Plausibility of a Discrete Point Manifold

  • #1
612
23

Main Question or Discussion Point

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##?
 

Answers and Replies

  • #2
806
23
Can we just consider each point's neighborhood to be a set containing only that point?
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.
 
  • #3
WannabeNewton
Science Advisor
5,803
530
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}##).
 
  • #4
22,097
3,277
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}##).
And to add to this: yes, the manifold will be orientable.
 
  • #5
612
23
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}##).
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:
  • #6
WannabeNewton
Science Advisor
5,803
530
That all looks good! Cheers.
 

Related Threads for: Plausibility of a Discrete Point Manifold

  • Last Post
Replies
6
Views
547
Replies
1
Views
2K
Replies
63
Views
5K
Replies
19
Views
5K
Replies
7
Views
3K
Replies
7
Views
3K
Replies
4
Views
6K
Top