Question about topological manifolds

  • Thread starter Laur
  • Start date
3
0
Does every (continous and second countable) topological manifold have an Euclidean neighbourhood around each of its points whose closure equals the whole manifold?
 
153
0
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:
3
0
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.
 
153
0
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.
 
3
0
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.
 

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
7
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.
 
709
0
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.
 

gel

533
5
Does every (continous and second countable) topological manifold have an Euclidean neighbourhood around each of its points whose closure equals the whole manifold?
Yes!

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.
 
709
0
Yes!

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.
I like that idea
 

gel

533
5
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:
709
0
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.
 

gel

533
5
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.
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.
 
709
0
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.
 

Related Threads for: Question about topological manifolds

Replies
3
Views
2K
Replies
6
Views
1K
Replies
1
Views
2K
Replies
4
Views
1K
Replies
2
Views
969
Replies
9
Views
2K
Replies
20
Views
1K
Top