2

Could anyone point out which line of the following reasoning is incorrect?

Let $\mathbb R$ be equipped with the indiscrete topology, $\{\emptyset, \mathbb R\}$.

  1. Then, every subset of $\mathbb R$ is compact.
  2. The sequence of sets $\{(n, \infty)\mid n \in \mathbb N\}$ is decreasing and all compact.
  3. According to Cantor's intersection, the intersection of the sequence of sets should be non-empty.

However, 3. is obviously false. Is Cantor's intersection theorem only applied to metric space? Is there any generalized version to only get the non-empty intersection result?

Jimmy R.
  • 35,868
seunglim
  • 71
  • Possible duplicate: http://math.stackexchange.com/questions/247645/intersection-of-nested-compact-subspaces-in-non-hausdorff-space – Michael M Jan 07 '16 at 16:36

1 Answers1

2

The proof uses the fact that the complement of a compact set is open, which as we see is not true in every topological space. It's true in any Hausdorff space.

David C. Ullrich
  • 89,985