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\title{\bf Contractible edges\\ in 2-connected locally finite graphs}

% input author, affilliation, address and support information as follows;
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% the street address 

\author{Tsz Lung Chan\\
\small Mathematisches Seminar\\[-0.8ex]
\small Universit\"at Hamburg\\[-0.8ex] 
\small Hamburg, Germany\\
\small\tt tlchantl@gmail.com\\
}

%\author{Dagwood Remifa\thanks{Supported by NASA grant ABC123.}\\
%\small Department of Inconsequential Studies\\[-0.8ex]
%\small Solatido College\\[-0.8ex] 
%\small North Kentucky, U.S.A.\\
%\small\tt remifa@dis.solatido.edu\\
%\and
%Forgotten Second Author \qquad  Forgotten Third Author\\
%\small School of Hard Knocks\\[-0.8ex]
%\small University of Western Nowhere\\[-0.8ex]
%\small Nowhere, Australasiaopia\\
%\small\tt \{fsa,fta\}@uwn.edu.ao
%}

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% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{May 30, 2014}{May 25, 2015}{Jun 15, 2015}\\
\small Mathematics Subject Classifications: 05C40, 05C45, 05C63}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
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\begin{abstract}
In this paper, we prove that every contraction-critical 2-connected infinite graph has no vertex of finite degree and contains uncountably many ends. Then, by investigating the distribution of contractible edges in a 2-connected locally finite infinite graph $G$, we show that the closure of the subgraph induced by all the contractible edges in the Freudenthal compactification of $G$ is 2-arc-connected. Finally, we characterize all 2-connected locally finite outerplanar graphs nonisomorphic to $K_3$ as precisely those graphs such that every vertex is incident to exactly two contractible edges as well as those graphs such that every finite bond contains exactly two contractible edges.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} contractible edge, Hamilton cycle, outerplanar, infinite graph
\end{abstract}

\section{Introduction}

Since the pioneering work of Tutte \cite{tutte} who proved that every 3-connected finite graph nonisomorphic to $K_4$ contains a contractible edge, a lot of research has been done on contractible edges in finite graphs. One may consult the survey paper by Kriesell \cite{survey} for details.

%Thomassen \cite{thomassen} proved that every $k$-connected triangle-free finite graph has a $k$-contractible edge. Dean \cite{dean} generalized Thomassen's result as follows.

%\begin{theorem}[Dean \cite{dean}]\label{dean}
%Let $G$ be a $k$-connected finite graph $(k\geq 2)$ which is triangle-free or has minimum degree at least $\lfloor \frac{3k}{2} \rfloor$. Then the subgraph $G_C$ induced by all the $k$-contractible edges in $G$ is $2$-connected.
 %\end{theorem}

%\begin{theorem}[Dean \cite{dean}]\label{dean}
%Let $G$ be a $k$-connected finite graph $(k\geq 2)$ which is triangle-free or has minimum degree at least $\lfloor \frac{3k}{2} \rfloor$. Then

%\begin{enumerate}[\rm (a)]
%\item the subgraph $G_C$ induced by all the $k$-contractible edges in $G$ is $2$-connected.
 
%\item every vertex of $G$ is incident to at least two $k$-contractible edges.
%\end{enumerate}

%For $k=2$, we only need the condition that $G$ is nonisomorphic to $K_3$, and (b) holds for $2$-connected locally finite graphs.

%The condition that $G$ is triangle-free or has minimum degree at least $\lfloor \frac{3k}{2} \rfloor$ can be dropped when $k=2$. Also, (b) holds for $2$-connected locally finite graphs.
%\end{theorem}

%For $k=2$, the above theorem is true for all finite graphs nonisomorphic to $K_3$. This follows from the well-known fact that every edge in a 2-connected graph nonisomorphic to $K_3$ can either be deleted or contracted so that the resulting graph remains 2-connected.

For any 2-connected graph nonisomorphic to $K_3$, we have the well-known fact that every edge can either be deleted or contracted so that the resulting graph remains 2-connected. This immediately leads to the following result.

\begin{theorem}\label{dean2}
Let $G$ be a $2$-connected finite graph nonisomorphic to $K_3$. Then the subgraph induced by all the contractible edges in $G$ is $2$-connected.
 \end{theorem}

Wu \cite{wu} investigated the distribution of contractible elements in matroids and extended Theorem \ref{dean2} to simple 2-connected matroids. He also characterized all simple 2-connected matroids $M$ having exactly $r(M)+1$ contractible elements (where $r(M)$ is the rank of $M$) as those matroids isomorphic to a graphic matroid of an outerplanar Hamiltonian graph.

%Combining this result with Theorem \ref{dean2}, we obtain the following characterization of 2-connected outerplanar finite graphs.

\begin{theorem}[Wu \cite{wu}]\label{wu}
Let $G$ be a $2$-connected finite graph nonisomorphic to $K_3$. Then every vertex of $G$ is incident to exactly two contractible edges if and only if $G$ is outerplanar.
\end{theorem}

On the other hand, only a few results were known for contractible edges in infinite graphs. For example, Mader \cite{mader} showed that every contraction-critical locally finite infinite graph has infinitely many triangles. Kriesell \cite{kriesell} provided a method of constructing contraction-critical $k$-connected infinite graphs $(k\geq 2)$. In Section 3, we will prove that every contraction-critical 2-connected infinite graph contains vertices of infinite degree only and has uncountably many ends.

A natural way to extend Theorems \ref{dean2} and \ref{wu} is to consider locally finite infinite graphs. Notice that Theorem \ref{dean2} is no longer true as demonstrated by the infinite double ladder (the cartesian product of a double ray and $K_2$). The subgraph $G_C$ induced by all the contractible edges is the disjoint union of two double rays and is not even connected. Interestingly, the situation changes dramatically by looking at the graph from a topological viewpoint as introduced by Diestel and K{\"u}hn \cite{cycles1,cycles2,tst}. By adding the two ends of the double ladder to $G_C$, the resulting closure $\overline{G_C}$ is a circle and is 2-arc-connected. In Section 4, we will prove that for every 2-connected locally finite infinite graph $G$, $\overline{G_C}$ is 2-arc-connected.

Returning to Theorem \ref{wu}, the backward direction is straightforward. For the forward direction, by Theorem \ref{dean2}, $G_C$ is spanning and 2-connected. Since every vertex is incident to exactly two contractible edges, $G_C$ is a Hamilton cycle. Then it is easy to see that $G$ is outerplanar. When extending to locally finite infinite graphs, we now need the non-trivial statement that if $G$ is a 2-connected locally finite infinite graph such that every vertex is incident to exactly two contractible edges, then $\overline{G_C}$ is a Hamilton circle. This will be proved in Section 5. We will use it to prove an infinite analog of Theorem \ref{wu} for any 2-connected locally finite graph $G$ nonisomorphic to $K_3$. Also we will show that $G$ is outerplanar if and only if every finite bond of $G$ contains exactly two contractible edges.

%Returning to Theorem \ref{wu}, the backward direction can be proved using Chartrand and Harary's characterization of outerplanar graphs \cite{outerplanar}. For the forward direction, by Theorem \ref{dean2}, $G_C$ is spanning and 2-connected. Since every vertex is incident to exactly two contractible edges, $G_C$ is a Hamilton cycle. Then it is easy to see that $G$ is outerplanar. When extending to locally finite infinite graphs, we now need the non-trivial statement that if $G$ is a 2-connected locally finite infinite graph such that every vertex is incident to exactly two contractible edges, then $\overline{G_C}$ is a Hamilton circle. This will be proved in Section 5. We will use it to prove an infinite analog of Theorem \ref{wu} for any 2-connected locally finite graph $G$ nonisomorphic to $K_3$ . Also we will show that $G$ is outerplanar if and only if every finite bond of $G$ contains exactly two contractible edges.

%As an immediate corollary, we obtain one of the main results of this paper.

%\begin{theorem}\label{outerplanar}
%Every $2$-connected outerplanar locally finite graph is Hamiltonian.
%\end{theorem}

\section{Definitions}

All basic graph-theoretical terminology can be found in Diestel \cite{diestel}. Unless otherwise stated, all graphs considered in this paper can be finite or infinite. An edge of a $k$-connected graph is said to be $k$-\emph{contractible} if its contraction results in a $k$-connected graph. Otherwise, it is called $k$-\emph{non-contractible}. A $k$-connected graph in which every edge is $k$-non-contractible is called \emph{contraction-critical} $k$-connected. We simply write 2-contractible as \emph{contractible}. Let $G=(V,E)$ be a 2-connected graph. Denote the set of all contractible edges in $G$ by $E_C$ and the subgraph induced by all the contractible edges by $G_C:=(V,E_C)$. Let $X$ and $Y$ be two disjoint subsets of $V$. An \emph{$X$-$Y$ path} $P$ is a path such that only the starting vertex of $P$ lies in $X$ and only the ending vertex of $P$ lies in $Y$. Denote the set of all edges between $X$ and $Y$ by $E_G(X,Y)$. If $X$ and $Y$ form a partition of $V$, then $E_G(X,Y)$ is called a \emph{cut}. A minimal non-empty cut is a \emph{bond}. Denote the set of all edges incident to a vertex $x$ by $E_G(x)$ and the set of all neighbors of $x$ by $N_G(x)$. Define $N_G(X):=(\bigcup_{x \in X}N_G(x))\setminus X$. A set $S$ of $k$ vertices is called an \emph{$k$-separator} if $G-S$ is not connected.

Let $G$ be a locally finite graph. A \emph{ray} is a $1$-way infinite path, a \emph{double ray} is a $2$-way infinite path, and the subrays of a ray or double ray are its \emph{tails}. An \emph{end} is an equivalence class of rays where two rays are \emph{equivalent} if no finite set of vertices separates them. Denote the set of the ends by $\Omega(G)$. We define a topological space, denoted by $|G|$, on $G$ together with its ends, which is known as the \emph{Freudenthal compactification} of $G$ as follows. View $G$ as a 1-complex. Thus, every edge is homeomorphic to the unit interval. The basic open neighborhoods of a vertex $x$ consists of a choice of half-open half edges $[xz)$, one for each incident edge $xy$, where $z$ is any interior point of $xy$. For an end $\omega\in\Omega(G)$, we take as a basic open neighborhood the set of the form: $\hat{C}(S,\omega):=C(S,\omega)\cup\Omega(S,\omega)\cup\mathring{E}(S,\omega)$, where $S\subseteq V$ is a finite set of vertices, $C(S,\omega)$ is the component of $G-S$ in which every ray from $\omega$ has a tail, $\Omega(S,\omega)$ is the set of all ends whose rays have a tail in $C(S,\omega)$, and $\mathring{E}(S,\omega)$ is the set of all interior points of edges between $S$ and $C(S,\omega)$. Let $H$ be a subgraph of $G$. Then the closure of $H$ in $|G|$ is called a \emph{standard subspace} and is denoted by $\overline{H}$. We say $H$ \emph{contains} a point $x$ of $|G|$ if $x\in\overline{H}$.

Let $X$ and $Y$ be two topological spaces. A continuous map from the unit interval $[0,1]$ to $X$ is a \emph{path} in $X$. A homeomorphic image of $[0,1]$ in $X$ is called an \emph{arc} in $X$. This induces an ordering $<$ for the points in the arc. The images of $0$ and $1$ are the \emph{endpoints} of the arc. An arc in $X$ with endpoints $x$ and $y$ is called an \emph{$x$-$y$ arc}. A homeomorphic image of the unit circle in $X$ is called a \emph{circle} in $X$. A (\emph{path-})\emph{component} of $X$ is a maximal (path-)connected set in $X$. $X$ is \emph{$2$-connected} (\emph{$2$-arc-connected}) if for all $x\in X$, $X\setminus x$ is connected (arc-connected). We say $X$ can be \emph{embedded} in $Y$ if there exists an injective continuous function $\phi: X\rightarrow Y$ such that $X$ is homeomorphic to $\phi(X)$ in the subspace topology of $Y$. Then $\phi$ is called an \emph{embedding} of $X$ in $Y$. Take $Y$ to be $\mathbb{R}^2$. A component of $\mathbb{R}^2\setminus\phi(X)$ is called a \emph{face} of $\phi(X)$ in $\mathbb{R}^2$. A graph $G$ is \emph{planar} if $G$ can be embedded in $\mathbb{R}^2$. A graph $G$ is \emph{outerplanar} if there exists an embedding $\phi$ of $G$ in $\mathbb{R}^2$ such that there is a face $f$ of $\phi(G)$ in $\mathbb{R}^2$ whose boundary $\partial f$ contains all the vertices of $G$. Chartrand and Harary \cite{outerplanar} characterized outerplanar finite graphs as precisely those graphs that do not contain a $K_{2,3}$- or $K_4$- subdivision.
%A continuous map from the unit interval $[0,1]$ to $|G|$ is a \emph{path} in $|G|$. A homeomorphic image of $[0,1]$ in $|G|$ is called an \emph{arc}. This induces an ordering $<$ for the points in the arc. The images of $0$ and $1$ are the \emph{endpoints} of the arc. Let $A$ be an arc in $|G|$ with endpoints $x$ and $y$. Then $A$ is called an \emph{$x$-$y$ arc}. We say an arc is an \emph{$\omega$-arc} if the end $\omega$ is one of its endpoints and unless otherwise stated, it corresponds to the image of $1$. A homeomorphic image of the unit circle in $|G|$ is called a \emph{circle}. An edge between two non-adjacent vertices in a circle is called a \emph{chord}. Two chords $x_1x_2$ and $y_1y_2$ of a circle $C$ are \emph{overlapping} if $x_1,y_1,x_2,y_2$ appear in the cyclic order of $C$. A \emph{Hamilton} circle in $|G|$ is a circle in $|G|$ that contains every vertex of $G$. Consider an arc $A$ in $|G|$. Let $x$ be a point in $A$ corresponding to a vertex of $G$. Define $x^-$ and $x^+$ to be the vertex immediately before and after $x$ in $A$ if they exist.

%Suppose $x_1Px_2$ and $y_1Qy_2$ are two disjoint $A$-$A$ paths in $G$ with $x_1<x_2$ and $y_1<y_2$. Then $P$ and $Q$ are \emph{overlapping} if $x_1<y_1<x_2<y_2$ or $y_1<x_1<y_2<x_2$. Otherwise, $P$ and $Q$ are \emph{nesting}.

Suppose $A$ is an arc in $|G|$ and $x$ is a vertex in $A$. Then the vertex immediately before $x$ in $A$ if exists is denoted by $x^-$ and the vertex immediately after $x$ in $A$ if exists is denoted by $x^+$. An arc in $|G|$ is an \emph{$\omega$-arc} if the end $\omega$ is one of its endpoints and unless otherwise stated, it corresponds to the image of $1$. Following Bruhn and Stein \cite{enddeg}, we define the \emph{end degree} of an end $\omega$ in $G$ as the supremum over the cardinalities of sets of edge-disjoint rays in $\omega$, and denote this number by $deg_G(\omega)$. In fact, they proved that this is equal to the supremum over the cardinalities of sets of edge-disjoint $\omega$-arcs in $|G|$. For a subgraph $H$ of $G$, define the degree of $\omega$ in $H$ as the supremum over the cardinalities of sets of edge-disjoint $\omega$-arcs in $\overline{H}$ which is denoted by $deg_H(\omega)$.
  
%, and this holds true for infinite graphs as well.


\section{Contraction-critical 2-connected infinite graphs}

It is well-known that the only contraction-critical 2-connected finite graph is $K_3$. However, there are infinitely many contraction-critical 2-connected infinite graphs as shown by the following construction due to Kriesell \cite{kriesell}. Define $G_0:=\emptyset$ and let $G_1$ be any 2-connected finite graph. Suppose we have constructed $G_n$ such that $G_{n-1}\subsetneq G_n$. For each edge $xy$ in $E(G_n)\setminus E(G_{n-1})$, add a new $x$-$y$ path of length at least 2. The resulting graph is $G_{n+1}$. Repeat the process inductively. Then the graph $G:=\bigcup_{i\geq 1}G_i$ is a contraction-critical 2-connected infinite graph. Note that $G$ has no vertex of finite degree and has uncountably many ends. We will show that this holds in general for any contraction-critical 2-connected infinite graph. First, we state a fundamental fact about contractible edges in 2-connected graphs.

\begin{lemma}\label{fund1}
Let $G$ be a $2$-connected graph nonisomorphic to $K_3$ and $e$ be an edge of $G$. Then $G-e$ or $G/e$ is $2$-connected.
\end{lemma}

Now, we can develop some tools that will be used for the rest of the paper.

\begin{lemma}\label{fund2}
Let $G$ be a $2$-connected graph nonisomorphic to $K_3$, and $e$ and $f$ be two non-contractible edges of $G$. Then $f$ is a non-contractible edge of $G-e$.
\end{lemma}

\noindent{\em Proof.}
By Lemma \ref{fund1}, $G-e$ is 2-connected. Since $V(f)$ is a 2-separator of $G$, $V(f)$ is also a 2-separator of $G-e$ and $f$ is a non-contractible edge of $G-e$.
\hfill $\Box$

\begin{lemma}\label{fund3}
Let $G$ be a $2$-connected graph nonisomorphic to $K_3$ and $F$ be a finite subset of $E(G)$.

\begin{enumerate}[\rm (a)]
\item If $G-F$ is disconnected, then $F$ contains at least two contractible edges.
 
\item If $G-F$ is connected but not $2$-connected, then $F$ contains at least one contractible edge.
\end{enumerate}
\end{lemma}

\noindent{\em Proof.}
For (a), suppose $F$ contains at most one contractible edge. Then by Lemma \ref{fund1} and \ref{fund2}, we can delete all the non-contractible edges in $F$ and the resulting graph is still 2-connected, a contradiction.

For (b), suppose all edges in $F$ are non-contractible. Then by Lemma \ref{fund1} and \ref{fund2}, we can delete all edges in $F$ and $G-F$ is still 2-connected, a contradiction.
\hfill $\Box$

\begin{lemma}\label{contexist}
Let $G$ be a $2$-connected graph nonisomorphic to $K_3$. Let $\{x,y\}$ be a $2$-separator of $G$ and $C$ be a component of $G-x-y$. If $|E_G(x,C)|$ is finite, then $E_G(x,C)$ contains a contractible edge.
\end{lemma}

\noindent{\em Proof.}
Note that $y$ is a cutvertex of $G-E_G(x,C)$. By Lemma \ref{fund3}(b), $E_G(x,C)$ contains a contractible edge.
\hfill $\Box$

\begin{lemma}\label{infnoncont}
Let $G$ be a $2$-connected graph nonisomorphic to $K_3$ and $x$ be a vertex of $G$. Suppose all edges incident to $x$ are non-contractible. Then 

\begin{enumerate}[\rm (a)]
\item $x$ has infinite degree.

\item For any edge $xy$ incident to $x$, every component of $G-x-y$ contains infinitely many neighbors of $x$.
\end{enumerate}
\end{lemma}

\noindent{\em Proof.}
For (a),  Suppose $x$ has finite degree. By applying Lemma \ref{fund3}(a) to $E_G(x)$, $x$ is incident to at least two contractible edges, a contradiction.

For (b), let $C$ be a component of $G-x-y$. By Lemma \ref{contexist}, $E_G(x,C)$ contains infinitely many edges.
\hfill $\Box$

%\begin{lemma}\label{critical}
%Let $G$ be a contraction-critical $2$-connected infinite graph. Then, for each vertex $x$ in $G$, we can construct a ray $R_x$ such that $x$ dominates $R_x$ directly at $x_0,x_1,x_2,\ldots$ with $x_0,x_1,x_2,\ldots$ appearing in that order in $R_x$, and $\{x,x_i\}$ separates $\{x_0,\ldots,x_{i-1}\}$ from $\{x_{i+1},x_{i+2},\ldots\}$ in $G$ for $i=1,2,\ldots$.
%\end{lemma}

%\noindent{\em Proof.}
%We will construct the ray $R_x$ inductively. Let $x_0$ be a neighbor of $x$. Since $xx_0$ is non-contractible, $\{x,x_0\}$ is a 2-separating set. Let $C_1$ be a component of $G-x-x_0$, $x_1$ be a neighbor of $x$ in $C_1$, and $y_1$ be a neighbor of $x_0$ in $C_1$. Then there is a path $Q_1$ between $y_1$ and $x_1$ in $C_1$. Define $P_0:=x_0$ and $P_1:=x_0y_1Q_1x_1$. As $xx_1$ is non-contractible, there exists a component $C_2$ of $G-x-x_1$ disjoint from $P_1$. Suppose we have obtained $k+1$ neighbors of $x$: $x_0,x_1,\ldots,x_k$, and a $x_0-x_k$ path $P_k$ which is disjoint from $x$ and contains $x_0,\ldots,x_k$ appearing in this order. Note that $P_{k-1}\subseteq P_k$. Since $xx_k$ is non-contractible, there exists a component of $C_{k+1}$ of $G-x-x_k$ disjoint from $P_k$. Let $x_{k+1}$ be a neighbor of $x$ in $C_{k+1}$ and $y_{k+1}$ be a neighbor of $x_k$ in $C_{k+1}$. Then there is a path $Q_{k+1}$ between $y_{k+1}$ and $x_{k+1}$ in $C_{k+1}$. Now, $\bigcup_{i\geq 1}P_i$ is the desired ray $R_x$.
%\hfill $\Box$
%\\

\begin{theorem}\label{uncountend}
Let $G$ be a contraction-critical $2$-connected infinite graph. Then every vertex of $G$ has infinite degree and $G$ has uncountably many ends.
\end{theorem}

\noindent{\em Proof.} By Lemma \ref{infnoncont}(a), every vertex of $G$ has infinite degree.

Next, we will construct a rooted binary infinite tree $T$ in $G$ together with edges incident to each vertex of $T$ with the following properties:

\begin{enumerate}[\rm (1)]

\item The root of $T$ is denoted by $x$.
%\item The root of $T$ is denoted by $x_{\emptyset}$.

\item The vertices of $T$ are denoted by $x_{n_1 n_2 \ldots n_k}$ where $k\in\mathbb{N}$ and $n_i\in\{0,1\}$ for $1\leq i\leq k$. For $k=0$, define $x_{n_1 n_2 \ldots n_k}:=x$.

\item Each vertex $x_{n_1 n_2 \ldots n_k}$ of $T$ is adjacent to two vertices $x_{n_1 n_2 \ldots n_k 0}$ and $x_{n_1 n_2 \ldots n_k 1}$ in $T$.

\item For each vertex  $x_{n_1 n_2 \ldots n_k}$ of $T$, there exists an edge $x_{n_1 n_2 \ldots n_k}y_{n_1 n_2 \ldots n_k}$ in $G$ such that $y_{n_1 n_2 \ldots n_k}$ does not lie in $T$.

\item The subtree of $T$ rooted at $x_{n_1 n_2 \ldots n_k}$ is defined as 

$T_{n_1 n_2 \ldots n_k}:=T[\bigcup_{i=0}^{\infty}\bigcup_{(m_1,m_2,\ldots,m_i)\in\{0,1\}^i} x_{n_1 n_2 \ldots n_k m_1 m_2 \ldots m_i}]$.

For fixed $n_1,n_2,\ldots,n_k$, $\bigcup_{j=0}^k \{x_{n_1 n_2 \ldots n_j},y_{n_1 n_2 \ldots n_j}\}$ separates $T_{n_1 n_2 \ldots n_k 0}$ and $T_{n_1 n_2 \ldots n_k 1}$ in $G$.
\end{enumerate}

Each ray in $T$ starting at $x$ is of the form: $xx_{n_1} x_{n_1n_2} x_{n_1n_2n_3}\ldots$. Let $R:=xx_{n_1} x_{n_1n_2}\newline x_{n_1n_2n_3}\ldots$ and $Q:=xx_{m_1} x_{m_1m_2} x_{m_1m_2m_3}\ldots$ be two distinct rays in $T$. Then there exists a smallest $k$ such that $n_i=m_i$ for all $i\leq k$ and $n_{k+1}\neq m_{k+1}$. By property (5) above, $\bigcup_{j=0}^k \{x_{n_1 n_2 \ldots n_j},y_{n_1 n_2 \ldots n_j}\}$ separates $R$ and $Q$ in $G$. Therefore, each ray in $T$ starting at $x$ belongs to a unique end of $G$, and $G$ has uncountably many ends.

Now, it remains to construct $T$ inductively. Let $x$ be any vertex in $G$. Define $T_0:=(\{x\},\emptyset)$. Choose any edge incident to $x$ in $G$, say $xy$. Let $C_0$ and $C_1$ be any two components of $G-x-y$. Let $x_0$ be a neighbor of $x$ in $C_0$ and $x_1$ be a neighbor of $x$ in $C_1$. Define $T_1:=(\{x,x_0,x_1\},\{xx_0,xx_1\})$. Note that $N_G(C_0)\subseteq\{x,y\}$ and $N_G(C_1)\subseteq\{x,y\}$. Also, $G-C_0$ and $G-C_1$ are both connected.

%Consider $x_{n_1}$ where $n_1\in\{0,1\}$. Since $x_{n_1}$ has infinite degree and $N_G(x_{n_1})\setminus C_{n_1}\subseteq N_G(C_{n_1})\subseteq \{x,y\}$ is finite, all but finitely many neighbors of $x_{n_1}$ lie in $C_{n_1}$. Let $z$ be any neighbor of $x_{n_1}$ in $C_{n_1}$. If $C_{n_1}-x_{n_1}-z$ is connected, then $C_{n_1}-x_{n_1}-z$ and $G-C_{n_1}$ are the only two components of $G-x_{n_1}-z$. By Lemma \ref{infnoncont}(b), $G-C_{n_1}$ contains infinitely many neighbors of $x_{n_1}$ contradicting $N_G(x_{n_1})\setminus C_{n_1}\subseteq N_G(C_{n_1})\subseteq \{x,y\}$ which is finite. Therefore, $C_{n_1}-x_{n_1}-z$ is not connected.

%Note that at least one component of $C_{n_1}-x_{n_1}-z$ is adjacent to $x_{n_1}$. If not, then, by the 2-connectedness of $G$, each component of $C_{n_1}-x_{n_1}-z$ has a neighbor in $\{x,y\}$. This implies $G-x_{n_1}-z$ is connected, a contradiction. Suppose there are two components of $C_{n_1}-x_{n_1}-z$, say $D$ and $D'$, that are both adjacent to $x_{n_1}$. Then choose $y_{n_1}:=z$, $C_{n_10}:=D$ and $C_{n_11}:=D'$. Suppose only one component of $C_{n_1}-x_{n_1}-z$ is adjacent to $x_{n_1}$, say $C$. Each component of $C_{n_1}-x_{n_1}-z$ other than $C$ is adjacent to $z$ by the connectedness of $C_{n_1}$ and has a neighbor in $\{x,y\}\subseteq G-C_{n_1}$ by the 2-connectedness of $G$. Since $G-x_{n_1}-z$ is not connected, $N_G(C)=\{x_{n_1},z\}$. Let $z'$ be a neighbor of $x_{n_1}$ in $C$. Then one component $D$ of $C_{n_1}-x_{n_1}-z'$ contains $z$ and the union of components of $C_{n_1}-x_{n_1}-z$ other than $C$. Since $G-x_{n_1}-z'$ is not connected, $D$ cannot be the only component of $C_{n_1}-x_{n_1}-z'$. Let $D'$ be any component of $C_{n_1}-x_{n_1}-z'$ other than $D$. Then $D'$ lies in $C$ and $N_G(D')\subseteq \{x_{n_1},z'\}\cup (N_G(C)-z)=\{x_{n_1},z'\}$. Now, choose $y_{n_1}:=z'$, $C_{n_10}:=D$ and $C_{n_11}:=D'$.

%In both cases, $x_{n_1}y_{n_1}$ lies in $C_{n_1}$, and $C_{n_10}$ and $C_{n_11}$ are two components of $C_{n_1}-x_{n_1}-y_{n_1}$ that are adjacent to $x_{n_1}$. Let $x_{n_10}$ be a neighbor of $x_{n_1}$ in $C_{n_10}$ and $x_{n_11}$ be a neighbor of $x_{n_1}$ in $C_{n_11}$. Define $T_2:=(\{x,x_0,x_1,x_{00},x_{01},x_{10},x_{11}\},\{xx_0,xx_1,x_0x_{00},x_0x_{01},x_1x_{10},x_1x_{11}\})$. Note that $N_G(C_{00})\subseteq N_G(C_0)\cup\{x_0,y_0\}\subseteq\{x,y,x_0,y_0\}$, $N_G(C_{01})\subseteq N_G(C_0)\cup\{x_0,y_0\}\subseteq\{x,y,x_0,y_0\}$, $N_G(C_{10})\subseteq N_G(C_1)\cup\{x_1,y_1\}\subseteq\{x,y,x_1,y_1\}$ and $N_G(C_{11})\subseteq N_G(C_1)\cup\{x_1,y_1\}\subseteq\{x,y,x_1,y_1\}$. Also, $G-C_{00}$, $G-C_{01}$, $G-C_{10}$ and $G-C_{11}$ are all connected.

Suppose we have constructed the rooted binary tree $T_k$ where

$V(T_k):=\bigcup_{i=0}^{k}\bigcup_{(n_1,n_2,\ldots,n_i)\in\{0,1\}^i} x_{n_1 n_2 \ldots n_i}$ and

$E(T_k):=\bigcup_{i=0}^{k-1}\bigcup_{(n_1,n_2,\ldots,n_i)\in\{0,1\}^i}\{x_{n_1 n_2 \ldots n_i}x_{n_1 n_2 \ldots n_i 0}, x_{n_1 n_2 \ldots n_i}x_{n_1 n_2 \ldots n_i 1}\}$ such that

\begin{enumerate}[\rm (i)]

\item each vertex $x_{n_1 n_2 \ldots n_i}$ $(0\leq i\leq k)$ lies in a connected subgraph $C_{n_1n_2\ldots n_i}$ of $G$ (for $i=0$, $x_{n_1 n_2 \ldots n_i}:=x$, $y_{n_1 n_2 \ldots n_i}:=y$ and $C_{n_1 n_2 \ldots n_i}:=G$),

\item for each vertex $x_{n_1 n_2 \ldots n_i}$ $(0\leq i<k)$, we have found an edge $x_{n_1 n_2 \ldots n_i}y_{n_1 n_2 \ldots n_i}$ that lies in $C_{n_1n_2\ldots n_i}$ such that $C_{n_1n_2\ldots n_i0}$ and $C_{n_1n_2\ldots n_i1}$ are two components of $C_{n_1n_2\ldots n_i}-x_{n_1n_2\ldots n_i}-y_{n_1n_2\ldots n_i}$ that are adjacent to $x_{n_1 n_2 \ldots n_i}$,

\item for fixed $n_1,n_2,\ldots,n_i$  $(1\leq i\leq k)$, $N_G(C_{n_1 n_2 \ldots n_i})\subseteq\bigcup_{j=0}^{i-1} \{x_{n_1 n_2 \ldots n_j},y_{n_1 n_2 \ldots n_j}\}$,

\item for fixed $n_1,n_2,\ldots,n_i$  $(1\leq i\leq k)$, $G-C_{n_1 n_2 \ldots n_i}$ is connected.

\end{enumerate}

Now, for each vertex $x_{n_1 n_2 \ldots n_k}$ in $T_k$, since it has infinite degree and $N_G(x_{n_1 n_2 \ldots n_k})\setminus C_{n_1 n_2 \ldots n_k}\subseteq N_G(C_{n_1 n_2 \ldots n_k})$ is finite by (iii), all but finitely many neighbors of $x_{n_1 n_2 \ldots n_k}$ lie in $C_{n_1 n_2 \ldots n_k}$. Let $z$ be a neighbor of $x_{n_1 n_2 \ldots n_k}$ in $C_{n_1 n_2 \ldots n_k}$ and $B:=C_{n_1 n_2 \ldots n_k}-x_{n_1 n_2 \ldots n_k}-z$. Suppose $B$ is connected. Since $B':=G-C_{n_1 n_2 \ldots n_k}$ is connected by (iv), $B$ and $B'$ are the only two components of $G-x_{n_1 n_2 \ldots n_k}-z$. By Lemma \ref{infnoncont}(b), $B'$ contains infinitely many neighbors of $x_{n_1 n_2 \ldots n_k}$ contradicting $N_G(x_{n_1 n_2 \ldots n_k})\setminus C_{n_1 n_2 \ldots n_k}\subseteq N_G(C_{n_1 n_2 \ldots n_k})$ which is finite by (iii). Therefore, $B$ is not connected.

Note that at least one component of $B$ is adjacent to $x_{n_1 n_2 \ldots n_k}$. If not, then, by the 2-connectedness of $G$, each component of $B$ has a neighbor in $N_G(C_{n_1 n_2 \ldots n_k})\subseteq G-C_{n_1 n_2 \ldots n_k}$. By (iv), this implies $G-x_{n_1 n_2 \ldots n_k}-z$ is connected, a contradiction. Suppose there are two components of $B$, say $D$ and $D'$, that are both adjacent to $x_{n_1 n_2 \ldots n_k}$. Then choose $y_{n_1 n_2 \ldots n_k}:=z$, $C_{n_1 n_2 \ldots n_k0}:=D$ and $C_{n_1 n_2 \ldots n_k1}:=D'$. Suppose only one component of $B$ is adjacent to $x_{n_1 n_2 \ldots n_k}$, say $C$. Each component of $B$ other than $C$ is adjacent to $z$ by the connectedness of $C_{n_1 n_2 \ldots n_k}$ and has a neighbor in $N_G(C_{n_1 n_2 \ldots n_k})\subseteq G-C_{n_1 n_2 \ldots n_k}$ by the 2-connectedness of $G$. Denote the union of components of $B$ other than $C$ by $C'$. Since $G-C_{n_1 n_2 \ldots n_k}$ is connected by (iv), $C'':=G[(G-C_{n_1 n_2 \ldots n_k})\cup C']$ is connected. Hence, $C$ and $C''$ are the only two components of  $G-x_{n_1 n_2 \ldots n_k}-z$ and $N_G(C)=\{x_{n_1 n_2 \ldots n_k},z\}$. Let $z'$ be a neighbor of $x_{n_1 n_2 \ldots n_k}$ in $C$. Then one component $D$ of $C_{n_1 n_2 \ldots n_k}-x_{n_1 n_2 \ldots n_k}-z'$ contains $z$ and $C'$. Since $G-x_{n_1 n_2 \ldots n_k}-z'$ is not connected, $D$ cannot be the only component of $C_{n_1 n_2 \ldots n_k}-x_{n_1 n_2 \ldots n_k}-z'$. Let $D'$ be any component of $C_{n_1 n_2 \ldots n_k}-x_{n_1 n_2 \ldots n_k}-z'$ other than $D$. Then $D'$ lies in $C$ and $N_G(D')\subseteq \{x_{n_1 n_2 \ldots n_k},z'\}\cup (N_G(C)-z)=\{x_{n_1 n_2 \ldots n_k},z'\}$. Now, choose $y_{n_1 n_2 \ldots n_k}:=z'$, $C_{n_1 n_2 \ldots n_k0}:=D$ and $C_{n_1 n_2 \ldots n_k1}:=D'$.

In both cases, $x_{n_1 n_2 \ldots n_k}y_{n_1 n_2 \ldots n_k}$ lies in $C_{n_1 n_2 \ldots n_k}$, and $C_{n_1 n_2 \ldots n_k 0}$ and $C_{n_1 n_2 \ldots n_k 1}$ are two components of $C_{n_1 n_2 \ldots n_k}-x_{n_1 n_2 \ldots n_k}-y_{n_1 n_2 \ldots n_k}$ that are adjacent to $x_{n_1 n_2 \ldots n_k}$. Let $x_{n_1 n_2 \ldots n_k 0}$ be a neighbor of $x_{n_1 n_2 \ldots n_k}$ in $C_{n_1 n_2 \ldots n_k 0}$ and $x_{n_1 n_2 \ldots n_k 1}$ be a neighbor of $x_{n_1 n_2 \ldots n_k}$ in $C_{n_1 n_2 \ldots n_k 1}$. Since $C_{n_1 n_2 \ldots n_k 0}\subseteq C_{n_1 n_2 \ldots n_k}$ and $C_{n_1 n_2 \ldots n_k 1}\subseteq C_{n_1 n_2 \ldots n_k}$, $N_G(C_{n_1 n_2 \ldots n_k 0})\subseteq N_G(C_{n_1 n_2 \ldots n_k}) \cup  \{x_{n_1 n_2 \ldots n_k},y_{n_1 n_2 \ldots n_k}\} \subseteq \bigcup_{j=0}^k \{x_{n_1 n_2 \ldots n_j},y_{n_1 n_2 \ldots n_j}\}$ and $N_G(C_{n_1 n_2 \ldots n_k 1})\subseteq N_G(C_{n_1 n_2 \ldots n_k}) \cup  \{x_{n_1 n_2 \ldots n_k},y_{n_1 n_2 \ldots n_k}\} \subseteq \bigcup_{j=0}^k \{x_{n_1 n_2 \ldots n_j},y_{n_1 n_2 \ldots n_j}\}$ by (iii). By the connectedness of $C_{n_1 n_2 \ldots n_k}$, every component of $C_{n_1 n_2 \ldots n_k}-x_{n_1 n_2 \ldots n_k}-y_{n_1 n_2 \ldots n_k}$ has a neighbor in $ \{x_{n_1 n_2 \ldots n_k},y_{n_1 n_2 \ldots n_k}\}$. For $n_{k+1}\in\{0,1\}$, denote the union of the components of $C_{n_1 n_2 \ldots n_k}-x_{n_1 n_2 \ldots n_k}-y_{n_1 n_2 \ldots n_k}$ other than $C_{n_1 n_2 \ldots n_kn_{k+1}}$ by $U_{n_{k+1}}$. Then $G[x_{n_1 n_2 \ldots n_k}y_{n_1 n_2 \ldots n_k}\cup U_{n_{k+1}}]$ is connected. Since $ x_{n_1 n_2 \ldots n_{k-1}}\in G-C_{n_1 n_2 \ldots n_k}$, $x_{n_1 n_2 \ldots n_{k-1}}x_{n_1 n_2 \ldots n_k}\in E(G)$ and $G-C_{n_1 n_2 \ldots n_k}$ is connected by (iv), $G-C_{n_1 n_2 \ldots n_kn_{k+1}}:=G[(G-C_{n_1 n_2 \ldots n_k})\cup x_{n_1 n_2 \ldots n_k}y_{n_1 n_2 \ldots n_k}\cup U_{n_{k+1}}]$ is connected.

Define $T_{k+1}$ where $V(T_{k+1}):=\bigcup_{i=0}^{k+1}\bigcup_{(n_1,n_2,\ldots,n_i)\in\{0,1\}^i} x_{n_1 n_2 \ldots n_i}$ and

$E(T_{k+1}):=\bigcup_{i=0}^{k}\bigcup_{(n_1,n_2,\ldots,n_i)\in\{0,1\}^i}\{x_{n_1 n_2 \ldots n_i}x_{n_1 n_2 \ldots n_i 0}, x_{n_1 n_2 \ldots n_i}x_{n_1 n_2 \ldots n_i 1}\}$.

Finally, define $T:=\bigcup_{k=0}^{\infty} T_k$. It is easy to see that $T$ satisfies properties (1) through (4). Let $z_0$ and $z_1$ be any vertices in $T_{n_1 n_2 \ldots n_k 0}$ and $T_{n_1 n_2 \ldots n_k 1}$ respectively. Then $z_0$ is of the form $x_{n_1 n_2 \ldots n_k 0 p_1 p_2 \ldots p_i}$ while $z_1$ is of the form $x_{n_1 n_2 \ldots n_k 1 q_1 q_2 \ldots q_j}$. We have $z_0=x_{n_1 n_2 \ldots n_k 0 p_1 p_2 \ldots p_i}\in C_{n_1 n_2 \ldots n_k 0 p_1 p_2 \ldots p_i}\subseteq C_{n_1 n_2 \ldots n_k 0 p_1 p_2 \ldots p_{i-1}}\subseteq \ldots \subseteq  C_{n_1 n_2 \ldots n_k 0}$ and $z_1=x_{n_1 n_2 \ldots n_k 1 q_1 q_2 \ldots p_j}\in C_{n_1 n_2 \ldots n_k 1 q_1 q_2 \ldots q_j}\subseteq C_{n_1 n_2 \ldots n_k 1 q_1 q_2 \ldots q_{j-1}}\subseteq \ldots \subseteq  C_{n_1 n_2 \ldots n_k 1}$. Therefore, $T_{n_1 n_2 \ldots n_k 0}\subseteq C_{n_1 n_2 \ldots n_k 0}$ and $T_{n_1 n_2 \ldots n_k 1}\subseteq C_{n_1 n_2 \ldots n_k 1}$. Since $\bigcup_{j=0}^k \{x_{n_1 n_2 \ldots n_j},y_{n_1 n_2 \ldots n_j}\}$ contains both $N_G(C_{n_1 n_2 \ldots n_k 0})$ and $N_G(C_{n_1 n_2 \ldots n_k 1})$, it separates $C_{n_1 n_2 \ldots n_k 0}$ and $C_{n_1 n_2 \ldots n_k 1}$ in $G$ and thus separates $T_{n_1 n_2 \ldots n_k 0}$ and $T_{n_1 n_2 \ldots n_k 1}$ in $G$. Hence, Property (5) holds for $T$ and the proof is complete.
\hfill $\Box$

%\noindent{\em Proof.}
%By Lemma \ref{critical}, every vertex $x$ of $G$ is of infinite degree since it dominates the ray $R_x$ directly. Consider any vertex $u$ in $G$. Denote $R_u$ to be the ray dominated by $u$ as constructed in Lemma \ref{critical}, and $u_0,u_1,u_2,\ldots$ to be the neighbors of $u$ in $R_u$ appearing in that order in $R_u$. We write $(u_{n_1,n_2,\ldots,n_{i-1}})_{n_i}$ as $u_{n_1,n_2,\ldots,n_{i-1},n_i}$. Now, consider a vertex $x$ in $G$ and the ray $R_x$ as constructed in Lemma \ref{critical}. Define $G_1:=G[x\cup x_0R_xx_2]$. Suppose we have constructed the rays $R_{x_{n_1,n_2,\ldots,n_{i-1}}}$ for $(n_1,n_2,\ldots,n_{i-1})\in\{0,2\}^{i-1}$ and $G_i$ such that $R_{x_{n_1,n_2,\ldots,n_{i-1}}}\cap G_{i-1}=\emptyset$ and $R_{x_{n_1,n_2,\ldots,n_{i-1}}}\cap R_{x_{m_1,m_2,\ldots,m_{i-1}}}=\emptyset$ for $(n_1,n_2,\ldots,n_{i-1})\neq (m_1,m_2,\ldots,m_{i-1})$.

%For each $(n_1,n_2,\ldots,n_i)\in\{0,2\}^i$, we construct a ray $R_{x_{n_1,n_2,\ldots,n_i}}$ using Lemma \ref{critical}. Since $G_i$ is finite, without loss of generality, we can assume that $R_{x_{n_1,n_2,\ldots,n_i}}\cap G_i=\emptyset$ for all $(n_1,n_2,\ldots,n_i)\in\{0,2\}^i$. Consider any two rays $R_{x_{n_1,n_2,\ldots,n_i}}$ and $R_{x_{m_1,m_2,\ldots,m_i}}$ where $(n_1,n_2,\ldots,n_i)\neq (m_1,m_2,\ldots,m_i)$. Let $k$ be the first integer such that $n_k\neq m_k$. Then $\{x_{n_1,n_2,\ldots,n_{k-1}}, x_{n_1,n_2,\ldots,n_{k-1},1}\}$ separates $R_{x_{n_1,n_2,\ldots,n_i}}$ and $R_{x_{m_1,m_2,\ldots,m_i}}$. Therefore, $R_{x_{n_1,n_2,\ldots,n_i}}\cap R_{x_{m_1,m_2,\ldots,m_i}}=\emptyset$ for $(n_1,n_2,\ldots,n_i)\neq (m_1,m_2,\ldots,m_i)$. Define $G_{i+1}:=G[G_i\cup\bigcup_{(n_1,n_2,\ldots,n_i)\in\{0,2\}^i}x_{n_1,n_2,\ldots,n_i,0}R_{x_{n_1,n_2,\ldots,n_i}}x_{n_1,n_2,\ldots,n_i,2}]$.

%Define $G_{\infty}:=\bigcup_{i\geq 1} G_i$. Consider any two distinct rays in $G_{\infty}$, $R_1:=xx_{n_1}x_{n_1,n_2}x_{n_1,n_2,n_3}\ldots$ and $R_2:=xx_{m_1}x_{m_1,m_2}x_{m_1,m_2,m_3}\ldots$ where $n_i,m_i\in\{0,2\}$ for $i\geq 1$. Let $k$ be the first integer such that $n_k\neq m_k$. Then $\{x_{n_1,n_2,\ldots,n_{k-1}},x_{n_1,n_2,\ldots,n_{k-1},1}\}$ separates $R_1$ and $R_2$ in $G$. Hence, $R_1$ and $R_2$ belongs to two different ends. Therefore, $G$ has uncountably many ends.
%\hfill $\Box$


\section{Subgraph induced by all the contractible edges}

In this section, we will extend Theorem \ref{dean2} to any 2-connected locally finite infinite graph $G$ and prove that $\overline{G_C}$ is 2-arc-connected. Note that Lemma \ref{fund3}(a) implies that every vertex is incident to at least two contractible edges. Hence, $G_C$ is spanning. Using the following two lemmas, it is easy to see that $\overline{G_C}$ is arc-connected.

\begin{lemma}[Diestel \cite{diestel}]\label{subspaceconn}
Let $G$ be a locally finite graph. Then a standard subspace of $|G|$ is connected if and only if it contains an edge from every finite cut of $G$ of which it meets both sides.
\end{lemma}

\begin{lemma}[Diestel and K{\"u}hn \cite{tst}]\label{connimplyaconn}
If $G$ is a locally finite graph, then every closed connected subspace of $|G|$ is arc-connected.
\end{lemma}

\begin{theorem}\label{G_Cconn}
Let $G$ be a $2$-connected locally finite infinite graph and $G_C$ be the subgraph induced by all the contractible edges in $G$. Then $\overline{G_C}$ is arc-connected.
\end{theorem}

\noindent{\em Proof.}
Let $F$ be any finite cut of $G$. By Lemma \ref{fund3}(a), $F$ contains at least two edges in $G_C$. Hence, $\overline{G_C}$ is connected by Lemma \ref{subspaceconn}. By Lemma \ref{connimplyaconn}, $\overline{G_C}$ is arc-connected.
\hfill $\Box$
\\

%\begin{lemma}\label{contexist}
%Let $G$ be a $2$-connected graph nonisomorphic to $K_3$. Let $\{x,y\}$ be a $2$-separator of $G$ and $C$ be a component of $G-\{x,y\}$. If $|E_G(x,C)|$ is finite, then $E_G(x,C)$ contains a contractible edge.
%\end{lemma}

%\noindent{\em Proof.}
%Note that $y$ is a cutvertex of $G-E_G(x,C)$. By Lemma \ref{fund3}(b), $E_G(x,C)$ contains a contractible edge.
%\hfill $\Box$

%\begin{lemma}[Dean \cite{dean}]\label{contexist}
%Let $G$ be a $2$-connected graph nonisomorphic to $K_3$. If $F$ is a finite subset of $E_G(G)$ such that $G-F$ is not $2$-connected, then $F$ contains a contractible edge.
%\end{lemma}

Next, we prove that $\overline{G_C}$ is 2-connected.

\begin{lemma}\label{key}
Let $G$ be a $2$-connected locally finite infinite graph and $x$ be a point of $|G|$. Suppose there is a partition $(X,X')$ of $V(G\setminus x)$ such that $E_G(X,X')$ is non-empty and all edges in $E_G(X,X')$ are non-contractible. Then $G$ contains a subdivision of a $1$-way infinite ladder $L$ consisting of two disjoint rays: $R:=x_0P_1x_1P_2x_2\ldots$ and $R':=x_0'P_1'x_1'P_2'x_2'\ldots$ with the rungs of the ladder being $x_0x_0',x_1x_1',x_2x_2',\ldots$, all of which are $X$-$X'$ edges such that $x\notin\overline{L}$.

%\begin{enumerate}[\rm (1)]
%\item $P_i$ is a path between $x_{i-1}$ and $x_i$, and $P_i'$ is a path between $x_{i-1}'$ and $x_i'$.

%\item $x_0,x_1,x_2,\ldots\in X$.

%\item $x_0',x_1',x_2',\ldots\in X'$.

%\item For $i\geq 1$, $\{x_i,x_i'\}$ separates $\{x_0,x_0'\ldots,x_{i-1},x_{i-1}'\}$ from $\{x_{i+1},x_{i+1}',x_{i+2},x_{i+2}',\ldots\}$ in $G$.

%\item $x\notin\overline{L}$.
%\end{enumerate}

\end{lemma}

\noindent{\em Proof.}
Since $G$ is 2-connected, $|E_G(X,X')|\geq 2$ unless $x$ is a vertex and $|E_G(X,X')|=1$. Consider any $X$-$X'$ edge $x_0x_0'$ that does not contain $x$. Let $C$ be the component of $G-x_0-x_0'$ containing $x$ and $C_1$ be a component of $G-x_0-x_0'$ not containing $x$.

%Since $G$ is 2-connected, $|E_G(X,X')|\geq 2$ unless $x$ is a vertex and $|E_G(X,X')|=1$. Consider any $X$-$X'$ edge $x_0x_0'$ that does not contain $x$. Let $C$ be the component of $G-x_0-x_0'$ containing $x$ and $C_1$ be a component of $G-x_0-x_0'$ not containing $x$. By applying Lemma \ref{contexist} to $E_G(x_0,C_1)$ and $E_G(x_0',C_1)$, there exist contractible edges $x_0y_1$ and $x_0'y_1'$ where $y_1\in C_1$ and $y_1'\in C_1$. Since $x\notin C_1$ and all edges in $E_G(X,X')$ are non-contractible, $y_1\in X$ and $y_1'\in X'$. Choose a path $Q_1$ in $C_1$ between $y_1$ and $y_1'$. Then there exists an $X$-$X'$ edge $x_1x_1'$ on $Q_1$ such that $V(y_1Q_1x_1)\subseteq X$ and $x_1'\in X'$. Define $P_1:=x_0y_1\cup y_1Q_1x_1$, $P_1':=x_0'y_1'\cup y_1'Q_1x_1'$ and $L_1:=x_0x_0'\cup P_1\cup P_1'\cup x_1x_1'$. Note that $L_1\subseteq G[C_1\cup x_0\cup x_0']$ and $G[C\cup L_1 - x_1 - x_1']$ is connected.


Suppose we have constructed the finite ladder $L_k$ consisting of two disjoint paths $R_k:=x_0P_1x_1P_2x_2\ldots x_{k-1}P_kx_k$ and $R_k':=x_0'P_1'x_1'P_2'x_2'\ldots x_{k-1}'P_k'x_k'$ with the rungs of the ladder being $x_0x_0',x_1x_1',\ldots x_kx_k'$, all of which are $X$-$X'$ edges such that $L_k\subseteq G[C_1\cup x_0\cup x_0']$ and $G[C\cup L_k - x_k - x_k']$ is connected. Let $C_{k+1}$ be a component of $G-x_k-x_k'$ not containing $x$. Then $C_{k+1}\subseteq C_1$ and $C_{k+1}\cap L_k=\emptyset$.  By applying Lemma \ref{contexist} to $E_G(x_k,C_{k+1})$ and $E_G(x_k',C_{k+1})$, there exist contractible edges $x_ky_{k+1}$ and $x_k'y_{k+1}'$ where $y_{k+1}\in C_{k+1}$ and $y_{k+1}'\in C_{k+1}$. Since $x\notin C_{k+1}$ and all edges in $E_G(X,X')$ are non-contractible, $y_{k+1}\in X$ and $y_{k+1}'\in X'$. Choose a path $Q_{k+1}$ in $C_{k+1}$ between $y_{k+1}$ and $y_{k+1}'$. Then there exists an $X$-$X'$ edge $x_{k+1}x_{k+1}'$ on $Q_{k+1}$ such that $V(y_{k+1}Q_{k+1}x_{k+1})\subseteq X$ and $x_{k+1}'\in X'$. Define $P_{k+1}:=x_ky_{k+1}\cup y_{k+1}Q_{k+1}x_{k+1}$, $P_{k+1}':=x_k'y_{k+1}'\cup y_{k+1}'Q_{k+1}x_{k+1}'$, $R_{k+1}:=R_k\cup P_{k+1}$, $R_{k+1}':=R_k'\cup P_{k+1}'$ and $L_{k+1}:=L_k\cup P_{k+1}\cup P_{k+1}'\cup x_{k+1}x_{k+1}'$. Note that $L_{k+1}\subseteq G[C_1\cup x_0\cup x_0']$ and $G[C\cup L_{k+1} - x_{k+1} - x_{k+1}']$ is connected.

Define $R:=\bigcup_{k\geq 0}R_k$, $R':=\bigcup_{k\geq 0}R_k'$ and $L:=\bigcup_{k\geq 0}L_k$. Then $L\subseteq G[C_1\cup x_0\cup x_0']$ and $x\notin\overline{L}$.
%We can repeat the above process indefinitely and obtain the desired subdivision of the 1-way infinite ladder.
\hfill $\Box$


\begin{theorem}\label{G_C2conn}
Let $G$ be a $2$-connected locally finite infinite graph and $G_C$ be the subgraph induced by all the contractible edges in $G$. Then $\overline{G_C}$ is $2$-connected.
\end{theorem}

\noindent{\em Proof.}
Suppose $\overline{G_C}$ is not 2-connected. Then there exists a point $x$ in $\overline{G_C}$ such that $\overline{G_C}\setminus x$ is not connected. Let $U$ and $U'$ be two disjoint non-empty open sets in $|G|$ such that $\overline{G_C}\setminus x\subseteq U\cup U'$, $(\overline{G_C}\setminus x)\cap U\neq\emptyset$ and $(\overline{G_C}\setminus x)\cap U'\neq\emptyset$. Define $X:=(\overline{G_C}\setminus x)\cap U\cap V(G)$ and $X':=(\overline{G_C}\setminus x)\cap U'\cap V(G)$. Since $G_C$ is spanning, $X\cup X'=V(G\setminus x)$. Suppose $U$ contains an interior point $a$ of an edge $bc$ of $G_C$. Then $\overline{G_C}\setminus x$ contains half edges $[ba]$ or $[ca]$ of $bc$. By the connectedness of half edge, $U$ contains $b$ or $c$. Suppose $U$ contains an end $\omega$ of $|G|$. Then $U$ contains a basic open neighborhood of $\omega$, say $\hat{C}(S,\omega)$, and thus contains infinitely many vertices. The same arguments hold for $U'$. Therefore, both $X$ and $X'$ are non-empty. Since $G\setminus x$ is connected, $E_G(X,X')$ is non-empty.

Suppose $x$ is a vertex or an end of $G$. Then all edges in $E_G(X,X')$ are non-contractible and $(X,X')$ is a partition of $V(G\setminus x)$. Suppose $x$ is an interior point of an edge $e$. Then all edges in $E_G(X,X')$ are non-contractible unless $e\in E_G(X,X')\cap E_C$. Note that, $E_G(X,X')-e$ is non-empty as $G$ is 2-connected and every edge in $E_G(X,X')-e$ is non-contractible. Let $e=yy'$ where $y\in X$ and $y'\in X'$. Suppose $X=\{y\}$. By Lemma \ref{fund3}(a), since $y$ is incident to at least two contractible edges, there is a contractible $X$-$X'$ edge other than $e$, which is impossible. Therefore, $|X|\geq 2$. Now, all edges in $E_G(X-y,X')$ are non-contractible and $(X-y,X')$ is a partition of $V(G\setminus y)$. In both cases, by Lemma \ref{key}, $G$ contains a subdivision of a $1$-way infinite ladder $L$ such that $x\notin\overline{L}$.


%and we can use $y$ instead of $x$ and $(X-y,X')$ as the required partition of $V(G\setminus y)$ in Lemma \ref{key}.

%In both cases, by Lemma \ref{key}, $G$ contains a subdivision of a $1$-way infinite ladder $L$ consisting of two disjoint rays: $R:=x_0P_1x_1P_2x_2\ldots$ and $R':=x_0'P_1'x_1'P_2'x_2'\ldots$ with the rungs of the ladder being $x_0x_0',x_1x_1',x_2x_2',\ldots$, all of which are $X$-$X'$ edges such that $x\notin\overline{L}$.

%\begin{enumerate}[\rm (1)]
%\item $P_i$ is a path between $x_{i-1}$ and $x_i$, and $P_i'$ is a path between $x_{i-1}'$ and $x_i'$.

%\item $x_0,x_1,x_2,\ldots\in X$.

%\item $x_0',x_1',x_2',\ldots\in X'$.

%\item $x\notin\overline{L}$.
%\end{enumerate}

Let $\omega$ be the end of $|G|$ containing $R$ and $R'$. Note that $\omega\neq x$. Since $G_C$ is spanning, $\overline{G_C}\setminus x$ contains all the ends of $|G|$ except possibly $x$. Without loss of generality, assume $\omega\in U$. Since $U$ is open, there exists a basic open neighborhood $\hat{C}(S,\omega)\subseteq U$. Since $x_0',x_1',x_2',\ldots\in X'\subseteq U'$ converge to $\omega$, all but finitely many of them lie in $\hat{C}(S,\omega)$, contradicting $U\cap U'=\emptyset$.
\hfill $\Box$
\\

Finally, we prove the main result of this section, namely, $\overline{G_C}$ is 2-arc-connected. This follows from a theorem by Georgakopoulos \cite{connsubsp} concerning connected but not path-connected subspaces of locally finite graphs. Note that since $|G|$ is Hausdorff, path-connectedness is equivalent to arc-connectedness.

\begin{theorem}[Georgakopoulos \cite{connsubsp}]\label{connsubsp}
Given any locally finite connected graph $G$, a connected subspace $X$ of $|G|$ is path-connected unless it satisfies the following assertions:

\begin{enumerate}[\rm (1)]
\item $X$ has uncountably many path-components each of which consists of one end only;

\item $X$ has infinitely many path-components that contain a vertex; and

\item every path-component of $X$ contains an end.
\end{enumerate}

\end{theorem}

\begin{theorem}\label{G_C2arcconn}
Let $G$ be a $2$-connected locally finite infinite graph and $G_C$ be the subgraph induced by all the contractible edges in $G$. Then $\overline{G_C}$ is $2$-arc-connected.
\end{theorem}

\noindent{\em Proof.}
Suppose $\overline{G_C}$ is not 2-arc-connected. Then there exists a point $x$ in $\overline{G_C}$ such that $\overline{G_C}\setminus x$ is not arc-connected. Note that $\overline{G_C}\setminus x$ is connected by Theorem \ref{G_C2conn}. By Theorem \ref{connsubsp}, $\overline{G_C}\setminus x$ has uncountably many path-components each of which consists of one end only. Let $\omega$ and $\omega'$ be two such path-components of $\overline{G_C}\setminus x$. Since $\overline{G_C}$ is arc-connected by Theorem \ref{G_Cconn}, there exists an arc $A$ joining $\omega$ and $\omega'$ in $\overline{G_C}$. Now, $x$ must lie in $A$ for otherwise $\omega$ and $\omega'$ would lie in the same path-component of $\overline{G_C}\setminus x$. But the path-component of $\overline{G_C}\setminus x$ containing $\omega$ also contains $[\omega A x)$, a contradiction.
\hfill $\Box$


\section{Outerplanarity of 2-connected locally finite graphs}

As mentioned in the introduction, in order to extend Theorem \ref{wu} to locally finite infinite graphs, we would like to prove that for any 2-connected locally finite infinite graph $G$, if every vertex is incident to exactly two contractible edges, then $\overline{G_C}$ is a Hamilton circle. This requires several lemmas listed below.

\begin{lemma}\label{arcvertex}
Let $G$ be a locally finite graph. Then every arc in $|G|$ whose two endpoints are ends contains a vertex.
\end{lemma}

\noindent{\em Proof.}
Suppose $A$ is an arc in $|G|$ whose two endpoints are ends $\omega_1$ and $\omega_2$. Then there exists a finite set $S$ of vertices such that $\hat{C}(S,\omega_1)$ and $\hat{C}(S,\omega_2)$ are distinct. By the connectedness of $A$, $A$ contains a vertex of $S$.
\hfill $\Box$

\begin{lemma}\label{arcray}
Let $G$ be a locally finite graph and $\omega$ be an end in $|G|$. Then every $\omega$-arc $A$ in $|G|$ contains a vertex, say $z$, and $zA$ contains a ray starting with $z$.
\end{lemma}

\noindent{\em Proof.}
Denote the starting point of $A$ by $a$. First, we show that $A$ contains a vertex. If $a$ is a vertex, then we are done. If $a$ is an end, then it is true by Lemma \ref{arcvertex}. If $a$ is an interior point of an edge $xy$, then by the connectedness of $A$, $A$ contains $x$ or $y$.

Let $z$ be a vertex in $A$. By the connectedness of $zA$, $zA$ contains an interior point of an edge incident to $z$, say $zz_1$. Then the connectedness of $zA$ implies $zz_1$ lies in $zA$. Repeat the above argument for $z_1A$ and so on. We obtain a ray that starts with $z$ and lies in $zA$.
\hfill $\Box$

\begin{lemma}\label{arcarcpath}
Let $G$ be a locally finite graph and $\omega$ be an end in $|G|$. Let $A_1$ and $A_2$ be two $\omega$-arcs in $|G|$ that are disjoint except at $\omega$. Then, for all finite subset $S$ of $V(G)$, $\hat{C}(S,\omega)$ contains a subarc $A_1'\omega A_2'$ of $A_1\omega A_2$ and there is an $A_1'$-$A_2'$ path in $C(S,\omega)$.
\end{lemma}

\noindent{\em Proof.}
Let $x_1$ be the last point of $A_1$ that lies in $S$ and $x_2$ be the last point of $A_2$ that lies in $S$. By Lemma \ref{arcray}, $x_1A$ contains a ray $R_1$ starting with $x_1$ and $x_2A$ contains a ray $R_2$ starting with $x_2$. Let $y_1$ be the neighbor of $x_1$ in $R_1$ and $y_2$ be the neighbor of $x_2$ in $R_2$. Then $y_1A_1\omega A_2y_2$ lies in $\hat{C}(S,\omega)$. Also there is a $y_1$-$y_2$ path in $C(S,\omega)$ which automatically contains a $y_1A_1$-$y_2A_2$ path.
\hfill $\Box$
\\

%\begin{lemma}[Diestel and K{\"u}hn \cite{cycles2}]\label{arctopath}
%Let $G$ be a locally finite graph. Let $X$ be a closed subset of $|G|$ and $A$ be an $a$-$b$ arc in $|G|$ disjoint from $X$ with $a,b\in V(G)$. Then $G$ contains an $a$-$b$ path disjoint from $X$.
%\end{lemma}

We also need a result on the characterization of a topological circle in $|G|$ in terms of its vertex and end degrees.

\begin{lemma}[Bruhn and Stein \cite{enddeg}]\label{circle}
Let $C$ be a subgraph of a locally finite graph $G$. Then $\overline{C}$ is a circle if and only if $\overline{C}$ is connected and every vertex and end of $|G|$ in $\overline{C}$ has degree two in $\overline{C}$.
\end{lemma}

Now, we can proceed with the proof.

\begin{theorem}\label{hamilton}
Let $G$ be a $2$-connected locally finite infinite graph and $G_C$ be the subgraph induced by all the contractible edges in $G$. If every vertex of $G$ is incident to exactly two contractible edges, then $\overline{G_C}$ is a Hamilton circle.
\end{theorem}

\noindent{\em Proof.}
Since $G_C$ is spanning, $\overline{G_C}$ contains all vertices and ends of $|G|$. By Theorem \ref{G_Cconn}, $\overline{G_C}$ is arc-connected. Obviously, every vertex of $G$ has degree two in $\overline{G_C}$. Therefore, it remains to prove that every end of $|G|$ has degree two in $\overline{G_C}$.

\begin{claim}\label{claim1}
Let $A$ be an arc in $\overline{G_C}$ and $x$ be a vertex in $A$. Suppose that both $x^-$ and $x^+$ exist in $A$. Let $y$ be any neighbor of $x$ other than $x^-$ and $x^+$. Then every $x^-$-$x^+$ arc in $|G|$ intersects $\{x,y\}$.
\end{claim}

\noindent{\em Proof.}
Since $xx^-$ and $xx^+$ are the only contractible edges incident to $x$, $xy$ is non-contractible. Lemma \ref{contexist} implies that $G-x-y$ has exactly two components, and $x^-$ and $x^+$ lie in different components. By the connectedness of an arc, every $x^-$-$x^+$ arc in $|G|$ intersects $\{x,y\}$.
\hfill $\Box$

%\begin{claim}\label{claim0}
%Let $xy$ be a non-contractible edge in $G$. Then $G-x-y$ has exactly two components, and for any component $C$ of $G-x-y$, $E_G(x,C)$ contains exactly one contractible edge.
%\end{claim}

%\noindent{\em Proof.}
%Let $C$ be any component of $G-x-y$. By Lemma \ref{contexist}, $E_G(x,C)$ contains a contractible edge. Since $x$ is incident to exactly two contractible edges, $G-x-y$ has exactly two components and $E_G(x,C)$ contains exactly one contractible edge.
%\hfill $\Box$

%\begin{claim}\label{claim1}
%Consider an arc $A$ in $\overline{G_C}$. Let $xy$ be a non-contractible edge such that $x,y\in A$, $x<y$, and $x^-, x^+, y^-, y^+\in A$. Then every $(Ax^-\cup y^+A)$-$x^+Ay^-$ path in $G$ intersects $\{x,y\}$.
%\end{claim}

%\noindent{\em Proof.}
%Since $xx^-$ and $xx^+$ are the only contractible edges incident to $x$, and $yy^-$ and $yy^+$ are the only contractible edges incident to $y$, $x^+Ay^-$ being connected lies in a component of $G-x-y$ and by Claim \ref{claim0}, $Ax^-\cup y^+A$ lies in the other component of $G-x-y$. Therefore, every $(Ax^-\cup y^+A)$-$x^+Ay^-$ path intersects $\{x,y\}$.
%\hfill $\Box$

%\begin{claim}\label{claim2}
%Consider an arc $A$ in $\overline{G_C}$. Let $xy$ be a non-contractible edge where $x\in A$, $y\notin A$, and $x^-, x^+\in A$. Then every $Ax^-$-$x^+A$ path in $G$ intersects $\{x,y\}$.
%\end{claim}

%\noindent{\em Proof.}
%Since $xx^-$ and $xx^+$ are the only contractible edges incident to $x$, by Claim \ref{claim0}, $Ax^-$ and $x^+A$ lie in different components of $G-x-y$, and the result follows.
%\hfill $\Box$


%Let $A$ be an arc in $\overline{G_C}$.  Suppose $x_1Px_2$ and $y_1Qy_2$ are two disjoint $A-A$ paths in $G$ with $x_1<x_2$ and $y_1<y_2$. Then $P$ and $Q$ are \emph{overlapping} if $x_1<y_1<x_2<y_2$ or $y_1<x_1<y_2<x_2$. Otherwise, $P$ and $Q$ are \emph{nesting}.

%\begin{claim}\label{claim3}
%Let $A$ be an arc in $\overline{G_C}$. Then there are no disjoint overlapping $A$-$A$ paths in $G$. 
%\end{claim}

%\noindent{\em Proof.}
%Suppose there are two disjoint overlapping $A$-$A$ paths $x_1Px_2$ and $y_1Qy_2$ in $G$. Without loss of generality, assume $x_1<y_1<x_2<y_2$. Let $y$ be the vertex adjacent to $y_1$ in $Q$. Note that $y_1y$ is non-contractible. If $y=y_2$, then $P$ contradicts Claim \ref{claim1}. If $y\neq y_2$, then $P$ contradicts Claim \ref{claim2}.
%\hfill $\Box$

%\begin{claim}\label{claim4}
%Let $\omega$ be an end in $|G|$. Suppose $A_1$ and $A_2$ are two $\omega$-arcs in $\overline{G_C}$ that are disjoint except at $\omega$. Then there are no disjoint overlapping $A_1$-$A_2$ paths in $G$ with respect to the arc $A_1\omega A_2$.
%\end{claim}

%\noindent{\em Proof.}
%This immediately follows from Claim \ref{claim3}.
%\hfill $\Box$

\begin{claim}\label{claim2}
Let $\omega$ be an end in $|G|$. Suppose $A_1$ and $A_2$ are two edge-disjoint $\omega$-arcs in $\overline{G_C}$. Then $A_1$ and $A_2$ can intersect only at the ends of $|G|$ with the only possible exception being that the starting points of $A_1$ and $A_2$ are the same vertex.
\end{claim}

\noindent{\em Proof.}
Obviously, $A_1$ and $A_2$ cannot intersect at an interior point of an edge. Suppose $A_1$ and $A_2$ intersect at a vertex $x$. If $x$ is not the starting point for both $A_1$ and $A_2$, then the degree of $x$ in $\overline{G_C}$ is at least three, a contradiction.
\hfill $\Box$

\begin{claim}\label{claim3}
Let $\omega$ be an end in $|G|$. Suppose $A_1$ and $A_2$ are two edge-disjoint $\omega$-arcs in $\overline{G_C}$ such that the starting points of $A_1$ and $A_2$ are distinct vertices. Then there exists an end $\omega'$ in $|G|$ such that there are three $\omega'$-arcs in $\overline{G_C}$ that are disjoint except at $\omega'$ unless $A_1\cap A_2=\{\omega\}$.
\end{claim}

\noindent{\em Proof.}
Suppose $A_1\cap A_2\neq\{\omega\}$. Let $\omega'$ be the first point of $A_2$ that intersects $A_1$. By Claim \ref{claim2}, $\omega'$ is an end in $|G|$ different from $\omega$. Then $A_1\omega'$, $A_2\omega'$ and $\omega A_1\omega'$ are the required three $\omega'$-arcs.
\hfill $\Box$

\begin{claim}\label{claim4}
Let $\omega$ be an end in $|G|$. Suppose there are three edge-disjoint $\omega$-arcs in $\overline{G_C}$. Then there exists an end $\omega'$ in $|G|$ such that there are three $\omega'$-arcs in $\overline{G_C}$ that are disjoint except at $\omega'$.
\end{claim}

\noindent{\em Proof.}
Let $A_1, A_2, A_3$ be three edge-disjoint $\omega$-arcs in $\overline{G_C}$. By Lemma \ref{arcray}, for each $i\in\{1,2,3\}$, $A_i$ contains a ray $R_i$. Denote the first edge of $R_i$ by $x_iy_i$. By Claim \ref{claim2}, $y_1, y_2, y_3$ are all distinct. Therefore, without loss of generality, we can assume that the starting points of $A_1, A_2, A_3$ are all distinct vertices. Consider $A_1$ and $A_2$. If $A_1\cap A_2\neq\{\omega\}$, then the claim follows from Claim \ref{claim3}. Suppose $A_1\cap A_2=\{\omega\}$. If $A_2\cap A_3\neq\{\omega\}$, then again the claim follows from Claim \ref{claim3}. Therefore, suppose $A_2\cap A_3=\{\omega\}$. But then, $A_1, A_2, A_3$ are the desired three $\omega$-arcs.
\hfill $\Box$

%\begin{claim}\label{claim7}
%Let $A_1,A_2$ be two $\omega$-arcs in $\overline{G_C}$ that are disjoint except at $\omega$. Then there are infinitely many disjoint nesting $A_1-A_2$ paths in $G$ with respect to the arc $A_1\omega A_2$.
%\end{claim}

%\noindent{\em Proof.}
%By Lemma \ref{arcvertex}, let $x$ be a vertex in $A_1$ and $y$ be a vertex in $A_2$. By applying Lemma \ref{arctopath} with $X=\emptyset$ and $A=xA_1\omega A_2y$, there exists an $x-y$ path  and hence an $A_1-A_2$ path, say $x_1P_1y_1$. Now apply Lemma \ref{arctopath} with $X=xA_1x_1\cup P_1\cup y_1A_2y$ and $A=x_1^+ A_1\omega A_2 y_1^+$, 
%\hfill $\Box$

%\noindent{\em Proof.}
%Take a basic open neighborhood $\hat{C}(S,\omega)$ of $\omega$. By Lemma \ref{arcvertex} and the connectedness of $C(S,\omega)$, there exists an $A_1-A_2$ path $P_1$ in $\hat{C}(S,\omega)$. Now consider $\hat{C}(S\cup P_1,\omega)$. As before, there exists an $A_1-A_2$ path $P_2$ in $\hat{C}(S\cup P_1,\omega)$. By Claim \ref{claim4}, $P_1$ and $P_2$ are nesting with respect to $A_1\omega A_2$. Suppose we have constructed nesting $A_1-A_2$ paths: $P_1,\ldots,P_k$. Consider $\hat{C}(S\cup P_1\cup\ldots\cup P_k,\omega)$.  By Lemma \ref{arcvertex} and the connectedness of $C(S\cup P_1\cup\ldots\cup P_k,\omega)$, there exists an $A_1-A_2$ path $P_{k+1}$ in $\hat{C}(S\cup P_1\cup\ldots\cup P_k,\omega)$. By Claim \ref{claim4}, $P_1,\ldots,P_k,P_{k+1}$ are nesting. By induction, we can construct infinitely many disjoint nesting $A_1-A_2$ paths.
%\hfill $\Box$

\begin{claim}\label{claim5}
For each end $\omega$ in $|G|$, $deg_{G_C}(\omega)\leq 2$.
\end{claim}

\noindent{\em Proof.}
Suppose there are three edge-disjoint $\omega$-arcs in $\overline{G_C}$. By Claim \ref{claim4}, there exists an end $\omega'$ in $|G|$ such that there are three $\omega'$-arcs in $\overline{G_C}$ that are disjoint except at $\omega'$. Denote these three $\omega'$-arcs by $A_1, A_2, A_3$. By Lemma \ref{arcvertex}, without loss of generality, we can assume $A_1, A_2, A_3$ start with vertices $a_1, a_2, a_3$ respectively.

By applying Lemma \ref{arcarcpath} to $A_1$ and $A_2$ with $S=\{a_1,a_2\}$, we obtain an $a_1^+A_1$-$a_2^+A_2$ path $P$.  Let $x_1=P\cap A_1$, $x_2=P\cap A_2$ and $x$ be the neighbor of $x_1$ in $Q$. If $P$ intersects $A_3$, then interchange $A_2$ and $A_3$. Therefore, without loss of generality, there is an $a_1^+A_1$-$a_2^+A_2$ path $P$ that does not intersect $A_3$.

Now, apply Lemma \ref{arcarcpath} to $x_2A_2$ and $A_3$ with $S=V(P)$. We obtain an $x_2^+A_2$-$A_3$ path $Q$ not intersecting $P$.  Let $y_2=Q\cap x_2^+A_2$ and $y_3=Q\cap A_3$. By Claim \ref{claim1}, $Q$ cannot intersect $A_2x_2^-$, and $Q$ cannot intersect both $A_1x_1^-$ and $x_1^+A_1$. Suppose $Q\cap A_1x_1^-\neq\emptyset$. Let $y$ be the first vertex of $Q$ that lies in $A_1x_1^-$. Then $x_1^-A_1yQy_2A_2\omega' A_1x_1^+$ is an $x_1^-$-$x_1^+$ arc not intersecting $\{x_1,x\}$, contradicting Claim \ref{claim1}. Suppose $Q\cap x_1^+A_1\neq \emptyset$. Then there is an $x_1^+A_1$-$A_3$ subpath in $Q$ not intersecting $A_2$, and we interchange $A_1$ and $A_2$. Therefore, without loss of generality, we can assume that there is an $x_2^+A_2$-$A_3$ path $Q$ that does not intersect $P\cup A_1\cup A_2x_2^-$. Let $u_2$ be the neighbor of $y_2$ in $Q$ and $u_3$ be the neighbor of $y_3$ in $Q$.

Finally, apply Lemma \ref{arcarcpath} to $x_1A_1$ and $y_2A_2$ with $S=V(P\cup Q)$. We obtain an $x_1^+A_1$-$y_2^+A_2$ path $R$ not intersecting $P\cup Q$.  Let $z_1=R\cap x_1^+A_1$ and $z_2=R\cap y_2^+A_2$. By Claim \ref{claim1}, $R$ cannot intersect $A_1x_1^-$ and $R$ cannot intersect $A_2y_2^-$. Also, $R$ cannot intersect both $A_3y_3^-$ and $y_3^+A_3$. Suppose $R\cap A_3y_3^-\neq\emptyset$. Let $z$ be the last vertex of $R$ that lies in $A_3y_3^-$. Then $y_3^-A_3zRz_2A_2\omega' A_3y_3^+$ is an $y_3^-$-$y_3^+$ arc not intersecting $\{y_3,u_3\}$, contradicting Claim \ref{claim1}. Suppose $R\cap y_3^+A_3\neq\emptyset$. Let $z'$ be the first vertex of $R$ that lies in $y_3^+A_3$. Then $y_2^-A_2x_2Px_1A_1z_1Rz'A_3\omega' A_2y_2^+$ is an $y_2^-$-$y_2^+$ arc not intersecting $\{y_2,u_2\}$, contradicting Claim \ref{claim1}. Therefore, $R\cap (A_1x_1^-\cup A_2y_2^-\cup A_3\cup P\cup Q)=\emptyset$. But, $y_2^-A_2x_2Px_1A_1z_1Rz_2A_2y_2^+$ is an $y_2^-$-$y_2^+$ arc not intersecting $\{y_2,u_2\}$, contradicting Claim \ref{claim1}.
\hfill $\Box$

%By applying Lemma \ref{arcarcpath} to $A_1$ and $A_2$ with $S=\emptyset$, we obtain an $A_1$-$A_2$ path $P$.  Let $x_1=P\cap A_1$, $x_2=P\cap A_2$ and $x$ be the neighbor of $x_1$ in $P$. If $P$ intersects $A_3$, then interchange $A_2$ and $A_3$. Therefore, without loss of generality, there is an $A_1$-$A_2$ path $P$ that does not intersect $A_3$.

%Now, apply Lemma \ref{arcarcpath} to $x_2A_2$ and $A_3$ with $S=V(P)$. We obtain an $x_2A_2$-$A_3$ path $Q$ not intersecting $P$.  Let $y_2=Q\cap x_2A_2$, $y_3=Q\cap A_3$ and $y$ be the neighbor of $y_2$ in $Q$. We claim that $Q\cap A_1\subseteq x_1A_1$. For otherwise, let $y'$ be the first vertex of $Q$ that lies in $A_1x_1$. Note that $y'\neq x_1$, and $x_1^-$ and $x_1^+$ exist in $A_1$. But, $x_1^-A_1y'Qy_2A_2\omega' A_1x_1^+$ is an $x_1^-$-$x_1^+$ arc not intersecting $\{x_1,x\}$, contradicting Claim \ref{claim1}. Suppose $Q\cap A_1\neq \emptyset$. Then there is an $x_1A_1$-$A_3$ subpath in $Q$ not intersecting $A_2$, and we interchange $A_1$ and $A_2$. Therefore, without loss of generality, we can assume that there is an $x_2A_2$-$A_3$ path $Q$ that does not intersect $P\cup A_1$.

%Finally, apply Lemma \ref{arcarcpath} to $x_1A_1$ and $y_2A_2$ with $S=V(P\cup Q)$. We obtain an $x_1A_1$-$y_2A_2$ path $R$ not intersecting $P\cup Q$.  Let $z_1=R\cap x_1A_1$ and $z_2=R\cap y_2A_2$. Note that $y_2^-$ and $y_2^+$ exist in $A_2$. Suppose $R\cap A_2y_2^-=\emptyset$. Then $y_2^-A_2x_2Px_1A_1z_1Rz_2A_2y_2^+$ is an $y_2^-$-$y_2^+$ arc not intersecting $\{y_2,y\}$, contradicting Claim \ref{claim1}. Suppose $R\cap A_2y_2^-\neq\emptyset$. Let $z$ be the last vertex in $R$ that lies in $A_2y_2^-$. Then $y_2^-A_2zRz_2A_2y_2^+$ is an $y_2^-$-$y_2^+$ arc not intersecting $\{y_2,y\}$, contradicting Claim \ref{claim1}.


\begin{claim}\label{claim6}
For each end $\omega$ in $|G|$, $deg_{G_C}(\omega)=2$.
\end{claim}

\noindent{\em Proof.}
Let $x$ be a vertex in $G_C$. Since $\overline{G_C}$ is arc-connected, there is an $\omega$-arc $A$ in $\overline{G_C}$ joining $x$ to $\omega$. Let $y$ be the neighbor of $x$ in $A$ and $a$ be an interior point of $xy$. Since $\overline{G_C}$ is 2-connected, $\overline{G_C}\setminus a$ is connected. Suppose $\overline{G_C-xy}$ is not connected. Then there exist two disjoint nonempty open sets $U$ and $V$ in $|G|$ such that $\overline{G_C-xy}\subseteq U\cup V$, $U\cap\overline{G_C-xy}\neq\emptyset$ and $V\cap\overline{G_C-xy}\neq\emptyset$. If $x,y\in U$, then $U\cup [x,a)\cup [y,a)$ and $V$ are two disjoint open sets in $|G|$ both intersecting $\overline{G_C}\setminus a$, and their union contains $\overline{G_C}\setminus a$, which is impossible. If $x\in U$ and $y\in V$, then $U\cup [x,a)$ and $V\cup [y,a)$ are two disjoint open sets in $|G|$ both intersecting $\overline{G_C}\setminus a$, and their union contains $\overline{G_C}\setminus a$, which is also impossible. Therefore, $\overline{G_C-xy}$ is connected and is arc-connected by Lemma \ref{connimplyaconn}. Let $A'$ be an $x$-$\omega$ arc in $\overline{G_C-xy}$. If $yA\omega \cap A'$ contains a vertex $u$, then $u$ has degree at least three in $G_C$, a contradiction. Let $\omega'$ be the first point in $yA\omega \cap A'$ which is an end. If $\omega'\neq \omega$, then $deg_{G_C}(\omega')\geq 3$ contradicting Claim \ref{claim5}. Therefore, $\omega'=\omega$ and we have $deg_{G_C}(\omega)\geq 2$. By Claim \ref{claim5}, $deg_{G_C}(\omega)=2$.
\hfill $\Box$
\\

%Let $x$ be a vertex in $G_C$. Since $\overline{G_C}$ is arc-connected, there is an $\omega$-arc $A$ in $\overline{G_C}$ joining $x$ to $\omega$. Let $xy$ be the starting edge in $A$ and $z$ be the neighbor of $x$ in $G_C$ other than $y$. Then $z\notin A$, for otherwise, $z$ will have degree at least three in $G_C$. Let $a$ be an interior point of $xz$. Since $\overline{G_C}$ is 2-connected, $\overline{G_C}\setminus a$ is connected. Suppose $\overline{G_C-xz}$ is not connected. Then there exist two disjoint nonempty open sets $U$ and $V$ in $|G|$ such that $\overline{G_C-xz}\subseteq U\cup V$, $U\cap\overline{G_C-xz}\neq\emptyset$ and $V\cap\overline{G_C-xz}\neq\emptyset$. If $x,z\in U$, then $U\cup [x,a)\cup [z,a)$ and $V$ are two disjoint open sets in $|G|$ both intersecting $\overline{G_C}\setminus a$, and their union contains $\overline{G_C}\setminus a$, which is impossible. If $x\in U$ and $z\in V$, then $U\cup [x,a)$ and $V\cup [z,a)$ are two disjoint open sets in $|G|$ both intersecting $\overline{G_C}\setminus a$, and their union contains $\overline{G_C}\setminus a$, which is also impossible. Therefore, $\overline{G_C-xz}$ is connected and is arc-connected by Lemma \ref{connimplyaconn}. Let $A'$ be an $z$-$\omega$ arc in $\overline{G_C-xz}$. If $A\cap A'$ contains a vertex $u$, then $u$ has degree at least three in $G_C$, a contradiction. Hence, $A$ and $A'$ are two edge-disjoint $\omega$-arcs in $\overline{G_C}$. We have $deg_{G_C}(\omega)\geq 2$. By Claim \ref{claim5}, $deg_{G_C}(\omega)=2$.
%Let $y$ be the neighbor of $x$ in $\overline{G_C}\setminus A$. Since $\overline{G_C}$ is 2-connected, $\overline{G_C}-xy$ is connected and by Lemma \ref{connimplyaconn}, $\overline{G_C}-xy$ is arc-connected. Let $A'$ be an $y-A$ arc in $\overline{G_C}-xy$ and $z=A\cap A'$. Then $z$ is an end of $G$. By Claim \ref{claim6}, $z=\omega$. Hence, $deg_{G_C}(\omega)\geq 2$ and $deg_{G_C}(\omega)=2$ by Claim \ref{claim6}.


We are now ready to prove the infinite analog of Theorem \ref{wu}.

\begin{theorem}\label{wuinfinite}
Let $G$ be a $2$-connected locally finite graph nonisomorphic to $K_3$. Then the following are equivalent:

\begin{enumerate}[\rm (1)]

\item Every vertex of $G$ is incident to exactly two contractible edges.

\item Every finite bond of $G$ contains exactly two contractible edges.

\item $G$ is outerplanar.

\end{enumerate}

\end{theorem}

\noindent{\em Proof.}

$(2)\Rightarrow(1)$ Trivial.

$(1)\Rightarrow(3)$ By Theorem \ref{wu}, this is true for finite $G$. Therefore, assume $G$ is infinite. Suppose every vertex of $G$ is incident to exactly two contractible edges. By Theorem \ref{hamilton}, $\overline{G_C}$ is a Hamilton circle. All edges in $E(G)\setminus E_C$ are chords of $\overline{G_C}$ and are non-contractible. Consider any chord $xy$ of $\overline{G_C}$. Since every vertex of $G$ is incident to exactly two contractible edges, by Lemma \ref{contexist}, $G-x-y$ consists of exactly two components $C_1$ and $C_2$. Without loss of generality, assume that $x^+\overline{G_C}y^-\subseteq\overline{C_1}$ and $y^+\overline{G_C}x^-\subseteq\overline{C_2}$. Then there is no chord of $\overline{G_C}$ between $x^+\overline{G_C}y^-$ and $y^+\overline{G_C}x^-$. Hence, no chords of $\overline{G_C}$ are overlapping. Embed $\overline{G_C}$ in a circle of $\mathbb{R}^2$ and denote the embedding by $\phi$. Now, draw every chord $xy$ of $\overline{G_C}$ as a straight line segment between $\phi(x)$ and $\phi(y)$ in $\mathbb{R}^2$. This shows that $G$ is outerplanar.

$(3)\Rightarrow(2)$ Let $B$ be a finite bond of $G$ between two components $X$ and $Y$ of $G-B$. By Lemma \ref{fund3}(a), $B$ contains at least two contractible edges. Suppose $B$ contains three contractible edges $x_1y_1, x_2y_2, x_3y_3$ such that $x_1, x_2, x_3\in X$ and $y_1, y_2, y_3\in Y$. Since $X$ and $Y$ are connected, there exists a path $P$ in $X$ joining $x_1$ to $x_2$ and a path $Q$ in $Y$ joining $y_1$ to $y_2$. Let $C:=P\cup x_1y_1\cup Q\cup x_2y_2$. Obviously, $x_3y_3\notin E(C)$. Take any $x_3$-$P$ path $P'$ in $X$ joining $x_3$ to $P$ at $x$ and any $y_3$-$Q$ path $Q'$ in $Y$ joining $y_3$ to $Q$ at $y$. Let $R':=P'\cup x_3y_3 \cup Q'$. If $R'=x_3y_3$, then $x_3y_3$ is a chord of $C$. Since $x_3y_3$ is contractible, the two components of $C-x_3-y_3$ are joined by a path, say $R$. Then $C\cup R \cup R'$ is a $K_4$-subdivision and $G$ is not outerplanar. Suppose $R'\neq x_3y_3$. If both $x$-$y$ paths in $C$ are not edges, then $C\cup R'$ is a $K_{2,3}$-subdivision and $G$ is not outerplanar. If one of the two $x$-$y$ paths in $C$ is an edge, then without loss of generality, assume $x_2=x$ and $y_2=y$. Since $x_2y_2$ is contractible, the two components of $(C\cup R')-x_2-y_2$ are joined by a path, say $R$. Then $C\cup R\cup R'$ is a $K_4$-subdivision and $G$ is not outerplanar.
\hfill $\Box$
\\
%there exists a cycle $C$ containing $x_1y_1$ and $x_2y_2$ such that $x_1Cx_2\subseteq X$ and $y_1Cy_2\subseteq Y$. Obviously, $x_3y_3\notin E_G(C)$. There exist a path $P_X$ in $X$ joining $x_3$ to $C$ at $x$ and a path $P_Y$ in $Y$ joining $y_3$ to $C$ at $y$. Let $P=P_X \cup x_3y_3 \cup P_Y$. If $P=x_3y_3$, then $x_3y_3$ is a chord of $C$. Since $x_3y_3$ is contractible, the two components of $C-x_3-y_3$ are joined by a path, say $Q$. Then $C\cup x_3y_3 \cup Q$ is a $K_4$-subdivision and $G$ is not outerplanar. Suppose $P\neq x_3y_3$. If none of the two $x-y$ paths in $C$ are edges, then $P\cup C$ is a $K_{2,3}$-subdivision and $G$ is not outerplanar. If one of the two $x-y$ paths in $C$ is an edge, then without loss of generality, assume $x_2=x$ and $y_2=y$. Since $x_2y_2$ is contractible, the two components of $P\cup C-x_2-y_2$ are joined by a path, say $R$. Then $P\cup C\cup R$ is a $K_4$-subdivision and $G$ is not outerplanar.


%\begin{theorem}\label{wuinfinite}
%Let $G$ be a $2$-connected locally finite infinite graph. Then every vertex of $G$ is incident to exactly two contractible edges if and only if $G$ is outerplanar.
%\end{theorem}

%\noindent{\em Proof.}
%$(\Rightarrow)$ Suppose every vertex of $G$ is incident to exactly two contractible edges. By Theorem \ref{hamilton}, $\overline{G_C}$ is a Hamilton circle. All edges in $E_G(G)\setminus E_C$ are chords of $\overline{G_C}$ and are non-contractible. Consider any chord $xy$ of $\overline{G_C}$. Since every vertex of $G$ is incident to exactly two contractible edges, by Lemma \ref{contexist}, $G-x-y$ consists of exactly two components $C_1$ and $C_2$. Without loss of generality, assume that $x^+\overline{G_C}y^-\subseteq\overline{C_1}$ and $y^+\overline{G_C}x^-\subseteq\overline{C_2}$. Then there is no chord of $\overline{G_C}$ between $x^+\overline{G_C}y^-$ and $y^+\overline{G_C}x^-$. Hence, no chords of $\overline{G_C}$ are overlapping. Now, embed $\overline{G_C}$ as a unit circle in $\mathbb{R}^2$ with the embedding denoted by $\phi$, and draw every chord $xy$ of $\overline{G_C}$ as a straight line segment between $\phi(x)$ and $\phi(y)$ in $\mathbb{R}^2$. This implies that $G$ is outerplanar.

%$(\Leftarrow)$ From Theorem \ref{dean}(b), we know that every vertex of $G$ is incident to at least 2 contractible edges. Suppose there is a vertex $x$ of $G$ that is incident to at least 3 contractible edges. Let $xa, xb, xc$ be three such contractible edges. Since $G$ is 2-connected, $xa$ and $xb$ lie on a finite cycle of $G$, say $C$. Suppose $xc$ is not a chord of $C$. As $G-x$ is connected, there is a path $P$ in $G-x$ joining $c$ to $C$. If $P\cap C\cap \{a,b\}=\emptyset$, then $C\cup P\cup xc$ is a $K_{2,3}$-subdivision and $G$ is not outerplanar. If $P\cap C\cap \{a,b\}\neq\emptyset$, without loss of generality, suppose $a\in P\cap C$. Since $xa$ is contractible, $G-x-a$ is connected and there exists a path $Q$ joining $P-a$ and $C-x-a$. Now, $C\cup P\cup xc\cup Q$ is a $K_4$-subdivision and $G$ is not outerplanar. Suppose $xc$ is a chord of $C$. As $xc$ is contractible, the two components of $C-x-c$ are joined by a path $Q$. Then $C\cup xc\cup Q$ is a $K_4$-subdivision and $G$ is not outerplanar.
%\hfill $\Box$
%\\

%As a corollary, we prove that every 2-connected outerplanar locally finite graph is Hamiltonian.
%\\

%\noindent{\bf Theorem 3.}
%{\em Every $2$-connected outerplanar locally finite graph is Hamiltonian.}
%\\

%\noindent{\em Proof.}
%This follows from Theorem \ref{wuinfinite} and Theorem \ref{hamilton}. 
%\hfill $\Box$


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
The author would like to thank Henning Bruhn for suggesting Theorem \ref{uncountend}, and the referees for comments that greatly improve the accuracy and presentation of the paper.

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