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Let $X,Y,Z$ be subsets of $\textbf{R}$. 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 $X$.

MY ATTEMPT

Since both $g$ is uniformly continuous, for every $\varepsilon > 0$, there is a $\delta_{1} > 0$ such that for every $w,z\in Y$ we have that \begin{align*} |w - z| \leq \delta_{1} \Longrightarrow |g(w) - g(z)| \leq \varepsilon \end{align*} On the other hand, $f$ is uniformly continuous too. Then for every $\delta_{1} > 0$, there is a $\delta > 0$ such that for every $x,y\in X$ we have that \begin{align*} |x - y| \leq \delta \Longrightarrow |f(x) - f(y)| \leq \delta_{1} \end{align*}

Hence, if we make the substitution $w = f(x)$ and $z = f(y)$, we conclude that for every $\varepsilon > 0$, there is a $\delta > 0$ such that \begin{align*} |x - y| \leq \delta \Longrightarrow |g(f(x)) - g(f(y))| \leq \varepsilon \end{align*} thence we conclude that $g\circ f$ is uniformly continuous as well.

Could someone provide any comments on the proposed solutions?

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  • $\begingroup$ If function maps cauchy sequence to cauchy sequence doesn't imply function is uniformly continuous $\endgroup$
    – Math_user
    Commented May 1, 2020 at 18:48
  • $\begingroup$ Indeed, you are right. Uniformly continuous functions map Cauchy sequence onto Cauchy sequences, but the converse in not true in general. Thanks for the comment. I have edited my answer. $\endgroup$
    – user0102
    Commented May 1, 2020 at 18:52

1 Answer 1

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I would just phrase it a bit differently.

Say $\delta_g$ and $\varepsilon_g$ in the first line, and $\delta_f$ and $\varepsilon_f$ in the second.

Then you only need to consider $\varepsilon_f\leq \delta_g$ and take, as you say, $w=f(x),z=f(y)$ to conclude.

I'm just pointing out the fact that the $\epsilon_f$ has to meet some requirements in order for $g$ to do as you desire. You can't say "for every $\delta$" everywhere because it's not really true. They are related.

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