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A dumb question:

If $\{O\}$ is a singleton, then we can think at it like a smooth manifold with the atlas given by the chart $( \{ O \}, \eta: \{O\}\to \{0\} )$.

My definition of differentiable function is:

$F:M\to N$ is differentiable in $p\in M$ if there are $(U,\phi)$ chart of $M$ in $p$, $(V,\psi)$ chart of $N$ such that:

1) $F(U)\subseteq V$

2) $\psi \circ F \circ \phi^{-1}:\phi(U)\to \psi(V)$ is $C^\infty$ in the sense of Analysis.

If I have $F:M\to \{O\}$, using the above defition i would chose any $(U,\phi)$ chart on $M$, then observe that $\eta \circ F \circ \phi^{-1}:\phi(U)\to\{0\}$ is costant, so $C^\infty$.

If I have $F:\{O\}\to M$, then using the above defition i would chose any $(U,\phi)$ chart on $M$ such that $F(O)\in U$, then consider $\phi \circ F \circ \eta^{-1}:\{0\}\to \phi(U)$. But in this case, in what sense I should say that $\phi \circ F \circ \eta^{-1}$ is $C^{\infty}$ in the sense of Analysis? Because in Analysis i consider functions defined in open subsets of $\mathbb{R}^n$.

I think i could think at $\phi \circ F \circ \eta^{-1}$ as the restricion of the function $g: (-\epsilon, \epsilon)\to \phi(U)$ given by $g(x)=\phi(F(O))$, so i can give sense at the sentence: " $\phi \circ F \circ \eta^{-1}$ is $C^{\infty}$ in the sense of Analysis".

But is correct to consider $\{0\}$ as a subset of $\mathbb{R}$? Because i think at $\{0\}=\mathbb{R}^0$, and considering that (for example) $\mathbb{R}$ is not subset of $\mathbb{R}^2$, in the same way should $\mathbb{R}^0$ non be subset of $\mathbb{R}$.

Maybe I'm confusing things beyond what is necessary. Thank you for anything that can clarify me the situation.

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    $\begingroup$ It has dimension zero. Otherwise, a one-point set is not open in any euclidean space. $\endgroup$ – Randall Jun 20 '18 at 16:24
  • $\begingroup$ Also, when you impose Hausdorff-ness, zero-dim manifolds must be discrete, therefore not interesting. $\endgroup$ – Randall Jun 20 '18 at 16:25

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