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Let $F$ be nonempty closed set in $\mathbb{R}^n$ and $F=\bigcup\limits_{m=1}^{\infty}F_m$, where $F_m$ are closed sets. Prove, that there exists a number $m_{_0}$ and a closed ball $\bar{U}\subset\mathbb{R}^n$ such that $F\cap\bar{U}=F_{m_{\ _0}}\cap\bar{U}$ and $F\cap U\ne\varnothing$.

There is a proof in the book the problem is from, by contradiction and using a decreasing sequence of nested balls. But there is a place in the proof, that I find dubious. So, I should like to see, how the community solves the problem.

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This follows from the Baire Category Theorem.

Since $F$ is a closed subset of a complete metric space it is complete.

By the Baire Category Theorem, one of the $F_m$ is NOT nowhere dense, say $F_{m_0}$.

But $F_{m_0}$, is closed so $F_{m_0}$ is NOT nowhere dense means it has non-empty interior. Take $U$ to be some open ball in the interior of $F_{m_0}$.

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  • $\begingroup$ Thank you, but in the standard calculus course, on which the book is based, there was no Baire theorem and even no dense set definition. Does there exist more elementary proof ? $\endgroup$ Dec 23, 2017 at 19:50
  • $\begingroup$ The problem is essentially the statement of the Baire Category Theorem (BCT3 on wikipedia) in $R^n$ so I am doubtful that a "simple" solution exists. The sequence of decreasing balls is a standard way to prove the BCT. $\endgroup$
    – Mohit
    Dec 23, 2017 at 22:37
  • $\begingroup$ I corrected what was wrong with the proof in the book. So, yes, decreasing balls do work. Now I can accept your answer. Thank you once again. $\endgroup$ Dec 23, 2017 at 23:05

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