Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone give me a short sketch of a proof or a space that serves as a counterexample to the fact that every first countable space is characterized by being compact and Hausdorff (or, stronger than Hausdorff, metrizable) ?

share|cite|improve this question
You have your answer. I am really wondering how did you come to this conclusion that first countable are necessarily compact and Hausdorff (since second countable implies first countable, it would mean that all second countable spaces are compact Hausdorff... which is obviously false.) – Asaf Karagila Oct 28 '11 at 10:47
You need to consider examples and definitions more closely, probably. You are surely familiar with $\mathbb R$ as a topological space, and you probably know that it is first countable, Hausdorff and not compact. – Mariano Suárez-Alvarez Oct 28 '11 at 11:19
@AsafKaragila Since you are wondering, I didn't come to this conclusion by any specific problem - I just came to it, thinking "Couldn't it be that...?". And if you are new to topology, it isn't at all obvious, like you said, that second countable space aren't always compact and Hausdorff. – resu Oct 31 '11 at 15:24
@resu another point of view would be that it is not obvious at all that there should be any connection between first countability and either Hausdorfness or compactness. – Olivier Bégassat Oct 31 '11 at 17:21
up vote 19 down vote accepted

Every infinite discrete space is first countable and Hausdorff (in fact metrizable) but not compact. The space $\omega_1$ of countable ordinals with the order topology is first countable and Hausdorff but neither compact nor metrizable. The integers with the cofinite topology are first countable and compact but not Hausdorff. The space $\langle\mathbb{N},\mathscr{T}\rangle$, where $$\mathscr{T} = \left\{\{0,\dots,n\}:n\in\mathbb{N}\right\}\cup\{\varnothing,\mathbb{N}\},$$ is first countable but neither compact nor Hausdorff (nor even $T_1$).

Metrizability is certainly not stronger than compactness plus Hausdorff: the real line is metrizable but not compact.

share|cite|improve this answer
Great answer, thanks (though the word "stronger" in my question just referred to "Hausdorff" not "Hausdorff + compact") – resu Oct 31 '11 at 15:25

Brian M Scott's answer was exhaustive, but let me add something, addressing a fundamental problem with the characterization your are looking for (or a similar one):

First countability, Hausdorffness, and metrizability are all hereditary with respect to subspaces. I.e., every subspace of a first countable (Hausdorff, metrizable) space is first countable (Hausdorff, metrizable). But compactness is not! The open unit interval $(0,1)$ is a noncompact subspace of the compact space $[0,1]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.