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I'm having problems with understanding why is it the case. Suppose $D\subset\mathbb{R}^n$ is an open, connected subset and for $p\in D$ define the tangent space $T_pD$ to be the set of the velocities of all curves $(-\epsilon,\epsilon)\to D$ which pass through $p$. The lecture notes I've got claim that for any $p$ $$ T_PD\cong \mathbb{R}^n $$ The proof goes as follows: let $c:I \to D$ be a curve such that $c(0)=p$, then the velocity of this curve is $c'(0)\in\mathbb{R}^n$.

Conversely, given $v\in \mathbb{R}^n$, let $c(t)=p+vt$ be the curve, then its velocity is $c'(0)=v$, so $v\in T_p D$. But $c$ is a straight line through $p$, so it may not even belong to $D$ at all, how is this not a contradiction?

Also, intuitively, suppose we have a sphere $S^2\subset\mathbb{R}^3$, then surely the tangent space at a point is going to be a plane, not the whole space itself?

Jimmy R
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    As you're only considering the tangent space at a point, I don't think the connectedness assumption is necessary here. – Michael Albanese Jan 26 '13 at 14:55
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    No, of course it's not necessary here. – Jimmy R Jan 26 '13 at 14:57
  • Think about the intuitive picture of an open set in $\mathbb{R}^n$...it is one where at each point you are able to travel some distance in ANY direction and not leave the set. So the possible "tangents" allowed would seem to be the whole of $\mathbb{R}^n$. – fretty Jan 26 '13 at 15:28

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Regarding your third paragraph: $c$ need only be defined on an interval around $0$ (or may be trivially extended beyond such an interval).

And yes, the tangent space to $S^2$ is isomorphic to $\mathbb R^2$ at each point. But $S^2\subset \mathbb R^3$ has nothing to do wih the situation for an open subset $D\subset \mathbb R^3$.

Hagen von Eitzen
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As $D$ is open and $p \in D$, there is $r > 0$ such that $B(p, r) \subseteq D$. Now note that you are free to choose $\epsilon$ in the definition of $c$, so you just choose $\epsilon$ small enough to ensure the range of $c$ is contained in $B(p, r)$ and hence $D$. You don't really need to introduce $B(p, r)$, but it is sometimes helpful to visualise the interior of a circle as opposed to some arbitrary subset of $\mathbb{R}^n$.

As for your second question, note that $S^2$ is not an open connected subset of $\mathbb{R}^3$ so you can't apply the result. $S^2$ is connected, but it is not an open subset of $\mathbb{R}^3$.

Michael Albanese
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