Question

Let \(X, Y, Z\) be metric spaces. \(f: X \rightarrow Y\) is called uniformly continuous if
\(\forall \epsilon > 0 \quad \exists \delta > 0 \quad \forall x, y \in X : d_{x}(x,y) \leq \delta \Rightarrow d_{Y}(f(x),f(y))\leq \epsilon \)
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a) \(f:X \rightarrow Y\) \(and\) \(g:Y\rightarrow Z \)are uniformly continuous \(\Rightarrow g o f : X \rightarrow Z\) is uniformly continuous

Answer 1

i think the profe is same as for continuity of gof

Answer 2

thx Myko i am not so good at Math and i need complete solution, actually i am late by delivering this homework, i am even not sure if my mentor will accept it because i am late... but any help would be apreciated.

Answer 3

et e > 0, then there is d1> 0 so that
1) d_Y(u,v) < d1 implies d_Z( g(u), g(v)) < e for all u and v
there is d> 0 such that d_X( x,y) < d implies that d_Y( f(x)) ,f(y)) <d1
by 1) Take u =f(x) and v=f(y)
to conclude that
d_X( x,y) < d implies that d_Y( f(x)) ,f(y)) <d1 implies d_Z( g(f(x)), g(f(y))) < e

Answer 4

e is suppose to be \[\epsilon\] right?

Answer 5

yes

Answer 6

ok Mr Elias and this end solution right or additional things need to be add?

Answer 7

the d's are also
\[
\delta's
\]

Answer 8

No That is it.

Answer 9

thank you very much Mr Elias you saved my life this week, here there is not much people who can solve this kind of question. thank you

Answer 10

Did you see my proof that x^2 is not uniformly continuous on R?

Answer 11

yw

Answer 12

You mean by my other question you solved before ?

Answer 13

I solve 30 minutes ago.

Answer 14

yes i ve..

Answer 15

sorry i have seen now what you exactly mean, i think your answer is better but i have already given best answer for that it woulnt be ethic to undo it but thanks anyway very much..