Basically I have tried to examine the fact. Suppose , $C=\{x_n\}$ be countable $G_\delta$ set.
Hence, $C=\cap V_n$ where each $V_n$ is open.
Thus, $\Bbb{R}=C\cup( \Bbb{R}\setminus C)=\cup \{x_n\} \cup\\ (\cup \Bbb{R}\setminus V_n)$
If I can show int$($cl $\Bbb{R}\setminus V_n)=\emptyset$, then it will prove that $\Bbb{R}$ is first category set contradicting the Baire Category Theorem. But to do this $V_n$ should be dense.
I can't proceed the proof even I am not getting any counter example of it.
Can anybody give such example or can prove it? Thanks for assistance in advance.

  • $\begingroup$ $\Bbb R$ is a "perfectly normal" space, that is, every closed set is $G_{\delta}.$ So any countable closed set (e.g. $\Bbb Z$ or $\{0\}\cup \{1/n: n\in \Bbb N\}$) will do $\endgroup$ Commented Aug 24, 2019 at 1:59

1 Answer 1


Let $V_n=\cup_{k=1}^{\infty} (k-\frac{1}{n},k+\frac{1}{n}).$

Then, each $V_n$ is open, and $\cap_{n\in \mathbb{N}} V_n=\mathbb{N},$ which is countable.


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