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This is in Cantor space ($2^\omega$ with the usual topology). In the course of trying to prove something else, I've found myself wanting to show that whenever $X$ is a countable $G_\delta$ set, the closure $\overline{X}$ is also countable.

This seems perfectly reasonable (keep in mind that countable $G_\delta$ sets are nowhere dense), but I can't seem to prove it (note that the closure of a countable nowhere dense set can easily have size continuum - take the endpoints of a Cantor-like set). Even worse, I vaguely remember having this problem on an analysis exam years ago and getting it right.

I'm sure I'm just having a silly moment (I was unsuccessful this morning in my quest to secure coffee), but: is the closure of a countable $G_\delta$ subset of Cantor space, itself countable?

Noah Schweber
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  • I suppose it's equivalent to ask whether there is a... co-countable?... $F_\sigma$ set in the cantor space whose interior is not... co-countable. Somehow, I thought that might help before I wrote it out. – Ben Grossmann Sep 08 '16 at 16:47
  • @NateEldredge Hang on, I'm not convinced anymore. Work in $\mathbb{R}$ rather than Cantor space for the moment, for simplicity. Let $X$ be the set of midpoints of the deleted intervals in the construction of the usual Cantor set (that is, the usual Cantor set is $\bigcap C_i$, each $C_{i+1}$ is $C_i\setminus{\bigcup_{n=1}^{2^i} I^i_n}$, and we're letting $X$ be the set of midpoints of the $I^i_n$s). Then $X$ is countable and (unless I'm missing something) $G_\delta$, but its closure contains the Cantor set. Thoughts? (Re: your hint, I'm unclear why $\partial X$ is meager in $\overline{X}$.) – Noah Schweber Sep 08 '16 at 17:29
  • @NoahSchweber: You're right, that was bogus. – Nate Eldredge Sep 08 '16 at 17:58
  • The other standard example is something like $X = {(k/n, 1/n) : 0 \le k \le n, n \in \omega}$ in $[0,1]^2$, which is countable, but $\overline{X} = X \cup [0,1] \times {0}$. Then in fact $[0,1] \times {0}$ is closed and nowhere dense in $\overline{X}$, and $X$ is even open in $\overline{X}$. So if this is true, it has to use something specific about the fact that we're working in $2^\omega$. – Nate Eldredge Sep 08 '16 at 18:04
  • @NateEldredge Re: your deleted answer, if $X$ is $G_\delta$ in $\overline{X}$, isn't it also $G_\delta$ in $2^\omega$ (that is, aren't $G_\delta$-in-closed sets $G_\delta$)? – Noah Schweber Sep 08 '16 at 18:57
  • @Noah: It appears to me that his $X$ is $G_\delta$ in $\Bbb R^2$, even. Enumerate it as ${p_n:n\in\omega}$, and let ${B(p_n,r_n):n\in\omega}$ be a pairwise disjoint family of open balls centred at the points $p_n$. For $k\in\omega$ let $U_k=\bigcup_{n\in\omega}B(p_n,2^{-k}r_n)$; then $X=\bigcap_{k\in\omega}U_k$. – Brian M. Scott Sep 08 '16 at 19:22
  • @NoahSchweber: Yeah, I had to rush off, and realized that about 3 seconds after I left my computer. Will fix and undelete now. – Nate Eldredge Sep 08 '16 at 19:55

2 Answers2

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It's not true.

Let me think of $C = 2^\omega$ as the Cantor set in $[0,1]$. It's homeomorphic to its square, so I'll actually work in $C^2 \subset [0,1]^2$.

Let $E_n$ be the endpoints of all the intervals remaining at the $n$th step of the construction of $C$, so $|E_n| = 2^{n+1}$. Let $X = \bigcup_n E_n \times \{1/3^n\}$ which is countable, and indeed, discrete.

Now for every $x \in E = \bigcup_n E_n$, clearly $(x,0) \in \overline{X}$. But $E$ is dense in $C$ so we have $C \times \{0\} \subset \overline{X}$, and that's uncountable.

Indeed, we can see that $\overline{X} = X \cup C \times\{0\}$. So back in $C^2$, we can write $X = \overline{X} \cap (C \times \{0\})^c$, which is the intersection of a closed (hence $G_\delta$) set and an open set. Thus $X$ is $G_\delta$ in $C^2$.

Nate Eldredge
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  • +1 Awesome! Given the number of times I've been wrong about this, I want to hold off on accepting for a bit until I'm sure; but this looks great! – Noah Schweber Sep 08 '16 at 20:36
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Another example: The set X of mid points of the open interval removed during the construction of the ternary Cantor set $C$ is countable and $G_{\delta}$ and its closure is $X \cup C$.

Mcnulty
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  • +1. Note that I mention this example in my comment to Nate Eldredge above; also, this is really an example in $\mathbb{R}$ (although it's easy to port it over to Cantor space). – Noah Schweber Sep 09 '16 at 23:03