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Set theory/ topology question

  1. Mar 16, 2004 #1
    Let f be a real-valued function defined and continuous on the set of real numbers R. Which of the following must be true of the set S = {f(c): 0<c<1}?

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

    The answer is I and III only. I understand why I is true. But, why is is bounded, and why is it not an open subset?

  2. jcsd
  3. Mar 16, 2004 #2

    matt grime

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    it is bounded because it is continuous on the compact set [0,1] and the continuous image of a compact (closed and bounded set) is closed and bounded, the image of (0,1) is a subsert of this bounnded set and is hence bounded.

    define f(x) = 0 for all x. the image of an open set is then closed.
  4. Mar 16, 2004 #3
    How can you reach a conclusion by simply considering the case f(x) = 0?
  5. Mar 16, 2004 #4

    matt grime

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    Because the question asks if it MUST be true that the image of an open set is open. I just showed that it isn't necessarily true. To disprove a statement it suffices to provide ONE counter example.

    The negation of the statement 'for all continuous f (on R) the restriction to (0,1) is an open map (ie the image is open)' is 'there exists A continuous map on r R whose restriction to (0,1) is not an open map'.
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