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?
Best Answer
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.