Hi all

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- April 1st 2008, 09:03 AMglen-cullenAlgebra
Hi all

- April 1st 2008, 10:17 AMteuthid
It's been a long time since I did anything like this, but I'll give it a try. I'm confident in the basic approach of the proof, but I feel like there might be something missing between the first and second step...

Let be a limit point of gr(f).

because f(x) is continuous.

gr(f) is a closed subset of - April 1st 2008, 12:57 PMOpalg
See this thread.

- April 1st 2008, 01:06 PMglen-cullenalgebra
Thank you again for your help

- April 1st 2008, 02:24 PMPlato
Suppose that that implies .

Let . From continuity we get

As a sort of lemma, any point in the ball centered at (a,b), .

Can you show that? Hint: .

So suppose some point of the graph is such that .

Can you explain the following contradiction?

That shows that the complement of the is open. - April 2nd 2008, 07:00 AMglen-cullenalgebra
Thank you Plato so much for your help. I understand that method much better. Thanks again