Can someone check my work? I've attached it. Thanks!

Printable View

- Nov 9th 2007, 07:34 AMfifthrapiersReal Analysis Proof
Can someone check my work? I've attached it. Thanks!

- Nov 9th 2007, 07:37 AMThePerfectHacker
It is wrong. Becauase $\displaystyle y\not = f(x)$, $\displaystyle y$ is some other point in the set. So $\displaystyle |x-y|<\delta$ are all points in the set so that this is true and

**not**$\displaystyle |x-f(x)|<\delta$. - Nov 9th 2007, 07:43 AMfifthrapiers
- Nov 9th 2007, 07:59 AMfifthrapiers
TPH, how does it look now? Attached the new one.

- Nov 9th 2007, 08:32 AMThePerfectHacker
Good proof.

In the second part where you show $\displaystyle (0,1]$ is not unfiormly continous, trying using the definition, i.e. show (without Cauchy sequences) that you can violate the definition of uniform continuity.