A result about trees in topology.

Question

before I begin, I would like to provide some definitions and theorems.

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.

Definition. A tree is a dendrite which is a finite union of arcs.

Lemma. Let $X$ be a dendrite, $B$ a subcontinuum of $X$, and $a$ a point. Then there exists a unique point $b \in B$ such that $ab \cap B = \{b\}$, and $b$ has the property that $b \in ay$ for every $y \in B$.

Theorem 3.3 (Brouwer's Reduction Theorem) Let $Y$ be a second countable space and $\mathcal{K}$ a non-empty family of closed subsets of $Y$ with the property that for every increasing sequence $K_0 \subset K_1 \subset K_2 \subset \ldots$ of elements of $\mathcal{K}$, there exists $K \in \mathcal{K}$ such that $K_n \subset K$ for all $n \geq 0$. Then $\mathcal{K}$ contains a maximal element in $\mathcal{K}$; that is, an element of $\mathcal{K}$ that is not contained in any other element of $\mathcal{K}$.

I need to prove the following statement:

Theorem. Let $X$ be a tree, $F$ a finite subset of $X$, and define $\mathcal{A}= \{ A \in X: A \text{ is a maximal free arc respect the property } F \cap A \subseteq E(A) \}$. Then $X = \bigcup\mathcal{A}$.

The statement looks none too simple, but I don´t get any proof.

My attempts:

My first idea to prove the equality was the following: take a point $x \in X \setminus (R(X) \cup E(X) \cup F)$, noting that this set is open. Consider the connected component of $X \setminus (R(X) \cup E(X) \cup F)$ that contains $x$, and takes its closure in $X$. My claim is that this closure is an arc. To prove this, I tried to use the characterization that an arc has exactly two non-cut points. In this regard, I looked for an alternative expression for the closure of the component, attempting to show that it is equal to a set $A \cup \{a,b\}$ where $a$ and $b$ are precisely the non-cut points. I tried to apply the Lemma above to find these points $a$ and $b$, but without success.

In parallel, I attempted a proof by contradiction, assuming that the closure of the component is not an arc. Since $X$ is a tree, then the closure is also a tree. From this assumption, two cases arise: one where some branch point is an interior point of the closure, and one where it is not. At this point, I got stuck.

I also tried to prove it by contraposition: suppose that the union is not equal to $X$, and from this show that $X$ is not a tree. If $X$ is not a tree, then it either contains simple closed curves, or is not a finite union of arcs, or it is not locally connected. I assumed that $X$ is locally connected and tried to prove that it contains simple closed curves. To this end, I took a point $x \in X$ that does not belong to the family $\mathcal{A}$, and tried to construct a family of free arcs having $x$ as an endpoint, but I did not achieve any conclusive result.

I tried to apply Brouwer Reduction Theorem to construct maximal arcs, applying this theorem to the family of free arcs whose endpoints lie in the set $F \cup E(X) \cup R(X)$. However, the key question arises: how do we know this family is not empty? it's a free arc?.

Can someone help me please?

Answer 1

First note that a dendrite $X$ is uniquely arcwise connected: it is arcwise connected, since it is a locally connected continuum, and between any pairs of distinct points $x,y\in X$ there exists a unique arc $[x,y]$, since if there were two distinct ones their union would contain a simple closed curve. Let $(x,y)=[x,y]\setminus\{x,y\}$.

Note also that an arc $[a,b]$ in a dendrite $X$ is free if and only if $(a,b)\cap R(X)=\varnothing$.

Now let $X$ be a tree, $F\subseteq X$ finite and $\mathcal A$ defined as above. Let $x\in X$ be arbitrary and we want to construct an arc $[a,b]\in\mathcal A$ with $x\in[a,b]$. Note that arcs $[a,b]\in\mathcal A$ are characterized by the properties that $(a,b)\cap (R(X)\cup F)=\varnothing$ and $\{a,b\}\subseteq E(X)\cup R(X)\cup F$.

If $x\in E(X)\cup R(X)$ let $y\in F\cup R(X)$ be such that $[x,y]\cap(F\cup R(X))\subseteq\{x,y\}$, where such an $y$ exists since $F\cup R(X)$ is finite. Take $a=x$, $b=y$ and verify that the arc $[a,b]$ is in $\mathcal A$.

If instead $x\in O(X)$ let $a,b\in E(X)\cup R(X)\cup F$ be such that $x\in[a,b]$ and $(a,b)\cap (R(X)\cup F)=\varnothing$. Once again verify that the arc $[a,b]$ is in $\mathcal A$ by the characterization above.