I need to prove that the infinite intersection of open sets may [must] not be open. I can show through examples that this is true, but this is not sufficient for a proof. - Can somebody give a formal proof ? Thanks.

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    $\begingroup$ A formal proof consists in finding one counter example to the statement: the infinite intersection of open sets is open $\endgroup$
    – marwalix
    Oct 2, 2015 at 9:36
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    $\begingroup$ to disprove something it's enough to give a counterexample $\endgroup$
    – user300
    Oct 2, 2015 at 9:37
  • $\begingroup$ There is no proof that the intersection is not open (since that is not true), so the best you can do is give a counterexample, which you presumably have. $\endgroup$ Oct 2, 2015 at 9:37
  • $\begingroup$ @GAVD: The intersection you have is the empty set, which is open. $\endgroup$ Oct 2, 2015 at 9:38
  • $\begingroup$ Thanks! It should be $\cap_{n=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)$. $\endgroup$
    – GAVD
    Oct 2, 2015 at 9:45

3 Answers 3


All you need is a counterexample to show a statemnt is not true: $\cap_{n=1}^\infty (-1/n,1/n) = \{0\}$ which is an intersection of open intervals resulting in a (non-open) closed set in the usual topology on $\mathbb{R}$ .

Obviously, there are infinite collections of open sets whose intersection is open. For example, $\cap_{n=1}^\infty (n,n+1) = \emptyset$ which is always open.

A term for countable intersections of open sets is a $G_\delta$ set. You can find $G_\delta$ sets which are neither open nor closed.

Thus, infinite intersections of open sets may be closed, open or neither.

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    $\begingroup$ The relevant fact is that $\{0\}$ is not open. Not that it's closed (as in general a set can be both open and closed). The fact that the intersection is closed proves nothing. The fact that it isn't open, does. $\endgroup$ Oct 2, 2015 at 9:50
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    $\begingroup$ Thanks, fixed. Though in the usual topology on $\mathbb{R}$, you could just add the fact that the only clopen sets are $\emptyset, \mathbb{R}$. $\endgroup$
    – Batman
    Oct 2, 2015 at 11:04
  • $\begingroup$ Sure, but that goes into connectedness which is non-trivial $\endgroup$ Oct 2, 2015 at 11:10
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    $\begingroup$ Because $0$ is in every interval of the form $(-1/n,1/n)$ for any $n \geq 1$. $\endgroup$
    – Batman
    Apr 25, 2017 at 12:43
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    $\begingroup$ No, $1/n$ is never zero. It gets close to zero, but is never zero. $\endgroup$
    – Batman
    Apr 25, 2017 at 15:11

The O.P here is not asking for an example or counter-example here. He is confused with the logic involved in disproving something.

Consider an example,

Proposition: I went to gymnasium everyday in August.

If you need to prove the statement either you have to check for everyday of august you have gone to gymnasium (or what mathematicians generally do is- take an arbitrary day of august and prove that you have gone to gym on that day, hence you have gone to gym on everyday of august.

But if you need to prove that the proposition is false, it is enough to find a particular such that you haven't gone to gymnasium on that day.

Similarly, to disprove infinite intersection of open sets is open, it is enough to give a particular collection of open sets such that intersection is not open.


1) To disprove the statement infinite intersection of open sets is open is not same as proving infinite intersection of open set is not open.

2) Your question was "prove that the infinite intersection of open sets may [must] not be open". Now there is a problem in the question. "infinite intersection of open sets may [must] not be open"- this is not a proposition or a statement. A proposition or statement is something which is either false or true, cannot be false and true together. The 'must' or 'may' creates a problem to be a proposition or a statement. So logically you cannot ask to prove something which is not a proposition. You can only ask to prove or disprove a proposition. Therefore, the correct statement should have been- Prove or Disprove, infinite intersection of open sets is a open set.

I hope this helps.


Take for instance $\cap_{n\in\mathbb{N}} K\left(0,\frac{1}{n}\right)$ where $K\left(x,\epsilon\right)$ denotes the open ball of radius $\epsilon$ around $x$.

Clearly the intersection of all these Sets is $\left\{0\right\}$ which is not open.

  • $\begingroup$ I think that the intersection in question is the empty set. Which is closed and open (in the topology induced by the standard metric). $\endgroup$
    – Loic
    Oct 25, 2021 at 13:42

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