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The relationship between con't function and a compact set

  1. Oct 3, 2008 #1
    suppose f:R^m -> R^n is a map such that for any compact set K in R^n, the preimage set f^(-1) (K)={x in R^m: f(x) in K} is compact, is f necessary continuous? justify.

    The answer is no.
    given a counterexample,

    function f:R->R

    f(x):= log/x/ if x is not equal to 0
    f(x):= 0 if x=0

    note, /x/ is the absolute value of x.


    I dont quite get how to draw the image log/x/
    and anyone can explain why ?
     
  2. jcsd
  3. Oct 4, 2008 #2

    morphism

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    What exactly are you having trouble with?
     
  4. Oct 4, 2008 #3
    let me try to plug in some number to the fn in the solution tomorrow...too late tonight, gonna sleep.. thx for asking :)
     
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