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 $E$?
Best Answer
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 $\inf \{f(x_1), f(x_2)... f(x_n)\} - \epsilon$