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So I'm following Calculus by Spivak and a particular part of a proof is troubling me. First off, the theorem that Spivak is trying to prove is the following: If f is continuous on $[a,b]$ then it is bounded above on $[a,b]$

To make this proof, the following set is considered:

$$A = \{x \mid a\leq x\leq b\land f\;\text{is bounded above on}\;[a,x]\}$$

Now, the author first shows that $\alpha = b$ (where $\alpha$ is the supremum) through contradiction. Suppose that $\alpha < b$. It's also easy to see that $\alpha > a$. Therefore, $a < \alpha < b$. Since $[a,b]$ is continuous, it's already proven that there exists a delta such that $(\alpha-\delta,\alpha+\delta)$ is bounded above.

Now, here's the statement that troubles me: "Since $\alpha$ is the least upper bound (supremum) of $A$, there is some $x_0\in A$ satisfying $x_0\in(\alpha-\delta,\alpha)$". Things that follow after this statement seem logical to me. Of course if this statement is true, you can easily conclude that $f$ is bounded above $[a,x_0]$ since that's the only way $x_0\in A$. But why is that quoted statement true in the first place? I don't quite understand the justification for it. Why must there exist an $x_0$ in $(\alpha-\delta,\alpha)$ with the property that it is also in $A$? I'm guessing this is something to do with the supremum.

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  • $\begingroup$ Yes, $\alpha$ is the least upper bound of $A$. Since $\alpha - \delta$ is less than $\alpha$, it (referring to $\alpha-\delta$) cannot be an upper bound of $A$: by definition of upper bound this is because some $x_0 \in A$ exceeds it. The only thing left to justify is that we can choose $x_0 < \alpha$: one way to see this is that the definition of $A$ means you can always pick a smaller element (unless it takes you below $a$). Alternatively, the proof probably goes through with the weaker assumption that $x_0 \in (\alpha -\delta, \alpha]$, which we got immediately. $\endgroup$
    – Erick Wong
    Commented Jun 19 at 3:36
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    $\begingroup$ If there were no such $x_0$, then $\alpha - \delta$ also would be an upper bound for $A$. But $\alpha - \delta \lt \alpha$, contradicting the statement that $\alpha$ is the least upper bound for $A$. $\endgroup$
    – Robert Shore
    Commented Jun 19 at 3:37
  • $\begingroup$ @RobertShore Ahh makes sense. Thanks a lot $\endgroup$
    – Aryaan
    Commented Jun 19 at 3:40
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    $\begingroup$ A far more direct proof would be to notice that $f^{-1}((-n,n))$ forms an open cover of $[a,b]$, which is compact and hence has a finite subcover. It follows that $f$ is bounded on $[a,b]$. $\endgroup$
    – copper.hat
    Commented Jun 19 at 5:21

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