Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to show that if $f$ is a uniformly continuous real valued function on the bounded set $E \subseteq \mathbb{R}$, then $f$ is bounded on $E$.

I want to define an open cover of $E$, then say that cover holds true for the closure of $E$. Then $E$ is compact and by continuity, $f(E)$ is compact - and so $f$ is bounded on $E$, by the Heine-Borel theorem.

This is problem 4.8 in Rudin's Principles of Mathematical Analysis.

My concern is the first part: Is it true that I can define the same cover for $E$ as the cover for $E$ closure?


share|cite|improve this question
What about $f:(0,1)\to\mathbb{R}$ defined by $f(x)=\frac{1}{x}$? – Clayton Dec 9 '12 at 3:35
Are you assuming that $f[E]\subseteq X$? Because if not, it’s not clear what you mean by $f$ is bounded on $f[E]$. – Brian M. Scott Dec 9 '12 at 3:39
Yes - or, I would assume so. I just edited the question, to exactly what it states. – user5262 Dec 9 '12 at 3:46
up vote 2 down vote accepted

Your method seems okay, but theres no need to mention covers. And you're missing a crucial component of the proof, you would need to show that you can extend $f$ to a uniformly continuous function $\tilde{f}$ on $\overline{E}$ (which crucially requires that $\mathbb{R}$ is complete). If you have already proved this than your proof will go through fine.

If you have not already proved this, I suggest proving it differently because proving that you can extend the function is more work than is needed.

Heres an alternative:

Suppose not, then there are $x_k\in E$ with $f(x_k)\geq k$ for every $k\in\mathbb{N}$.

Since $x_k \in E$ and $E$ is bounded in $\mathbb{R}$, it has a convergent subsequence $x_{k_j} \rightarrow x$. (This is called Bolzano- Weierstrass)

So then since $f$ is uniformly continuous $f(x_{k_j})$ is a Cauchy sequence. And since $\mathbb{R}$ is complete this sequence converges. But by construction this sequence diverges to infinity!

share|cite|improve this answer
@JoshuaBunce I'm not sure what you mean $j \geq k$, I think your formatting may have screwed up – Deven Ware Dec 9 '12 at 7:00
Yea, it did - what I meant, I had decided was wrong - just deleted the comment, since I couldn't fix the formatting and it wouldn't let me edit again - I'm good now - thanks for the proof. I like this one much better - and will also still show the extension - For prob. 13, if anything. Thx! Appreciate the help! – user5262 Dec 9 '12 at 7:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.