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 $|f(x)| \leq 1 + M$


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
    $\begingroup$ This question was asked and partially answered already on MO. $\endgroup$
    – t.b.
    Dec 4, 2011 at 13:36
  • $\begingroup$ Kristian, welcome to math.SE. Next time you post on several different fora please do mention that, in order to avoid duplication of effort. $\endgroup$
    – t.b.
    Dec 4, 2011 at 13:36
  • $\begingroup$ Yes, but there we assumed that [a,b] was closed. Here we do not. $\endgroup$
    – Kristian
    Dec 4, 2011 at 13:36
  • $\begingroup$ 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. $\endgroup$
    – Kristian
    Dec 4, 2011 at 13:37
  • 2
    $\begingroup$ Related question: math.stackexchange.com/questions/87770/… $\endgroup$ Dec 4, 2011 at 13:50

1 Answer 1


$f$ is presumably a real-valued function (the value of $f$ at any single point is finite); so, the $f(z_i)$ are fixed and, in particular, bounded numbers (infinities are not real numbers).

There are a finite number of the $z_i$; so, $M$ is the maximum of a finite set of numbers and is, thus, finite.

  • $\begingroup$ Thanks, David. That makes sense :) $\endgroup$
    – Kristian
    Dec 4, 2011 at 13:39

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