I was studying the proof of the boundedness theorem of continuous functions on a closed interval from Apostol's Calculus Volume 1. I am unable to understand a crucial step in the proof.
First the theorem :-
"If a function $f$ is continuous on a closed interval [$a,b$] then it is bounded on that interval "
The proof provided is as follows:-
"Suppose that $f$ is not bounded on the interval [$a,b$].
Let $c$ be the mid point of [$a,b$] . $f$ will be unbounded in at least one of the two intervals [$a,c$] and [$c,b$] . We choose the interval on which it is unbounded (in case it is unbounded on both, we choose the left interval). We call this interval as [$a_1,b_1$].
This process of bisection is carried out indefinitely so that the interval [$a_{n+1},b_{n+1}$] denotes that half of [$a_n,b_n$] in which $f$ is unbounded. In case it is unbounded on both halves, the left half is selected.
The length of the $n$th interval is $(b-a)/2^n$.
Let $A$ denote the set of leftmost endpoints $a,a_1,a_2,a_3…$ so obtained. Let $\alpha$ denote the supremum of $A$. Then $\alpha$ lies in $[a,b]$.
Since $f$ is continuous at $\alpha$ , there exists a $\delta > 0 $ such that
$$
|f(x)| < 1 + |f(\alpha)|
$$
in the interval $(\alpha -\delta,\alpha + \delta)$
(In case $\alpha = a$ , the interval should be $[a,a + \delta)$.
In case $\alpha = b$ , the interval should be $(b – \delta,b]$.)
However , the interval [$a_n,b_n$] lies inside the interval $(\alpha -\delta,\alpha + \delta)$, provided $(b-a)/2^n < \delta$.
Therefore, f is bounded in $(b-a)/2^n$ , which is a contradiction , hence completing the proof.
"
My trouble is that I am unable to understand how it is guaranteed that the interval [$a_n,b_n$] lies inside the interval $(\alpha -\delta,\alpha + \delta)$. I can somewhat understand (but not too sure about it)that if any one point of [$a_n,b_n$] lies inside the interval $(\alpha -\delta,\alpha + \delta)$ , then it is guaranteed that entire interval lies inside $(\alpha -\delta,\alpha + \delta)$ , as it's length is less than $\delta$. But how is it guaranteed that at least one point from it lies inside $(\alpha -\delta,\alpha + \delta)$. I think it is something to do with the way the intervals were selected. But I am unable to link the two crucial aspects of the proof.
Also I am unable to visualise why we pick the left interval when both intervals are such that $f$ is unbounded over them.
Thanks.
Best Answer
Note that the sequence $\{a_i\}$ is monotone increasing. So $\alpha$, as the supremum of the $a_i$, implies that $(\alpha-\delta,\alpha]$ contains a tail of $\{a_i\}$.
Alternatively, note that each interval $[a_n,b_n]$ is a subinterval of the preceding interval. You have $a_{n+1}\in[a_n,b_n]$ for each $n$. (In fact, $a_{n+1}=a_n$ or $a_{n+1}={b_n−a_n\over2}$ for each $n$.)
So, noting $\alpha\ge a_n$, if the length of $[a_n,b_n]$ is less than $\delta$, we see that $[a_n,b_n]\subseteq(\alpha-\delta,\alpha+\delta)$.
The left intervals are picked simply so that the process is well-defined.