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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.

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1  
Note $\{a_i\}$ is monotone increasing. So $\alpha$, as the supremum of the $a_i$, implies that $(\alpha-\delta,\alpha]$ contains a tail of $\{a_i\}$. The left intervals are picked to avoid ambiguity. –  David Mitra Dec 15 '12 at 23:23
    
Thanks for the reply. {${a_i}$} is monotone increasing. Is this so because of the kind of construction involved. I.e once we have zeroed in on a half interval over which $f$ is unbounded, at a given stage , we won't ever go to the other half interval ,over which f is bounded, at a later stage. But I am unable to state this rigorously. Another thing is that Apostol doesn't introduce the notion of sequences at this stage in the book. It comes in quite a later chapter. Anyways, thanks again for the reply, it helped clear some fog. –  ameyask86 Dec 16 '12 at 5:38
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It's simply because each interval $[a_n,b_n]$ is a subinterval of the preceeding interval. You have $a_{n+1}\in [a_n, b_n]$ for each $n$ (in fact $a_{n+1}=a_n$ or $a_{n+1}={b_n-a_n\over2}$. –  David Mitra Dec 16 '12 at 14:18
    
Thanks. That has made the matter crystal clear. Is there a way to mark the question as answered , even if the answers were in comment form ? –  ameyask86 Dec 17 '12 at 9:23
    
I'll add the comments as an answer. –  David Mitra Dec 17 '12 at 12:50

1 Answer 1

up vote 2 down vote accepted

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.

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Marked the question as answered. :) –  ameyask86 Dec 17 '12 at 17:00

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