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Topic: GRE paractice question (Read 447 times) |
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knightfischer
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GRE paractice question
« on: Apr 10th, 2008, 3:56pm » |
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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?
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pex
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Re: GRE paractice question
« Reply #1 on: Apr 10th, 2008, 11:29pm » |
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on Apr 10th, 2008, 3:56pm, knightfischer wrote:| couldn't the function be 1/x-1, which is unbounded as c approaches 1? |
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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.
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| « Last Edit: Apr 10th, 2008, 11:30pm by pex » |
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knightfischer
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Re: GRE paractice question
« Reply #2 on: Apr 11th, 2008, 3:34am » |
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OK, thanks.
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