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If a metric space is complete, we know that the Cantor's Intersection Theorem holds. Does the converse also hold? And if not, what is a suitable counterexample for the same?

Also, if the converse does not hold, does the Cantor's Intersection Theorem hold only for complete metric spaces, or does it hold for more general spaces too?

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Yes, Cantor's Intersection Theorem fails to hold in all incomplete spaces. For example, take any non-convergent Cauchy sequence $x_n$, and consider the sets: $$C_k = \{x_n : n \ge k\}.$$ Because $x_n$ is Cauchy, this implies $\operatorname{diam} C_k \to 0$.

If these sets were not closed, then there would be some accumulation point $x \in C_k$ for some $k$, which would suggest a convergent subsequence of $x_n$ to $x$, which would imply $x_n \to x$, since $x_n$ is Cauchy. Hence, no such accumulation point exists, and the sets must be closed.

Finally, if the sets had a point in the intersection, then the sets would have to have a constant subsequence, which again would imply convergence.

Hence, the $C_k$s form a counterexample to Cantor's Intersection Theorem, when the space is incomplete.

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  • $\begingroup$ Seems simpler just to let $D_n=\overline{C_n}$ and note that $\bigcap D_n$ is empty... $\endgroup$ May 17, 2019 at 14:14
  • $\begingroup$ @DavidC.Ullrich I don't see it. Surely you'd have to replace the argument that $C_n$ is closed with an argument that $D_n$ contains no extra points in common? $\endgroup$ May 17, 2019 at 14:19
  • $\begingroup$ I don't know what an "extra point" is, but it doesn't matter. If $x\in\bigcap D_n$ it's clear that $x_n\to x$. $\endgroup$ May 17, 2019 at 14:28

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