Uniqueness of smooth structure on a zero-dimensional smooth manifold

In John Lee's Book "Introduction to Smooth Manifolds" on page 17, Example 1.12, the author states that the smooth structure on any zero - dimensional manifold is unique. That confuses me, suppose for example the $M = \{ p \}$ is a point. Then I can for example give this manifold a smooth structure by taking $(U, \psi)$ to be given by \begin{equation} U = \{ p \}, \quad \psi(p) = 1 \end{equation} or, I could also give it the structure \begin{equation} U = \{ p \}, \quad \psi(p) = 2 \end{equation} From what I understand, these are different smooth structures. What am I missing ?

• $\mathbf R^0$ is just a point, isn't it? Any transition maps are going to send that point to itself. – Dylan Moreland Apr 18 '12 at 20:18
• @Dylan: Suppose M is the set $\{A,C,T,G\}$ the bases for DNA. I can identify this set with any discrete set consisting of four real numbers. What I don't understand is, to what extend is any such identification unique ? – harlekin Apr 18 '12 at 20:57
• You're off by one dimension. the charts should map to $R^0$, not $R^1$. – David Roberts Dec 13 '15 at 15:31

As Dyland Moreland points out $\mathbb{R}^0=\{p\}$. Thus, there is for each discrete space $X$ a unique zero-dimensional manifold structure with the charts $\psi_x:\{x\}\to\mathbb{R}^0$ being the unique such maps.
• @Youcis: In what sense do you mean $\mathbb{R}^0 = \{p\}$ ? I thought that $R^0$ stands for any discrete set, not necessarily a one - point set ? – harlekin Apr 18 '12 at 21:01
• Well, then the symbol $\mathbb{R}^0$ wouldn't make any sense. No, $\mathbb{R}^0$ is the $0^{\text{th}}$-dimensional Euclidean space which is just a point. – Alex Youcis Apr 18 '12 at 21:04
• @harlekin If you mean how does it allow you to locally think about Euclidean space that's simple. To think about something locally being $\mathbb{R}^0$ means it locally looks like a point, which means that it's discrete. If you meant coordinate in the sense that you can do interesting things, like calculus, you really can't. While you can define a zero-dimensional manifold $\mathbb{R}^0$ doesn't carry nearly enough structure to do complicated things like calculus. Does that help? – Alex Youcis Apr 18 '12 at 21:12
• Part of the definition of being a smooth manifold is that the transition maps are smooth. In what sense are these transition maps smooth? In what sense is the (only) map $\Bbb R^0\to\Bbb R^0$ smooth? – Alex Ortiz Nov 11 '17 at 6:10