Does every cut-point space embed into the plane?

Question

Let $X$ be a connected separable metrizable topological space. Call it a cut-point space if $X\setminus \{x\}$ is disconnected for every $x\in X$. Then does $X$ embed into the plane?

My thoughts:

(0) It is not difficult to see that $X$ must have dimension $1$, and therefore embeds into $\mathbb R ^3$.

(1) Every separable cut-point space has only a countable set of cut points of order greater than 2 ($x$ is a cut point of order $n\in \{0,1,2,...\}\cup \{\infty\}$ if $X\setminus \{x\}$ has $n$ components). Coincidentally, or perhaps not, there is no uncountable collection of disjoint "tacks" in the plane: How many tacks fit in the plane?.

(2) Every dendrite embeds into the plane. A dendrite is a locally connected continuum $X$ such that for every two points $x,y\in X$, there is a third point $z\in X$ such that $x$ and $y$ belong to different components of $X\setminus \{z\}$. It is not clear to me whether every cut-point space can be embedded into a dendrite.

Answer 1

Here's an example that contradicts (0).

Take $n\in\mathbb{N}$ and let $f:[0,1]\to[0,1]^n$ be a function with the following property: whenever $F\subseteq[0,1]^{n+1}$ is closed and $\pi_1[F]$ is uncountable then the graph of $f$ intersects $F$ ($\pi_1$ is the projection onto the first coordinate). Such an $f$ is constructed by enumerating the family of closed sets as $\langle F_\alpha:\alpha<\mathfrak{c}\rangle$ and then choosing, at every stage $\alpha$, a point $x_\alpha\in[0,1]\setminus\{x_\beta:\beta<\alpha\}$ and a point $y_\alpha\in[0,1]^n$ such that $\langle x_\alpha,y_\alpha\rangle\in F_\alpha$. If one uses a well-order of $[0,1]$ of type $\mathfrak{c}$ and always lets $x_\alpha$ be the first point with the desired property then $\{\langle x_\alpha,y_\alpha\rangle:\alpha<\mathfrak{c}\}$ is de desired function.

Note that every point $\langle x_\alpha,y_\alpha\rangle$ is a cut point of $f$.

Since $f$ intersects every continuum that connects $\{0\}\times[0,1]^n$ and $\{1\}\times[0,1]^n$ it follows that $\dim f\ge n$.

Also $f$ is connected: if $U$ and $V$ are open and disjoint in $[0,1]^{n+1}$ such that $f\subseteq U\cup V$ then $F=[0,1]^{n+1}\setminus(U\cup V)$ is closed and disjoint from $f$, hence $K=\pi_1[F]$ is countable (and closed, by compactness). For $x\notin K$ we have $\{x\}\times[0,1]^n\subseteq U$ or $\{x\}\times[0,1]^n\subseteq V$. Assuming that $U\cap f$ and $V\cap f$ are both nonempty the sets $U_1=\{x:\{x\}\times[0,1]^n\subseteq U\}$ and $V_1=\{x:\{x\}\times[0,1]^n\subseteq V\}$ are nonempty, open (by compactness), and disjoint. Because $K$ is countable the union $U_1\cup V_1$ is dense and hence there is an $x\in\operatorname{cl}U_1\cap\operatorname{cl}V_1$. Say $\langle x,f(x)\rangle\in U$; there is an $\varepsilon>0$ such that $(x-\varepsilon,x+\varepsilon)\times (f(x)-\varepsilon,f(x)+\varepsilon) \subseteq U$, but then $U\cap V\neq\emptyset$. Contradiction.