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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, 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 $m=1,\ldots, n$.
  • $\begingroup$ 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

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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