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?

  • 1
    $\begingroup$ 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. $\endgroup$ – tomasz Nov 22 '18 at 10:47
  • $\begingroup$ @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? $\endgroup$ – user198044 Nov 22 '18 at 11:15
  • $\begingroup$ Yes. For a smooth manifold you need some further restrictions. But every smooth manifold is in particular a topological manifold. $\endgroup$ – tomasz Nov 22 '18 at 12:12
  • $\begingroup$ @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. $\endgroup$ – user198044 Nov 24 '18 at 3:28
  • $\begingroup$ 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 $n$. $\endgroup$ – 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 $C_\alpha$

  • $\begingroup$ 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? $\endgroup$ – user198044 Nov 22 '18 at 11:09
  • 1
    $\begingroup$ 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. $\endgroup$ – Enkidu Nov 22 '18 at 12:30
  • $\begingroup$ Enkdiu, the empty set and the one-point sets are the only connected subsets of a totally disconnected set by the definition. Thank you! $\endgroup$ – user198044 Nov 22 '18 at 14:21
  • 1
    $\begingroup$ Ah, ok thank you, then yes, precisely! $\endgroup$ – Enkidu Nov 22 '18 at 14:31

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.

  • $\begingroup$ 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. $\endgroup$ – user198044 Nov 22 '18 at 10:34
  • $\begingroup$ Well this is my definition. Do you consider the set of 1/n together with 0 to be a manifold ? Isn't it a counterexample to your question? $\endgroup$ – Tom Nov 24 '18 at 23:37
  • $\begingroup$ 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. $\endgroup$ – user198044 Nov 25 '18 at 8:44

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