## Closed infinite intersection of open sets

For the discussion of math. Duh.

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### Closed infinite intersection of open sets

I was told that the infinite intersection of the open sets (-1/1, 1,1), (-1/2, 1/2), (-1/3, 1/3), ... (-1/n, 1/n) as integer n -> infinity is just {0}, a closed set.
However, I'm not sure this makes sense to me.

Since for any two sets in the intersection, one is a proper subset of the other, the "smallest" set in the intersection must be the intersection, and since all the sets are open, the intersection must therefore also be open. Am I missing something?

jestingrabbit
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### Re: Closed infinite intersection of open sets

>-) wrote:Since for any two sets in the intersection, one is a proper subset of the other, the "smallest" set in the intersection must be the intersection, and since all the sets are open, the intersection must therefore also be open. are we missing something?

Yes, you are missing that the intersection of an infinity of sets is the set of points contained in all of the sets. Your argument about pairs of sets is valid in that any finite intersection of open sets is open, but irrelevant as we are not performing a finite intersection.

Can you describe a point that is not 0 that is in the intersection of all these sets? Is 0 in all these sets? What is the intersection of all these sets?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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### Re: Closed infinite intersection of open sets

I see that the intersection appears to be {0}, but what breaks down with my reasoning when the intersection is infinite?

DR6
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### Re: Closed infinite intersection of open sets

Your argument is perfectly valid for the intersection of two sets: the intersection of two sets is open. Just in that case. Technically, you haven't even proved it for, say, three sets: just for two. To prove that it works for more than two sets, you need to use induction: so your argument can be extended to any collection with a natural number of elements, i.e. finite sets. For infinite sets your argument doesn't work anymore, because induction only extends to natural numbers.

Tirian
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### Re: Closed infinite intersection of open sets

>-) wrote:Since for any two sets in the intersection, one is a proper subset of the other, the "smallest" set in the intersection must be the intersection, and since all the sets are open, the intersection must therefore also be open. are we missing something?

You already see where your argument breaks down, because it is the word that you put quotations around. There is no smallest real open interval containing 0, just like there is no largest integer.

phlip
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### Re: Closed infinite intersection of open sets

Do you know the formal definition of an open set? Do you know the proof that the intersection of two open sets is open? Try to take that proof and extend it to an infinite intersection, and see where it breaks down...

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### Re: Closed infinite intersection of open sets

DR6 wrote: For infinite sets your argument doesn't work anymore, because induction only extends to natural numbers.

Tirian wrote:You already see where your argument breaks down, because it is the word that you put quotations around. There is no smallest real open interval containing 0, just like there is no largest integer.

Ah, that explains it. Thank you!

thefargo
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### Re: Closed infinite intersection of open sets

Sorry to ask a dumb question, but is there a difference between (0) and {0} (that is to say 0 as an open set vs closed set?). I tried to google around for an answer, but couldn't find anything

Tirian
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### Re: Closed infinite intersection of open sets

Yes. (0) = {real x | 0 < x < 0} is the empty set.

jestingrabbit
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### Re: Closed infinite intersection of open sets

thefargo wrote:Sorry to ask a dumb question, but is there a difference between (0) and {0} (that is to say 0 as an open set vs closed set?). we tried to google around for an answer, but couldn't find anything

Sets are different only if they contain different points. That is to say, if A != B, either there is an element of A that is not an element of B, or there is an element of B that is not an element of A. You don't get to talk about "A as an open set" or "A as a closed set". A is either open or closed, and it doesn't make any difference how you want to talk about it.

{0} is the set which contains one element only, that element being 0. I have no idea what you mean when you write (0). "0 as an open set" is also not something that makes any sense to me. You might be meaning to use interval notation, but to do that you need two numbers, like so: (a, b). (0, 0) isn't something that I would find acceptable interval notation, but if it means anything, it means the empty set. [0, 0] is also not something that I like the look of, but again, if it signifies anything, it is {0}.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

Qaanol
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### Re: Closed infinite intersection of open sets

It may be worth mentioning that “open” and “closed” are extrinsic properties of a set: they depend on the underlying space being considered.

{0} as a subset of ℝ is closed and not open
{0} as a subset of ℤ is closed and open

For a more interesting example, consider the half-open interval in the Cartesian plane, call it h = [0, 1)×{0}. In other words, h is the set of all points (x, y) in ℝ² with 0≤x<1 and y=0. Symbolically, h = {(x, y) | 0≤x<1, y=0}.

Now consider, is h open, and is h closed, as a subset of…
• ℝ²
• [0, 1)×ℝ
• [0, 1]×{0}
• [0, 1)×ℤ
?
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flownt
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### Re: Closed infinite intersection of open sets

Qaanol wrote:{0} as a subset of ℤ is closed and open

Depends on the topology on ℤ. While ℝ has a "default" topology, on ℤ several are sensible. Offhand, one might consider ℤ as a subspace of ℝ, this leads to the discrete topology and is probably what you intend. On the other hand, one can also consider ℤ a subspace of hat ℤ, the profinite completion.This topology is the one used in Fuerstenberg's proof that the number of primes is not finite. Thirdly one can consider the cofinite (or even cocompact) topology which makes all finite sets closed. Under this topology {0} is not open.