Let $X, Y, Z$ be subsets of $\mathbb R$. Let $f : X \rightarrow Y$ be a function which is uniformly continuous on $X$, and let $g : Y \rightarrow Z$ be a function which is uniformly continuous on Y. Show that the function $g \circ f : X \rightarrow Z$ is uniformly continuous on $X$.

  • $\begingroup$ What have you tried? As a tip: the proof is very similar to proving the same thing of continuous functions. $\endgroup$ – J. David Taylor May 5 '16 at 6:39

We want to find $\delta$ such that for any $x,y\in X$ and $\epsilon>0$, $|x-y|<\delta$ implies $|g\circ f(x)-g\circ f(y)|<\epsilon$.

But since $g$ is uniformly continuous on $Y$, there is $\delta_1$ such that if two points in $Y$ (which are equal to some $f(x),f(y)$ where $x,y\in X$) satisfy $|f(x)-f(y)|<\delta$, then $|g(f(x))-g(f(y))|<\epsilon$.

Now we need to make sure that $f(x)$ and $f(y)$ are sufficiently close; I'll let you fill in the details.

Overview: we can pick $x,y\in X$ making $f(x),f(y)\in Y$ sufficiently close to make $g(f(x)), g(f(y))$ arbitrarily close.

  • $\begingroup$ not sure i am stuck $\endgroup$ – Ben May 5 '16 at 18:17

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