# In any metric space: is the intersection of a countable sequence of dense, open sets always non-empty and dense?

I am self studying Principles of Mathematical Analysis (Rudin). There is an exercise which asks to prove Baire's theorem:

Prove that the intersection of a sequence of open, dense subsets, $G_n$, of a complete metric space, $X$, is not empty. In fact, it will be dense.

With my approach, $X$ being complete seems irrelevant... Is it?

My approach was the following (assuming that {$G_n$} is a countable sequence):

The intersection of an open set and a dense set is not empty.

Proof:

Let $G$ be a dense set and $B$ an open set.

If $G \cap B = \emptyset$, then $\forall p \in B$ is an accumulation point of $G$. As such, a neighbourhood of $p$ is contained by $G$, i.e., $N_p \subset G$.

But $B$ is open, as such $N_p \subset B$. Which would mean that $N_p \subset B \cap G$, which is a contradiction and the intersection cannot be empty.

From the above, $\bigcap^\infty_1 G_n$ is not empty.

I went on to show that $\bigcap^\infty_1 G_n$ is also dense.

Proof:

Suppose $p \in G_i$ and $p \notin G_k$.

Then $N_p \subset G_k$. But the $G$'s are open; then $N_p \subset G_i$.

It follows that $N_p \subset G_i \cap G_k$ ad $p$ is an accumulation point of it.

As the intersection of two open sets is also open, I consider, by induction, that any given $p_k \in G_k$ | $p_k \notin \bigcap^\infty_1 G_n$, will be an accumulation point of such intersection.

Now, if $x \notin G_n, \forall n$, it follows that $N_x \subset G_n, \forall n$. Then $N_x \subset \bigcap^\infty_1 G_n$, which makes $x$ an accumulation point of the intersection.

Hence $\bigcap^\infty_1 G_n$ is dense.

The fact that the metric space $X$ is complete seems irrelevant... Is that so?

Consider the metric space $\Bbb Q$ with the usual metric. For each $q\in\Bbb Q$ the set $\Bbb Q\setminus\{q\}$ is open and dense, but the intersection of these sets is empty.
I can’t make any sense of your first argument. The fact that a dense set has non-empty intersection with every non-empty open set certainly does not allow you to conclude that $\bigcap_nG_n\ne\varnothing$. You do of course know that $\bigcap_{k\le n}G_k$ is dense and open for each $n$, but you can’t generalize that to the infinite intersection.