How does this follow from the Baire category theorem?

Question

The book says that statement 2 is a direct consequence of statement 1. I don't see how they prove statement 2 directly from statement 1, can you please help me?

Statement 1:

A complete metric space $(\Omega,\rho)$ is not the union of a countable collection of nowhere dense sets.

Statement 2:

Let $\{G_n\}_{n=1}^\infty$ be a sequence of dense open subsets of a complete metric space. Then $\cap_{n=1}^\infty G_n$ is dense.

So I struggle to see $1\rightarrow 2$. What I do know is that since each $G_n$ is dense, then $G_n^c$ is nowhere dense.(A set is nowhere dense iff the complement og its closure is dense, an $G_n^c$ is already closed, since $G_n$ is open). Hence I do know that $\cap_{n=1}^\infty G_n$ is nonempty, since it's complement is a countable union of nowhere dense sets, and from statement 1, this can't be the whole set.

But I only get that it has atleast 1 element, that is a long way from proving that it is dense.

Answer 1

Use the fact that an open subset of a complete metric space is homeomorphic to a complete metric space. Let $U$ be a non-empty open subset of $\Omega$. Then the subspace $U$ has a complete metric $d$ that generates the subspace topology. For $n\in\Bbb N$ let $H_n=G_n\cap U$; each $H_n$ is a dense, open subset of $U$, so by your argument $\bigcap_{n\in\Bbb N}H_n\ne\varnothing$. Thus,

$$\varnothing\ne\bigcap_{n\in\Bbb N}H_n=\bigcap_{n\in\Bbb N}(G_n\cap U)=U\cap\bigcap_{n\in\Bbb N}G_n\;,$$

and $\bigcap_{n\in\Bbb N}G_n$ is dense in $\Omega$.