A corollary of Baire Category Theorem

Question

I have learnt that the theorem says:

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

I also know that

A subset $Y$ of $X$ is nowhere dense if $\bar Y$ (closure of Y) has no interior points, and also
$Y$ is nowhere dense $\iff$ int($\bar Y$) = $\emptyset$ $\iff X \setminus \bar Y$ is a dense open subset of X.

I have trouble figuring out a corollary:

Let $\{E_n\}_{n=1}^\infty$ be a sequence of nowhere dense subsets of a complete metric space X.
Then $\cup_{n=1}^\infty E_n$ has empty interior.

My thoughts:

$X\setminus \bar E_n$ are dense open sets.
Therefore $\displaystyle\cap_{n=1}^\infty (X \setminus \bar E_n)$ is also dense in X,
and $\cap_{n=1}^\infty (X \setminus \bar E_n)$ = $X \setminus \cup_{n=1}^\infty \bar E_n$

but I have no idea where to go next.

Answer 1

You’re basically there. You’ve concluded that:

$$X\setminus\bigcup_{n\ge1}\overline{E_n}$$

Is dense in $X$. That means the complement is hollow, or in your words:

$$\bigcup_{n\ge1}\overline{E_n}$$Has no interior points.

In particular that’ll imply any subset will have no interior points! So the subset $\bigcup_{n\ge1}E_n$ also has no interior points (is hollow).

The dual, but much less commonly stated, result is this:

If $X$ is a complete metric space and $(G_n)_{n\in\Bbb N}$ is a sequence of (dense) sets each having dense interior, then: $$G:=\bigcap_{n\in\Bbb N}G_n$$Is dense.