Exercise 4.9

Question

\def\eps{\varepsilon} Exercise 9: Show that the requirement in the definition of uniform continuity can be rephrased as follows, in terms of diameters of sets: To every $\varepsilon>0$ there exists a $\delta>0$ such that $\mathop{\rm diam}f(E)<\varepsilon$ for all $E\subset X$ with $\mathop{\rm diam}E<\delta$.

analambanomenos · Accepted Answer

\def\eps{\varepsilon} Suppose $f$ is a uniformly continuous function from the metric space $X$ to the metric space $Y$. Let $\varepsilon>0$. Then there is a $\delta>0$ such that $d_Y\bigl(f(p), f(q)\bigr)<\varepsilon$ if $d_X(p,q)<\delta$. Let $E\subset X$ with $\mathop{\rm diam}E<\delta$. If $p,q\in E$, then $d_X(p,q)\le\mathop{\rm diam}E<\delta$, so $d_Y\bigl(f(p),f(q)\bigr)<\varepsilon$. Hence $\mathop{\rm diam}f(E)<\varepsilon$. Conversely, let $f$ be a function from the metric space $X$ to the metric space $Y$ with the diameter property. Let $\varepsilon>0$ and $\delta>0$ such that $\mathop{\rm diam}f(E)<\varepsilon$ for all $E\subset X$ with $\mathop{\rm diam}E<\delta$. Let $p,q\in X$ such that $d_X(p,q)<\delta$. Letting $E=\{p,q\}$ we have $\mathop{\rm diam}E<\delta$, so $\mathop{\rm diam}f(E)=d_Y\bigl(f(p),f(q)\bigr)<\varepsilon$. Hence $f$ is uniformly continuous.

Exercise 4.9

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Exercise 4.9

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