Baire's Theorem Rudin's RCA

Question

Now I'm reading Rudin's RCA.

There I've readen Baire's Theorem.

There it is:

If $X$ is a complete metric space, the intersection of every coutanble collection of dense open subsets of $X$ is dense in $X$ .

Why should these dense subsets of $X$ be open?

What will change if we consider dense closed subsets of $X$?

Answer 1

If those sets are not open, then claim simply doesn't hold.

For example $\mathbb R$ is complete metric space space, and intersection of $\mathbb Q$ and $\mathbb R \setminus \mathbb Q$ (which are dense, but not open) is not dense. You could take countably many of the same set to see that claim explicitly does not hold.

Set is dense when its closure is whole space, so only dense closed set is whole space $X$. It does not make much sense to talk about dense closed set when you could just say it's whole space. And here it becomes trivial.