Uniformly continuous functions

[Bachelier]

Question

Let (S; d) and (T;D) be metric spaces. A function f : X -> Y is said to be
uniformly continuous if ( for all epsilons > 0)(there exists a sigma > 0) such that d(x; y) < sigma => D(f(x); f(y)) < epsilon
a. Show that a uniformly continuous function maps Cauchy sequences to Cauchy seq.
b. Show that f(x) = 1/x is uniformly continuous on [1 , + infinity) but not on (0 , 1).

Answer:

a. let Sn be Cauchy.
then ( for all epsilons > 0)(there exists a sigma > 0) such that d(Sn; Sn) < sigma => D(f(Sn); f(Sn)) < epsilon
therefore f(Sn) is Cauchy since it is bounded (inside the ball with radius epsilon).

b. reason: D(f(x),f(y)) < epsilon fails on (0, 1) because
let x= 1/10^100 and y= 1/2, both are in (0,1) yet D(10^100,2) >>>>>>> epsilon

what do you guys think?

thanks

 

[LCKurtz]

Question

Let (S; d) and (T;D) be metric spaces. A function f : X -> Y is said to be
uniformly continuous if ( for all epsilons > 0)(there exists a sigma > 0) such that d(x; y) < sigma => D(f(x); f(y)) < epsilon
a. Show that a uniformly continuous function maps Cauchy sequences to Cauchy seq.
b. Show that f(x) = 1/x is uniformly continuous on [1 , + infinity) but not on (0 , 1).

Answer:

a. let Sn be Cauchy.
then ( for all epsilons > 0)(there exists a sigma > 0) such that d(Sn; Sn) < sigma => D(f(Sn); f(Sn)) < epsilon
therefore f(Sn) is Cauchy since it is bounded (inside the ball with radius epsilon).

b. reason: D(f(x),f(y)) < epsilon fails on (0, 1) because
let x= 1/10^100 and y= 1/2, both are in (0,1) yet D(10^100,2) >>>>>>> epsilon

what do you guys think?

thanks

What I think is that you need to write your general ideas up much more carefully using the appropriate definitions to make a genuine valid argument.