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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 $B$.

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B is the closed interval and b is one of the endpoint, then B\{b} is path connected. And if b is a midpoint of the interval, then it is not path connected. –  lee Dec 9 '12 at 15:02
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In fact, it may be the case that $B\setminus\{b\}$ is path connected, but $B$ itself is not: Consider a two-point set $B$. –  Hagen von Eitzen Dec 9 '12 at 15:13
    
@lee : question edited so as to take your example into account. –  Ewan Delanoy Dec 9 '12 at 16:15
    
@HagenvonEitzen : question edited so as to take your example into account. –  Ewan Delanoy Dec 9 '12 at 16:15
    
Can a single comment contain notifications to several users? –  Ewan Delanoy Dec 9 '12 at 16:17

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up vote 2 down vote accepted

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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The definition of your set $S$ looks strange to me. Since $\gamma_1$ avoids $b_1=\gamma_0(t_0)$, must not $S$ be empty ? –  Ewan Delanoy Dec 9 '12 at 17:42
    
In general it is not at all simple to extract a simple path from an arbitrary path (which can be fractal, Peano-like, nowhere differentiable, you name it). –  Ewan Delanoy Dec 9 '12 at 17:45
    
Since any path connected subset of the plane is also arc connected, however, one can assume that $\gamma_0$ and $\gamma_1$ are arcs. –  Ewan Delanoy Dec 9 '12 at 19:27

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