Spiral points on the sphere Let $(X,dist)$ a compact metric space and $\phi$ a flow on $X$. We say that $x\in X$ is a spiral point if there is a continuous function $h:R\to R$ and $τ>0$ such that $h(t) − t > τ$ for all $t ≥ 0$ and $dist(\phi_t(x), \phi_{h(t)}(x)) → 0$ as $t → +∞$
I would appreciate it if you could help me prove the following statement: If $\phi$ is a continuous flow on the two-dimensional sphere then every point is spiral.
 A: Choose any $x\in S^2$. Poincaré-Bendixson tells us that if $\omega(x)$ (the $\omega$-limit set of $x$) is either (1) a fixed point, (2) a closed orbit (which in the case of the 2-sphere is necessarily periodic), or (3) a finite chain of fixed points connected by trajectories. In the first and second case, we let $\omega(x)$ be an orbit of period $T > 0$ (in the case that $\omega(x)$ consists of a single fixed point, this can be any positive number). Then, we note that $\phi(x, t)$ approaches $\omega(x)$ in the sense that $d(\phi(x, t), \omega(x))\to 0$, as otherwise, there is some $\{t_n\}_{n=1}^{\infty}\subset \{t > 0 : d(\phi(x, t), \omega(x)) > \epsilon\}$ for some $\epsilon > 0$ such that $t_n\to \infty$, and so we have some subsequence $\{t_{n_k}\}$ such that $\phi(x, t_{n_k})\to y'$ and $d(y', \omega(x)) > 0$. This implies $y'\notin \omega(x)$, which contradicts the definition of $\omega(x)$. Then, we let $h(t) = t+T$, so $h(t)-t = T > \tau$ for all $t\geq 0$ for any $\tau < T$, and we note that $$d(\phi(x, t), \phi(x, h(t))) = d(\phi(x, t), \phi(x, t+T)) = d(\phi(x, t), \phi(\phi(x, t), T))\to 0$$ as $d(\cdot, \cdot)$ and $\phi(\cdot, T)$ are continuous (and therefore uniformly continuous on $S^2\times S^2$ and $S^2$ respectively, as these are compact) and $d(y, \phi(y, T)) = 0$ for all $y\in \omega(x)$.
Now, we will handle the case where $\omega(x)$ is a chain of fixed points.   If anybody has a nicer or less hand-wavy way of doing this, let me know (although, this logic theoretically also works in the case of the closed orbit). As before, $d(\phi(x, t), \omega(x))\to 0$. We choose a transversal $T$ to $\omega(x)$ and let $\{t_n\}_{n=1}^{\infty}\subset \mathbb{R}^+$ be the times at which $\phi(x, t)$ crosses $T$. Then, we let $h : \mathbb{R}\to \mathbb{R}$ be a continuous, increasing map such that $h(t_n) = t_{n+1}$ and for $S_{\epsilon}(t) := \{y\in \omega(x) : d(\phi(x, t), y) < \epsilon\}$ and some $\{\epsilon_n\}_{n=1}^{\infty}\subset \mathbb{R}$ decreasing to $0$ such that $d(\phi(x, t), \omega(x)) < \epsilon_n$ for all $t\in [t_n, t_{n+1}]$, we choose $h$ so that $S_{\epsilon_n}(t)\cap S_{\epsilon_n}(h(t))\neq \emptyset$ (basically, $h$ is chosen so that $\phi(x, h(t))$ "follows" $\phi(x, t)$ around $\omega(x)$, one pass ahead). Now, we note that $h(t)-t$ is bounded below, as otherwise, $\phi(x, t)$ travels around $\omega(x)$ arbitrarily fast, which implies that $\phi(\cdot, t)$ is not continuous near $\omega(x)$. Thus, we have some $\tau > 0$ such that $h(t)-t > \tau$. Furthermore, $d(\phi(x, t), \phi(x, h(t))) < 2\epsilon_n$ for $t\in [t_n, t_{n+1}]$, so $d(\phi(x, t), \phi(x, h(t)))\to 0$.
