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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. $(A \setminus E(A)) \cap R(X)= \emptyset$ then $A$ is an free arc in $X$.

My attempt: In order to prove it, 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 $\mid \operatorname{fr}_X(V_0) \mid \leq 2$. But how can we find an open set $B$ in $X$ such that $p \in B \subseteq A \setminus E(A)$?

I have been working in this result for a long time, but I have not been able to prove it or disprove it. I hope can you help me.

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Suppose $p \in A \setminus E(A)$ and choose open $U$ such that $p \in U \subseteq cl(U) \subseteq X \setminus \{a,b\}$ and $\mid \operatorname{fr}_X(U) \mid \leq 2$. $U$ must have two boundary points within $A \setminus E(A)$, say $c$ and $d$. $p$ must be an ordinary point of the arc $cd$.

Suppose $q \in U \setminus A$. Cover $cd$ with open sets which have only two boundary points and which do not contain $a$, $b$ or $q$. (Their boundaries will all be within $A$.) Take a finite subcovering and call its union $V$. The boundary of $V$ is contained in the union of the boundaries of the members of the subcovering, and so is contained in $A \setminus cd$. So $q \in cl(U) \setminus V = U \setminus cl(V)$, which is clopen, contradicting $X$ is connected.Hence there cannot be such a $q$, $U \setminus A$ is empty and $U \subset A \setminus E(A)$. Therefore $A \setminus E(A)$ is open.

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  • $\begingroup$ It's not clear to me why $U$ must have two boundary points within $A\setminus E(A)$ $\endgroup$ Commented Aug 29 at 19:11
  • $\begingroup$ As $U \cap A$ is a non-empty open subset of $A$ not containing the endpoints, it must have at least two boundary points in $A$ (in the subspace topology). But these are also boundary points of $U$ in $X$, which only has two. Therefore they must both be in $A$. However, as I had already realised, that didn't exclude them being $a$ or $b$. I've edited the answer to require $U$ is chosen so that $cl(U)$ doesn't include those points, which ensures the two boundary points are in $A \setminus E(A)$. $\endgroup$ Commented Aug 30 at 2:33
  • $\begingroup$ Now, I don't see why $V$ has two points in its boundary. $\operatorname{fr}_X(V) = \overline{V} \setminus V$, but $V$ is a finite union of open sets, so $\operatorname{fr}_X(V) = \left(\bigcup_{i=1}^k \overline{U_i}^X\right) \setminus \left(\bigcup_{i=1}^k U_i \right) \subseteq \bigcup_{i=1}^k \operatorname{fr}_X(U_i)$ and it is bounded by $2k$ for some $k$. I would like to prove it was bounded by $2$ and conclude its boundary has two points with a similar argument as at the beginning. $\endgroup$ Commented Aug 30 at 20:07
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    $\begingroup$ No.Just add a sequence of isolated points converging on a point inside the arc. E.g. $[0,1] \times \{0\} \cup \{(1/2,1/n)\}| n=1,2,. . \}$. $\endgroup$ Commented Sep 3 at 9:05
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    $\begingroup$ An arc $A$ in a space $X$ is free if, and only if, each point of $A$ other than the end-points has order 2 in $X$ and has a connected neighbourhood in X. $\endgroup$ Commented Sep 4 at 0:35

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