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I was trying to prove that a collection of sets is a topological space over a set and I came over this question which I can't figure out now:

Can the infinite intersection of infinite sets (all subsets of a given set) be finite but not empty?

If it is possible can anyone construct any example?


So the actual question was: Let $X$ be a an infinite set. Does the following collection of subsets of $X$ given by $T=\{U \in X : X \backslash U$ is infinite or empty $\}$ defines a topology? Answer : No.

Which now by your help I can prove. Thnx everyone :)

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up vote 8 down vote accepted

Take the sets $S_n = \{n^k \colon k \ge 0\}$, $\bigcap_{n \ge 1} S_n = \{1\}$.

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If $X = \mathbf{R}$ and $I_r = \left[-\frac{1}{r},\frac{1}{r}\right]$, then $$\bigcap_{r=1}^\infty I_r = \{0\}.$$ This is called the nested interval property.

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+1) "nested interval property" – Paul Apr 10 '13 at 1:20

This may be helpful for you:

Let $X=\mathbb N$, and let $\mathcal U=\{\{1,3, 5\}, \{1,3, 5, 0\}, \{1,3, 5, 0, 2 \}, \{1,3, 5, 0, 2 , 4\},....\}$. Clearly, the infinite intersection of $\mathcal U$ is always $\{1,3,5\}$, hence finite.

I may introduce a definition of $calibre-{\aleph_1}$ which between Countable chain condition and separable.

A topological space $X$ is called $calibre-{\aleph_1}$ if for any uncountable nonempty open sets of $X$, there is an uncountable subcollection such that $\bigcap\{U_\xi: \xi < \aleph_1\}\not=\emptyset$.

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