# Intersection of open dense subsets in a complete metric space is dense

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 $$\delta_1 gives $$\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?

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 $$0. Now, $$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
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.
• Yes, but I need help with the fact that $\bigcap\limits_{1}^{n}G_n$ is dense. Jan 17, 2020 at 20:35
• I'm sorry. I did not think about it properly. So correct me if I am wrong. By the last argument, since we can prove that $U\cap \bigcap\limits_nG_n$ is not empty for any arbitrary $x\in X,\therefore, \forall x\in X, U$ intersects $\bigcap\limits_n G_n \implies G_n$ is dense in $X$ Jan 17, 2020 at 23:31
• You take an arbitrary open set $U$ and show that it intersects $\bigcap\limits_n G_n$ and that means $\bigcap\limits_n G_n$ is dense, right? Jan 20, 2020 at 2:00