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Infinite metric space

  1. Jul 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Let M be an infinite metric space. Prove that M contains an open set U s.t. both U and its complement are infinite.


    2. Relevant equations



    3. The attempt at a solution

    For Euclidean spaces it is easy. You take (among other sets) [tex] R^{+} [/tex]. However, I do not think that you have that +/- directionality in general so that does not work.
     
    Last edited: Jul 29, 2007
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  3. Jul 29, 2007 #2

    morphism

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    Here's an idea. I haven't tried it out, but I think it works. Consider the collection [itex]\mathcal{C}[/itex] of all triples (U, V, f) such that U is open in X, V is a subset of X that is disjoint from U, and f:U->V is a bijection. Show that [itex]\mathcal{C}[/itex] is nonempty. Now partially order [itex]\mathcal{C}[/itex] by 'extension', i.e. (U1, V1, f1) <= (U2, V2, f2) iff U1 [itex]\subset[/itex] U2, V1 [itex]\subset[/itex] V2, and f2 restricted to U1 is f1. Then show that any chain in [itex]\mathcal{C}[/itex] has an upperbound, and so by Zorn's lemma [itex]\mathcal{C}[/itex] has a maximal element (U*, V*, f*). Show that U* is the open set we're after.
     
  4. Jul 29, 2007 #3
    Take a point P and take and let B(R,P) denote the open ball of radius R around it. I think we can always make R small enough so that there are an infinite number of points outside of B(R,P). I think if B(R,P) is in X and there are an infinite number of points in X outside of B(R,P), one can make a bijection between B(R,P) and some set in the complement of B(R,P) but I cannot think of a specific way to do this.

    Does letting U be the metric space itself give an upper bound? Or does the upper bound need to be finite?


    We need to show U* and/or equivalently V* in the maximal element cannot be finite (U*, V*, f*). If they are finite there are an infinite number of points left in the infinite metric space, so let U' be the union of U* and another open set (disjoint from U* and V*) in the metric space, then I think we can find a suitable V' and a bijection f' (with the method in first paragraph if that is correct).
     
  5. Jul 29, 2007 #4

    Dick

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    Consider the set of all points x such that B(R,x) is finite for sufficiently small R. These are the discrete points. If they are infinite in number then your job is pretty easy. Each {x} is open since B(R,x)={x} for sufficiently smaller R. If they are finite in number then the non-discrete points are infinite in number and your job is even easier (for a different reason).
     
    Last edited: Jul 29, 2007
  6. Jul 29, 2007 #5
    If the discrete points are infinite, we can take them as our infinite open set U. Now, the completement of U consists either of the null set or a union of open sets of non-discrete points which must be infinite, since any open set with a non-discrete point is infinite. However, in the case of the complement being the null set, we need to divide those discrete points somehow?

    If the discrete points are finite, there must be an infinite number of non-discrete points, any one of which could serve as U.
     
  7. Jul 29, 2007 #6

    Dick

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    Yes, if there are no non-discrete points you will need to divide the discrete points into two infinite sets. I didn't want to deprive you of all of the fun of solving the problem. It really helps to be able to write down explicit examples.
     
    Last edited: Jul 29, 2007
  8. Jul 29, 2007 #7

    Dick

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    Now you've got me wondering. How do I know I can find an infinite subset of an infinite set whose complement is also infinite? It's easy if it has a countable subset. Do I need the axiom of choice to show this? I'm confused about my foundations. Help!
     
  9. Jul 29, 2007 #8
    Take the set of lattice points in the real number as an explicit example. Let P be a point in this set, take the set of points that are farther away from a point adjacent to point (call it P+ 1) than to P. Similarly, take the set of points that are farther away from the other point adjacent to P (P-1) than to P.

    Generalize this to any metric space by using the two points closest to P instead of simply P-1 and P+1.

    How is that?
     
  10. Jul 30, 2007 #9
    Isn't that precisely what we are trying to prove? How is the statement

    "Prove that there exists an infinite subset of an infinite set whose complement is infinite."

    different from the statement in my thread-opening post

    "Let M be an infinite metric space. Prove that M contains an open set U s.t. both U and its complement are infinite." ?
     
  11. Jul 30, 2007 #10

    Dick

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    No. This problem is just dealing with sets not metric spaces. I'm not sure I understand your solution. I suppose one way to deal with the problem is to take a countable subset of the discrete points and then a bijection of that to the integers. Now take the subset to be all of the points that map to even numbers. Seems a little awkward though.
     
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