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

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.

• This is another version of the completeness of $\mathbb{R^n}$. See Baire's Catagory Theorem. – Arjun Banerjee Oct 25 '18 at 11:33
• @ArjunBanerjee: That would be if the union was countable rather than finite. For a finite union it holds in arbitrary spaces. – Henning Makholm Oct 25 '18 at 12:01