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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: \begin{equation} K \subsetneq O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n} \end{equation} 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: \begin{equation} x \in A_1 \cap K \subsetneq O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n} \end{equation} or: \begin{equation} x \in B_1 \cap K \subsetneq O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n} \end{equation} Does this count as a valid proof? I feel like I'm missing something, but am not sure what it is.

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  • $\begingroup$ 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." $\endgroup$ – Michael Aug 27 '14 at 10:53
  • $\begingroup$ @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. $\endgroup$ – Hunter Aug 27 '14 at 10:57
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    $\begingroup$ If both $A\cap K$ and $B\cap K$ have finite subcovers, then taking the two subcovers together covers the whole of $K$, since $K\subset A\cup B$. $\endgroup$ – Vincent Boelens Aug 27 '14 at 11:05
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    $\begingroup$ And no, your proof is not valid, since it assumes the conclusion. $\endgroup$ – Vincent Boelens Aug 27 '14 at 11:06
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    $\begingroup$ @Michael because, for contradiction, let us assume that no finite subcover exists for $K$ (this is what is stated in the book). $\endgroup$ – Hunter Aug 27 '14 at 11:50
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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 $x\in B\cap K$.

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  • $\begingroup$ Thanks! That was indeed the idea behind the proof, but I'm still very new to the idea of proving something rigorously, and then also writing it down so that other people understand it. $\endgroup$ – Hunter Aug 27 '14 at 12:26
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    $\begingroup$ @Hunter You're welcome. Your last edit was already a step in the right direction. The only thing you needed to add in the end was the final sentence I wrote. $\endgroup$ – Vincent Boelens Aug 27 '14 at 12:28

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