(Theorem 3.3.8)[Heine-Borel Theorem] Let \(K\) be a subset of \(\mathbf{R}\). All of the following statements are equivalent in the sense that any one of them implies the other two
  1. \(K\) is compact.
  2. \(K\) is closed and bounded.
  3. Every open cover for \(K\) has a finite subcover.


Proof

We proved that (i) and (ii) are equivalent in theorem 3.3.4. Next we will show that (iii) implies (ii) and thus (i) as well. So assume that every open cover for (K) has a finite subcover. We’ll show that (K) is bounded and closed.

To show that \(K\) is bounded, construct an open cover for \(K\) by defining \(O_x\) to be an open interval of radius 1 around each element of \(x \in K\) so that \(O_x = V_1(x)\). The open cover is defined then as \(\{O_x: x \in K\}\). Since we assume that (iii) is true, then this means there exists a finite subcover \(\{O_{x_1}, O_{x_2}, O_{x_3}, ...O_{x_n}\}\) that contains \(K\). Since the subcover is a finite collection of bounded sets, then \(K\) must be bounded as well.

To show that \(K\) is closed, suppose that \((y_n)\) is a Cauchy sequence contained in \(K\) such that \((y_n) \rightarrow y\). We need to show that \(y \in K\) since a closed set must contain all of its limit points. To show this, suppose for the sake of contradiction that \(y \notin K\). This means that for all \(x \in K\), \(x\) is some positive distance away from \(y\). So construct an open cover by taking \(O_x\) to be an interval of radius \(|x-y|/2\) around each element. Since we assumed (iii) is true, then there exists a finite subcover \(\{O_{x_1}, O_{x_2}, O_{x_3}, ...O_{x_n}\}\) such that it contains \(K\). But this is impossible. To see why, set

$$ \begin{align*} \epsilon_0 = \min \left\{\frac{x_i - y}{2}: 1 \leq i \leq n \right\}. \end{align*} $$

Now, since \((y_n) \rightarrow y\), then for all \(\epsilon > 0\), there exists an \(N \in \mathbf{N}\) such that when \(n \geq N\), we have \(|y_n - y| < \epsilon_0\). But this implies that \(y_n\) is not in any \(O_x\) meaning that

$$ \begin{align*} y_n \notin \bigcup_{i=1}^{n} O_{x_i}. \end{align*} $$

This means that the finite cover doesn’t actually contain all of \(K\) and therefore this is a contradiction and we must have \(y \in K\) as required to show that \(K\) is closed.

Next, we need to assume (ii) and prove (iii). So suppose that \(K\) is closed and bounded. We will prove that for every open cover of \(K\), there exists a finite subcover. [TODO] \(\blacksquare\)

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