(Baby Rudin Chapter 4 Exercise 8)
I am trying to prove:
$f$ is a real, uniformly continuous function on the bounded subset $E$ in $\mathbb{R}^1 \implies f$ is bounded on $E$.
My attempt:
Suppose $E \subset \mathbb{R}$ is bounded and $f$ is a real, uniformly continuous function on $E$. For a contradiction, assume $f$ is not bounded on $E$. Then there exists a non-finite $f(p)$ for some $p \in E$. Let $r\in E$ be distinct from $p$. The boundedness of $E$ guarantees the existence of the real number $M$ such that $|p-r|< M$. Due to the usual properties of the absolute value function (and $p\ne r$), we conclude that $0<|p-r|<M$, that is, $M$ is positive. By the definition of uniform continuity,
\begin{equation}\tag{8.1}
|p-r|<M \implies |f(p)-f(r)| <\epsilon
\end{equation}
for some $\epsilon>0$. (We utilize the uniform continuity of $f$ in asserting that the $\delta$ in (8.1) would hold for our choice of $p$, given an arbitrary $r \in E$.) Note that the consequent (8.1) births a contradiction since the absolute difference of the non-finite $f(p)$ and $f(x)$ cannot be made smaller than $\epsilon$.
My question: Is my proof correct in general? In particular, I am not sure if I should have taken $r \in \mathbb{R}$ instead of taking $r \in E$ above. The issue is that the definition of a bounded set mandates that $r$ be some real number (not necessarily in $E$) and the definition of uniform continuity requires that $r$ be in $E$. Is my proof totally incorrect? Can someone suggest how this proof can be improved / completed?
Definition of a bounded set:
Let $X$ be a metric space. $E$ is bounded if there is a real number $M$ and a point $q \in X$ such that $d(p, q)<M$ for all $p \in E$.
Definition of uniform continuity:
Let $f$ be a mapping of a metric space $X$ into a metric space $Y$. We say that $f$ is uniformly continuous on $X$ if for every $\epsilon>0$ there exists a $\delta>0$ such that $d_Y(f(p), f(q))$ for all $p$ and $q$ in $X$ for which $d_X(p, q)<\delta$.
Best Answer
If $f$ is not bounded there exists a sequence $(x_n)$ in $E$ such that $|f(x_n)|\to \infty$. Since $E$ is bounded so is the sequence $(x_n)$. Hence there is a convergent subsequence $(x_{n_i})$. Note that $|x_{n_i}-x_{n_j}| \to 0$ as $i,j \to \infty$. Now use the definition of uniform continuity to show that $|f(x_{n_i})-f(x_{n_j})| \to 0$ as $i,j \to \infty$. But then $(f(x_{n_i}))$ is Cauchy sequence. Any Cauchy sequence of real numbers is convergent, hence bounded. We have arrived at a contradiction to the fact that $|f(x_{n_i})|\to \infty$.