I'm reading about flows and there is a proof that I don't understand. Here we come:
Let $\psi$ be a flow of $V\in\mathfrak{X}(M^{n}).$ If $\phi_{t}(p)=p$ for a sequence of $t-$values approaching $0,$ then $V_{p}=0$ (hence $\psi_{t}(p)=p$ for all $t.$)
The proof given is this: Let $f:M\rightarrow\mathbb{R}$ be a smooth function and $\alpha(t)=f\circ\psi_{t}(p).$ Then $0=\alpha^{'}(0)=V_{p}(f).$
I don't understand the reason of such equality and why this finishes the proof.
Here $\mathfrak{X}(M^{n})$ is the set of all vector fields of smooth manifold $M$ of dimension $n.$
Any kind of help is thanked in advanced.
The chain rule tells us that $$\alpha'(0) = df\left(\frac{d}{dt}\Big|_{t=0}\psi_t(p)\right)=df(V_p)=V_p(f).$$
On the other hand, we know $\alpha$ is a smooth function (since it is the composition of a smooth function and a smooth curve), and we have $\alpha(t_k)= f(p)$ for a sequence $t_k \to 0$; so we must have $\alpha'(0) = 0.$ (If this isn't clear to you, apply the sequential definition of the limit of a function to the definition of the derivative.)
Finally, since we have shown $V_p(f) = 0$ for all functions $f$, we can conclude $V_p = 0$.