Contents

Statement

Lemma

(Michael's theorem, Michael 53, theorem 1)

Let $X$ be a topological space such that

1. $X$ is regular;

2. every open cover of $X$ has a refinement by a union of a countable set of locally finite sets of open subsets (not necessarily covering).

Then $X$ is paracompact topological space.

Proposition

(second-countable regular spaces are paracompact)

Let $X$ be a topological space which is

Then $X$ is paracompact topological space.

Proof

Let $\{U_i \subset X\}_{i \in I}$ be an open cover. By Michael's theorem (lemma 1) it is sufficient that we find a refinement by a countable cover.

But second countability implies precisely that every open cover has a countable subcover:

Every open cover has a refinement by a cover consisting of base elements, and if there is only a countable set of these, then the resulting refinement necessarily contains at most this countable set of distinct open subsets.

References

Last revised on August 9, 2017 at 04:58:41. See the history of this page for a list of all contributions to it.