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Property: Strongly zero-dimensional spaces

Open Moniker1998 opened this issue 2 years ago • 0 comments

A property that isn't on pi-base is that of strongly zero-dimensional spaces.

Those are spaces $X$ for which the Lebesgue covering dimension or equivalently large inductive dimension is $0$.

Equivalently, and I think this should be the definition on pi-base, a space $X$ is strongly zero-dimensional or ultranormal if for any closed disjoint $E, F\subseteq X$ there is clopen $U\subseteq X$ with $E\subseteq U, F\subseteq U^c$.

Contrast with zero-dimensional spaces, for which small inductive dimension of $X$ is $0$.

If we add this property the theorem [regular + P-space => zero-dimensional] on pi-base could be strengthened by replacing it with [regular + P-space => strongly zero-dimensional] as in this paper https://www.sciencedirect.com/science/article/pii/0016660X72900268

We could also add [strongly zero-dimensional + $T_1$ => zero-dimensional] and [strongly zero-dimensional + regular => zero-dimensional] See Charalambous's Dimension Theory

If we also add the property of ultrametrizable spaces, we could add those theorems: [metrizable + strongly zero-dimensional => ultrametrizable] [ultrametrizable => strongly zero-dimensional] [ultrametrizable => metrizable] A space is called ultrametrizable if it can be given an ultrametric i.e. metric $d$ such that $d(x, y)\leq \max(d(x, z), d(z, y))$

The following converse to strong zero-dimensionality could be added: [zero-dimensional + Lindelof => strongly zero-dimensional] Proof: If $X$ is Lindelof then $\text{dim }X \leq \text{ind } X$ where $\text{dim}$ is the Lebesgue covering dimension. If $\text{ind }X = 0$ then $\text{dim }X = 0$ so $X$ is strongly zero-dimensional (as I mentioned before, $\text{dim }X = 0$ iff $\text{Ind }X = 0$ where $\text{Ind}$ is the large inductive dimension).

Another one is [normal + extremally disconnected => strongly zero-dimensional]

Moniker1998 avatar Jul 22 '23 18:07 Moniker1998