# Does every cut-point space embed into the plane?

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.

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.