# If $f:[a,b]\to\mathbb{R}$ is continuous but unbounded, then on which subintervals is $f$ unbounded?

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.

• "or" in mathematics is not exclusive or. Dec 14, 2015 at 10:09
• 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. Dec 14, 2015 at 10:10
• You need not adhere the exclusiveness of 'or'. The proof works either the word 'or' is exclusive or not. Dec 14, 2015 at 10:13
• @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? Dec 14, 2015 at 10:16
• I am completely sure that this post will help you a lot. :) Dec 14, 2015 at 10:32

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 $[(a+b)/2,b]$.
• 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. Dec 14, 2015 at 10:39