# Real Analysis Proof

• Nov 9th 2007, 07:34 AM
fifthrapiers
Real Analysis Proof
Can someone check my work? I've attached it. Thanks!
• Nov 9th 2007, 07:37 AM
ThePerfectHacker
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 AM
fifthrapiers
Quote:

Originally Posted by ThePerfectHacker
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 $\displaystyle |x-f(x)|<\delta$.

Ok, change all the f(x)'s to y's. Then, I think, it is correct.
• Nov 9th 2007, 07:59 AM
fifthrapiers
TPH, how does it look now? Attached the new one.
• Nov 9th 2007, 08:32 AM
ThePerfectHacker
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.