Let $\{G_n\}$ be a sequence of dense open subsets of a complete metric space $X$. Show that $ \bigcap_{n \geq 1}G_n$ is nonempty.
Proof:
Consider some point $p_1 \in G_1$. There exist real numbers $0< r_1 <r<1$ such that
$$N_1 = N_{r_1} (p_1) \subset N_r (p_1) \subset G_1.$$
Let $q$ be a limit point of $N_1$. For any $\epsilon > 0$ we can find $x \in X$ such that $x \in N_1$ and $x \in N_\epsilon (q)$. Hence
$$d(p_1, q) \leq d(p_1, x) + d(x, q)<r_1 + \epsilon.$$
Since $\epsilon$ was arbitrary it follows that
$$d(p_1, q) \leq r_1.$$
Thus, if $q$ is any limit point of $N_1$, $q$ is an interior point of $G_1$ (to see this, simply consider the neighbourhood of radius $0<\epsilon<r-r_1$) and so we can write
$\overline{N_1} \subset G_1.$
Since the $G_n$ are dense, either $p_1$ is a point in $G_2$ or a limit point of $G_2$. In either case, arguing as above, we can find a point $p_2 \in G_2 $ and a neighbourhood of $p_2$ with radius $r_2 < \min\big(r_1, \frac{1}{2}\big)$ such that
$$\overline{N_2} \subset \overline{N_1}$$
and, since the $G_n$ are all open, we also choose $r_2$ small enough that
$$\overline{N_2} \subset G_2.$$
Continuing this process, we can construct the sequence of closed subsets $\Big\{\overline{N_n}\Big\}$. Further, by the way we have chosen the $r_n$ (i.e., $0<r_n<\frac{1}{n}$), it follows $r_n \to 0$, so we have actually constructed a nested sequence of closed bounded sets $\Big\{\overline{N_n}\Big\}$ such that $diam \overline{N_n} \to 0.$ Since $X$ is a complete metric space, the intersection of this sequence is nonempty and since each $$\overline{N_n} \subset G_n$$ the result follows. $\qquad \square$
Assuming this proof is correct, I am wondering if it can be extended to show that the intersection of the $G_n$ is actually dense, or if that requires a completely different line of argument?
Best Answer
That the intersection of the $G_n$ is dense, is a small modification of the above argument: let $O$ be any non-empty open set in $X$, and start with an open ball $N_0 \subseteq O$ and stay inside $N_0$ with all subsequent steps. This hardly takes any effort at all, but does show that the $x \in \cap G_n$ is also in $N_0$ hence in $O$. So the intersection of the $G_n$ intersects every non-empty open set, hence is dense.
The construction of the $N_n$ can be a bit simplified:
No distinguishing limit points etc. Just go straight to the goal.
Then the $\overline{N_n}$ $n \ge 1$ form the required nested family that the Cantor intersection theorem can be applied to. The promised $p \in \bigcap_{n \ge 1} \overline{N_n} \subseteq O \cap \bigcap_{n \ge 1}O_n$ witnesses the denseness of $\bigcap_{n \ge 1} G_n$.