Kenneth Falconer's Fractal Geometry gives a proof that the Julia set of a polynomial $f$ is the closure of the repelling periodic points of $f$. To show that $J(f)$ is contained in the closure of the repelling periodics, he gives the following argument, which I understand except for the claim that $w$ must be in the closure of repelling periodics.
Let $E=\{w\in J(f) : \exists v\neq w \text{ with } f(v)=w \text{ and } f'(v)\neq 0\}$. Suppose that $w\in E.$ Then there is an open neighborhood $V$ of $w$ on which we may find a local analytic inverse $f^{-1}: V\rightarrow \mathbb{C}\setminus V$ so that $f^{-1}(w)=v\neq w$ (just choose values of $f^{-1}(z)$ in a continuous manner). Define a family of analytic functions $\{h_k\}$ on $V$ by \begin{equation*} h_k(z) = \frac{f^k(z)-z}{f^{-1}(z)-z}. \end{equation*} Let $U$ be any open neighbourhood of $w$ with $U\subset V.$ Since $w\in J(f)$ the family $\{f_k\}$ and thus, from the definition, the family $\{h_k\}$ is not normal on $U.$ By Montel's theorem 14.5, $h_k(z)$ must take the value 0 or 1 for some $k$ and $z\in U.$ In the first case $f^k(z)=z$ for some $z\in U$; in the second case, $f^k(z)=f^{-1}(z)$ so $f^{k+1}(z)=z$ for some $z\in U.$ Thus $U$ contains a periodic point of periodic point of $f$, so $w$ is in the closure of the repelling periodic points for all $w\in E.$ Since $f$ is a polynomial, $E$ contains all of $J(f)$ except for a finite number of points. Since $J(f)$ contains no isolated points, by Proposition 14.9, $J(f)\subset \overline{E}$ is a subset of the closure of the repelling periodic points.
Why must $w$ be in the closure of the repelling periodics? Didn't we only show that there is a periodic point (not necessarily repelling) in every neighborhood of $w$? Is the periodic point we found in $U$ repelling, or is there another point that we must find? I feel like the point $v$ is an important piece here that I don't understand.
