The Cantor Intersection Theorem is that

Let $\{S_1,S_2,S_3,...\}$ be a countable collection of nonempty sets in $\mathbb R$ such that:

  1. $S_{k+1} \subset S_k$ for $k=1,2,3...$

  2. Each $S_k$ is closed and $S_1$ is bounded,

then the intersection $\bigcap_{k=1}^\infty S_k$ is closed and nonempty.

My question is that, what if for some $S_k$ is not closed, how does it fail?

What is the counterexample?


1 Answer 1


The most trivial one would be

$$S_k= (0,\frac{1}{k}).$$

Note that $\bigcap_{k=1}^\infty S_k$ is empty.

If only finitely many of them are not closed, then the result still hold.

  • $\begingroup$ I think you meant to say "If only finitely many are not closed$\dots$" $\endgroup$ Commented Dec 6, 2014 at 8:47
  • $\begingroup$ @MikeEarnest: Thanks, edited. $\endgroup$
    – user99914
    Commented Dec 6, 2014 at 9:46

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