm not entirely in agreement (SE is a place to ask questions, after all, and I personally don't mind answering common questions), however I think I can certainly agree to do a bit more homework before posting something new. – Michael Stachowsky Oct 31 '17 at 12:24

1 Answers1

1

Assume nonzero x in the intersection. By the Archimedean property of
numbers, there is some n in N with 1/n < x if x is positive
or x < -1/n if x is negative. Thus x is not in A_n and hence
not in the intersection. So none of the nonzero numbers are in
the intersection, leaving only 0 which clearly is in the intersection.

William Elliot
  • 17,598