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Plausibility of a Discrete Point Manifold

  1. Jul 3, 2013 #1
    Consider a discrete set of ##k## points.

    First, is it a manifold? I know that a manifold is a topological space that contains a neighborhood homeomorphic to Euclidean space for each point. Can we just consider each point's neighborhood to be a set containing only that point?

    Second, would the structure be orientable for ##k>2##?
     
  2. jcsd
  3. Jul 3, 2013 #2
    You can, but then where is the homeomorphism to an open subset of [itex]\mathbb{R}^n[/itex]? All singleton sets in [itex]\mathbb{R}^n[/itex] are closed.
     
  4. Jul 3, 2013 #3

    WannabeNewton

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    Every countable discrete space is a topological 0-manifold; the converse holds as well. Here a discrete space is a set equipped with the discrete topology. This is a trivial exercise so try to show it yourself. Define a topological n-manifold as a second countable Hausdorff space that is locally Euclidean of dimension n (i.e. every point has a neighborhood that is homeomorphic to an open subset of ##\mathbb{R}^{n}##).
     
  5. Jul 3, 2013 #4

    micromass

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    And to add to this: yes, the manifold will be orientable.
     
  6. Jul 4, 2013 #5
    I'm not very confident with this material, so...here I go:

    Lemma #1: If the space is a countable discrete space, then it is also a second countable space.
    //Proof: If the space is countable and discrete, then there are a countable number of singletons. The singletons can, therefore, form a countable basis for the discrete space.##\halmos##​

    Lemma #2: If the space is a countable discrete space, then it is a Hausdorff space.
    //Proof: If the space is a discrete space, then the points are necessarily disjoint. ##\halmos##​

    Lemma #3: Each point in a discrete space is homeomorphic to ##\mathbb{R}^0##.
    //Proof: Homeomorphism is the identity map. ##\halmos##?​

    Considering lemmas 1, 2, and 3, we have established that a countable discrete space is a second countable Hausdorff space of dimension 0 with each neighborhood homeomorphic to Euclidean space of dimension 0, and therefore a 0-manifold. ##\blackhalmos##

    I'm not sure about the third lemma. It feels too...blunt.

    @Micromass: How is it orientable? Could you, perhaps...nudge me in the right direction?
    (Also...you've not responded to my personal message about operators. It's been 3 weeks. :frown:)
     
    Last edited: Jul 4, 2013
  7. Jul 4, 2013 #6

    WannabeNewton

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    That all looks good! Cheers.
     
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