Plausibility of a Discrete Point Manifold

[Mandelbroth]

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 $k>2$?

 

[Number Nine]

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 \mathbb{R}^n? All singleton sets in \mathbb{R}^n are closed.

 

[WannabeNewton]

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 $\mathbb{R}^{n}$).

 

[micromass]

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.

 

[Mandelbroth]

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. )

 

[WannabeNewton]

That all looks good! Cheers.