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I am struggling to understand the proof of the Heine-Borel theorem in "Measure, Integration and Real Analysis" by Sheldon Axler.

The Theorem states: Every open cover of a closed bounded subset of $\mathbb{R}$ has a finite subcover.

The part of the proof that I am concerned with goes as follows:

Suppose $F$ is a closed bounded subset of $\mathbb{R}$ and $\mathcal{C}$ is an open cover of $F$.

First consider the case where $F=[a,b]$ for some $a,b \in \mathbb{R}$ with $a<b$. Thus $\mathcal{C}$ is an open cover of $[a,b]$. Let \begin{align*} D = \{ d \in [a,b] : [a,d] \text{ has a finite subcover from }\mathcal{C} \} \end{align*}

Note that $a\in D$ (because $a\in G$ for some $G \in \mathcal{C}$). Thus $D \neq \emptyset$. Let $s = \sup{D}$. Thus $s\in [a,b]$. Hence there exists an open set $G \in \mathcal{C}$ such that $s\in G$. Let $\delta >0$ be such that $(s-\delta, s+\delta) \subset G$. Because $s = \sup{D}$, there exist $d \in (s - \delta, s ]$ and $n \in \mathbb{Z}^+$ and $G_1, \ldots, G_n \in \mathcal{C}$ such that \begin{align*} [a,d] \subset G_1 \cup \ldots \cup G_n. \end{align*} Now \begin{align*} [a,d'] \subset G \cup G_1 \cup \ldots \cup G_n \end{align*} for all $d' \in [s, s+\delta)$. Thus $d' \in D$ for all $d' \in [s, s+\delta) \cap [a,b]$. This implies that $s=b$.

I am able to follow the proof up until the last line of what I have written. My question is: how does this imply that $s=b$?

user12345
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  • I think the idea here is that we can keep applying the same argument again and again otherwise. If $s+ \delta \le b$ then there exists another $G^*$ which covers $s+ \delta$, which since it's an open set will also cover a smidge more of the interval and so on. Can't think of how that makes an airtight argument. Also seems that that is a bit complicated for the authors to have left unstated... – Chessnerd321 Oct 26 '22 at 02:16

1 Answers1

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Suppose that $a \leqslant s<b.$ Then we can pick $t$ so that $s<t<b.$ We can pick it so that also $s<t<s+\delta.$ So now our $t$ qualifies to be one of the $d'$ values. Concretely:

$$t \in [s, s+\delta) \cap [a,b].$$

But this is absurd; $t$ belongs to $D$, but is strictly greater than $\sup D.$

311411
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  • Does this make sense to you, user12345? If so you can vote up and click the grey check mark. Or maybe you want me to clarify something? Welcome to Math Stack. – 311411 Oct 29 '22 at 13:59