# Thread: prove that the composition gof:A-> C is uniformly continuous

1. ## prove that the composition gof:A-> C is uniformly continuous

Let A,B,C be nonempty sets of real numbers, and let f:A->B and g:B->C be uniformly continuous functions. Prove that the composition gof:A->C is also unfiormly continuous,

Here's what I have. Is it correct?

Let$\displaystyle \epsilon >0$ then there is $\displaystyle \delta>0$ s.t for all x1,y1 $\displaystyle \in A$ if $\displaystyle |x_1-y_1|<\delta$ then |$\displaystyle f(x_1)-f(y_1)|<\epsilon$ so f(x1),f(y1)$\displaystyle \in$B. Now let $\displaystyle \epsilon$>0 then there exists $\displaystyle \delta$>0 s.t for all x2,y2$\displaystyle \in$B if |x2-y2|<$\displaystyle \delta$ then |g(x2)-g(y2)|<$\displaystyle \epsilon$. So |(gof)(x1)-(gof)(y1)|=|g(f(x1))-g(f(y1))|<$\displaystyle \epsilon$.

2. No, that won't do. You've given the definitions of uniform continuity for f and g, but you haven't actually shown how they combine to produce the end result.

Let $\displaystyle \epsilon_1>0$. The uniform continuity of $\displaystyle g$ gives you a special $\displaystyle \delta_1>0$. Then let $\displaystyle \epsilon_2=\delta_1$, and the uniform continuity of $\displaystyle f$ gives you $\displaystyle \delta_2>0$. Show that if $\displaystyle |x-y|<\delta_2$ then $\displaystyle |(g\circ f)(x)-(g\circ f)(y)|<\epsilon_1$.

3. Originally Posted by alice8675309
Let A,B,C be nonempty sets of real numbers, and let f:A->B and g:B->C be uniformly continuous functions. Prove that the composition gof:A->C is also unfiormly continuous
Start with $\displaystyle \epsilon>0$ pick a $\displaystyle \delta>0$ for $\displaystyle \mathbf{g}$. Then pick a $\displaystyle \gamma$ with respect to $\displaystyle \delta$ for $\displaystyle \mathbf{f}$

PS. I did not see reply #2 before posting.
BUT you need to work on your LaTeX
I tried to edit your post. Double click on the code to see how it works.
In particular, how subscripts work.

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# f,g are uniform continuous then its composition is

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