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If the tangent space $T_{x_0}\mathbb{R}^n$ is the tangent space to $\mathbb{R}^n$ at the point $x_0$, as the set of all vectors applied to $x_0$ and so it is isomorphic to $\mathbb{R}^n$, that is the set of all vectors applied in $0$: $$\textbf{what is the meaning of $T_{x_0}\mathbb{R}$, and why it is isomorphic to $\mathbb{R}$?}$$

I mean in $\mathbb{R}$ there is no a notion of vector so I can't imagine what is the tangent space in this case. Can you help me?

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If you define tangent spaces for any differentiable manifold, you'll notice that when your manifold is a (finite-dimensional) real vector space $V$ you get to identify $T_p V$ with $V$ at any point $p$ (to see why, see for example this question).

The case of $\Bbb R$ is simply the one-dimensional case of this phenomenon, since $\Bbb R$ is (the) one-dimensional real vector space.
(I don't really get why you would say "there's no notion of vector in $\Bbb R$", but you can easily "visualize" what a vector in $\Bbb R$ is as you can do in any other $\Bbb R^n$: as an arrow going from the origin, i.e. $0$, to any element of the space. That is, you're looking at every real number $t$ as a vector of length $|t|$, going left or right depending on the sign of $t$, respectively negative or positive sign.)

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    $\begingroup$ Really thanks for the explanation! $\endgroup$
    – pawel
    Mar 6, 2021 at 13:46

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