# Exercise 3.3.8 from Understanding Analysis by Stephen Abbott

Motivation: trying to prove that if $K \subseteq \mathbb{R}$ is compact (and thus, by the Heine-Borel theorem, closed and bounded), then this implies that any open cover for $K$ has a finite subcover.

Exercise: In order to prove the above, I first need to show the following:

Let $\{ O_\lambda \mid \lambda \in \Lambda \}$ be an open cover for $K$ and, for contradiction, let us assume that no finite subcover exists for $K$. Let $I_0$ be a closed interval containing $K$, and bisect $I_0$ into two closed intervals $A_1$ and $B_1$. Why must either $A_1 \cap K$ or $B_1 \cap K$ (or both) have no finite subcover consisting of sets from $\{O_\lambda \mid \lambda \in \Lambda \}$?

I'm not sure how to prove the above. Since by assumption $K$ has no finite subcover, this means that: $$K \subsetneq O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n}$$ Now, if we cut $I_0 \supseteq K$ into two intervals, then surely there will be a point $x \in K \subseteq I_0$, with $x \notin O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n}$, such that: $$x \in A_1 \cap K \subsetneq O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n}$$ or: $$x \in B_1 \cap K \subsetneq O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n}$$ Does this count as a valid proof? I feel like I'm missing something, but am not sure what it is.

• Can you clarify the question? You can either prove that if set in $\mathbb{R}$ is closed and bounded, then every open cover has a finite subcover. Or you can prove the opposite direction. It is not clear which direction interests you in your "motivation." – Michael Aug 27 '14 at 10:53
• @Michael I'm trying to prove that if a set in $\mathbb{R}$ is closed and bounded, then every open cover has a finite subcover. – Hunter Aug 27 '14 at 10:57
• If both $A\cap K$ and $B\cap K$ have finite subcovers, then taking the two subcovers together covers the whole of $K$, since $K\subset A\cup B$. – Vincent Boelens Aug 27 '14 at 11:05
• And no, your proof is not valid, since it assumes the conclusion. – Vincent Boelens Aug 27 '14 at 11:06
• @Michael because, for contradiction, let us assume that no finite subcover exists for $K$ (this is what is stated in the book). – Hunter Aug 27 '14 at 11:50

Your proof idea works, but the phrasing is imprecise and crucial details are left out. What you're trying to do is to show that for any finite subcollection $O_{\lambda_1},\ldots O_{\lambda_n}$,the following holds: $A\cap K\subsetneq O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$ or $B\cap K\subsetneq O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$. This amounts to showing that there is an $x\in A\cap K$ or an $x\in B\cap K$ such that $x\not \in O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$.
Now, since by assumption $K\subsetneq O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$, there is an $x\in K$ such that $x\not \in O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$. Furthermore, since $K\subset A\cup B$, we must have $x\in A$ or $x\in B$ (or both), so $x\in A\cap K$ or $x\in B\cap K$.