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Is con't fn maps compact sets to compact sets converse true?

  1. Oct 2, 2008 #1
    Is "con't fn maps compact sets to compact sets" converse true?

    The question is here,
    Suppose that the image of the set S under the continuous map f: s belongs to R^n ->R is compact, does it follow that the set S is compact? Justify your ans.

    I already know how to prove the original thm, it requires us using another thm: Given S belongs to R^n, a belongs to S, and f: S->R^m, the following are equivalent:
    a. f is con't at a.
    b. For any {x_k}sequence in S that converges to a, the sequence {f(x_k)} converges to f(a).

    If I need to prove the question on the top, I have to get the converse of this thm first.

    And I see someone post a similar question before, please take a look as well:

    "That f is continuous and that there is a continuous inverse, g, say.

    So all we're doing is using the more basic fact that the continuous image of a compact set is compact.

    Ie K compact implies f(K) compact, and f(K) compact imples gf(K)=K is compact."

    Here, I agree this method, but I doubt this is not enough to prove my question, isn't it?
     
  2. jcsd
  3. Oct 3, 2008 #2
    Re: Is "con't fn maps compact sets to compact sets" converse true?

    Indeed it is not enough to prove your assertion, as f may not have a continuous inverse. It may not have an inverse at all!

    Do you think the converse is true or false? Have you tried looking for counter-examples?
     
  4. Oct 3, 2008 #3

    HallsofIvy

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    Re: Is "con't fn maps compact sets to compact sets" converse true?

    Think about the function f(x)= 0 for x any member of Rn.
     
  5. Oct 3, 2008 #4
    Re: Is "con't fn maps compact sets to compact sets" converse true?

    you are right, but i still dont 100% get it.

    my professor gave me a similar example, f(x)=c.

    as you said, x can be any number, assume S={x_k}, f(S)=c, right?

    but if a set is made up by the points on a line, say y=c, then this is not compact because it's not closed?
    i am not sure here.
    if the set of points on a line is counted to be a compact set, then you are right cuz {x_k} is not compact.
     
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