I found this problem while solving Baby Rudin. The problem states
$X$ is a complete metric space. $\{G_n\}$ is a sequence of dense open subsets of $X$. Prove that $\bigcap\limits_{1}^{\infty}G_n$ is not empty. Prove that it is dense in $X$.
Hint: Find a sequence of shrinking neighborhoods $E_n$ such that $\overline{E_n}\subset G_n$.
My attempt:
$G_1$ is an open set. Construct a neighborhood $N_{r_1}(p_1)\subseteq G_1$ for some $p_1\in G_1$. Constructing a smaller neighborhood $E_1$ of $p_1\in G_1$ of radius $\delta_1<r_1$ gives $\overline{E_1}\subset G_1$.
$p_1\in G_1\subseteq X$. Hence either $p_1\in G_2$ OR $p_1$ is a limit point of $G_2$. In either case, neighborhood of $p_1$ contains points of $G_2$. Hence some point $p_2\in E_1$ and $p_2\in G_2$. Hence, constructing a similar neighborhood sequence with $p_2$ gives $\overline{E_2}\subset\overline{E_1}$.
Repeating the similar construction, we get a sequence of sets $\{E_n\}$ such that $\overline{E_1}\supset\overline{E_2}\supset…$
It can be concluded that all $\overline{E_n}$ are closed and bounded, hence, compact. Also, according to a theorem stated in Rudin, if $K_n$ is a sequence of compact sets such that $K_1\supset K_2\supset K_3…$, then $\bigcap\limits_{1}^{\infty}K_n$ is not empty. Hence, that part is proven. As far as I can tell, this argument is correct. Do point out if it is not.
I have hence proved that the intersection is not empty and contains at least one point. How do I go from this point to proving that every point in the intersection or its limit point is in X? i.e.
How do I prove that the intersection of sets is also dense?
Best Answer
This is Baire's Theorem: let $U$ be open in $X$. Then, $U_1:=U\cap G_1$ is open and non-empty and so contains an $x_1\in B_{r_1}(x_1)$ such that $\overline{B_{r_1}(x_1)}\subseteq U_1$ and $0<r_1<1$. Now, $U_2:=B_{r_1}(x_1)\cap G_2\neq \emptyset$ so we can find $x_2\in B_{r_2}(x_2)\subseteq \overline{B_{r_2}(x_2)}\subseteq U_2$ such that $r_2<1/2.$
We get a sequence of sets $\overline{B_{r_1}(x_1)}\supseteq\cdots \supseteq \overline{B_{r_n}(x_n)} \supseteq ...$ such that $0<r_n<1/n.$
It's easy to show that $(x_n)$ is Cauchy and so has a limit point $x$ in $X$, which is then contained in $\overline{B_{r_n}(x_n)}$ for each $n$ because this set is closed. It follows that $x\in G_n\cap U$ for each $n$ which means that $U$ meets $\bigcap_n G_n$ non voidly.