Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 9

Problem

Show that the requirement in the definition of uniform continuity can rephrased as follows, in terms of diameters of sets: To every $\epsilon > 0$ there exists a $\delta > 0$ such that $\text{diam}f(E)<\epsilon$ for all $E \subset X$ with $\text{diam}E < \delta$.

Definition 4.18

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 $\delta >0$ such that $d_Y(f(p),f(q))<\epsilon$ for all $p$ and $q$ in $X$ for which $d_X(p,q) < \delta$.

Answer

Let us denote Definition 4.18 and the new definition as Definition A and Definition B. We will show that Definition A implies Definition B and vice versa.

A $\Rightarrow$ B: For $\epsilon>0$, there exists $\delta_A>0$ such that $d_Y(f(p),f(q))<\epsilon/2$ for all $p$ and $q$ in $X$ for which $d_X(p,q) < \delta_A$. Let $\delta:=\delta_A$ and consider $E \subset X$ satisfying $\text{diam}E < \delta$. Then for all $x,y \in E$, $d_Y(f(x),f(y)) < \epsilon/2$ since $d_X(x,y) \leq \text{diam}E < \delta$. So, we can conclude that $\text{diam}f(E) \leq \epsilon/2 < \epsilon$.

B $\Rightarrow$ A: For $\epsilon>0$, there exists $\delta_B>0$ such that $\text{diam}f(E)<\epsilon$ for all $E \subset X$ with $\text{diam}E < \delta_B$. Let $\delta:=\delta_B/4$ and consider $x,y \in X$ satisfying $d_X(p,q)<\delta$. Since $\text{diam}N_{\delta}(p) = \delta_B/2 < \delta_B$, $\text{diam}f(N_{\delta}(p)) < \epsilon$, which implies $d_Y(f(x),f(y)) < \epsilon$.

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