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Motivation: I read that compact subsets of metric spaces have this property, and I was wondering why this is not true for all topological spaces.

If $\bigcap_{i=1}^\infty A_i$, then there is at least one pair of sets $A_k,A_l$ such that $A_k\bigcap A_l=\emptyset$. If there is no such pair of disjoint sets, then a point $x$ is present in every set $A_i$, and hence in $\bigcap_{i=1}^\infty A_i$.

By the finite intersection property, $A_k\bigcap A_l\neq\emptyset$ for any pair of sets. Hence, $\bigcap_{i=1}^\infty A_i\neq\emptyset$.

Thanks in advance!

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    $\begingroup$ No, if $\bigcap_{i=1}^\infty A_i = \varnothing$, there need not be two disjoint sets in the family. Consider $A_i = (0,\, 1/i) \subset \mathbb{R}$. $\endgroup$ – Daniel Fischer Aug 15 '13 at 12:01
  • $\begingroup$ Sorry this was an exceedingly stupid question. I just found a flaw in my own argument. If there are no disjoint pairs, that does not imply one single point $x$ is included in all sets. There may be different points in intersections of different pairs of sets. So the intersection of all sets need not contain that one point. Thanks. $\endgroup$ – fierydemon Aug 15 '13 at 12:07
  • $\begingroup$ Wikipedia article on Cantor intersection's theorem mentions that: On the other hand, both the sequence of open bounded sets $C_k = (0, 1/k)$ and the sequence of unbounded closed sets $C_k = [k, \infty)$ have empty intersection. All these sequences are properly nested. $\endgroup$ – Martin Sleziak Aug 15 '13 at 14:37
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Take the collection $A_i = (-\frac1{i},\frac1{i})\backslash \{0\}$. Its intersection is empty, but it has the required property...

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Let $$A_n=\{ n, n+1, n+2,... \} = \{ m \in \mathbb Z | m \geq n \}$$

Then, for any finite collection, the intersection is one of the sets (the one with the largest index), but the infinite intersection is the emptyset.

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