# Continuity and Boundedness

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?

Thanks.

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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

## 1 Answer

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!

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@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