# Intersection of closed and bounded subsets contained within open set

$\textbf{Problem statement}:$ Suppose that $\{F_j\}$ are closed and bounded subsets of $\mathbb{R}$, and $G$ is an open subset. Show that if $\cap_{j=1}^{\infty} F_j \subset G,$ then there is a finite subset $\{j_1,\dots,j_k \}$ such that $\cap_{i=1}^k F_{j_i} \subset G$.

Let the open set $G$ be $(a,b)$, and I have showed that the countable intersection of closed sets is closed already, and since each $F_j$ is bounded, the total intersection is some $[a',b'] \subset (a,b)$.

Now assume that there is no such finite subset $\{j_1,\dots,j_k \}$ such that $\cap_{i=1}^k F_{j_i} \subset G$. We know that for each finite intersection of closed and bounded sets, the intersection is closed and bounded, and the lower bound is $sup(inf(F_j))$ which converges to $a'$, and the upper bound is $inf(sup(F_j))$ which converges to $b'$. Since we have this convergence, $\vert sup(inf(F_j)) - a'\vert < \epsilon$ and $\vert inf(sup(F_j)) - b' \vert < \epsilon$. Then we can find an $N_1$ such that $\vert sup(inf(F_{N_1}) - a'\vert < \frac{a' - a}{2}$ equal to $min(\frac{a' - a}{2},\frac{b - b'}{2})$, $N_2$ so $\vert inf(sup(F_{N_2})) - b' \vert < \frac{b - b'}{2})$ and take $N = max(N_1,N_2)$ which is a finite $N$ such that the intersection of those $N$ subsets is contained in $(a,b)$ and thus a contradiction.

I doubt the rigor of this proof to be honest, and feels messy. If anyone can correct me or provide something cleaner I'd appreciate it. A hint on generalization to $\mathbb{R}^n$ would also be helpful.

## 1 Answer

Your proof assumes that the intersection is a segment in $\mathbb{R}$, which does not have to be true.

Assume there does not exist a finite $I \subseteq \mathbb{N}$ such that $\bigcap_{i \in I} F_i \subseteq G$.

Then in particular, $F_1 \not\subseteq G$ so there exists $x_1 \in F_1 \setminus G$.

Also, $F_1 \cap F_2 \not\subseteq G$ so there exists $x_2 \in (F_1 \cap F_2 )\setminus G$.

Inductively, we arrive at a sequence $(x_n)_{n=1}^\infty$ such that $x_n \in \left(\bigcap_{i=1}^n F_i\right) \setminus G$ for all $n \in \mathbb{N}$.

Notice that $(x_n)_{n=1}^\infty$ is a sequence in $F_1 \setminus G$, which is closed and bounded and hence compact. Therefore, there exists a convergent subsequence of $(x_n)_{n=1}^\infty$ with the limit in $F_1 \setminus G$. WLOG we can assume that $x_n \xrightarrow{n\to\infty } x \in F_1 \setminus G$.

Note that for any $k \in \mathbb{N}$ we have that $(x_n)_{n \ge k}^\infty$ is a sequence in $\left(\bigcap_{i=1}^k F_i\right) \setminus G$ which is a closed set so its limit $x \in \left(\bigcap_{i=1}^k F_i\right)\setminus G$.

Thus, $x \in \left(\bigcap_{i=1}^\infty F_i\right)\setminus G$, which is a contradiction with the assumption that $\bigcap_{i=1}^\infty F_i \subseteq G$.

• Thank you so much! This was amazing. Can you just explain why $F_1 - G$ has to be closed? What if $F_1$ is $[-1,2]$ and $G$ is $(0,1)$? Wouldn't $F_1 \ G$ be $[-1,0) \cup (1,2]$? – blanchey Apr 1 '18 at 21:59
• @blanchey If $G$ is open then $G^c$ is closed so $F_1 \setminus G = F_1 \cap G^c$ is closed as an intersection of two closed sets. In your example, $[-1, 2] \setminus (0,1) = [-1,0] \cup [1,2]$, which is closed. – mechanodroid Apr 1 '18 at 22:03