Question about topological manifolds

In summary: I think.In summary, every topological manifold has an Euclidean neighbourhood around each of its points whose closure equals the whole manifold.
  • #1
Laur
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?
 
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  • #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.
 
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  • #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.
 
  • #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.
 
  • #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.
 
  • #6
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.
 
  • #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.
 
  • #8
Laur said:
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.
 
  • #9
gel said:
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
 
  • #10
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.
 
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  • #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.
 
  • #12
wofsy said:
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.
 
  • #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.
 

1. What is a topological manifold?

A topological manifold is a mathematical concept that describes a space that locally resembles Euclidean space, but may have a more complicated global structure. It is a type of topological space that is locally homeomorphic to Euclidean space.

2. How is a topological manifold different from a smooth manifold?

A smooth manifold is a type of topological manifold that has the additional structure of a smooth atlas, which allows for a notion of differentiability. Topological manifolds do not have this additional structure, and therefore cannot be differentiated or integrated in the traditional sense.

3. What types of objects can be represented as topological manifolds?

Topological manifolds can represent a wide range of objects, including surfaces in 3D space, curves in higher-dimensional spaces, and abstract mathematical objects such as the set of real numbers or complex numbers.

4. What is the importance of topological manifolds in mathematics?

Topological manifolds are important in mathematics because they provide a framework for studying and understanding geometric and topological properties of various objects. They also have applications in fields such as physics, engineering, and computer science.

5. Are there any real-world examples of topological manifolds?

Yes, there are many real-world examples of topological manifolds. For example, the surface of the Earth can be approximated as a topological manifold, with each point on the surface having a neighborhood that resembles a portion of a flat plane. Additionally, road maps, subway systems, and networks of blood vessels can all be modeled as topological manifolds.

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