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Prove or disprove: If the metric $X$ has the property that every open set in $X$ can be written as union of countably many closed sets then $X$ must be locally compact.

I think this statement is true because $\mathbb{R}$ is locally compact and every open set can be written in the manner described. But how do I prove this? Help!

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  • $\begingroup$ A normal space for which every open set is an $F_{\sigma}$ set ( a countable union of closed sets) is called perfectly normal. All metric spaces are perfectly normal. $\endgroup$ – DanielWainfleet Jan 5 '17 at 16:50
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This property is equivalent to that every closed subset of $X$ is countable intersection of open sets. It holds for every metric space. In fact, for a subset $A$ of $X$, $A$ is closed if and only if $$A = \bigcap_{n=1}^\infty B(A, 1/n)$$ where $B(A,r) = \{x\in X\mid \exists a \in A: d(a,x)<r\}$.

However, not every metric space is locally compact; e.g. an infinite-dimensional Banach space.

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This is not true. Consider $\mathbb{Q}$ with the usual metric. Every open set can be written as a union of countable closed sets as is the case in $\mathbb{R}$, however $\mathbb Q$ is not locally compact.

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  • $\begingroup$ My edit was to change "it" to "$\mathbb Q$" in the second sentence, as I felt that "it" was uncomfortably close to "$\mathbb R$". $\endgroup$ – DanielWainfleet Dec 29 '16 at 1:24

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