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) ?
first countable $\Leftrightarrow$ compact and Hausdorff?
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1@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*. – 2011-10-31
2 Answers
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.
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0Great answer, thanks (though the word "stronger" in my question just referred to "Hausdorff" not "Hausdorff + compact") – 2011-10-31
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 $[0,1]$.