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Cantor's intersection theorem states

Let $S$ be a topological space. Given a decreasing nested sequence of bounded nonempty compact, closed subsets of S satisfying $$C_1 \supset C_2 \supset C_3 \supset ...$$ it follows that $$ \left(\bigcap _{k}C_{k}\right)\neq \emptyset$$

Wouldn't the intersection of all $C_k$ simply just be the subset $C_N$ with the largest index?

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  • $\begingroup$ What if there are infinitely many such intervals, though? What exactly is the "largest" index? $\endgroup$ – Arturo Magidin Nov 20 '19 at 21:59
  • $\begingroup$ @Gae.S. yes, of course. I corrected my comment. $\endgroup$ – uniquesolution Nov 20 '19 at 22:05
  • $\begingroup$ @Gae.S. yes, you are right again. $\endgroup$ – uniquesolution Nov 20 '19 at 22:11
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    $\begingroup$ It does not make sense, as far as I know, to classify a subset of a topological space as "bounded" or otherwise. $\endgroup$ – Gae. S. Nov 20 '19 at 22:18
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It is true that for finite $N \in \mathbb{N}$, $\cap_{k=1}^{N} C_{k} = C_{N}$. But consider $C_{n} = \left[ -\frac{1}{n}, \frac{1}{n} \right] \subset \mathbb{R}$. Taking an infinite intersection, we have that $\cap_{k \in \mathbb{N}} C_{k} = \{ 0 \}$. Here, the notion of largest index does not quite make sense. In a general topological space, $\cap_{k} C_{k}$ is non-empty because one can construct a sequence which converges in the intersection by the closed property of the intersection.

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  • $\begingroup$ Or consider the Cantor set, where every $C_n$ is a finite union of intervals and the intersection contains no interval at all... $\endgroup$ – Henno Brandsma Nov 20 '19 at 22:39
  • $\begingroup$ Or the intersection is nonempty because it is a family of closed subsets of a compact set, and every finite subfamily has nonempty intersection; hence the entire family has nonempty intersection. $\endgroup$ – Arturo Magidin Nov 20 '19 at 23:04

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