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Continuous functions

  1. Aug 8, 2006 #1
    Let [tex]f: D \rightarrow \mathbb{R}[/tex] be continuous.

    Is there an easier function that counterexamples;
    if D is closed, then f(D) is closed
    than D={2n pi + 1/n: n in N}, f(x)=sin(x) ?????


    Plus, these counterexamples are all the same with the domain changed, just correct me if I'm wrong.

    If D is not closed, then f(D) is not closed.
    CE: D = (0, 1) and f(x) = 5
    If D is not compact, then f(D) is not compact.
    CE: We use same CE as above
    If D is infinite, then f(D) is infinite.
    CE: D = all real numbers and f(x) = 5
    If D is an interval, then f(D) is an interval
    CE: Use same CE as first
     
    Last edited: Aug 8, 2006
  2. jcsd
  3. Aug 8, 2006 #2

    berkeman

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    "If D is not closed, then f(D) is not closed.
    CE: D = (0, 1) and f(x) = 5"

    Why is D = (0,1) not closed? I think you mean more like
    CE: D = N, and f(x) = 5
     
  4. Aug 8, 2006 #3
    ???

    [tex]D \subseteq \mathbb{R}[/tex]

    an open set is one that doesn't contain it's bounds. The closure of (0, 1) is [0, 1]....umm...i'm not sure what to say...(0, 1) is an open set.
     
  5. Aug 8, 2006 #4

    berkeman

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    Fair enough, I guess I misunderstood the notation. I thought you were saying that the domain of the function was all the numbers between 0 and 1.
     
  6. Aug 8, 2006 #5
    that's exactly what i meant....am i misunderstanding something?
     
  7. Aug 9, 2006 #6
    (0, 1) is indeed (extremely) standard notation for the set {x; 0 < x < 1}, which is not closed.

    What about D = the natural numbers and f(x) = 1/x?

    But f(D) = {5} = [5, 5]. The intermediate value theorem garantuees that you can't find a counterexample.
     
    Last edited: Aug 9, 2006
  8. Aug 9, 2006 #7

    mathwonk

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    the continuous image of a compacts et is always compact, so you must use a closed set that is not bounded. say x-->1/[1+x^2] applied to the (closed) real line.

    or the exponential function applied to the real line. or arctan applied to the real line.

    etc......
     
  9. Aug 9, 2006 #8
    Oh yeah...haha..I typed that one up wrong. It was supposed to be "if D is an interval that is not open, then f(D) is an interval that is not open"

    my bad....
     
  10. Aug 9, 2006 #9
    how about D = (0, 1] and f(x) = (sin(1/x))/x
     
  11. Dec 4, 2007 #10
    That function would have f(D)=(-infinity,infinity) or, in other words, the reals. This interval is both open and closed so it is not a counterexample.
     
  12. Dec 4, 2007 #11
    I got it!

    This is for: if D is an interval that is not open, then f(D) is an interval that is not open.

    Take f:D->Reals

    f(x)=(x*sin(x))+x+(1/x)

    D=[1,infinity) which is not open

    f(D)=(0,infinity) which is open

    Woot! Woot!

    Patrick
    South Dakota State University - Real Analysis I
     
    Last edited: Dec 4, 2007
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