Exercise 3.3.9

Question

Follow these steps to prove that being compact implies every open cover has a finite subcover.

Assume $K$ is compact, and let $\left\{O_{\lambda}: \lambda \in \Lambda\right\}$ be an open cover for $K$. For contradiction, let's assume that no finite subcover exists. Let $I_{0}$ be a closed interval containing $K$.
  \begin{enumerate}[label=(\alph*)] 
  \item Show that there exists a nested sequence of closed intervals $I_{0} \supseteq I_{1} \supseteq I_{2} \supseteq$ $\cdots$ with the property that, for each $n, I_{n} \cap K$ cannot be finitely covered and $\lim \left|I_{n}\right|=0$.
  \item Argue that there exists an $x \in K$ such that $x \in I_{n}$ for all $n$.
  \item Because $x \in K$, there must exist an open set $O_{\lambda_{0}}$ from the original collection that contains $x$ as an element. Explain how this leads to the desired contradiction.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Bisect $I_0$ into two intervals, and let $I_1$ be the interval where $I_1 \cap K$ cannot be finitely covered. Repating in this fashion we have $\lim |I_n| = \lim |I_0|(1/2)^n = 0$.
  \item The nested compact set property with $K_n = I_n \cap K$ gives $x \in \bigcap_{n=1}^\infty K_n$ meaning $x \in K$ and $x \in I_n$ for all $n$.
  \item Since $x \in O_{\lambda_0}$ and $|I_n| 	o 0$ with $x \in I_n$ for all $n$, there exists an $N$ where $n > N$ implies $I_n \subseteq O_{\lambda_0}$ contradicting the assumption that $I_n \cap K$ cannot be finitely covered since $\{O_{\lambda_0}\}$ is a finite subcover for $I_n \cap K$.
   \end{enumerate}

Exercise 3.3.9

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Exercise 3.3.9

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