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 ?
[Math] Uniqueness of smooth structure on a zero-dimensional smooth manifold
differential-geometrymanifolds
Related Solutions
It took me a while to understand the great idea proposed by Anthony Carapetis since I think that other people may have the same doubt that I had, I decided to write a more detailed answer using his idea.
First of all, remember the Proposition $1.17$ (this proposition is straightforward to prove) of John Lee's Introduction to smooth manifold:
and note that $F_s: B(0,1) \rightarrow B(0,1)$ is an homeomorphism that isn't a diffeomorphism $\forall s>0$ and $s\neq 1$, and $F_s: B(0,1)\setminus \{0\} \rightarrow B(0,1)\setminus \{0\}$ is a smooth diffeomorphism $\forall s>0$ ( where $B(y,r) = \{x\in \mathbb{R}^n;$ $|x-y|<r$ $\}$).
Let $\mathcal{A} = (\varphi_i,U_i)_{i \in I}$ be a smooth atlas of $M$. We will construct a smooth atlas $\mathcal{B}$ such that $\mathcal{A} \cup \mathcal{B}$ is not a smooth atlas.
Note that for every $x$ $\in$ $M$, there exists a chart $(\varphi_x,U_x)$ $\in$ $\mathcal{A}$, satisfying $x$ $\in$ $U_x$. Using that $\varphi(U_x) \subset \mathbb{R}^n$ is open, $\exists$ $\delta_x >0$, such that, $B(\varphi(x), \delta_x) \subset \varphi (U_x)$.
So, we can define a smooth atlas $$\mathcal{C} = \left\{(\varphi_x, \varphi^{-1}_x \left( B(\varphi(x),\delta_x) \right) )\right\}_{x \in M},$$ which has the same smooth structure of $\mathcal{A}$.
Moreover, note that for every $x\in M$, there is a function $\xi_x : B(\varphi(x), \delta_x) \rightarrow B(0,1)$ such that $\xi_x $ is a smooth diffeomorphism between $B(\varphi(x), \delta_x)$ and $B(0,1)$ (in fact we cand define $\xi_x(y) = \frac{1}{\delta_x}(y-\varphi(x) )$ ).
Consequently, we are able to define a new smooth atlas $$\mathcal{D} = \left\{\left(\xi_x \circ \varphi_x, \varphi_x^{-1}\left(B(\varphi(x), \delta_x) \right)\right)\right\}_{x \in M},$$ which has the same smooth structure of $\mathcal{A}$ and $B(0,1) = \xi_x \circ \varphi^{-1}_x(B(\varphi(x),\delta_x ))$, $\forall$ $x$ $\in$ $M$.
Now, fixed $x_0$ $\in$ $M$, and using that $M$ is Hausdorff, for every $y$ $\in$ $M$ ($y$ $\neq$ $x$), exists a neighborhood $V_y$ of $y$, such that $x$ $\notin$ $V_y$.
Therefore, $$\mathcal{E} = \left\{(\xi_{x_0} \circ \varphi_{x_0}, \varphi_x^{-1}(B(\varphi(x_0), \delta_{x_0}) )\right\} \cup \left\{(\xi_y \circ \varphi_y, \varphi_y^{-1}(B(\varphi(y), \delta_y) \cap V_y )\right\}_{y \in M\setminus \{x_0\}}$$ is a smooth atlas which has same smooth structure of $\mathcal{A}$.
So, we can finally use Carapetis' idea. Define $$\mathcal{B} = \left\{(F_s \circ \xi_{x_0} \circ \varphi_{x_0}, \varphi_x^{-1}(B(\varphi(x_0), \delta_{x_0}) )\right\} \cup \left\{ (\xi_y \circ \varphi_y, \varphi_y^{-1}(B(\varphi(y), \delta_y) \cap V_y )\right\}_{y \in M\setminus \{x_0\}},$$
Now, we need to prove that $\mathcal{B}$ is a smooth atlas, the only nontrivial property that needs to be checked is: if, $\forall$ $y$ $\in$ $M\setminus \{x_0\}$
$$F_s \circ \xi_{x_0} \circ \varphi_{x_0} \circ (\xi_{y} \circ \varphi_y)^{-1}: \xi_y \circ \varphi_y (Z_y) \rightarrow F_s \circ \xi_{x_0}\circ \varphi_{x_0} (Z_y), $$
( where $Z_y = \left( V_y \cap \varphi_y^{-1}(B(\varphi(y), \delta_{y})) \right) \cap \varphi_{x_0}^{-1}(B(\varphi(x_0), \delta_{x_0}))$ ) is a smooth diffeomorphism.
This follows directly from the fact that $\xi_{x_0} \circ \varphi_{x_0} \circ (\xi_{y} \circ \varphi_y)^{-1}$ and $F_s\vert_{ \xi_{x_0} \circ \varphi_{x_0} \circ (\xi_{y} \circ \varphi_y)^{-1}(Z_y)}$ are diffeomorphisms, because $0$ $\notin \xi_{x_0} \circ \varphi_{x_0} \circ (\xi_{y} \circ \varphi_y)^{-1}(Z_y)$, sinse $x_0$ $\notin$ $U_y$.
Then $\mathcal{B}$ is a smooth atlas, but $\mathcal{B} \cup \mathcal{D}$ isn't a smooth atlas, because
$$F_s = F_s \circ \xi_{x_0} \circ \varphi_{x_0} \circ ( \xi_{x_0} \circ \varphi_{x_0})^{-1} : B(0,1) \rightarrow B(0,1) $$ is not a smooth diffeomorphism.
Using Proposition 1.17, we conclude that the smooth structure determined by $\mathcal{B}$ is different of the smooth structure determined by $\mathcal{A}$ (because of the smooth structure determined by $\mathcal{A}$ $=$ smooth structure determined by $\mathcal{D}$ $\neq$ smooth structure determined by $\mathcal{B}$ ). Since this result holds for all $s>0$, we can construct uncountable many differents smooth atlas $\mathcal{B}$, which all have different smooth structures among them, which completes the proof.
The distinction to be made is that a differentiable structure is a choice of maximal smooth atlas $\mathcal A$, but two different choices $\mathcal A$ and $\mathcal A'$ can lead to isomorphic smooth structures. As an example, the canonical smooth structure $\mathcal A$ on $\mathbb R$ that contains the smooth function ${\rm id}:\mathbb R\longrightarrow \mathbb R$ is isomorphic to the smooth structure $\mathcal A'$ that contains the smooth function $x\mapsto x^3$, although $\mathcal A'\neq \mathcal A$. Thus, although a manifold admits uncountably many different smooth structures, it may have finitely many isomorphism classes of such structures.
Best Answer
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.