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- Apr 13, 2013

- 3,718

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?

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?

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