Prove that a uniformly continuous function of a uniformly continuous function is uniformly continuous.

In other words, is it saying that for a given uniformly continuous function $f$ and $g$ such that $f: X \to Y$ and $g: Y\to Z$ then $g \circ f$ is also uniformly continuous?

If it is then since $f$ is uniformly continuous then for $\delta >0$ and for all $x,y$ we have $|x-y| < \delta$ implies $|f(x) - f(y)| <\epsilon$. And for $\delta'>0$ we have for all $y,z$ $|y-z|<\delta'$ then $|g(y)-g(z)|<\epsilon$. So now we have to prove that $g\circ f$ is uniformly continuous.

So I have to prove $|g(f(x)) - g(f(y))| <\epsilon$. How can I show that? I know that I can make $f(x) = y$ since it is defined there but can I make $f(y) = z$ even though it is not defined for $z$? I don't think I can.


Fix $\varepsilon>0$. Since $g$ is uniformly continuous there is a $\delta'>0$ such that $|g(x')-g(y')|<\varepsilon$ whenever $|x'-y'|<\delta'$. But since $f$ is uniformly continuous as well, there is a $\delta>0$ such that $|f(x)-f(y)|<\delta'$ whenever $|x-y|<\delta$. By putting $x'=f(x)$ and $y'=f(y)$ we conclude: $|g(f(x))-g(f(y))|<\varepsilon$ whenever $|x-y|<\delta$, and this shows, that even $g\circ f$ is uniformly continuous.

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  • $\begingroup$ So $y'$ is an element of $Z$? $\endgroup$ – Tom Oct 28 '13 at 4:45
  • $\begingroup$ No, $x,y\in X$ and $x',y'\in Y$. $\endgroup$ – sranthrop Oct 28 '13 at 11:23

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