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Hello forumites! I got another question (I have a lots of them, maybe

  1. Jun 11, 2010 #1
    Hello forumites! I got another question (I have a lots of them, maybe I should change my textbook haha).

    Suppose you have a continuous real-valued function f on the metric space X and a closed set E in X. Is f(E) closed in the reals? By definition, I know that if f(A) is closed, so is the inverse image of f(A) for a set A in X if f is continuous. But if we know that E is closed, would it be correct to say that f(E) is too?

    Thanks a lot!
     
  2. jcsd
  3. Jun 11, 2010 #2
    Re: Continuity

    No, let X = [0, 1), E = X, and f:X->R be the identity function. Then E is closed, but f(E) = [0, 1) as a subset of R is not closed.
     
  4. Jun 11, 2010 #3
    Re: Continuity

    I could be wrong, but suppose the sequence 0.9, 0.99, 0.999, etc. Then this is a sequence of elements of E that does not converge to an element of E. Thus E isn't closed.
     
  5. Jun 11, 2010 #4
    Re: Continuity

    Let me try something (and I'm surely wrong again) : Suppose a continuous real-valued function on a metric space X and a closed set E in X. Then there is a set V in the reals such that V=f(E) (Because f maps every points in X). If V is open, so is the inverse image of V, that is, E. But E is closed so V must be closed.
     
    Last edited: Jun 11, 2010
  6. Jun 11, 2010 #5
    Re: Continuity

    You are forgetting that sets can be neither open nor closed. Take this as your counterexample. X = R. E = [1,∞). f(x) = 1/x. Then, f(E) = (0,1] is not closed. However, it is true that the image of a compact set over a continuous function is compact.
     
  7. Jun 11, 2010 #6
    Re: Continuity

    No, that is "completeness" you are describing.

    "Closed" just requires for any *convergent* sequence in X such that every element of the sequence is in E, it's limit must also be in E. But in the example I gave 1 was not in the metric space [0, 1), therefore that's not a convergent sequence. (Similar to your example, the sequence 9, 99, 999, 9999, ... does not converge in the reals, but the entire real line is still a closed set).
     
  8. Jun 11, 2010 #7
    Re: Continuity

    Isn't that's weird? That the metric space is't closed but that a subset with the exact same points is? Is it because we define "closure" in respect to a specific metric space instead of, for exemple, the reals (in our case)?
     
  9. Jun 11, 2010 #8
    Re: Continuity

    Well in your exemple a "neither open nor closed" is mapped into a "neither open nor closed" set. But I'm talking about a mapping from a closed set into another one (where we don't know if it is open, closed or neither). I'm sorry if I just didn't understand where you were trying to get!
     
  10. Jun 11, 2010 #9
    Re: Continuity

    No! The metric space [0, 1) is closed! It's the entire space, so it must be open AND closed. You can also see this by applying any of the definitions of closure in a metric space.
     
  11. Jun 11, 2010 #10
    Re: Continuity

    Regarding some_dude's example: I didn't even notice that. Most of the analysis I've done deals with complete metric spaces, so I took it for granted. But the example is valid. As for my example, the set [1,∞) is closed in R.
     
  12. Jun 11, 2010 #11
    Re: Continuity

    _DJ_british_?, try playing around with metrics in more generality, you're taking for granted the stuff that holds for the reals holds everywhere.

    I could have also given you the counter example:

    (R, d) metric space where d(x, y) = 1 if x = y, and d(x, y) = 0 otherwise. Then let f be the identity function. Then f( (-1, 1) ) = (-1, 1), an open and NOT closed set, but (-1, 1) is closed (and open) with respect to the metric I gave above.
     
  13. Jun 11, 2010 #12
    Re: Continuity

    Gah! The last three posts were very insightful! That cleared that up for me. Many thanks guy! And yeah, I may have a bit of difficulty trying to see the reals only as a specific metric space, since I've used them for so long. I'll get used to it, I guess. Anyway, thank you guys.
     
  14. Jun 12, 2010 #13

    Landau

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    Science Advisor

    Re: Continuity

    Openness and closedness are actually relative concepts.
    [0,1) is closed in [0,1), but not closed in R, nor in [0,1].
    Every metric (topological) space is both closed and open in itself.
     
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