# How to find a curve passing through a sequence of points?

Here's a (slightly paraphrased) excerpt from Complex Analysis by Stein & Shakarchi. The function $$F:\mathbb D\to P$$ is conformal where $$P$$ is a polygonal region (open, bounded, simply connected, polygonal boundary).

Lemma 4.4: Let $$z_0$$ be a point on the unit circle. Then $$F(z)$$ tends to a limit as $$z$$ approaches $$z_0$$ within the unit disc.

Proof. If not, there are two sequences $$\{z_n\},\{z_n'\}$$ in the unit disc that converge to $$z_0$$ and are so that $$\{F(z_n)\},\{F(z_n')\}$$ converge to two distinct points $$\zeta,\zeta'\in\bar P$$. ... We may therefore choose two disjoint discs $$D,D'$$ centered at $$\zeta,\zeta'$$, respectively. For all large $$n$$, $$F(z_n)\in D, F(z_n')\in D'$$. Therefore, there exist two continuous curves $$\Lambda$$ and $$\Lambda'$$ in $$D\cap P$$ and $$D'\cap P$$, respectively, with $$F(z_n)\in\Lambda,F(z_n')\in\Lambda'$$ for all large $$n$$, and with the end-points of $$\Lambda$$ and $$\Lambda'$$ equal to $$\zeta,\zeta'$$, respectively.

I wonder why the curves in the highlighted sentence would exist. I tried (by path-connectedness) to connect the $$z_n$$'s one by one, but the curve may not even have finite length (say if $$|z_n-z_{n+1}|=1/n$$). Does this use the assumption of $$P$$ being polygonal in an essential way? Or would simply connected with a "nice" boundary suffice? I think we are just working inside the disc $$D$$, but I'm not sure--maybe if $$P$$ has a weird shaped boundary, its intersection with $$D$$ would be too small to guarantee such curves.