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What is the source of the claim that "Strongly zero-dimensional and zero-dimensional is equivalent for separable metrizable spaces"? My university lecture told me that this is true, but I fail to find it anywhere in literature. Could you provide a source for this?

Definitions

A Hausdorff topological space $X$ is zero dimensional if for every point $x$ of $X$ and every neighborhood $U$ of $x$ in $X$, there exists a nonempty clopen subset $V$ of $X$ such that $x \in V \subset U$. The clopen basis of any zero-dimensional space is a collection of clopen sets that is closed under complements and finite intersections.

A Hausdorff topological space $X$ is said to be strongly zero-dimensional whenever for every closed subset $A$ of $X$ and every open subset $U$ of $X$ such that $A \subseteq U$, there exists a clopen subset $V$ of $X$ such that $A \subseteq V \subseteq U.$

Arthur
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Tereza Tizkova
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  • If $\text{ind}$ and $\text{Ind}$ denote the small and large inductive dimensions, then $\text{ind} = \text{Ind}$ for separable metric spaces. Zero dimensional space is zero-dimensional with respect to $\text{ind}$, while strongly zero-dimensional, with respect to $\text{Ind}$ (or equivalently Lebesgue covering dimension) – Jakobian May 14 '23 at 17:56

2 Answers2

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What you call "strongly zero-dimensional" is usually called ultranormal. In this common notation a Tychonoff space is called strongly zero-dimensional, if its Stone-Cech compactification is zero-dimensional. A space is ultranormal, iff it is normal and strongly zero-dimensional. In particular, for metric spaces, ultranormality is the same as strong zero-dimensionality. See, for instance here.

In the paper cited in the above link, you also find a proof for "Every Lindelof zero-dimensional space is ultraparacompact" (hence ultranormal), thus answering your question, since a separable metric space is, of course, Lindelof. See also Engelking, General topology 6.2.7.

Ulli
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  • Thank you! But isn´t it enough to say 1) Every metric & separable space is Lindelof and 2) Every Lindelof zero-dimensional space is strongly zero dimensional and from 1) and 2) together conclude the result? I don´t see how the ultranormality etc. is needed. – Tereza Tizkova Jul 15 '22 at 14:58
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    @TerezaTizkova: According to your defintion, you are not asking for strong zero-dimensionality, but for ultranormality. Of course, a (regular) Lindelof space is normal, hence a Lindelof zero-dimensional space is ultranormal, or, according to your definition "strongly zero-dimensional". But for non-Lindelof, non-metric spaces, strong zero-dimensionality is strictly weaker than ultranormailty. – Ulli Jul 15 '22 at 15:12
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It´s easy to see that every strongly zero-dimensional space is a zero-dimensional space, but conversely it is not necessarily true.

Theorem: Every zero-dimensional Lindelöf space is strongly zero-dimensional. You can see this result in Engelking, in the theorems 6.2.2 to 6.2.7.

Answering your question for metrizable spaces. I think you should ask for the space to be metrizable and separable, in order to have a Lindelöf space and use the previous result. I do not have at the moment, a counterexample on a zero-dimensional, metrizable and non-separable space that is not strongly zero-dimensional

AmottX
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