Show that gof is uniformly continuous

[evinda]

Hi! :)
I am given the following exercise:
$f:A \to B,g:B \to R$
If $f$ is uniformly continuous at $A$ and $g$ is uniformly continuous at $B$,show that gof is uniformly continuous.
That's what I have tried so far:
Let $\epsilon'>0$.Since $f$ is uniformly continuous at $A$ there is a $\delta=\delta(\epsilon')>0$ such that $\forall x,y \in A$ with $|x-y|<\delta \Rightarrow |f(x)-f(y)|< \epsilon'$.

Let $\epsilon>0$.Since $g$ is uniformly continuous at $B$,there is a $w=w(\epsilon)>0$ such that $\forall u,v \in B$ with $|u-v|<w \Rightarrow |g(u)-g(v)|< \epsilon$.

Am I right so far?How can I continue?

 

[ThePerfectHacker]

Hi! :)
I am given the following exercise:
$f:A \to B,g:B \to R$
If $f$ is uniformly continuous at $A$ and $g$ is uniformly continuous at $B$,show that gof is uniformly continuous.
That's what I have tried so far:
Let $\epsilon'>0$.Since $f$ is uniformly continuous at $A$ there is a $\delta=\delta(\epsilon')>0$ such that $\forall x,y \in A$ with $|x-y|<\delta \Rightarrow |f(x)-f(y)|< \epsilon'$.

Let $\epsilon>0$.Since $g$ is uniformly continuous at $B$,there is a $w=w(\epsilon)>0$ such that $\forall u,v \in B$ with $|u-v|<\epsilon \Rightarrow |g(u)-g(v)|< \epsilon$.

Am I right so far?How can I continue?

First you mean to say $f$ is uniformly continuous on $A$, $g$ is uniformly continuous on $B$.

Let $h=g\circ f: A\to \mathbb{R}$. We want to show this is uniformly continuous. What does the definition say? For any $\varepsilon > 0$ we can find $\delta > 0$ so that if $|x-y|<\delta \text{ and }x,y \in A$ then $|h(x)-h(y)| < \varepsilon $.

Now $|h(x)-h(y)| = |f(g(x)) - f(g(y))|$. Now the idea is that since $f$ is uniformly continuous if $g(x),g(y)$ are sufficiently close to each other then $|f(g(x)) - f(g(y))|$ can be made arbitrary small. So the question is how do you mean $g(x),g(y)$ close to each other? This is where you start to use uniform continuity of $g$.

 

[evinda]

First you mean to say $f$ is uniformly continuous on $A$, $g$ is uniformly continuous on $B$.

Let $h=g\circ f: A\to \mathbb{R}$. We want to show this is uniformly continuous. What does the definition say? For any $\varepsilon > 0$ we can find $\delta > 0$ so that if $|x-y|<\delta \text{ and }x,y \in A$ then $|h(x)-h(y)| < \varepsilon $.

Now $|h(x)-h(y)| = |f(g(x)) - f(g(y))|$. Now the idea is that since $f$ is uniformly continuous if $g(x),g(y)$ are sufficiently close to each other then $|f(g(x)) - f(g(y))|$ can be made arbitrary small. So the question is how do you mean $g(x),g(y)$ close to each other? This is where you start to use uniform continuity of $g$.

So,can I say it like that?

Let $\epsilon'>0$.Since $f$ is uniformly continuous at $A$ there is a $\delta=\delta(\epsilon')>0$ such that $\forall x,y \in A$ with $|x-y|<\delta \Rightarrow |f(x)-f(y)|< \epsilon'$.

Let $\epsilon>0$.Since $g$ is uniformly continuous at $B$,there is a $w=w(\epsilon)>0$ such that $\forall u,v \in B$ with $|u-v|<w \Rightarrow |g(u)-g(v)|< \epsilon$.

We want to show that $|h(x)-h(y)|=|g(f(x))-g(f(y))|< \epsilon$,and this is true,if $u=f(x),v=f(y),w=\epsilon'$?? Or do I have to say something else to show that gof is uniformly continuous?

 

[ThePerfectHacker]

So,can I say it like that?

Let $\epsilon'>0$.Since $f$ is uniformly continuous at $A$ there is a $\delta=\delta(\epsilon')>0$ such that $\forall x,y \in A$ with $|x-y|<\delta \Rightarrow |f(x)-f(y)|< \epsilon'$.

Let $\epsilon>0$.Since $g$ is uniformly continuous at $B$,there is a $w=w(\epsilon)>0$ such that $\forall u,v \in B$ with $|u-v|<w \Rightarrow |g(u)-g(v)|< \epsilon$.

We want to show that $|h(x)-h(y)|=|g(f(x))-g(f(y))|< \epsilon$,and this is true,if $u=f(x),v=f(y),w=\epsilon'$?? Or do I have to say something else to show that gof is uniformly continuous?

You basically got it. Here is a more polished version of what you are trying to say.

Given $\varepsilon > 0$ choose $\delta > 0$ so that $|g(a) - g(b)| < \varepsilon $ provided that $|a-b| < \delta$. Now choose $\eta > 0$ so that $|f(x) - f(y)| < \delta$ provided that $|x-y| < \eta$. If $|x-y|<\eta$ then $|f(x)-f(y)| < \delta$ and then $|g(f(x)) - g(f(y)) | < \varepsilon$.

 

[evinda]

You basically got it. Here is a more polished version of what you are trying to say.

Given $\varepsilon > 0$ choose $\delta > 0$ so that $|g(a) - g(b)| < \varepsilon $ provided that $|a-b| < \delta$. Now choose $\eta > 0$ so that $|f(x) - f(y)| < \delta$ provided that $|x-y| < \eta$. If $|x-y|<\eta$ then $|f(x)-f(y)| < \delta$ and then $|g(f(x)) - g(f(y)) | < \varepsilon$.

I understand... :) Thank you very much!