Is it possible to have an empty intersection of a set and its subsets?

[tinfoilhat]

question about the set [a,a)

If S=[a,a) Can such a set exist? It implies that a is in S and not in S, which doesn't make sense, but it seems a problem I'm trying to do requires it to be considered empty.

The question is:

Let I_n = [a_n,b_n) where

I_n+1 < I_n for all natural numbers n. [< denotes subset]

Give an example of those I_n for which the intersection of I_n (for all n) is empty.
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I can't see any other way to construct an empty set.

 

[Hurkyl]

(a, a] would be the set of all x such that a < x <= a.

However, I think the question intends you to limit yourself to intervals whose endpoints are distinct.


Recall that when you take a nested intersection of nonempty closed sets, you never get the empty set. Since you're searching for behavior that is not demonstrated by closed sets, I would suggest focusing at the open end of your intervals, where they resemble closed sets the least.

 

[tinfoilhat]

Hi. Thanks for the response. Based on your definition [a,a) is indeed empty. Although I'm not sure why you think the question requires the endpoints to be distinct.

If every set I_n+1 is a subset of I_n, then the only way the intersection of I_n for all n is empty is if one of those sets is empty. To my eyes (which admittedly are stupid at times), the only set of the form [a,b) that is empty is when a=b (or b<a)

 

[tinfoilhat]

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I just learned that the set [a,a) is not a proper subset of itself. So if I_n=[a,a), I_n+1 can't satisfy I_n+1 < I_n.

Now I have no idea what's going on. If no I_n can be empty, how can the intersection of all I_n be empty in any case! HELP