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Continuous Mappings Part 2

  1. Feb 16, 2006 #1

    JasonRox

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    Can there exist a continuous function from the real numbers to the rationals?

    Where we are considering the usually topology for the real numbers and the relative topology for the rationals with respect to the topology of real numbers.

    At first my intuition said no. Quickly going through random functions in my head, I couldn't create one.

    Is my intuition right?

    I've been thinking about a contradiction if a continuous function did exist, but can't find one that seals the deal.

    Can someone atleast confirm or correct my intuition without giving a solution?

    Hopefully I get a solution before tomorrow night. I'll have time at work to think about it some more, and try to use different approaches. I'll try looking at invariants and what not.
     
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  3. Feb 17, 2006 #2

    AKG

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    I suppose you mean a continuous function from the reals onto the rationals. The answer is "no." What are some important topological properties that the reals have and that the rationals don't?
     
  4. Feb 17, 2006 #3

    JasonRox

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    Yeah, I got it using the limit point definition of continuity.

    I'm reading two different books where one uses open sets as the main focus and the other using limit points.

    I used the idea that atleast one of the points in the rationals will have an inverse image of an uncountable number of elements in R. It must exist. And since that one point is closed, then the inverse image must also be closed by definition of continuity. Then I went from there.
     
  5. Feb 17, 2006 #4

    AKG

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    Where did you go from there? I was thinking about using connectedness.
     
  6. Feb 17, 2006 #5

    matt grime

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    Yep, me too, connectedness seemed to be the important thing, nay the only thing that sprang to mind, and I too would like to see the sequential proof.
     
  7. Feb 17, 2006 #6

    JasonRox

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    Here is the proof, if it's valid.

    Now assume a continuous function does exist.

    We know the inverse image of an element in Q (call it p) will map to an uncountable number of elements.

    Since f is continuous and p is closed, then f^-1(p) is closed. Now, f^-1(p) has uncountable many elements, and it is closed (call it set S in R). Now, we know that S will contain a limit point call it x. By definition of continuity, if x is a limit point of S, then f(x) is a limit point of f. But f(x)=p and f={p}, but p isn't a limit point to the set {p}, which contradicts the fact that f is continuous.

    Note: If it's not making any sense, let me know. I'll write it more formally.

    It could be false, but it seems fine.

    Note: I guess what is left to prove is that a set with uncountable many elements in R has atleast one limit point. Now, if the set (call it S) did not have a limit point, then for some interval for every point in S, there is a distance e such that an interval about that point where it does not contain a point in S. Take the minimum e amongst all the elements. So, all the elements are separated by atleast that much. That makes it countable now, contradicting the fact that it was uncountable.
     
    Last edited: Feb 18, 2006
  8. Feb 18, 2006 #7
    Hey Jason about your proof that there is a limit point. You must take into account that the points may repeat. In fact you may have points repeating for an entire sequence without having limit points. For example

    1,1,2,2,3,3,4,4,...

    Then your distance between points would be 0, but you'd have no limit point. Also you wouldn't be able to throw away any finite number of points to make your process work.

    What you're trying to prove though is still true. Remember all you need to prove is that there is some bounded subset of the real numbers which contains an infinite (countable suffices) number of elements from f^-1(p). Consider if there is no such bounded subset (ie ever bounded subset contains a finite number of elements from f^-1(p)) and what that would mean about f^-1(p).
     
  9. Feb 18, 2006 #8

    matt grime

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    Sorry it's not. In fact your proof indicates that there can never be a continuous map from the reals to any one point set, which surely you can see is wrong.

    this is unnecessary, just let f be a map and show f cannot be continuous, there is no need to go for a contradiction here.



    p is a limit point, the only one, of {p}, sequences are allowed to be constant. I think you're confusing this with subsequences where you can't pick the same element of the sequence repeatedly, ie x_1,x_1,x_1,... is not a subsequence of x_1,x_2,x_3,.....


    Note also how you say '{p} is closed' and then that {p} has no limit points. I thought your preferred definition of closed was that it contained all its limit points.
     
    Last edited: Feb 18, 2006
  10. Feb 18, 2006 #9

    JasonRox

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    Yeah, big mistake. :redface:

    Something else now. Using connectedness mentionned earlier.

    By construction of the topology of the rationals, it will contain open and closed sets. This is what I "see".

    Let a and b be irrational.

    The interval [itex](a,b) \cap Q[/itex] is an open set in Q call it S.

    The interval [itex][a,b] \cap Q[/itex] is a closed set in Q, which is precisely the set S from above. Therefore, S is closed and open.

    Is that right?

    That means Q is not connected.
     
    Last edited: Feb 18, 2006
  11. Feb 18, 2006 #10

    JasonRox

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    I'm not getting what your saying here.
     
  12. Feb 18, 2006 #11

    matt grime

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    Q is not connected because, well, it isn't. The open sets S={ s in Q : s<sqrt(2)} and T={t in Q :t> sqrt(2)} disconnect it. As do uncountably many other choices of irrational number.
     
  13. Feb 18, 2006 #12

    JasonRox

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    Yes, but is my example valid?
     
  14. Feb 19, 2006 #13

    matt grime

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    What do you think? Have you produced two open non-empty disjoint sets?
     
  15. Feb 19, 2006 #14
    You said to take "e" as being the minimum width of the intervals. I'm saying that that distance can be 0 even without having a limit point. Then you can't use that to conclude it's countable.
     
  16. Feb 19, 2006 #15

    matt grime

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    One more thing: you can't take the 'minimum of the e' since that will not exist. the inf will, but it is not guaranteed to be any of the e and there is nothing to suppose that the inf is not zero.
     
  17. Feb 19, 2006 #16

    JasonRox

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    Finding an open and closed set within the set other than Q and the null set is basically finding two open sets that are disjoint.

    So, I would say yes I did find two open sets that are disjoint.
     
  18. Feb 19, 2006 #17

    JasonRox

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    I see what's going on now.

    Good thing I posted the question and "proof".
     
  19. Feb 21, 2006 #18

    JasonRox

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    At first this got me confused, but in fact I am right.

    {p} does not have a limit point.

    Sequences can be constant, but can not "finish" as a contant of the point p.

    An equivalent statement is that every neighbourhood of p has a point in the set A pther than p. For the set {p}, that is impossible.

    http://mathworld.wolfram.com/LimitPoint.html
     
  20. Feb 21, 2006 #19

    JasonRox

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    I almost forgot.

    As an added note, any finite subset of a T1 space does not contain any limit points.
     
  21. Feb 21, 2006 #20

    mathwonk

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    do you know the intermediate value theoirem? it syas that the continuous image of any intervalk is another interval. so a continuous map from the interval of all reals to the rsati0onals would have image an interval of the reals which contains only rationals. what are msuch intervals?
     
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