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        riddles >> general problem-solving / chatting / whatever >> GRE paractice question
        
(Message started by: knightfischer on Apr 10^th, 2008, 3:56pm)

Title: GRE paractice question
Post by knightfischer on Apr 10^th, 2008, 3:56pm

Let f be a real-valued functiond efined and continuous on the set of real numbers R. Which of the following muct be true of the set S = {f(c): 0<c<1}?

I S is connected subset of R
II S is an open subset of R
III S is a bounded subset of R

Answer I and III.

Why is it bounded? couldn't the function be 1/x-1, which is unbounded as c approaches 1?

Can anyone help?

Title: Re: GRE paractice question
Post by pex on Apr 10^th, 2008, 11:29pm

on 04/10/08 at 15:56:17, knightfischer wrote:

couldn't the function be 1/x-1, which is unbounded as c approaches 1?

No; that function isn't defined (let alone continuous) at x=0.

Edit: I see you probably meant 1/(x-1). The same point now applies at x=1.

Title: Re: GRE paractice question
Post by knightfischer on Apr 11^th, 2008, 3:34am

OK, thanks.