A uniformly continuous function of a uniformly continuous function is uniformly continuous. State this more precisely and prove it.

Theorem: Let $E,X$ and $Y$ are metric spaces. Let $f$ maps $E$ into $X$, and $g$ maps the rang of $f$, $f(E)$ into $Y$. Let $h=g\circ f$. Then $h:E\to Y$ is uniformly continuous mapping.

Proof: Let $\varepsilon>0$ be given then $\exists \delta>0$ such that for all $z,t\in f(E)$ and $d_X(z,t)<\delta$ implies $d_Y(g(z),g(t))<\varepsilon$.

For $\delta>0$ $\exists \eta>0$ such that for all $x,y\in E$ and $d_E(x,y)<\eta$ implies $d_X(f(x),f(y))<\delta$.

Combining these two statements we get:

For given $\varepsilon>0$ $\exists \eta>0$ such that for all $x,y\in E$ and $d_E(x,y)<\eta$ implies $$d_Y(g(f(x)),g(f(y)))=d_Y(h(x),h(y))<\varepsilon.$$

Is this proof correct?

  • 2
    $\begingroup$ Yes, this is the standard argument. $\endgroup$ – Math1000 Sep 25 '15 at 21:36

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