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Let X be a compact Hausdorff space. If it has a countable basis, is it metrizable? And if it is metrizable, does it have a countable basis?

carmichael561
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    If it is metrizable and separable it has a countable basis. Let $a_n$ denumerate a dense subset. $ { B(a_n,q) | n \in \mathbb{N},q \in \mathbb{Q} } $ is a basis. This doesn't require compactness. – Neil May 25 '16 at 04:45
  • Does "metrizable" mean equipped with a positive definite billinear form? – marshal craft May 25 '16 at 05:07
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    @marshalcraft, please google for "metrizable space" (or even just "metrizable") – Mariano Suárez-Álvarez May 25 '16 at 05:08
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    See http://math.stackexchange.com/questions/234018/compact-metrizable-space-has-a-countable-basis-munkres-topology?rq=1 for metrizable $\Rightarrow$ countable basis. For the other direction, see Urysohn's theorem: https://en.wikipedia.org/wiki/Metrization_theorem (and note that compact Hausdorff implies regular). – Moishe Kohan May 25 '16 at 05:29

2 Answers2

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  • A compact Hausdorff space $X$ with a countable base is indeed metrisable. This is Urysohn's metrisation theorem. The standard proof embeds $X$ into the metric space $[0,1]^\mathbb{N}$.

  • If a compact space $X$ is metrisable (with metric $d$), it has a countable base (even Lindelöf is enough for this, i.e. every open cover has a countable subcover). For this, for every $n$, cover $X$ by the open balls $B_d(x, \frac{1}{n})$ and take a countable (or finite) subcover, and call these balls $\mathcal{B}_n$, and show that $\cup_n \mathcal{B}_n$ is a countable base for $X$.

So for compact spaces $X$, $X$ is metrisable iff $X$ is Hausdorff and has a countable base.

For non-compact spaces $X$ we have more complicated theorems: A space $X$ is metrisable iff $X$ is regular and $T_1$ (also known as $T_3$) and $X$ has a $\sigma$-locally finite base (see the Bing-Nagata-Smirnov metrisation theorem). This is quite non-trivial.

Henno Brandsma
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A metrisable uncoutable space X with the discrete meetric cannothave a countable base. If it does then it would have a countable dense subset A with closure A=A=couable=X!

Adelafif
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