# If a topological space is covered by finitely many closed sets, then one of them has nonempty interior [duplicate]

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I got this question, which I have to prove by induction:

Let $$X$$ be the topological space and $$A_1, \dotsc, A_n$$ closed subsets of $$X$$ with $$\bigcup_{k=1}^n A_k = X$$. Then there exists some $$i \in \{1,...,n\}$$ such that $$A_i^\circ \neq \emptyset$$.

I really don't know how to prove that, not even how to start... Would be nice if you could give me some tips.

## marked as duplicate by YuiTo Cheng, callculus, egreg, Jendrik Stelzner, jgonMay 22 at 1:23

Under the additional assumption that $$X\neq\emptyset$$ here's your induction:
1. For $$n=1$$ this is clear because $$A_1=X$$
2. Assume it holds for some $$n\geq 1$$. Consider $$n+1$$ and $$X=A_1\cup\cdots\cup A_{n+1}$$. If $$\text{int}(A_{n+1})\neq\emptyset$$ then we are done so assume that $$\text{int}(A_{n+1})=\emptyset$$. Consider $$U=X\backslash(A_1\cup\cdots\cup A_n)$$ Note that it is open. Since $$X=A_1\cup\cdots\cup A_{n+1}$$ then this implies that $$U\subseteq A_{n+1}$$. So if $$U$$ is nonempty then it contradicts $$\text{int}(A_{n+1})$$ being empty. It follows that $$U=\emptyset$$ and so $$X=A_1\cup\cdots\cup A_{n}$$ We can now apply our induction hypothesis to conclude that $$\text{int}(A_m)\neq\emptyset$$ for some $$m=1,\ldots, n$$.
• @Mangafreak13 to be 100% clear: the induction hypothesis is: "for some $n\geq 1$ and any $n$ closed subsets $A_1,\ldots, A_n\subseteq X$ such that $A_1\cup\cdots\cup A_n=X$ there is $m$ such that $\text{int}(A_m)\neq\emptyset$". What we are proving is that under this hypothesis the same holds for $n+1$. – freakish May 22 at 13:24
Suppose all $$A_n$$ are closed with empty interior. Then $$O_n=X\setminus A_n$$ are open and dense (as $$\overline{O_n}=X\setminus \operatorname{int}(A_n)= X$$ by a general formula relating closures, interiors and complements).
But a finite intersection of open and dense sets is open and dense so cannot be empty while $$\bigcap_{i=1}^n O_i = X\setminus \bigcup_{i=1}^n A_i= \emptyset$$ by de Morgan..