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.