Free arcs in the Universal dendrite

Question

Definition. A topological space $(X, \tau_X)$ is a continuum, if $X$ is a non-empty, metric, compact and connected space.

Definition. Let $X$ be a continuum and define $E(X)=\{p \in X: ord_X(p)=1\}$, $O(X)=\{p \in X: ord_X(p)=2\}$ and $R(X)=\{p \in X: ord_X(p) \geq 3\}$. $E(X)$, $O(X)$ and $R(X)$ denote the set of endpoints, ordinary points and branch points of $X$, respectively.

Definition. Let $(X, \tau_X)$ be a topological space, let $p \in X$, and let $\kappa$ be a cardinal number. We say that the order of $p$ in $X$, denoted by $\operatorname{ord}_X(p)$, is equal to $\kappa$ if the following conditions are satisfied:

1. The order of $p$ in $X$ is less than or equal to $\kappa$; that is, for every open subset $U$ of $X$ containing $p$, there exists an open subset $V$ of $X$ such that $p \in V \subseteq U$ and $|\operatorname{fr}_X(V)| \leq \kappa$.
2. $\kappa$ is the smallest cardinal number that satisfies condition (1); that is, for every cardinal number $\alpha < \kappa$, there exists an open subset $U_\alpha$ of $X$ containing $p$ such that for every open subset $W$ of $X$ with $p \in W \subseteq U_\alpha$, it holds that $|\operatorname{fr}_X(W)| > \alpha$.

Definition. An arc, is every topological space which is homeomorphic to the interval $[a,b]$. An arc with endpoints $a,b$ is denoted by $ab$.

Definition. Let $X$ be a topological space and $ab$ an arc, we said that $ab$ is a free arc if $ab \setminus \{a,b\}$ is open in $X$.

Definition. A continuum $X$ is called a dendrite if it is locally connected and contains no simple closed curves.

The Wazewski´s Universal dendrite is constructed as follows:

I would like understand why the set of branch points is dense, I don´t see it. I would appreciate your help.

Answer 1

As I understand, $D_\infty={\lim_\limits\leftarrow} \{D_i,f_i\}_{i=1}^\infty $ is the subspace of the Tychonoff product $\prod_{i=1}^\infty D_i$, consisting of sequences $(d_i)_{i=1}^\infty$ such that $f_i(d_{i+1})=d_i$ for each natural $i$. Moreover, for each natural $i$ let $\pi_i:D_\infty\to D_i$ be the restriction onto $D_\infty$ of the projection from $\prod_{j=1}^\infty D_j $ to $D_i$. On the other hand, adapting the construction of $D_\infty$ with the equality (1), we identify $D_i$ with the set $D_\infty\cap D_i^\infty$. Let $D=\bigcup_{i=1}^\infty D_i$. Then the set $D_\infty \setminus D$ is zero-dimensional, because it is contained in $C^\infty$. Let $A\subset D_\infty$ be any arc with distinct endpoints $x$ and $y$. Then $A\cap D$ is dense in $A$. So to show (3) we can assume that $x,y\in D$ and it suffices to show that the set $A \cap C_i$ is nonempty. There exists natural $N$ such that $x,y\in D_N$. Then $\pi_n(x)=x$ and $\pi_n(y)=y$ for each natural $n\ge N$. Let $h\colon[0,1]\to A$ be a homeomorphism such that $h(0)=x$ and $h(1)=y$. Then $\pi_nh$ is a continuous map from $[0,1]$ to $D_n$ such that $\pi_nh(0)=x\ne y=\pi_nh(1)$. I assume that the tree-like structure of $D_n$ ensures that there is a unique shortest path $A_n$ in $D_n$ from $x$ to $y$ (such that $A_n\subset g([0,1])$ for any continuous map $g\colon[0,1]\to D_n$ with $g(0)=x$ and $g(1)=y$) and, moreover, $A_n=A_N\subset D_N$ and $C_i\cap A_N$ is dense in $A_N$. All this looks intuitively true, but a rigorous proof can be not very short. Then for each $c\in C_i\cap A_N $, $((\pi_nh)^{-1}(c))_{n=N}^\infty$ is a nondecreasing sequence of closed subsets of a compact space $[0,1]$, there exists a point $c'\in \bigcap_{n=N}^\infty (\pi_nh)^{-1}(c)$. Then $h(c')=c\in C_i\cap A_N$.