Manifolds of zero dimension and $\mathbb R^0$?

Tu Manifolds Section 5.4

Example 5.13 (Manifolds of dimension zero). In a manifold of dimension zero, every singleton subset is homeomorphic to $$\mathbb R^0$$ and so is open. Thus, a zero-dimensional manifold is a discrete set. By second countability, this discrete set must be countable.

Why exactly is the manifold $$M$$ discrete? I actually proved that the singleton subsets are open in their components but was not able to show they are open in $$M$$ itself.

Here is what I have done thus far:

Let $$M$$ be a smooth manifold with dimension zero, which means by definition that all of the connected components of $$M$$'s topological manifold (see here) $$\{C_{\alpha}\}_{\alpha \in J}$$ have dimension zero.

Let $$\alpha \in J$$. $$C_{\alpha}$$ has dimension zero, which means by definition (see here) that $$\forall p \in C_{\alpha}, \exists$$ homeomorphism $$\varphi: U \to V$$ for some $$U$$, a neighborhood of $$p$$ in $$C_{\alpha}$$ and some $$V$$, an open subset of $$\mathbb R^0=\{0\}$$. $$V$$ is either $$\{0\}$$ or $$\emptyset$$. Since $$U$$ contains $$p$$, $$U \ne \emptyset$$. Hence, $$V \ne \emptyset$$ because from nothing comes nothing, so $$V=\mathbb R^0=\{0\}$$. Sets that are homeomorphic to singletons are singletons. Therefore, $$U$$ is a singleton containing p, so $$U=\{p\}$$.

Therefore, we have

• $$\forall p \in M, \exists$$ unique $$\alpha \in J: \{p\}$$ is open in $$C_{\alpha}$$.

I remember the connected components $$C_{\alpha}$$ are:

• closed in $$M$$

• not necessarily open in $$M$$.

• open in $$M$$ if $$J$$ is finite.

I know $$\{p\}$$ is open in one of the connected components of $$M$$. How do we arrive at the conclusion that $$\{p\}$$ is open in $$M$$ itself?

• I don't think you have the right definition of an $n$-dimensional manifold. Being an $n$-dimensional manifold means just that the space is locally homeomorphic to open subsets of ${\mathbf R}^n$. In your other question, you have the hypothesis of "locally Euclidean", which implies in particular that the connected components are open, in which case the two notions of $n$-dimensional coincide. – tomasz Nov 22 '18 at 10:47
• @tomasz I think in Tu, "space is locally homeomorphic to open subsets" is the definition for topological manifold instead of smooth manifold. By $n$-dimensional manifold, do you refer to smooth manifold or topological? – user198044 Nov 22 '18 at 11:15
• Yes. For a smooth manifold you need some further restrictions. But every smooth manifold is in particular a topological manifold. – tomasz Nov 22 '18 at 12:12
• @tomasz I will take it to mean that you refer to a smooth manifold. Tu's definition is different from yours, if I understand correctly. – user198044 Nov 24 '18 at 3:28
• That is my point: it is not (really) different. By the definition you gave 5.2, a manifold is a locally Euclidean space, which implies that connected components are open (because Euclidean spaces are locally connected). Hence, all connected components being $n$-dimensional is the same as being locally Euclidean of dimension $n$. – tomasz Nov 25 '18 at 14:58

You wrote down yourself that you now know that every point in your manifold is clopen (U={p} implies this, since the domain of charts has to be open). But the only clopen subsets of a connected space are the space itself and the empty set, hence $$\{p\}$$ is a maximal connected component, and hence all of $$C_\alpha$$
• Oh, we deduce $\{p\}$ is open in $C_{\alpha}$ and know $\{p\}$ is closed in $C_{\alpha}$ by Hausdorff and so conclude $M$ is totally disconnected, which implies discrete? – user198044 Nov 22 '18 at 11:09
• well, I do not know what totally disconnected means, but we conclude that $C_\alpha$ is just a point, and since $\alpha$ was arbitrary, all connected components are singletons, which are clopen, hence the union of them (i.e. M) is disrcete. – Enkidu Nov 22 '18 at 12:30
I don't know what you know but I would do it like this: Pick a point $$p\in M$$. It has an open neighborhood $$U$$ homeomorphic to $$\mathbf R^0$$. So $$U=\{p\}$$ (it has only one point!). Hence $$\{p\}$$ is open.
• How do you know $p$ has such an open neighborhood in M? I know $p$ has such an open neighborhood in one of the connected components of M, but I don't know how to extend this to $M$ itself. – user198044 Nov 22 '18 at 10:34
• Tom, are you actually tomasz and mistakenly replied here instead of in the comments on the question? If not, I don't understand your reply here. My problem is I know $\{p\}$ is open in a connected component of $M$ but don't know how to conclude $\{p\}$ is open in $M$ itself. – user198044 Nov 25 '18 at 8:44