confusion about the definition of Uniform Continutiy

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|<delta =>|f(x)-f(y)|<epsilon.


Here is what confuses me. How can there be a delta such that |x-y|<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 x-y is bigger than delta.

For example, x^2 is uniformly continuous on [-5,5] because for epsilon>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|<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.

tiny-tim

hi torquerotates!

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

(so you have to choose |x-y| < 1/10)

Landau

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?

lavinia

This line can be read

if |x-y| is less than delta then |f(x)-f(y)| is less than epsilon.