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Prove or disprove: If $f$ and $g$ are uniformly continuous on $\Bbb{R}$ then $f\circ g$ is uniformly continuous on $\Bbb{R}$. I think there's something crooked in my attempt. I would like to know what it is and would appreciate your replies.

$Attempt:$ Let $\epsilon>0$. $f$ is uniformly continuous and therefore $\exists \delta>0$ such that $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta$, $|f(x)-f(y)|<\epsilon$. Since $g$ is uniformly continuous, for this $\delta$ there exists $\delta _1$ such that $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_1$, $|g(x)-g(y)|<\delta$. Taking $\delta_{f\circ g}=\min\{\delta, \delta_1\}$, we get that $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_{f\circ g}, |f(x)-f(y)|<\epsilon $ and $|g(x)-g(y)|<\delta$. Therefore $|f(g(x))-f(g(y))|<\epsilon$ $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_{f\circ g}$, proving $f\circ g$ is uniformly continuous on $\Bbb{R}$.

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    $\begingroup$ The only crooked thing I see is that you misspelt "attempt". $\endgroup$ – David Mitra Jan 28 '15 at 23:32
  • $\begingroup$ Oh, Too tired(and not native lol). Thank you for your evaluation. $\endgroup$ – Donna Jan 28 '15 at 23:33
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The proof is basically good but this part is a bit hard to follow and possibly misleading as regarding what the argument is.

Taking $\delta_{f\circ g}=\min\{\delta, \delta_1\}$, we get that $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_{f\circ g}, |f(x)-f(y)|<\epsilon $ and $|g(x)-g(y)|<\delta$.

Also, you do not need the $\min$. The following works:

We get $\forall x,y\in \Bbb{R}$ fulfilling $|x-y|<\delta_1$ that $|g(x)-g(y)|<\delta$, and thus $|f(g(x))-f(g(y))|<\epsilon $.

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  • $\begingroup$ This is what made me so confused and scattered... Thank you. $\endgroup$ – Donna Jan 28 '15 at 23:48
  • $\begingroup$ Glad it was helpful. I also had to look twice to note that the $\min$ is not necessary. $\endgroup$ – quid Jan 28 '15 at 23:49

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