TITLE: Subset of the plane with many cut points
QUESTION [1 upvotes]: This is yet another question of mine related to this other stack overflow question.
Let $A$ be a path connected, dense subset of ${\mathbb R}^2$. Let $B$ denote the complement of $A$, and $C$ denote the set of cut points in $B$, i.e. the set of all $b\in B$ such that $B\setminus \lbrace b \rbrace$ is not path connected. Assume that $B$ contains at least three points and has no isolated points. 
Prove or find a counterexample :  $C$ is dense in $B$.

REPLY [2 votes]: Consider $b\in B$.
Let $a$ be another point of $B$.
Assume that $C$ does not intersect an open neighbourhood $U$ of $b$.
Select $b_0\in U\cap B\setminus \{a,b\}$ (using the fact that $b$ is not an isolated point).
Then $b_0\notin C$, hence there is a path $\gamma_0\colon[0,1]\to B$ from $a$ to $b$ avoiding $b_0$.
Select $b_1=\gamma_0(t_0)$ on $\gamma_0([0,1])\cap U\setminus\{a,b\}$.
Then $b_1\notin C$, hence there is a path $\gamma_1\colon[0,1]\to B$ from $a$ to $b$ avoiding $b_1$.
The set $T_>:=\{t\in[t_0,1]\mid \gamma_0(t)\in\gamma_1([0,1])\}$ is compact, we have $1\in T_>$, $t_0\notin T_>$. Let $t_1=\min T_>$.
The set $T_<:=\{t\in[0,t_0]\mid \gamma_0(t)\in\gamma_1([0,1])\}$ is compact, we have $0\in T_<$, $t_0\notin T_<$. Let $t_2=\max T_<$.
There exists $s_1\in [0,1]$ with $\gamma_1(s_1)=\gamma_0(t_1)$.
The set $S := \{s\in[0,1]\mid \gamma_1(s)=\gamma_0(t_0)\}$ is compact. Let $s_0$ be an element that minimizes the continuous function $s\mapsto |s-s_1|$.
Then we can glue together a simple closed curve $\gamma\colon[0,1]\to B$ per
$$\gamma(t)=\begin{cases}\gamma_0(t_0+2t(t_1-t_0))&\text{if }t<\frac12 \\
\gamma_1(s_1+(1-2t)(s_0-s_1))&\text{otherwise}\end{cases}$$
By the Jordan curve theorem, we obtain two open sets, the interior and te exterior of the curve. Since $A$ is dense, there are interior and exterioir points of $A$. Any path between such points intersects our Jordan curve $\subseteq B$, contradicting the path connectedness of $A$.
Therefore $C$ intersects every openneighbourhoodof $B$, i.e. $b\in\overline C$.
Since $b\in B$ was arbitrary, $C$ is dense in $B$.
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