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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?

Thomas Wening
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    Write down a good definition of differentiability for $\mathbb R^n$. See what it means for $\mathbb R^0$. Take into account the meaning of quantification over the empty set. I think you will conclude that the unique map $\mathbb R^0 \to \mathbb R^0$ is differentiable. – GEdgar May 30 '18 at 10:44
  • A function $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ is differentiable at $x_0\in\mathbb{R}^m$, if there exists a linear map $L:\mathbb{R}^m\rightarrow\mathbb{R}^n$, such that

    $\lim\limits_{h\rightarrow0}\frac{|f(x_0+h)+f(x_o)-L(h)|{\mathbb{R}^n}}{|h|{\mathbb{R}^m}}=0.$

    On $\mathbb{R}^0$ we can pick $L:\mathbb{R}^0\rightarrow\mathbb{R}^0,0\mapsto0$. But then I do not comprehend how to evaluate the limit, given that the only possible norm is $|0|=0$.

     – Thomas Wening May 30 '18 at 11:03
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    OK, your definition has $\lim_{h \to 0}$ in it. That involves quantifying over all $h$ satisfying some conditions, including $h \ne 0$. Since there are no such $h$, the relevant statement is true. – GEdgar May 30 '18 at 11:08

1 Answers1

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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 $\psi(h)$.

GEdgar
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    In fact, it is also vacuously true that $\lim_{h \to 0} \psi(h) = 17$, but that is irrelevant. – GEdgar May 30 '18 at 11:18