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I have come across a proof which I understand almost completely, except for one part:

THEOREM: If $f$ is uniformly continuous on a bounded interval $I$, then $f$ is also bounded on $I$.

PROOF: In this case we assume that $I$ is of the form $(a,b), (a,b], [a,b)$, or $[a,b]$, with $a,b \in \mathbb{R}$. Fix an $\epsilon > 0$, for instance $\epsilon = 1$. Since $f$ is uniformly continuous, there is a $\delta > 0$ such that:

$|f(x_1) - f(x_2)| < \epsilon = 1$ when $x_1, x_2 \in I$ and $|x_1 - x_2| < \delta$

Divide $I$ into $N$ intervals, $I_1, . . ., I_N$, where $N$ is chosen so that $\frac{b-a}{N} < \delta$.

Let $z_i$ be the center point of $I_i$. For each $i$ and $x \in I_i$, $|x - z_i| < \delta$, and then we have:

$|f(x)| = |f(x) - f(z_i) + f(z_i)| \leq |f(x) - f(z_i)| + |f(z_i)| \leq 1 + |f(z_i)|$. Then for $x \in I_i$,

$|f(x)| \leq 1 + \max_{1 \leq i \leq N}\{|f(z_i)|\}$.

Let $M = \max_{1 \leq i \leq N}\{|f(z_i)|\}$. Then $|f(x)| \leq 1 + M$

QED

OK, so the one thing I am a bit unsure of here, is when we write:

Let $M = \max_{1 \leq i \leq N}\{|f(z_i)|\}$.

How is it that we know for sure that each $|f(z_i)|$ is also bounded? I see how the presence of a maximum value completes the proof, but why is it not possible that we have an $|f(z_i)|$ which is unbounded?

If anyone could explain this to me I would greatly appreciate it!

  • 1
    This question was asked and partially answered already [on MO](http://mathoverflow.net/questions/82604/uniform-continuity-and-boundedness).2011-12-04
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    Kristian, welcome to math.SE. Next time you post on several different fora please do mention that, in order to avoid duplication of effort.2011-12-04
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    Yes, but there we assumed that [a,b] was closed. Here we do not.2011-12-04
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    t.b - will do! I realized I forgot to add that the given interval did not neccessarily have to be closed on MO. And I was directed to this forum for the type of question I had.2011-12-04
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    Yes, it was me who pointed you here :) I just added the links to the threads for the sake of transparency. That's all. As I said, welcome to this site and I hope you'll have a good experience here!2011-12-04
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    Thanks! I appreciate you pointing the way for me :). I'm a bit stressed these days since I have my real analysis exam this upcoming Thursday, so finding a forum like this is really of great help when I have questions!2011-12-04
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    Related question: http://math.stackexchange.com/questions/87770/if-fx-is-uniformly-continuous-at-0-1-then-is-it-bounded-at-0-12011-12-04
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    Thanks for the link. Yeah, I see I could also have used Cauchy sequences to prove this.2011-12-04
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    Hi what is the difference between this and normal continuity? Cos I know normal continuity does not imply boundedness? I don't really understand the proof2014-04-24
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    Why this proof doesn't work on normal continuity to imply boundedness, which i know is not true.2014-04-24

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