In metric spaces, every open set is a countable union of closed sets.
is the converse true?
A topological space with the property "every open set is a countable union of closed sets" has to be metrizable?
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No the Sorgenfrey line (a.k.a. $\mathbb{R}$ in the lower limit topology) is perfectly normal (as this property is called) but not metrisable.
A compact counterexample is the Double Arrow space, which is a related example.
answered May 3 '16 at 18:05
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$\begingroup$ Why is the Helly space perfectly normal? Steen and Seebach "Counterexamples in Topology" (1978), p. 211, Problem 128 invite to show that the Helly space is not even completely normal. $\endgroup$ – yada Jun 27 '16 at 12:03
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$\begingroup$ @yadaddy it's not (I used pi-base which had an error at the time), as Helly space contains the Sorgenfrey plane as a subspace, I think. $\endgroup$ – Henno Brandsma Jun 29 '16 at 11:20
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$\begingroup$ I have recently modified in pi-base the description for the wrongly asserted properties for "perfectly normal" and T5 for the Helly space, but I cannot delete the properties themselves, since I have no administration rights. The available information in pi-base is based on the 1st edition of Steen, Seebach "Counterexamples in Topology" (1970). The 2nd (and current) edition (1978) corrects some errors and extends some unknown properties. As an example, only in the 2nd edition it is mentioned that the Helly space is not completely normal (as an exercise). $\endgroup$ – yada Jun 29 '16 at 12:08
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No. Take the indiscrete topology as a counterexample.
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$\begingroup$ Using a pseudometrizable space as your counterexample seems almost like cheating. $\endgroup$ – bof Jun 29 '16 at 11:29
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If that were true, then any countable T$_1$-space would be metrizable. In fact there are lots of countable Hausdorff spaces that aren't even first countable. For example, topologize $\mathbb N$ so that $$S\text{ is closed }\iff1\in S\text{ or }\sum_{n\in S}\frac1n\lt\infty.$$
