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

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

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 $\epsilon>0$ pick a $\delta>0$ for $\mathbf{g}$. Then pick a $\gamma$ with respect to $\delta$ for $\mathbf{f}$

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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