# Is the intersection of a nested sequence of compact subsets of a metric space equal to the smallest subset?

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?

• What if there are infinitely many such intervals, though? What exactly is the "largest" index? – Arturo Magidin Nov 20 '19 at 21:59
• @Gae.S. yes, of course. I corrected my comment. – uniquesolution Nov 20 '19 at 22:05
• @Gae.S. yes, you are right again. – uniquesolution Nov 20 '19 at 22:11
• It does not make sense, as far as I know, to classify a subset of a topological space as "bounded" or otherwise. – Gae. S. Nov 20 '19 at 22:18

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.
• Or consider the Cantor set, where every $C_n$ is a finite union of intervals and the intersection contains no interval at all... – Henno Brandsma Nov 20 '19 at 22:39