I've been asked to prove that if $f$,$g$ are uniformly continuous functions, then their composition $g\circ f$ is also uniformly continuous.
This is my attempt:
Since $f$ and $g$ are uniformly continuous, we have,
For a given $\epsilon_1>0$, $\exists\ \delta_1>0$ such that $|x_1-x_2|< \delta_1\implies|f(x_1)-f(x_2)|< \epsilon_1$ (1)
Similarly, For a given $\epsilon_2>0$, $\exists\ \delta_2>0$ such that $|x_1-x_2|< \delta_2\implies|g(x_1)-g(x_2)|< \epsilon_2$ (2)
Now let, $\epsilon= \min(\epsilon_1,\epsilon_2)$ and let $\delta=\min(\delta_1,\delta_2)$
Then (1) becomes
For a given $\epsilon>0$, $\exists\ \delta>0$ such that $|x_1-x_2|< \delta \implies |f(x_1)-f(x_2)|< \epsilon$
Let $\delta=\epsilon$,for a particular choice of $\epsilon$
Then we have $|f(x_1)-f(x_2)|<\delta$ for some $\epsilon>0$ (3)
Then by (2) and (3) we have,
Hence $g\circ f$ is uniformly continuous
Any advice is welcomed!!