Prove sequence has limit in $\gamma (S^1)$ This is a seemingly interesting exercise from my topology notes, but I can't solve it for the life of me. It's like this: take a closed curve $\gamma : S^1\to \mathbb R^2$, and a sequence $\{p_n\}\subset \mathbb R^2\setminus\gamma (S^1)$ such that $\mu(\gamma,p_m)\neq\mu(\gamma,p_n)$, for all $m \neq n$. Here $\mu(\gamma, p)$ is the winding number of $\gamma$ relative to $p$.
Now I have to prove that $\{p_n\}$ has a limit point, or equivalently a convergent subsequence, to some point in $\gamma(S^1)$. I have honestly no ideas, so I can't say what I've tried. The only thing that occurs to me is maybe proving that the sequence is in some compact set containing the curve? Then prove by contradiction that the limit has to lie on the curve? I'm only barely beginning algebraic topology, so the hypotheses here are hard for me to work with, and I'd appreciate some help. Thanks.
 A: Your intuition is spot on! What follows is more of a hint than a complete answer.
Suppose first that $d(p_m, \gamma(S^1))$ (the distance between the point and the curve) is unbounded. Then eventually the winding number will be zero, which contradicts. So there's some closed ball $K$ (therefore compact) containing the curve and all the $p_n$.
By compactness, $(p_n)$ has a limit point $p$ lying in $K$. If $p$ doesn't lie on $\gamma(S^1)$, you can show that eventually, the winding number will be constant (equal to the winding number around $p$). Again, contradiction.
A: Show that $\gamma$ is homotopic to $0$. The image of the homotopy is a compact set. Show that for points $p$ outside this compact set, $\mu(\gamma,p) = 0$ and hence all but at most one point of the sequence must be inside.
If the limit is not on $\gamma$, choose a connected neighborhood not intersecting $\gamma$. Then the sequence must eventually lie in this neighborhood and the winding number on this connected set must be constant.
