Is an Hausdorff separable topological space that is uniform and complete necessarily a Polish space ?
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1$\begingroup$ What do uniform and complete mean here? I guess the latter is Čech-complete, does the former mean the topology is induced by a uniformity? $\endgroup$– Alessandro CodenottiCommented Nov 29, 2023 at 18:17
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6$\begingroup$ How about $\{0,1\}^{2^\omega}$? $\endgroup$– Emil JeřábekCommented Nov 29, 2023 at 18:26
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2$\begingroup$ @AlessandroCodenotti Perhaps “uniform and complete” means “induced by a complete uniformity”? That’s how I would read it. $\endgroup$– Emil JeřábekCommented Nov 29, 2023 at 18:30
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4$\begingroup$ A space with cardinality below the first uncountable measurable cardinal can be endowed with a complete uniformity if and only if it is realcompact. And any paracompact space can be endowed with a complete uniformity (and a space is paracompact if and only if it can be endowed with a supercomplete uniformity). These facts allow us to obtain counterexamples such as $\mathbb{Q},\beta\mathbb{N}$, the space of real numbers with the lower limit topology, the Sorgenfrey plane, and the product of up to continuumly many copies of the Sorgenfrey plane. $\endgroup$– Joseph Van NameCommented Nov 29, 2023 at 21:19
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3$\begingroup$ Functional analysis is awash with complete, separable, non metrisable locally convex spaces. $\endgroup$– terceiraCommented Nov 30, 2023 at 5:17
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2 Answers
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A simple counterexample is $\{0,1\}^{2^\omega}$:
It is a compact Hausdorff space, hence $T_{3\frac12}$, hence uniformizable, and any uniformity on a compact space is complete.
It is separable by the Hewitt–Marczewski–Pondiczery theorem.
It is not first-countable (it has character $2^\omega$), hence it is not metrizable.
answered Nov 29, 2023 at 20:26
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$\begingroup$ And not just uniformizable, it's a topological group. $\endgroup$– bofCommented Nov 30, 2023 at 1:38
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Questions like these are often answerable by a search of the pi-Base (noting that every Hausdorff paracompact space is completely uniformizable): https://topology.pi-base.org/spaces?q=%20hausdorff%2B%20separable%20%2B%20paracompact%2B%20~Metrizable
- Weak topology on separable Hilbert space (see comments)
- Arens-Fort Space
- Sorgenfrey line
- Double arrow space
- Appert space
- Continuum-power of closed unit intervals
- Stone-Cech compactification of the integers
- Single ultrafilter topology
- Arens space
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1$\begingroup$ The weak topology of an infinite dimensional Hilbert space is not complete. $\endgroup$ Commented Dec 4, 2023 at 16:56
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2$\begingroup$ So, does @JochenWengenroth remark imply that "every Hausdorff paracompact space is completely uniformizable" is incorrect? Or does it mean that there is some complete uniform structure on $l^2$, compatible with the weak topology, other than the usual weak uniform structure? $\endgroup$ Commented Dec 4, 2023 at 21:23
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2$\begingroup$ I think "Stone-Cech compactification of the integers" is the best example, here. It is compact, therefore completely uniformizable. It has $\mathbb Z$ as a dense subset, therefore it it separable. But it has power $2^{\mathfrak{c}}$, so it is not Polish. $\endgroup$ Commented Dec 4, 2023 at 21:28
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1$\begingroup$ I repeated the questionable formulation the weak topology [...] is not complete although I meant the weak uniformity of a Hilbert space. This uniformity is indeed incomplete. Whether there are complete uniformities generating the weak topology is a different property (which seems to be true for the weak topology of a separable Hilbert space). However, I did not find a reference for paracompactness. $\endgroup$ Commented Dec 5, 2023 at 8:13
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1$\begingroup$ The weak topology of a Hilbert (or reflexive Banach) space is $\sigma$-compact, hence Lindelöf and regular and thus paracompact. $\endgroup$ Commented Dec 5, 2023 at 8:44