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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.

Notation. If $X$ is a topological space and $A\subseteq X$, then we denoted the boundary of $A$ in $X$ by $fr_X(A)$.

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 $|fr_X(V)| \leq \kappa$.
  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 $|fr_X(W)| > \alpha$.

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 $X$.

I would like to prove: Let $X$ be a continuum and $A$ an arc. $A$ is an free arc in $X$ if and only if $(A \setminus E(A)) \cap R(X)= \emptyset$.

I haven’t been able neither to prove nor to disprove that result, my attempt:

Suppose that $A$ is a free arc, then $A\setminus E(A)$ is open in $X$. I wanna prove that for every $x \in A\setminus E(A)$, $ord_X(x) \leq 2$. The only fact we know is $ord_A(x)=2$, and that is all I could do.

I hope you can help me pls

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  • $\begingroup$ You already asked this question but apparently deleted it. Moreover, you did not define $\mathrm{fr}$. $\endgroup$ Commented Jul 20 at 12:05
  • $\begingroup$ Yes, I deleted it because the theorem I used to prove sufficiency was false. Here, $fr$ denotes the boundary of a set. $\endgroup$ Commented Jul 20 at 16:27

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If $A=ab$ is a free arc, then $A\setminus E(A)=ab\setminus\{a,b\}$ is open in $X$, and for all $p\in A\setminus E(A)$, for all open set $U$ in $X$ containing $p$, the intersection $U\cap(A\setminus E(A))$ is an open set in $X$ containing $p$ and its connected component containing $p$ is an open interval $I$ whose boundary has only two elements.

EDIT: Here are more details to prove the last sentence. The closure of $I$ in $A$ can be written $I\cup\{u,v\}$ where $u$ and $v$ are its (distinct) boundary points in $A$. But it is a closed subset of the compact set $A$ so it is compact, hence it is a closed subset of $X$. Therefore it contains the closure of $I$ in $X$. Conversely, the closure in $A$ must be included in the closure in $X$, so these closures are equal.
Finally, the boundary of $I$ in $X$ is the pair $\{u,v\}$.

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  • $\begingroup$ I thought I had seen it, but I was mistaken. How do we know that $K$ is an open set in $X$? All we know is that $K$ is a connected subset of the open set $U \cap (A \setminus E(A))$. Since $X$ is not locally connected, we can’t guarantee that $K$ is an open set in $X$. So I think we need that extra hypothesis. $\endgroup$ Commented Jul 20 at 21:36
  • $\begingroup$ An open subset of an open subset is an open subset. $\endgroup$ Commented Jul 21 at 3:56
  • $\begingroup$ In order to prove necessity, all I have been able to do is: Suppose that $(A \setminus E(A)) \cap R(X)$ is empty. We want to prove that $A$ is a free arc. Since, for each $y \in A\setminus E(A)$, $\operatorname{ord}_A(y)=2$, using our hypothesis we have that $\operatorname{ord}_X(y)= 2$ for each $y \in A\setminus E(A)$. Let $p \in A \setminus E(A)$. We know that $A\setminus E(A)$ is an open set in $A$, so there is an open set $V$ in $X$ such that $p \in V \cap A \subseteq A\setminus E(A)$. Also, we have that there is an open subset $V_0$ of $X$ such that $p \in V_0 \subseteq V$ and ... $\endgroup$ Commented Jul 21 at 20:45
  • $\begingroup$ ... $\mid \operatorname{fr}_X(V_0) \mid \leq 2$. But how can we find an open set $B$ in $X$ such that $p \in B \subseteq A \setminus E(A)$? $\endgroup$ Commented Jul 21 at 20:48
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    $\begingroup$ I suggest to open a new question now that this one has an accepted answer. $\endgroup$ Commented Jul 22 at 9:24

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