Prove that $\mathbb{R^n}$ is not union of a finite number of closed sets with empty interior

Question

I don't know how to finish this one. Here's what I've done so far:

We're trying to find a contradiction.

Suppose that $\mathbb{R^n}=C_1\cup C_2 \cup \dots \cup C_n$, with $C_i$ closed subsets (with empty interior) of $\mathbb{R^n}$.

It follows that $C_1^c \cap C_2^c \cap \dots \cap C_n^c = \emptyset$.

I've got to use somehow that $Int(C_i)=\emptyset$ but I can't figure out how.

EDIT I've been only defined the closed-open set concept, neighborhood, closure and interior.

I'd appreciate some hint to finish this one.

Thanks for your time.

Answer 1

Can you prove that the union of two closed sets with empty interior is itself a closed set with empty interior?

If so, then use induction to extend this to "finitely many".

(I find it slightly easier to think of the complements instead and prove that the intersection of two dense open sets is dense and open. Your mileage may vary, though).