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I am reading Theorem 3.11 from the book "Apostol calculus Vol 1". Let $f$ be continuous on a closed internval $[a,b]$. Then $f$ is bounded on $[a,b]$. That is, there is a number $C\ge 0$ such that $|f(x)|\le C$ for all $x$ in $[a,b]$.

The book called the theorem "Boundedness theorem for continuous functions" and used a method of "successive bisection" to prove the theorem.

The proof works like this: assume $f$ is not bounded on $[a,b]$. Let $c$ be the midpoint of $[a,b]$, at least one of the subintervals $[a,c]$ and $[c,b]$ will be unbounded. If $f$ is unbounded on $[a,c]$, then we choose $[a_1, b_1]$ be $[a,c]$. Otherwise we let $[a_1,b_1]$ be $[c,b]$.

This bisection process continues. Let $A$ denote the set of the $a, a_1, a_2, ..., a_n, ...$, then from the real number's least upper bound axiom, let $\alpha$ be the supremum of $A$. We know $\alpha$ lies in $[a,b]$. By continuity of $f$ at $\alpha$, there is an interval of $(\alpha - \delta, \alpha + \delta)$ in which $|f(x) - f(\alpha)| < 1$. This means $|f(x)|$ is bounded by $1 + |f(\alpha)|$.

However, the interval $[a_n, b_n]$ lies inside $(\alpha - \delta, \alpha + \delta)$ when n is so large that $(b-a)/2^n < \delta$. Therefore $f$ is also bounded in $[a_n, b_n]$ contradicting that $f$ is unbounded on $[a_n, b_n]$. Proof completes.

My confusion is that why are we so sure that $[a_n, b_n]$ lies inside $(\alpha - \delta, \alpha + \delta)$? Yes the subinterval can be as small as we want, but what if $|a_n - \alpha| > \delta$? The proof did not say anything about the distance between $a_n$ and $\alpha$. I find it confusing. Did I miss anything here?

Dachuan Huang
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Note that for each $n, [a_{n+1}, b_{n+1}] \subseteq [a_n, b_n]$. This implies $a_n \le a_{n+1}$. I.e., the sequence $\{a_n\}_{n \in \Bbb N}$ is increasing.

Because $\alpha$ is the supremum of $\{a_n\}_{n \in \Bbb N}$, for every $\delta > 0$, there has to be an $a_N$ such that $\alpha - \delta < a_N$. Otherwise $\alpha - \delta$ would be an upper bound of $A = \{a_n\}_{n \in \Bbb N}$, contradicting that $\alpha$ is the smallest upper bound. But for all $n \ge N,$ $$ \alpha - \delta < a_N \le a_n \le \alpha$$

Thus by choosing $n$ high enough, we are guaranteed that $|\alpha - a_n| < \delta$.

Paul Sinclair
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  • Nice, thanks! I think you used Theorem I.32 from the book. It's: Let $h$ be a given positive number and let $S$ be a set of real numbers, then if $S$ has a supremum, then for some $x$ in S, we have $x > \sup S - h$. Together with the fact that the $a_n$ is non-decreasing, we can say there is a number bigger than $\alpha - \delta$ and the neighborhood argument holds now. – Dachuan Huang Aug 01 '21 at 18:26
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    I don't know what is in your book, but that is the property that distinguishes the supremum from other upper bounds, so it is one you should become familiar with. – Paul Sinclair Aug 01 '21 at 19:57