Exercise 2.30

Question

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Exercise 30: If $\mathbb R^k = \bigcup^\infty F_n$ where each $F_n$ is closed, then at least one $F_n$ has non-empty interior.

ghostofgarborg · Accepted Answer

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Assume for contradiction that $\mathbb R^k = \bigcup^\infty F_n$ where each $F_n$ is closed and has non-empty interior. 
    Let $N_0$ be a ball of finite radius around a point $x_1 \in F_1$, so that $\bar N_0$ is compact. 
    Assume $N_{i-1}$ is open and does not contain any points of $F_1, \cdots F_{i-1}$. 
    This set must contain a point $x_i$ not in $F_i$, as it otherwise would belong to the interior of $F_i$. 
    Furthermore, $x_i$ must be contained in a neighborhood $N_i \subseteq N_{i-1}$ that does not intersect
    $F_i$, as $x_i$ otherwise would be a limit point of $F_i$ and therefore belong to $F_i$. 
    We can choose $N_i$ such that $\bar N_i \subseteq N_{i-1}$, and we observe that it does not contain any points of $F_1, \cdots, F_i$.

Now observe that each $\bar N_i$ is compact, and that $ \bar N_{i+1} \subseteq \bar N_{i} $, so that by the corollary to thm. 2.36, $I = \bigcap_i \bar N_i$ is non-empty.
    However, by construction, if $ x \in I$, then $x 
ot \in F_i$ for any $i$. But this implies that $x 
ot \in \bigcup F_i = \mathbb R^k$, a contradiction.

Amit Awasthi · Answer

\begin{proof}
      Assume all $F_n$ have empty interiors. We will construct a sequence $\{V_n\}$ of neighborhoods as follows. Let $V_1$ be any neighborhood of $x_1 
otin F_1$, where $V_1 \bigcap F_1 = \emptyset$. If $V_1$ consists of all $y \in \mathbb{R}^k$ such that $|y-x_1| < r$, the closure $\overline{V}_1$ of $V_1$ is the set of all $y \in \mathbb{R}^k$ such that $|y-x_1| \le r$. Suppose $V_n$ has been constructed so that $V_n \bigcap F_n = \emptyset$. Then, there is a neighborhood $V_{n+1}$ such that (i) $\overline{V}_{n+1} \subset V_n$, (ii) $x_n 
otin F_{n+1}$, (iii) $V_{n+1} \bigcap F_{n+1} = \emptyset$. By (iii), $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed. Then, $\bigcap^\infty_{n=1} \overline{V}_n 
eq \emptyset$ by our construction, and the points in $\overline{V}_n$ are not in $F_n$ for any $n$. However, $\overline{V}_n \subset \mathbb{R}^k$ for all $n$ which contradicts that all $F_n$ have empty interiors.
    \end{proof}

Exercise 2.30

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Exercise 2.30

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