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Question about topological manifolds

  1. Jun 1, 2009 #1
    Does every (continous and second countable) topological manifold have an Euclidean neighbourhood around each of its points whose closure equals the whole manifold?
  2. jcsd
  3. Jun 2, 2009 #2
    In what sense do you mean equals? Diffeomorphic?

    I would think do. The disjoint union of two manifold is a manifold, right? But any neighborhood has only one connected component but the whole manifold has two.

    ETA: Assuming both pieces are of the same dimension.
    Last edited: Jun 2, 2009
  4. Jun 4, 2009 #3
    I meant a manifold that is connected (wrote it as "continuous" by accident).

    By "equals", i mean that the neighbourhood is dense in the manifold.
  5. Jun 4, 2009 #4
    Since the product of two manifolds is a manifold, the torus which is just S1 x S1 is a manifold. But any point has neighborhoods that are contractible, while the whole torus isn't.
  6. Jun 4, 2009 #5
    But on the torus, every point has a neighbourhood (obtained by removing two orthogonal circles from the torus) whose closure is the whole torus, and which is homeomorphic to the plane (can be "flattened" into a rectangle). That's what I mean. By saying that the neighbourhood is "Euclidean", I mean that it is homeomorphic to R^n.
  7. Jun 5, 2009 #6


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    Obviously it's true for all 1-manifolds and all 2-manifolds. And it's true for S^3, T^3 and RP^3. So a counter-example, if it exists, will be found in dimension 3 or greater in a less common manifolds... I'm quite curious as to what the answer is as well.
  8. Jun 14, 2009 #7
    I don't know the full answer but for a Riemannian manifold the cut locus works. The manifold is the union of an open cell and its boundary which is the cut locus. For non-smooth manifolds I am not sure. apparently there are toplogical manifolds that do not have handlebody decompositions. They may be candidates.

    A sub question would be: if a manifold does have a handle body decomposition is it the union of an open cell and its boundary? This should not be too hard.
  9. Jun 14, 2009 #8


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    Consider the set S of all neighbourhoods of a point P which are homeomorphic to the open ball. Try showing that this set has a maximal element (for this you just need to show that the union of a chain in S is again in S). Then that the closure of such a maximal element is the whole manifold.

    I haven't worked through the details, but it seems like this argument should work.
  10. Jun 14, 2009 #9
    I like that idea
  11. Jun 14, 2009 #10


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    Thinking a bit more about my idea for a proof, the following statement can be proven

    - if S is a nonempty collection of subsets of the manifold which are each homeomorphic to Rn, and totally ordered (by inclusion), then the union of S is also homeomorphic to Rn.

    This can be done by applying the generalized http://en.wikipedia.org/wiki/Jordan%E2%80%93Sch%C3%B6nflies_theorem" [Broken] to 'glue' together the different homeomorphisms. It follows that any open subset homeomorphic to Rn is contained in a maximal such subset.

    The question of whether such a maximal euclidean neighbourhood has closure equal to the entire manifold still remains.
    Last edited by a moderator: May 4, 2017
  12. Jun 14, 2009 #11
    Interesting. One can have a nested chain of open disks in a manifold whose union does not have the whole manifold as its boundary.

    Take the height function on a torus standing on end on the the plane. From the bottom, f(x) < a, for a small enough is a a nested sequence of open disks that has a figure eight as its boundary and leaves out all of the torus above the first saddle point of f.
  13. Jun 14, 2009 #12


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    Well, the unit open ball in Rn is the nested union of the open balls {x:|x|<a} for a < 1. It follows that every Euclidean neighborhood is a nested union of such neighborhoods.
    Your example shows that the boundary of such a neighborhood can be a figure of eight, but it isn't maximal.

    Showing that a Euclidean neighborhood whose closure isn't the entire manifold can be enlarged sounds difficult. This is because the boundary of the neighbourhood can be very complicated, and fractal-like.
  14. Jun 14, 2009 #13
    Yeah but showing that you can always cross the boundary to create a larger open ball does not seem easy. Maybe its doesn't always work.
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