# confusion about the definition of Uniform Continutiy

by torquerotates
Tags: confusion, continutiy, definition, uniform
 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, |x-y||f(x)-f(y)|0, when delta=epsilon/10, |x^2-y^2|<|x+y||x-y|< or = 10|x-y|<10*delta=epsilon. Right here delta=epsilon/10. The definition states that for ANY x,y in domain f, |x-y|0. But that is impossible since I can pick epsilon=1. 4<(1/10) is not true.
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P: 25,514
hi torquerotates!

(have a delta: δ and an epsilon: ε )
 Quote by torquerotates 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, |x-y||f(x)-f(y)|
no, it states that if you choose |x-y| < δ, then |x2 - y2| < ε

(so you have to choose |x-y| < 1/10)
P: 904
 Quote by torquerotates 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, |x-y||f(x)-f(y)|
You should read the definition carefully. The delta should have the property that the implication "if |x-y|<delta then |f(x)-f(y)|<epsilon" holds for all x and y.
You are saying: the delta should have the property that "|x-y|<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?