Verifying the consistency of the statement

Question

The Baire Theorem says:

A countable union of closed sets with empty interior has empty interior in a complete metric space.

From this, I was asked to show that in a complete metric space a countable union of nowhere dense sets has empty interior. For a nowhere dense set I refer to a set $A$ in the metric space such that its closure has empty interior.

First of all, is this statement true? (if it's not true, please give an argument)

Finally, to show this result using the above theorem, one should add "...nowhere dense closed sets ..." Then closed empty interior set coincides with nowhere dense closed set.

Am I right?

Thanks in advance!

Answer 1

The Baire theorem already states the fact that a countable union of nowhere dense closed sets has empty interior. Because a closed set is nowhere dense iff it has empty interior, by definition.

Now this is slightly generalised by remving the "closed" part: suppose that $A_n$, $n \in \Bbb N$ are nowhere dense. By definition $\overline{A_n}$ is closed with empty interior, and

$$\operatorname{int}(\bigcup_n A_n) \subseteq \operatorname{int}(\bigcup_n \overline{A_n}) = \emptyset$$

where the latter is by the Baire theorem, and the former as interior is monotone ($A \subseteq B$ implies $\operatorname{int}(A) \subseteq \operatorname{int}(B)$). So the interior of the union of countably many nowhere dense sets is also empty, as claimed. As you can see, it's only slightly more general.

Answer 2

If the sets are already closed, the result would be trivial.

Hint: consider the union of the closures of your nowhere dense sets.

Answer 3

" Z is nowhere-dense" means "$int (\overline Z)=\emptyset$".

Let $Y=\overline X.$ Since $\overline Y=Y,$ we have $$ X \text { is nowhere-dense } \implies int (\overline X)=\emptyset \implies$$ $$\implies int (\overline Y)=int (Y)=int (\overline X)=\emptyset \implies$$ $$\implies int(\overline Y)=\emptyset\implies Y \text { is nowhere-dense}.$$ So if $F$ is a family of nowhere-dense sets then $G=\{\overline X: X\in F\}$ is a family of closed nowhere-dense sets. And $int (\cup F)\subset int (\cup G)$ because $\cup F\subset \cup G.$