Real Analysis Proof

**fifthrapiers** · November 9th 2007, 07:34 AM

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

**ThePerfectHacker** · November 9th 2007, 07:37 AM

It is wrong. Becauase $y\not = f(x)$ , $y$ is some other point in the set. So $|x-y|<\delta$ are all points in the set so that this is true and not $|x-f(x)|<\delta$ .

**fifthrapiers** · November 9th 2007, 07:43 AM

Originally Posted by ThePerfectHacker

It is wrong. Becauase $y\not = f(x)$ , $y$ is some other point in the set. So $|x-y|<\delta$ are all points in the set so that this is true and not $|x-f(x)|<\delta$ .

Ok, change all the f(x)'s to y's. Then, I think, it is correct.

**fifthrapiers** · November 9th 2007, 07:59 AM

TPH, how does it look now? Attached the new one.

**ThePerfectHacker** · November 9th 2007, 08:32 AM

Good proof.

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

Thread: Real Analysis Proof

LinkBack

Thread Tools

Display