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This may have a simple answer, but I couldn't find it so far either in textbooks or in math.stackexchange. Let $X$ be a metric space, and $$A=\bigcap^\infty_{n=1}A_n$$ a $G_\delta$ subset of $X$, where $A_n\subset X$ is open for each $n\in\mathbb{N}$. We assume for simplicity that $A_n\supset A_{n+1}$ for all $n\in\mathbb{N}$.

Question: Given any $B\subset X$ open such that $B\supset A$, is there an $n\in\mathbb{N}$ such that $B\supset A_n$?

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Not necessarily.

Let $X=\Bbb R\setminus\{1\}$ under the subspace topology, $B=(0,1),$ and each $A_n=X\cap\left(0,1+\frac1n\right).$ Then $A=B,$ but no $A_n$ is a subset of $B$.

It may be true if we require the metric space to be complete, but I'm not sure. I'll think on it.

Edit: Drat! I see that Ittay already posted the super-nice counterexample I was about to add. And now Asaf has posted another (which hadn't occurred to me).

The upshot is that if the $G_\delta$ set $A$ is open, and is a proper subset of the $A_n$s, then letting $B=A$ gives a counterexample.

Second Edit: It occurs to me (belatedly, of course!), that my original example can be easily adapted into a counterexample where $X$ is complete, connected, and $A$ is non-empty.

Instead, put $X=\Bbb R$ in the usual (completely metrizable and connected) topology, $B=(0,1),$ and $A_n=(0,1)\cup\left(1,1+\frac1n\right).$ (Aside from $X$, these are all exactly the same sets.)

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  • $\begingroup$ What if one assumes in addition that $X$ is complete (which is clearly false in your counter-example)? $\endgroup$ Jun 6, 2013 at 0:56
  • $\begingroup$ @Pedro: Note that $X$ here is completely metrizable, so there is a complete metric inducing that subspace topology, and the open sets are the same. So the same example works -- we just need to take a different metric. $\endgroup$
    – Asaf Karagila
    Jun 6, 2013 at 1:04
  • $\begingroup$ Yeah, I got it... Can the possibility that $A$ is open and a proper subset of $A_n$ for each $n\in\mathbb{N}$ be evaded by assuming completeness and (say) connectedness of $X$? The latter property seems to fail in both counter-examples... $\endgroup$ Jun 6, 2013 at 1:10
  • $\begingroup$ @Pedro: Not at all. Ittay's example shows that this won't work. $\endgroup$ Jun 6, 2013 at 1:11
  • $\begingroup$ @Pedro: This may amuse you (it certainly amused me). See my second edit. $\endgroup$ Jun 6, 2013 at 1:57
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The answer is not necessarily, even if we assume that $X$ is complete. The reason is that $A$ itself might be open, and a proper subset of any of the $A_n$'s.

Consider $X=\Bbb N$ with the discrete metric, then let $A_n=\{0\}\cup\{k\in\Bbb N\mid k\geq n\}$.


We can also require that $X$ is connected, take $X=\Bbb R$ and let $A_n=(0,1)\cup(2,2+\frac1n)$.

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  • $\begingroup$ I've deleted a pair of messed-up comments before I could see your reply to one of them... You're right, $X$ here is totally disconnected. $\endgroup$ Jun 6, 2013 at 1:36
  • $\begingroup$ It seems that disconnectedness of some of the sets involved plays a common role in all these counter-examples... My last move: what if $X$ is complete and connected, $A$ is connected and $A_n$ is connected for all $n\in\mathbb{N}$? $\endgroup$ Jun 6, 2013 at 1:43
  • $\begingroup$ Hm. That's a good question. I'll think about it a bit. It might be possible to come up with a convoluted counterexample in $\Bbb R^n$ for some $n>1$. $\endgroup$
    – Asaf Karagila
    Jun 6, 2013 at 1:46
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No. Consider $\mathbb R$ with the usual metric. Take $A_n=(0,\frac{1}{n})$. Then $A=\emptyset$, yet there are plenty of $B$ that do not contain any of the $A_n$.

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  • $\begingroup$ What if $X$ is assumed complete and connected as in your counter-example, but also that $A\neq\varnothing$? $\endgroup$ Jun 6, 2013 at 1:27
  • $\begingroup$ Never mind, Asaf provided a suitable counter-example... $\endgroup$ Jun 6, 2013 at 1:40

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