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Is $X$ is second countable implies or equivalent to Every closed(or open) set in $X$ is a countable intersection of open sets in $X$?

I know, metric space implies this by Closed set as a countable intersection of open sets, and Every open set in $\mathbb{R}$ is a countable union of closed sets. Also I know metric space is 1st countable but not 2nd countable.

If so How one can prove this statement?

phy_math
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  • In any metric space any closed set is a countable intersection of open sets. A metric space need not be second countable. So equivalence of these two is not true. – Kavi Rama Murthy Jul 30 '21 at 06:15
  • @Kavi Rama Murthy, yes I know metric space satisfies the above property. I wonder whether a similar thing holds for 2nd countable case or need to impose a further topological condition, such as normal. – phy_math Jul 30 '21 at 06:17
  • You definitely need something more, e.g., $X$ has at least 2 points, with only 3 open sets. Then the nontrivial closed set $F\subset X$ is not an intersection of opens. – user10354138 Jul 30 '21 at 06:24
  • i.e., Since $X$ regular+second countable implies $X$ is metrizable and separable, normal+second countable space is metric space, so for this case the above holds – phy_math Jul 30 '21 at 06:24

2 Answers2

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The natural numbers $\ge 2$ endowed with the topology of the subsets that contain the divisors of each of their elements is second-countable, but not $G_\delta$.

  • Off-topic: Your answer here still seems correct to me: Even though the series converges to an $L^2$ function a.e. according to WolframAlpha, there is no $L^2$ function with that series as its Fourier series. – Maximilian Janisch Jul 31 '21 at 15:05
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The Double Arrow space ($[0,1] \times \{0,1\}$ in the order topology induced from the lexicographer order) is not second countable but is compact, connected and every closed subset is a countable intersection of open sets (it’s perfectly normal). So it’s not equivalent.

Henno Brandsma
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