Let $X$ be a smooth vector field defined on the sphere $S^2$ such that $X$ has only one singular point. Prove that:
The $\alpha$ and $\omega$-limit set of every point is the singular point
Suppose that the north pole $N$ is the singular point $X(N)=0$ and fix $p\in S^2$ with integral curve $\gamma$ in $S^2$ such that $\gamma(0) = p$. Then taking the spherical projection $f: S^2\setminus \{N\} \rightarrow \mathbb R^2$ and denoting points in $\mathbb R^2$ which are in the image of $f$ by $\tilde a = f(a)$, we may define a vector field $\tilde X$ on $\mathbb R^2$ by $\tilde X(\tilde a) = df(f^{-1}(\tilde a))X(f^{-1}(\tilde a )).$ We note that $\tilde \gamma = f\circ \gamma$ is the solution for the vector field $\tilde X$ in $R^2$ such that $\tilde \gamma(0)= \tilde p$. Since $X$ has only $N$ as a singular point, the induced vector field $\tilde X$ on $\mathbb R^2$ has only regular points. By contradiction, suppose that $q\in \omega_X(p), q \neq N.$ Then $\tilde q \in \omega_{\tilde X}(\tilde p)$. We note that $\tilde \gamma^+ = \{\tilde \gamma(t):t\geq 0\}$ cannot be inside any compact set on $\mathbb R^2$, because since $\omega_{\tilde X}(\tilde p)$ contains only regular points, by Poincaré-Bendixson it would follow that $\omega_{\tilde X}(\tilde p)$ is a closed orbit. But in $\mathbb R^2$, any closed orbit must contain a singular point in its interior, which cannot happen with our vector field $\tilde X$.
How do I continue? I want to prove that $q\neq N$ in $S^2 \implies q\not\in \omega_X(p),$ because since $S^2$ is compact, $\omega(p)\neq\emptyset$ and hence $\omega(p)=\{N\}$.
Any help on how to continue? Pictorially it seems that the orbit of $\tilde \gamma$ as soon as it comes close to $\tilde q$, it must get back and get very far from it and then encircles again towards the point $\tilde q$ and again go very far and get back again...