Register to reply 
Confusion about the definition of Uniform Continutiy 
Share this thread: 
#1
Feb1911, 12:16 PM

P: 204

A function is uniformly continuous iff for every epsilon>0 there exists delta>0 such that for all x in the domain of f and for all y in the domain of f, xy<delta =>f(x)f(y)<epsilon.
Here is what confuses me. How can there be a delta such that xy<delta for ALL x and y. Since epsilon depends on delta, we can pick epsilon such that delta is small. Then we can surely pick x and y such xy is bigger than delta. For example, x^2 is uniformly continuous on [5,5] because for epsilon>0, when delta=epsilon/10, x^2y^2<x+yxy< or = 10xy<10*delta=epsilon. Right here delta=epsilon/10. The definition states that for ANY x,y in domain f, xy<delta. If we pick x=5 and y=1 we have 4<epsilon/10 for any epsilon>0. But that is impossible since I can pick epsilon=1. 4<(1/10) is not true. 


#2
Feb1911, 01:23 PM

Sci Advisor
HW Helper
Thanks
P: 26,158

hi torquerotates!
(have a delta: δ and an epsilon: ε ) (so you have to choose xy < 1/10) 


#3
Feb1911, 02:08 PM

Sci Advisor
P: 905

You are saying: the delta should have the property that "xy<delta" holds for all x and y. That's a completely different property. The implication "if it rains tomorrow, I will be wet" holds for all days. But surely it doesn't rain every day? 


#4
Feb2111, 12:55 PM

Sci Advisor
P: 1,716

Confusion about the definition of Uniform Continutiy
if xy is less than delta then f(x)f(y) is less than epsilon. 


Register to reply 
Related Discussions  
Confusion on the definition of a quotient map  Differential Geometry  2  
Confusion with Continuity Definition  Calculus & Beyond Homework  4  
Electric flux definition confusion  General Physics  1  
Showing continutiy  Calculus & Beyond Homework  9 