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?
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.