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Theorem: Let $(X,d)$ be a complete metric space, and let $D_n, n\in \mathbb N$ be open, dense subsets of $X$. Then also $\bigcap_{n\in\mathbb N} D_n$ is dense in $X$.

This statement is false if $X$ is not complete. Take $X=\mathbb Q=\{q_1,q_2,\dots\}$ and $D_n=X\setminus \{q_n\}$.

$\bullet $ $\mathbb Q$ is not complete

$\bullet$ $D_n$ is open since $X\setminus D_n=\{q_n\}$ and singeltons are closed sets.

$\bullet$ $D_n$ is dense in X. This is what I do not understand. The closure of $D_n$ is equal $\mathbb R$, but not $\mathbb Q$. But $\mathbb R$ is no subset of $X$. So is actually here the closure of $D_n=\mathbb Q$? And if, why?

$\bullet$ $\bigcap_{n=1}^\infty D_n=\emptyset$ which is not dense in $\mathbb Q$.

Could someone elaborate on the third bullet?

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  • $\begingroup$ $\mathbb{R}$ contains $X$. Density w.r.t. $\mathbb{R}$ means for any point $x$ in $\mathbb{R}$ we can find a converging sequence of points in $D_n$ which have limit $x$. In particular, we can choose any point of $X$. $\endgroup$ – Tony S.F. Jun 16 at 19:25
  • $\begingroup$ Whenever you talk about "closure" of a subset, it is with respect to some ambient metric space. While the closure of $D_n$ in $\mathbb{R}$ is $\mathbb{R}$, the closure of $D_n$ in $\mathbb{Q}$ is $\mathbb{Q}$, since every element of $\mathbb{Q}$ is a limit of points in $D_n$. $\endgroup$ – mathworker21 Jun 16 at 19:25
  • $\begingroup$ @TonyS.F. your comment seems irrelevant $\endgroup$ – mathworker21 Jun 16 at 19:26
  • $\begingroup$ @mathworker21 it's not irrelevant; i edited it anyway to be more precise but the point is that if you are dense w.r.t. some set you are also dense w.r.t. any subset of it. $\endgroup$ – Tony S.F. Jun 16 at 19:27
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The metric space $X$ under consideration here is the set of the rational numbers with the metric inherited from the real numbers. That is, given two rational numbers $p,q\in X$, the distance between them is $|p-q|$. Now if you take any rational number, $p$, then $\mathbb{Q}\backslash\{p\}$ is dense in $X$, because given any rational number $r$, there exists a sequence in $\mathbb{Q}\backslash\{p\}$ converging to $r$. This is of course obvious if $r\neq p$, and if $r=p$, then the sequence $\{p-\frac{1}{n}\}_{n=1}^{\infty}$ belongs to $\mathbb{Q}\backslash\{p\}$ and converges to $r$, so that $\mathbb{Q}\backslash\{p\}$ is dense.

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In that example, $X$ is $\mathbb Q$. So, it makes no sense to assert that $\overline{D_n}=\mathbb R$. Your universe here is $\mathbb Q$ and therefore $\overline{D_n}$ must be a subset of $\mathbb Q$. And, in fact, it is equal to $\mathbb Q$.

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