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We know that A closed set in a metric space is $G_\delta$

Is there any topological space where a closed set is not necessarily $G_\delta$?

I am thinking a space where singletons are well known to be closed, but cannot be represented as countable intersection of open sets.

Community
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Fraïssé
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  • I think something like the double origin line ($\Bbb R\cup{0^}$) would suffice. Roughly speaking, any open set containing either $0$ or $0^$ must contain the other, while the remaining topology is what you would expect. However, ${0}$ is a closed set. – Clayton Jun 08 '16 at 04:41

1 Answers1

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Example: The Sierpiński space $S=\{0,1\}$ with open sets $S,\ \emptyset,\ \{1\}.$ The set $\{0\}$ is closed but is not the intersection of any number of open sets.

Example: An uncountable set with the cofinite topology, i.e., the open sets are the cofinite sets and the empty set. Points are closed but are not $G_\delta$-sets. Of course this is not a Hausdorff space.

Example: The product of uncountably many compact Hausdorff spaces, each having at least two points. This is a compact Hausdorff space, and single points are not $G_\delta$-sets.

bof
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  • I like the last one. Similarly, points in the Stone-Cech remainder of a non-compact space are not $G_\delta$. – Forever Mozart Jun 08 '16 at 05:11
  • In your second example, points are closed because ${x}^c$ is co-finite and they are not $G_\delta$ since countable intersection of co-finite/infinite set is not finite, correct? Thanks – Fraïssé Jun 08 '16 at 05:12
  • @Lookbehindyou A related question which may be of interest: If every closed subset of $X$ is a $G_\delta$, then is $X$ metrizable? – Forever Mozart Jun 08 '16 at 05:14
  • @Lookbehindyou A countable union of finite sets is countable, so a countable intersection of cofinite sets is cocountable. – bof Jun 08 '16 at 05:20
  • @ForeverMozart In a countable T$1$-space, *every* subset is a $G\delta,$ and it's easy to give examples of such spaces which are not first-countable, therefore not metrizable. – bof Jun 08 '16 at 05:25
  • @bof ah, ok so maybe I need more assumptions. – Forever Mozart Jun 08 '16 at 05:36
  • @ForeverMozart Helly space is compact Hausdorff separable perfectly normal (i.e. normal and every closed set is a $G_\delta$) but not metrisable. The Sorgenfrey line (lower limit topology) is a non-compact example. – Henno Brandsma Jun 08 '16 at 13:04