# Infinite intersection of open sets

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.

• A formal proof consists in finding one counter example to the statement: the infinite intersection of open sets is open – marwalix Oct 2 '15 at 9:36
• to disprove something it's enough to give a counterexample – user300 Oct 2 '15 at 9:37
• 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. – Prahlad Vaidyanathan Oct 2 '15 at 9:37
• @GAVD: The intersection you have is the empty set, which is open. – Prahlad Vaidyanathan Oct 2 '15 at 9:38
• Thanks! It should be $\cap_{n=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)$. – GAVD Oct 2 '15 at 9:45

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.

• 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. – Henno Brandsma Oct 2 '15 at 9:50
• Thanks, fixed. Though in the usual topology on $\mathbb{R}$, you could just add the fact that the only clopen sets are $\emptyset, \mathbb{R}$. – Batman Oct 2 '15 at 11:04
• Sure, but that goes into connectedness which is non-trivial – Henno Brandsma Oct 2 '15 at 11:10
• Because $0$ is in every interval of the form $(-1/n,1/n)$ for any $n \geq 1$. – Batman Apr 25 '17 at 12:43
• No, $1/n$ is never zero. It gets close to zero, but is never zero. – Batman Apr 25 '17 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.

Note-

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 closed.