This is from Rudin Chapter 4, problem 9:
Show that the requirement in the definition of uniform continuity can be rephrased as follows, in terms of diameters of sets: To every $\epsilon > 0$ there exists a $\delta > 0$ such that diam $f(E)<\epsilon$ for all $E\subset X$ with diam $E<\delta$.
Here is the solution from Baby Rudin that I'm following:
Suppose $f$ is uniformly continuous on a metric space $X$. By the definition, we know that for any $\epsilon >0$, there is $\delta>0$ such that $d(f(x), f(y))<\frac{\epsilon}{2}$ for any $x, y \in X$ satisfying $d(x, y)<\delta$. Thus, for all $E$ with $\text{diam}(E)<\delta$, if $x, y \in E$,then $d(x, y)≤\text{diam}(E)<\delta$, so that $d(f(x), f(y))<\frac{\epsilon}{2}$. The last inequality implies $\text{diam}(f(E))\leq \frac{\epsilon}{2}<\epsilon$.
I don't get the last implication. Can smb explain it? Thanks.
By definition $$ \text{diam} (E)=\sup_{x,y\in E}d(x,y). $$ For the proof, $\epsilon/2$ is an upper bound of the set $\{d(f(x), f(y))\mid x,y\in E\}$. But the supremum is the least upper bound whence $$ \text{diam}(f(E))=\sup_{x,y \in E}d(f(x),f(y))\leq \epsilon/2<\epsilon $$