Exercise Baby Rudin 4.8 Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb{R}^1$. Prove that $f$ is bounded on $E.$
I am struggling to show that $E$ is compact (I know how to resolve the problem after that). I was originally going to say that it is a subset of a 1-cell since it's bounded but I know that $(0,1)$ isn't compact when $[0,1]$ is. Is it possible to make an argument about limit points?
 A: Why does $E$ need to be compact? You only know that it is bounded, it need not be closed. You can solve it using the Bolzano-Weierstrass Theorem like this.
Suppose otherwise: without loss of generality, let $f$ be not bounded above. Then for any $n\in\mathbb{N}$, there exists $x_n\in E$ such that $f(x_n)>n$. Because $(x_n)_{n\in\mathbb{N}}$ is a sequence in a bounded subset of $\mathbb{R}$, it has a convergent, and thus Cauchy, subsequence $(x_{n_k})_{k\in\mathbb{N}}$.
Since $f$ is uniformly continuous, we can fix some $\delta>0$ such that for all $x,y\in E$,
$$|x-y|<\delta\implies |f(x)-f(y)|<1.$$
As $(x_{n_k})_{k\in\mathbb{N}}$ is Cauchy, there exists some $N\in\mathbb{N}$ such that for all $l,m>N$, $|x_{n_k}-x_{n_l}|<\delta$. Then for all $l,m>N$, $|f(x_{n_k})-f(x_{n_l})|<1$.
This contradicts the unboundedness of $(f(x_{n_k}))_{k\in\mathbb{N}}$ (Why?) and therefore, $f$ is bounded.
A: As pointed out, compactness of $E$ is not needed.
Let $\delta > 0$ be such that $|f(x) - f(y)| < 1$ whenever $x, y \in E$ and $|x - y| < \delta$.
Since $E$ is bounded, you can cover $E$ with finitely many (say, $N$) intervals of length $\delta$. We may also assume that all these intervals actually intersect $E$.
Pick $x_1, \ldots, x_N \in E$ from the each interval. Put $M = \max\{|f(x_1)|, \ldots, |f(x_N)|\} + 2.$
Now, if $y \in E$, then $y$ is in the $\delta$ neighbourhood of some $x_k$. In that case $$|f(y)| \le |f(y) - f(x_k)| + |f(x_k)| < 1 + |f(x_k)| < M.$$
Thus, $M$ is an upper bound.
