Scratch that, I actually have no idea how to do this problem.
Prove that f(x)=1/x is not uniformly continuous on (0, inf)?
First of all I wanted to make sure that the negation of:
for every epsilon > 0 there exists delta > 0 such that x,y element S and |x-y| => |f(x)-f(y)|< epsilon
is:
there exists epsilon > 0 such that for every delta > 0 there exists x, y element S such that |x-y|< delta and |f(x)-f(y)|>= epsilon
So here is how I started:
We can set epsilon = 1 so |1/x - 1/y|>= 1. Then |x-y|<delta so |x-y|=delta/2 which means y = x + delta/2. Then we can plug in to get
|1/x - 1/(x+delta/2)|>=1. Simplifying we get delta/2 >= x(x+delta/2).
Here I don't know what x I need to use so that the statement is true. Actually, I don't think my way is possible. I was wondering if there is another way to approach this but still use the definition.
Thanks
Let sn = 1/n for n element N. Then (sn) is cauchy in (0,inf). Since f(sn) = n, (f(sn)) is not cauchy. Therefore f cannot be uniformly continuous on (0, inf).
I think that proof works. I was just wondering how you would do a proof using the precise definition uniformly continuous and a contradiction
for each eps >0 there exists delt>0 s. t. s, y ele S and |x-y| => |f(x)-f(y)|<eps
THanks