# Proof that continuous functions are bounded

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.

• 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. Commented Jun 19 at 3:36
• 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 $A$. Commented Jun 19 at 3:37
• @RobertShore Ahh makes sense. Thanks a lot Commented Jun 19 at 3:40
• 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 $[a,b]$. Commented Jun 19 at 5:21