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This question already has an answer here:

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.

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

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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 $m=1,\ldots, n$.
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  • $\begingroup$ Thanks for proof! I have 1 question: What exactly is the induction hypothesis? Thats the only part i dont understand. I think that the induction hypothesis is that X does have a int(Am)≠∅ for n, thats why you showed that X is the same for n and n+1 $\endgroup$ – Mangafreak13 May 22 at 13:06
  • $\begingroup$ @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 $n+1$. $\endgroup$ – freakish May 22 at 13:24
  • $\begingroup$ Thats what i meant, sorry that my english isnt good enough! Thanks i understand it now 100% $\endgroup$ – Mangafreak13 May 22 at 13:33
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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..

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