Given the discrete space $X$ and the atlas $A=\{(\{x\},\varphi_x)|x\in X\}$, where $$\varphi_x:\{x\}\rightarrow\mathbb{R}^0,x\mapsto 0,$$
in what sense is the transition map $\varphi_x\circ\varphi_x^{-1}=id_{\mathbb{R}^0}$ smooth? It eludes me how differentiability is defined for a one-point set.
Does it have to do with the fact that since $\mathbb{R}^0$ is 0-dimensional, any map on it has zero components and therefore no partial derivatives?
OK answering based on my comments.
I have some formula $\psi(h)$ involving $h \in \mathbb R^n$ with values $\psi(h) \in \mathbb R$. Any formula at all. I claim, when $n=0$, the statement $$ \lim_{h \to 0} \psi(h) = 0 $$ is vacuously true. Indeed, it means:
For every $\epsilon > 0$, there exists $\delta > 0$ such that for all $h$ with $0 < \|h\| < \delta$ we have $|\psi(h)| < \epsilon$.
This is vacuouslsy true, of course, since in $\mathbb R^0$ there are no $h$ satisfying $0 < \|h\| < \delta$. I don't even have to look at the formula $\psi(h)$.