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Problem 8-16 in Spivak's Calculus (fourteenth printing, 1993) begins exactly as follows:

Suppose $f$ were continuous on $[a,b]$, but not bounded on $[a,b]$. Then $f$ would be unbounded on either $[a,(a+b)/2]$ or on $[(a+b)/2,b]$. Why?

The point is that you go on to prove a contradiction using a bisection argument.

However, I don't see how this first sentence is justified. If $f$ is as described, then of course $f$ is unbounded on at least one of the two subintervals given. Can we show that $f$ is unbounded on at most one of the two subintervals?

I don't see how to do this without implicitly assuming (or just outright proving by another method) the ultimate conclusion, that is, the fact that continuity on a closed, bounded interval implies the function is bounded.

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    $\begingroup$ "or" in mathematics is not exclusive or. $\endgroup$ Dec 14, 2015 at 10:09
  • $\begingroup$ In my experience, "either ... or..." is exclusive or. Besides, the point is that you choose an interval and then use bisections; I don't see how to work a bisection argument where every subinterval needs to be bisected. $\endgroup$
    – Will R
    Dec 14, 2015 at 10:10
  • $\begingroup$ You need not adhere the exclusiveness of 'or'. The proof works either the word 'or' is exclusive or not. $\endgroup$
    – Hanul Jeon
    Dec 14, 2015 at 10:13
  • $\begingroup$ @HanulJeon: I find this surprising. So the point is that I just pick an interval and carry on? I guess that would work: a contradiction is a contradiction. Is this the right idea? $\endgroup$
    – Will R
    Dec 14, 2015 at 10:16
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    $\begingroup$ I am completely sure that this post will help you a lot. :) $\endgroup$ Dec 14, 2015 at 10:32

2 Answers 2

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You might say that from $f$ continuity follows that $f$ is bounded on $(a+\varepsilon , b-\varepsilon )$. But $f$ is unbounded on $[a,b]$ so $f$ is unbounded on $[a,a+\varepsilon )$ or on $(b-\varepsilon , b]$ which are respectively subsets from $[a,(a+2)/2]$ and $[(a+b)/2,b]$.

I hope I helped you.

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    $\begingroup$ That's also an interesting way of looking at it. It took me a moment to process your first sentence in the form "$f$ continuous at $(a+b)/2$ implies $f$ bounded on $((a+b)/2-\delta,(a+b)/2+\delta)$ for some $\delta>0$, which implies $f$ bounded on $(a+\epsilon,b-\epsilon)$ for some $\epsilon>0$." On that basis, +1. However, I will not "accept" this answer because, while you are providing an alternative argument for what I had stated in the question, you have not answered the question. I hope this is not unsatisfactory. $\endgroup$
    – Will R
    Dec 14, 2015 at 10:39
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The wording of the question is perhaps slightly misleading: some people interpret "either...or" to mean exclusive or, some people don't interpret it this way and just see it as "or".

The point is that it doesn't matter how many subintervals the function is continuous but unbounded on: just choose one and carry on with the argument. The goal is to get a contradiction, and it doesn't really matter how you get there (as long as all the steps in-between are valid, of course).

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