Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb{R^1}$. Prove that $f$ is bounded on $E$

Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb{R^1}$. Prove that $f$ is bounded on $E$

My (Attempted) Proof

Since $E$ is bounded, put $\alpha = \sup E$, $\beta = \inf E$. Now since $f$ is uniformly continuous on $E$, we only have to prove convergence of $f$ as $x \in E \to \alpha$ and $x \in E \to \beta$.

So let $\{x_n\}$ be a Cauchy sequence and fix $\epsilon > 0$. Let $\delta$ be such that $d(f(p), f(q)) < \epsilon$.

Now take $N$ so large such that $m, n > N \implies d(x_n, x_m) < \delta$. We then have $d(f(x_n), f(x_m)) < \epsilon$ (By uniform continuity of $f$)

Thus $\{f(x_n)\}$ is a Cauchy sequence, and thus converges to some point $L \in \mathbb{R^1}$

\begin{aligned}\therefore \text{As} \ \ x_n \to \beta \ , \ \ f(x_n) \to L_1\\ \ \ x_n \to \alpha \ , \ \ f(x_n) \to L_2\\\end{aligned}

(Comment: Since we let $\{x_n\}$ be an arbitrary Cauchy sequence, we can pick two different Cauchy sequences, one that converges to $\beta$, and one that converges to $\alpha$, and thus the above statement holds)

Now put $\gamma = \max\{L_1, L_2, f(x)\ | \ x \in E\}$ and $\eta = \min\{L_1, L_2, f(x)\ | \ x \in E\}$.

Then we have $\gamma = \sup f[E]$, and $\eta = \inf f[E]$, and thus $f[E]$ is bounded. $\ \square$

First of all if my proof rigorously correct? Secondly, if you have any criticism on my proof-writing skills, please let me know, as I'm always looking to improve.

Finally, I proved that $f$ is bounded on $E$ by proving that the sequence $\{f(x_n)\}$ was Cauchy, and would thus converge, but are there other cleaner/more efficient ways of proving that $f$ is bounded on $E$?

• The function $f$ is not necessarily uniformly continuous unless $E$ is closed as well. – Olivier Moschetta Oct 19 '16 at 8:17
• @OlivierMoschetta, $f$ is given as a uniformly continuous function. My apologies, I left the uniform part out of the title – Perturbative Oct 19 '16 at 8:18
• Why should the problem of not being bounded only appear at the supremum and infimum of $E$? $E$ does not have to be connected, does it? – Paul K Oct 19 '16 at 8:27
• @menag, no $E$ is not given to be connected. I thought that since $f$ is uniformly continuous on $E$ and $E$ is bounded, that we'd only need to prove that as $x \in E$ approached $\sup E$ or $\inf E$, that $f[E]$ would converge to some point in $\mathbb{R^1}$, because $\sup E$ and $\inf E$ need not be contained in $E$ for $E$ to be bounded. Since $f$ is uniformly continuous on $E$, it doesn't contain any discontinuities on $E$, but $f$ uniformly continuous on $E$ doesn't imply that $f[E]$ is bounded – Perturbative Oct 19 '16 at 8:35
• But there could be holes, i.e. $(-1,0) \cup (0,1)$. Why shouldn't $f$ go to infinite for $x \to 0$? – Paul K Oct 19 '16 at 8:38

2 Answers

Fix $\epsilon$. There is a $\delta$ such that $|x-y| < \delta$ implies $|f(x) - f(y)| < \epsilon$.

Since $E$ is bounded, there exists a sequence $x_1, ...x_n$ of elements of $E$ such that

$$\bigcup_{1 \leq i \leq {n}} (x_i - \delta, x_i+\delta)$$

is a finite open cover of $E$.

Note $\{f(x_1), f(x_2)... f(x_n)\}$ is finite hence bounded. Thus $f$ is bounded above by $\sup \{f(x_1), f(x_2)... f(x_n)\} + \epsilon$ and bounded below by $\inf \{f(x_1), f(x_2)... f(x_n)\} - \epsilon$

Take $\varepsilon = 1$. By uniform continuity there is $\delta > 0$ such that $\lvert x - y \rvert < \delta$ implies $\lvert f(x) - f(y) \rvert < \varepsilon$. Now by boundedness there are finitely many invertvals $(x_i - \delta, x_i + \delta)$ (proof!) which cover $E$. Now take a reference point $x \in E$ and try to bound the difference $\lvert f(x) - f(y) \rvert$ for all $y \in E$.