Exercise 3.3.10

Question

Here is an alternate proof to the one given in Exercise 3.3.9 for the final implication in the Heine-Borel Theorem.

Consider the special case where $K$ is a closed interval. Let $\left\{O_{\lambda}: \lambda \in \Lambda\right\}$ be an open cover for $[a, b]$ and define $S$ to be the set of all $x \in[a, b]$ such that $[a, x]$ has a finite subcover from $\left\{O_{\lambda}: \lambda \in \Lambda\right\}$.
  \begin{enumerate}[label=(\alph*)] 
  \item Argue that $S$ is nonempty and bounded, and thus $s=\sup S$ exists.
  \item Now show $s=b$, which implies $[a, b]$ has a finite subcover.
  \item Finally, prove the theorem for an arbitrary closed and bounded set $K$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $S$ is nonempty since $x=a$ has the finite subcover $\{O_{\lambda_0}\}$ for $a \in O_{\lambda_0}$. $S$ is bounded since $x \le b$ for all $x \in S$.
  \item Suppose for contradiction that $s < b$, letting $s \in O_{\lambda_0}$ implies $[a,s]$ is finitely coverable since we can take the finite cover of an $x \in O_{\lambda_0}$ with $x < s$. This is causes a contradiction however since there exist points $y > s$ with $y \in O_{\lambda_0}$ meaning $[a,y]$ is also finitely coverable. therefore the only option is $s = b$, since any $s < b$ doesn't work.
  \item (a) still works, for (b) we must also consider the case where $y$ does not exist / there is a gap. Let $y = \inf\;[s, b]\cap K$ and suppose $y 
e s$. since $y \in [s,b] \cap K$ we know
    $$[a,y] \cap K = ([a,s]\cap K) \cup ((s, y]\cap K) = [a,s]\cap K \cup \{y\}$$
    therefore if $\{O_{\lambda_1}, \dots, O_{\lambda_n}\}$
    covered $[a, s]$ then letting $y \in O_{\lambda_{n+1}}$ would give the finite cover $\{O_{\lambda_1}, \dots, O_{\lambda_{n+1}}\}$ contradicting the assumption that $s < b$, therefore $s = b$ is the only option, and so $K$ can be finitely covered.
   \end{enumerate}

Exercise 3.3.10

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Exercise 3.3.10

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