# Union of closed sets in a metric space

There is an exercise that asks to show that:

If a countable union of closed sets has nonempty interior in a complete metric space, then at least one set of the union has nonempty interior.

Is this result valid? Doesn't the complete metric space that should be the countable union itself?

How could I prove it?

So far, I have the Baire theorem in hand:

Let $$(E,d)$$ be a complete metric space. Then every intersection of countably many dense open sets is dense.

• It is preferable that countable means "finite or countably infinite" so that "uncountable" means "not countable". So your last sentence should say "Then the intersection of any non-empty countable family of dense open sets is dense." – DanielWainfleet Jul 7 at 5:53

## 2 Answers

Call the sets be $$A_n$$ and suppose they all have empty interior while the union does not. Then the complement of each, $$A_n^c$$, is dense. So by the Baire theorem, $$\cap_nA_n^c$$ is dense. But $$\cap_nA_n^c=(\cup A_n)^c$$, implying $$\cup A_n$$ has empty interior, a contradiction.

• Oh very nice! thank you. – Danmat Jul 7 at 1:52

We argue by contraction. We assume that all the closed set , say $$F_n$$, have no interior. Then we know $$(F_n)^c$$ is dense and open, hence by Baire category theorem, we get $$\bigcap\limits_{n=1}^\infty (F_n)^c$$ is dense, that is $$\bigcup\limits_{n=1}^\infty F_n$$ has empty interior, it’s a contraction.

• How did you conclude that the union of closed sets has empty interior? – Danmat Jul 7 at 2:26
• Ok, I already understood. You used the fact that the closure of complement is the complement of interior – Danmat Jul 7 at 2:29