Q: The boundary of a totally invariant set is also totally invariant Let $P$ be a degree-two polynomial. A subset $X \subset \mathbb{C}$ is totally invariant if for all $z \in X$ we have $P(z) \in X$ and $P^{−1}(z) = \{w \in \mathbb{C} | P(w) = z\} \subset X$.
I want to prove that if $X$ is totally invariant then its boundary, $\partial X$ is also totally invariant.
Let $Y = \mathbb{C} \setminus X$. We know that $Y$ is totally invariant as $X$ is totally invariant.
Suppose $z_0 \in \partial X$, we want to see that $P(z_0) \in \partial X$. That is, that $\forall U \text{ neigbourhood of } P(z_0)$ we have $U \cap X \neq \emptyset$ and $U \cap Y \neq \emptyset$. By continuity of P, $\exists V \text{ neigbourhood of } z_0$ such that $P(V) \subset U$. Now, as $z_0 \in \partial X$, we have that $V \cap X \neq \emptyset$ and $V \cap Y \neq \emptyset$ and taking into account that $X$ and $Y$ are totally invariant,
$$P(V \cap X) \subset U \cap X \neq \emptyset$$ and $$P(V \cap Y) \subset U \cap Y \neq \emptyset$$.
Now, how do you prove that if $w$ is such that $P(w) = z_0 \in \partial X$, then $w \in \partial X$.

A: Suppose $p:C \to C$ be a continuous map of degree $2$ and $X$ closed (resp. open.)
Let $p(w) \in \partial X$ for some $w$. We want to show that $w \in \partial X$.
Notice that $w \in \overline{X}$ (that is the closure of $X$) which follows trivially.
We now show that $w \notin int(X)$

Suppose For sake of contradiction $w \in int(X)$ and $p(w) \notin int(X)$ (trivial for when $p(w) \in int(X)$). Now here is a Lemma (Let me know if you need the proof in the comment section. We use the fact that the boundary contains the limit points

Lemma: Every continuous map $f:U \to U$ (where $U$ is any topological space) is sequentially continuous meaning if $x_n$ converges to $x$ ($x_n \to x$) then $f(x_n)$ converges to $f(x)$ $(f(x_n) \to f(x))$
