Dispersion points and rational curves

TL;DR

This paper constructs specific connected plane sets with unique properties, embedding them into rational curves, and addresses a question about dispersion points, contributing new examples to the study of rational curves.

Contribution

It introduces two novel connected plane sets with dispersion points and indecomposability, both embeddable into rational curves, answering a previously open question.

Findings

Constructed a biconnected set with a dispersion point

Created an indecomposable connected set

Both sets are completely metrizable

Abstract

We construct two connected plane sets which can be embedded into rational curves. The first is a biconnected set with a dispersion point. It answers a question of Joachim Grispolakis. The second is indecomposable. Both examples are completely metrizable.

Equations

$$0(\sigma )$$

$$1(\sigma )$$

$$\text{@text@daccent∇}X=\nabla [X]\cup \{⟨\frac{1}{2},0⟩\}.$$

$$f(c)=\sum _{\{n<\omega \text{mathchar}58d_n<c\}}{2}^{-n-1}.$$

$$\underset{c\to d_n^{-}}{\lim }f(c)=r_n\text{mathchar}58=f(d_n)\text{, whereas}$$

$$\underset{c\to d_n^{+}}{\lim }f(c)=s_n\text{mathchar}58=f(d_n)+2^{-n-1}.$$

$$\{⟨c,r⟩\in C\times [0,1]\text{mathchar}58(r>f(c)\text{ if }c\notin \{d_n\text{mathchar}58n<\omega \})\text{ or }(r>s_n\text{ if }c=d_n)\}$$

$$T_i^n(⟨c,r⟩)=⟨0({\sigma }_i^n)+\frac{c}{3^n},a_i^n+r(b_i^n-a_i^n)⟩$$

$$\pi_{1}\big{[}\textstyle B(\sigma)\times\mathbb{R}\cap\bigcup\{T^{k}_{i}[D]\mathrel{\mathop{\mathchar 58\relax}}k\leq n\text{ and }i<|\mathcal{R}_{k}|\}\big{]}=[-n,n+1];$$

$$d(I,T_j^k[D])\text{mathchar}58=\min \{\varrho (x,y)\text{mathchar}58x\in I\text{ and }y\in T_j^k[D]\}<\frac{1}{n+1}+\frac{1}{3^n}.$$

$$R_0^1$$

$$R_1^1$$

$$R_2^1$$

$$R_3^1$$

$$\{B(\sigma )\times [a,a+\frac{1}{2}]\text{mathchar}58\sigma =⟨0⟩,⟨1⟩\text{ and }a=-1,-\frac{1}{2},1,\frac{3}{2}\}.$$

$${\pi }_1[\{0(\sigma )\}\times \mathbb{R}\cap \bigcup \{T_i^k[D]\text{mathchar}58k<n\text{ and }i<|{\mathcal{R}}_k|\}];\text{ and}$$

$${\pi }_1[\{1(\sigma )\}\times \mathbb{R}\cap \bigcup \{T_i^k[D]\text{mathchar}58k<n\text{ and }i<|{\mathcal{R}}_k|\}]$$

$$-n+1\le x_0<y_0<x_1<y_1<...<x_m<y_m\le n$$

$$B(\sigma )\times [-n,x_0];$$

$$B(\sigma )\times [y_m,n+1]\text{; and}$$

$$B(\sigma )\times [y_j,x_{j+1}]\text{, }j<m,$$

$$Y=C\times \mathbb{R}\setminus \bigcup \{T_i^n[D\setminus M]\text{mathchar}58n<\omega \text{ and }i<|{\mathcal{R}}_n|\}.$$

$$P=C\times \mathbb{R}\setminus \bigcup \{T_i^n[D]\text{mathchar}58n<\omega \text{ and }i<|{\mathcal{R}}_n|\}$$

$$Q=\bigcup \{T_i^n[M]\text{mathchar}58n<\omega \text{ and }i<|{\mathcal{R}}_n|\}.$$

$$\Xi (⟨c,r⟩)=⟨c,\frac{1}{\pi }\arctan (r)+\frac{1}{2}⟩.$$

$$I\subseteq R_{i(n)}^n\cup R_{j(n)}^{k(n)}$$

$$0<d(I,T_{j(n)}^{k(n)}[D])<\frac{1}{n+1}+\frac{1}{3^n}$$

$$\overline{Y\cap (C\setminus {\pi }_0[Q])\times \mathbb{R}}=C\times \mathbb{R}.$$

$$\varnothing \ne (U\setminus {\pi }_0[Q])\times (r,s)\cap \overline{A}\cap \overline{B}\subseteq T_i^n[D].$$

$$X\cap \{c\}\times [a,b]=\{⟨c,r⟩\}.$$

$$[a,b]=V\cap [{\pi }_1(T_i^n(⟨d_m,r_m⟩)),{\pi }_1(T_i^n(⟨d_m,s_m⟩))].$$

$${\Pi }^0\text{mathchar}58=\{⟨n,i⟩\text{mathchar}58B({\sigma }_i^n)\times [a_i^n,b_i^n]\subseteq U\times (a,\frac{a+r}{2})\}.$$

$$\displaystyle W_{0}=\Big{\{}\langle e,s\rangle\in X\mathrel{\mathop{\mathchar 58\relax}}\big{(}\exists k<\omega\text{ and }\langle n,i\rangle\in\Pi^{0}_{k}\big{)}\big{(}e\in$$

$$\displaystyle s>\max\pi_{1}\big{[}\{e\}\times\mathbb{R}\cap T^{n}_{i}[D]\big{]}\big{)}\Big{\}}.$$

$$\displaystyle W_{1}=\Big{\{}\langle e,s\rangle\in X\mathrel{\mathop{\mathchar 58\relax}}\big{(}\exists k<\omega\text{ and }\langle n,i\rangle\in\Pi^{1}_{k}\big{)}\big{(}e\in$$

Full text

1. Introduction

Sur les ensembles connexes [8], published by B. Knaster and C. Kuratowski in 1921, was the first paper devoted to the study of connected sets in topology [16]. The most surprising feature of the article was its presentation of an ensemble biconnexe, known today as the “Knaster-Kuratowski fan” or “Cantor’s leaky tent”. The essence of the example is conveyed in [9]: “Cantor’s leaky tent is connected, but just barely: remove one point and the whole thing falls apart.”

The Knaster-Kuratowski fan has a certain countable structure. It is rational, meaning it has a basis of open sets with countable boundaries. However, it does not embed into any rational continuum [15, Example 1].111Tymchatyn’s argument in [15] shows that in order for a connected subset $X$ of the Cantor fan to embed into a rational continuum, the complement of $X$ must contain a dense union of arcs. Hints to [4, Exercise 1.4.C(c)] reveal, moreover, that each of its compactifications has a dense subspace homeomorphic to $\mathbb{P}\times(0,1)$, where $\mathbb{P}$ is the set of irrational numbers. In this sense, the Cantor fan’s uncountable structure is encoded in the Knaster-Kuratowski fan.

In the Houston Problem Book [1, Problem 79], Joachim Grispolakis asked: Is it true that no biconnected set with a dispersion point can be embedded into a rational continuum? The primary goal of this paper is to show the answer is “no”. We will construct a biconnected subset of the Cantor fan which has a dispersion point and can be embedded into a rational continuum.

By a technique in [11], our counterexample generates an indecomposable connected plane set which also can be embedded into a rational continuum. Indecomposable continua, by contrast, are known to locally resemble the Cantor set times the unit interval. And every separator of an indecomposable continuum contains a Cantor set.

For rational compactifications of our examples, we will rely on the following theorem by E.D. Tymchatyn and S.D. Iliadis: A separable metrizable space $X$ has a rational metrizable compactification if and only if $X$ has a basis of open sets with scattered boundaries. See [15, Theorem 1] and [6, Theorem 8].222The introduction to [6] indicates there was a gap in the original proof of [15, Theorem 1]. [6, Theorem 8] is a stronger result with a corrected proof. A shorter proof of [6, Theorem 8] is given in [5]. Our $X$’s will have bases of open sets with discrete boundaries. Suslinian compactifications, which are almost rational, will be described in detail near the end of the paper. Each will be the union of a zero-dimensional set and a null-sequence of Hawaiian earrings.

2. Preliminaries

2.1. Terminology

A connected topological space $X$ is biconnected provided $X$ cannot be written as the union of two disjoint non-degenerate connected subsets.

A point $x$ in a non-degenerate connected space $X$ is a dispersion point if $X\setminus\{x\}$ is hereditarily disconnected, i.e. if every connected subset of $X\setminus\{x\}$ is degenerate. Every connected set with a dispersion point is biconnected, but the converse is false under the Continuum Hypothesis [12].

A space X is rational (rim-discrete) provided $X$ has a basis of open sets whose boundaries are countable (discrete). If $X$ has a basis of clopen sets, then $X$ is zero-dimensional.

A continuum is a connected compact metrizable space with more than one point.

A continuum is Suslinian if there is no uncountable collection of pairwise disjoint subcontinua [10]. Suslinian continua are frequently called curves because they are $1$-dimensional.

Rational continua are Suslinian. In fact, by second countability, each rational continuum can be written as the union of a zero-dimensional subspace and a countable set. Every subcontinuum must intersect the countable set.

A connected space $X$ is indecomposable if $X$ cannot be written as the union of two proper closed connected subsets. This is equivalent to saying every proper closed connected subset of $X$ is nowhere dense. A connected space $X$ is widely-connected if every proper closed connected subset of $X$ is degenerate.

Biconnected and widely-connected spaces are known to be punctiform, meaning they contain no continua.

2.2. The Cantor set and Cantor fan

Let $C\subseteq[0,1]$ be the middle-thirds Cantor set. For each $\sigma\in 2^{<\omega}$ define

[TABLE]

where $n=\operatorname{dom}(\sigma)$. The points $0(\sigma)$ and $1(\sigma)$ are called the “endpoints” of $C$. More precisely, they are the endpoints of the connected components of $\mathbb{R}\setminus C$.

Let $B(\sigma)=[0(\sigma),1(\sigma)]\cap C$. The set of all $B(\sigma)$’s is the canonical clopen basis for $C$.

Let $\nabla\mathrel{\mathop{\mathchar 58\relax}}C\times[0,1]\to[0,1]^{2}$ be defined by $\langle c,y\textstyle\rangle\mapsto\langle\frac{y(2c-1)+1}{2},y\rangle,$ so that $\nabla\restriction C\times(0,1]$ is a homeomorphism and $\nabla^{-1}\{\langle\frac{1}{2},0\rangle\}=C\times\{0\}$. For any set $X\subseteq C\times(0,1]$ we put

[TABLE]

The Cantor fan is the set \nabla(C\times[0,1])=\textbf{\@text@daccent{\nabla}}(C\times(0,1]).

2.3. Null-sequences and Hawaiian earrings

Let $\varrho$ denote the Euclidean metric on $\mathbb{R}^{2}$. A sequence of sets $A_{0},A_{1},...\subseteq\mathbb{R}^{2}$ is a null-sequence provided the diameter (with respect to $\varrho$) of $A_{n}$ converges to [math] as $n\to\infty$. A continuum $H$ is called a Hawaiian earring if there is a null-sequence of simple closed curves $J_{0},J_{1},...\subseteq\mathbb{R}^{2}$ such that $H=\bigcup\{J_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ and $|\bigcap\{J_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}|=1$.

3. Main examples

Example 1**.**

Our goal is to construct a hereditarily disconnected $X\subseteq C\times(0,1)$ such that \textbf{\@text@daccent{\nabla}}X is rim-discrete and connected. To accomplish this, we will reduce $C\times(0,1)$ to a zero-dimensional set $P$ by deleting $\omega$-many pairwise disjoint copies of a compact set $D\subseteq C\times[0,1]$. Then $X$ will be obtained by adding back a countable point set $Q$.

Throughout this example, $\pi_{0}$ and $\pi_{1}$ are the coordinate projections in $C\times\mathbb{R}$ defined by $\pi_{0}(\langle c,r\rangle)=c$ and $\pi_{1}(\langle c,r\rangle)=r$.

We begin with a function which was introduced by Wojciech Dębski in [2]. Let $\{d_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}\subseteq C$ be a dense set of non-endpoints. Each $d_{n}$ is the limit of both increasing and decreasing sequences of points in $C$. Define $f\mathrel{\mathop{\mathchar 58\relax}}C\to[0,1]$ by

[TABLE]

Observe that $f$ is non-decreasing, and $\{d_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ is the set of (jump) discontinuities of $f$;

[TABLE]

Let $\textstyle D=\{\langle c,f(c)\rangle\mathrel{\mathop{\mathchar 58\relax}}c\in C\}\cup\bigcup\{\{d_{n}\}\times[r_{n},s_{n}]\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ be the graph of $f$ together with the vertical arcs from $\langle d_{n},r_{n}\rangle$ to $\langle d_{n},s_{n}\rangle$. Then:

• $\square$

$D$ is compact;

• $\square$

$\pi_{1}[D]=[0,1]$;

• $\square$

$C\times[0,1]\setminus D$ is the union of two disjoint open sets $\{\langle c,r\rangle\in C\times[0,1]\mathrel{\mathop{\mathchar 58\relax}}r<f(c)\}$ and

[TABLE]

consisting of all half-open intervals “below” $D$ and “above” $D$, respectively. In fact, $D$ extends to a Jordan arc which separates $[0,1]^{2}$ in this manner.

Say that a collection of sets $\mathcal{R}$ is a partial tiling of $C\times\mathbb{R}$ if: for every $R\in\mathcal{R}$ there exists $\sigma\in 2^{<\omega}$ and two numbers $a<b\in\mathbb{R}$ such that $R=B(\sigma)\times[a,b]$; and for every two elements $R_{1}\neq R_{2}$ of $\mathcal{R}$ there exists $\sigma\in 2^{<\omega}$ and $c\in\mathbb{R}$ such that $R_{1}\cap R_{2}\subseteq B(\sigma)\times\{c\}$.

We will recursively define a sequence $\mathcal{R}_{0},\mathcal{R}_{1},...$ of finite partial tilings of $C\times\mathbb{R}$ so that for each $n<\omega$:

• (i)

$\mathcal{R}_{n}$ consists of rectangles $R^{n}_{i}=B(\sigma^{n}_{i})\times[a^{n}_{i},b^{n}_{i}]$, where $\sigma^{n}_{i}\in 2^{n}$ and $i<|\mathcal{R}_{n}|<\omega$;

• (ii)

$0<b^{n}_{i}-a^{n}_{i}\leq\frac{1}{n+1}$ for all $i<|\mathcal{R}_{n}|$;

• (iii)

the linear transformations $T^{n}_{i}\mathrel{\mathop{\mathchar 58\relax}}C\times[0,1]\to R^{n}_{i}$ defined by

[TABLE]

are such that $T^{n}_{i}[D]\cap T^{k}_{j}[D]=\varnothing$ whenever $k<n$ or $i\neq j$;

• (iv)

for each $\sigma\in 2^{n}$,

[TABLE]

• (v)

for every arc $I\subseteq C\times[-n,n+1]\setminus\bigcup\{T^{k}_{i}[D]\mathrel{\mathop{\mathchar 58\relax}}k\leq n\text{ and }i<|\mathcal{R}_{k}|\}$ there are integers $i<|\mathcal{R}_{n}|$, $k\leq n$, and $j<|\mathcal{R}_{k}|$ such that $I\subseteq R^{n}_{i}\cup R^{k}_{j}$ and

[TABLE]

Here, $\varrho$ is the Euclidean metric on $\mathbb{R}^{2}$.

Note that $T^{n}_{i}$ is just the natural homeomorphism between $C\times[0,1]$ and $R^{n}_{i}$.

• $n=0$:

Let $\mathcal{R}_{0}=\{R^{0}_{0}\}=\{C\times[0,1]\}=\{B(\varnothing)\times[0,1]\}$, and $T^{0}_{0}=\operatorname{id}_{C\times[0,1]}$.

• $n=1$:

Tile $([0,\frac{1}{3}]\cap C)\times[f(\frac{1}{3}),1]$ and $([\frac{2}{3},1]\cap C)\times[0,f(\frac{2}{3})]$ with four rectangles:

[TABLE]

Let $R^{1}_{4},R^{1}_{5},...,R^{1}_{11}$ enumerate the set of eight rectangles

[TABLE]

Let $\mathcal{R}_{1}=\{R^{1}_{i}\mathrel{\mathop{\mathchar 58\relax}}i<12\}$.

• $n\geq 2$:

Assume $\mathcal{R}_{k}$ has been defined for all $k<n$.

Fix $\sigma\in 2^{n}$, and partially tile $B(\sigma)\times[-n,n+1]$ in the following way. The sets

[TABLE]

can be enumerated $x_{0},...,x_{m}$ and $y_{0},...,y_{m}$, respectively, such that

[TABLE]

and $x_{j}$ and $y_{j}$ belong to the same $T^{k}_{i}[D]$. Note that $B(\sigma)\times[x_{j},y_{j}]$ is contained in the closed $\frac{1}{3^{n}}$-neighborhood of the unique $T^{k}_{i}[D]$ it intersects. Tile each strip

[TABLE]

with finitely many rectangles of the form $B(\sigma)\times[a,b]$ where $0<b-a\leq\frac{1}{n+1}$.

Repeat for each $\sigma\in 2^{n}$ to obtain $\mathcal{R}_{n}$.

Note that conditions (i) through (iv) are satisfied. We have also guaranteed that for all $l,m<n$, $j<|\mathcal{R}_{l}|$, $k<|\mathcal{R}_{m}|$, and $\sigma\in 2^{n}$, there exists $i<|\mathcal{R}_{n}|$ such that $T^{n}_{i}[D]$ is between $T^{l}_{j}[D]\cap B(\sigma)\times\mathbb{R}$ and $T^{m}_{k}[D]\cap B(\sigma)\times\mathbb{R}$. Together with (iv), this implies (v).

Now let $M=\{\langle d_{n},\textstyle\frac{r_{n}+s_{n}}{2}\rangle\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}=\{\langle d_{n},f(d_{n})+2^{-n-2}\rangle\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ be the set of midpoints of vertical arcs in $D$. Note that $M$ is discrete. Define

[TABLE]

In other words, $Y=P\cup Q$ where

[TABLE]

and

[TABLE]

Let $\Xi\mathrel{\mathop{\mathchar 58\relax}}C\times\mathbb{R}\to C\times(0,1)$ be the homeomorphism defined by

[TABLE]

Finally, $X=\Xi[Y].$

Claim 1**.**

$X$* is hereditarily disconnected.*

Proof of Claim 1.

Suppose not. Then there is a vertical arc $I\subseteq Y$ of positive height. Let $N<\omega$ be such that $I\subseteq C\times[-N,N+1]$.

Apparently, $I\cap Q=\varnothing$ because each point $\langle c,r\rangle\in Q$ is isolated in its strand $Y\cap\{c\}\times\mathbb{R}$. Thus $I\subseteq P$. For each $n\geq N$ let $i(n)<|\mathcal{R}_{n}|$, $k(n)\leq n$, and $j(n)<|\mathcal{R}_{k(n)}|$ be given by condition (v), so that

[TABLE]

and $d(I,T^{k(n)}_{j(n)}[D])<\frac{1}{n+1}+\frac{1}{3^{n}}$. See Figure 3.

The total height of $R^{n}_{i(n)}\cup R^{k(n)}_{j(n)}$ is less than or equal to $\frac{2}{k(n)+1}$, so $\{k(n)\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ must be finite. On the other hand,

[TABLE]

implies $\{k(n)\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ is infinite. We have a contradiction.∎

Claim 2**.**

̣\nabla$$X** **is connected.

Proof of Claim 2.

Note that $\pi_{0}[Q]$ is countable, as is $\{c\}\times\mathbb{R}\setminus Y$ for every $c\in C\setminus\pi_{0}[Q]$. In particular, $\overline{Y\cap\{c\}\times\mathbb{R}}=\{c\}\times\mathbb{R}$ for each $c\in C\setminus\pi_{0}[Q]$, and

[TABLE]

Supposing \textbf{\@text@daccent{\nabla}}X is not connected, there exists a clopen partition $\{A,B\}$ of $Y$ and a point $c\in C\setminus\pi_{0}[Q]$ such that $A\cap\{c\}\times\mathbb{R}$ and $B\cap\{c\}\times\mathbb{R}$ are non-empty. Since $\{c\}\times\mathbb{R}\subseteq\overline{A}\cup\overline{B}$, we have $\overline{A}\cap\overline{B}\cap\{c\}\times\mathbb{R}\neq\varnothing$. Applying the Baire Category Theorem in $(C\setminus\pi_{0}[Q])\times\mathbb{R}\cap\overline{A}\cap\overline{B}$, we find there exists $n,i<\omega$ and a $C\times\mathbb{R}$-open set $U\times(r,s)$ such that

[TABLE]

Let $\langle c^{\prime},r^{\prime}\rangle$ be an element of this intersection.

The interval $\{c^{\prime}\}\times(r^{\prime},s)$ is contained in either $\overline{A}\setminus\overline{B}$ or $\overline{B}\setminus\overline{A}$, and likewise for $\{c^{\prime}\}\times(r,r^{\prime})$. But $\{c^{\prime}\}\times[(r,r^{\prime})\cup(r^{\prime},s)]$ is contained in neither $\overline{A}$ nor $\overline{B}$. Without loss of generality, $\{c^{\prime}\}\times(r^{\prime},s)\subseteq\overline{A}\setminus\overline{B}$ and $\{c^{\prime}\}\times(r,r^{\prime})\subseteq\overline{B}\setminus\overline{A}$.

In $(U\setminus\pi_{0}[Q])\times(r,s)$, all intervals above $T^{n}_{i}[D]$ with $C$-coordinates sufficiently close to $c^{\prime}$ belong to $\overline{A}\setminus\overline{B}$, and all lower segments eventually belong to $\overline{B}\setminus\overline{A}$. But there exists $m<\omega$ such that $\pi_{0}(T^{n}_{i}(\langle d_{m},0\rangle))$ is very close to $c^{\prime}$ and $T^{n}_{i}[\{d_{m}\}\times[r_{m},s_{m}]]\subseteq U\times(r,s)$. See Figure 4. We have $T^{n}_{i}(\langle d_{m},\frac{r_{m}+s_{m}}{2}\rangle)\in A\cap B$, a contradiction. ∎

Claim 3**.**

$Y$* is rim-discrete.*

Proof of Claim 3.

Let $U\times V$ be any $C\times\mathbb{R}$-neighborhood of a point $\langle c,r\rangle\in Y$.

Case 1: $c\notin\pi_{0}[Q]$. By Claim 1, $\langle c,r\rangle$ is the limit of increasing and decreasing sequences of points in $\{c\}\times\mathbb{R}$ from infinitely many $T^{n}_{i}[D]$’s. So there exist $n,m,i,j<\omega$ such that $R^{n}_{i}\cup R^{m}_{j}\subseteq U\times V$, $c\in B(\sigma^{n}_{i})\cap B(\sigma^{m}_{j})$, and $b^{n}_{i}<r<a^{m}_{j}$. Let $W$ be the set of points in $X$ between $T^{n}_{i}[D]$ and $T^{m}_{j}[D]$. Then $W$ is an open subset of $X$, and $\langle c,r\rangle\in W\subseteq U\times V$. Finally, $\partial_{Y}W$ is discrete because it is contained in the discrete set $T^{n}_{i}[M]\cup T^{m}_{j}[M]$.

Case 2: $c\in\pi_{0}[Q]$. There is an interval $[a,b]\subseteq V$ such that $a<r<b$ and

[TABLE]

For in this case there are integers $n,i,m<\omega$ such that $\langle c,r\rangle=T^{n}_{i}(\langle d_{m},\frac{r_{m}+s_{m}}{2}\rangle)$. Assume $V$ is a closed interval, and let

[TABLE]

Let $N$ be an integer such that $b^{n}_{i}-a^{n}_{i}<\frac{r-a}{2}$ whenever $n\geq N$. Only finitely many vertical arcs in $D$ have height at least $\frac{r-a}{2}$, and there are only finitely many $T^{k}_{i}[D]$’s with $k<N$. It follows that $F_{0}\mathrel{\mathop{\mathchar 58\relax}}=\{e\in C\mathrel{\mathop{\mathchar 58\relax}}\{e\}\times(a,\textstyle\frac{r+a}{2})\cap X\subseteq Q\}$ is finite. For if $e\in F_{0}$ then $\{e\}\times(a,\frac{r+a}{2})$ is covered by a countable collection of pairwise disjoint arcs and singletons. So $\{e\}\times(a,\frac{r+a}{2})$ must be contained in one of the arcs; a single vertical arc of some $T^{k}_{i}[D]$ with $k<N$.

Likewise, $F_{1}\mathrel{\mathop{\mathchar 58\relax}}=\{e\in C\mathrel{\mathop{\mathchar 58\relax}}\{e\}\times(\textstyle\frac{r+b}{2},b)\cap X\subseteq Q\}$ is finite.

Let $A_{0}\supseteq A_{1}\supseteq...$ be a decreasing sequence of $C$-clopen sets such that $A_{0}\subseteq U\setminus[(F_{0}\cup F_{1})\setminus\{c\}]$ and $\{c\}=\bigcap\{A_{k}\mathrel{\mathop{\mathchar 58\relax}}k<\omega\}$. Let

[TABLE]

By the case $c\notin\pi_{0}[Q]$ we have $A_{0}\setminus\{c\}\subseteq\bigcup\{B(\sigma^{n}_{i})\mathrel{\mathop{\mathchar 58\relax}}\langle n,i\rangle\in\Pi^{0}\}$. Moreover, for each $k<\omega$ there is a finite $\Pi_{k}^{0}\subseteq\Pi^{0}$ such that $A_{k}\setminus A_{k+1}\subseteq\bigcup\{B(\sigma^{n}_{i})\mathrel{\mathop{\mathchar 58\relax}}\langle n,i\rangle\in\Pi^{0}_{k}\}$. Let

[TABLE]

We may define $\Pi^{1}$ and finite $\Pi^{1}_{k}$’s similarly for $U\times(\frac{r+b}{2},b)$, and then let

[TABLE]

Observe that $W\mathrel{\mathop{\mathchar 58\relax}}=(W_{0}\cap W_{1})\cup\{\langle c,r\rangle\}$ is an $X$-open subset of $U\times V$ and $\partial_{Y}W$ is discrete.∎

Claim 4**.**

̣\nabla$$X** is rim-discrete.**

Proof of Claim 4.

By Claim 3 we only need to show ̣\nabla$$X has a basis of open sets with discrete boundaries at the point $\langle\frac{1}{2},0\rangle$. To that end, let $\varepsilon\in(0,1)$. We will show there is a \textbf{\@text@daccent{\nabla}}X-open set \textbf{\@text@daccent{\nabla}}U such that \langle\frac{1}{2},0\rangle\in\textbf{\@text@daccent{\nabla}}U\subseteq\textbf{\@text@daccent{\nabla}}[C\times(0,\varepsilon)] and \partial_{\textbf{\@text@daccent{\nabla}}X}\textbf{\@text@daccent{\nabla}}U is discrete.

Let $r=\tan(\pi(\varepsilon-\frac{1}{2}))$. The arguments in Claim 3 show there is a finite collection of $B(\sigma^{n}_{i})$’s which cover $C$ and satisfy $B(\sigma^{n}_{i})\times[a^{n}_{i},b^{n}_{i}]\subseteq C\times(-\infty,r)$. Let $W$ be the set of all points in $Y$ below the corresponding $T^{n}_{i}[D]$’s, and put $U=\Xi[W]$. Then \textbf{\@text@daccent{\nabla}}U is as desired. ∎

Note that $D\setminus M$ is $\sigma$-compact, so $Y$ is a $G_{\delta}$-subset of $C\times\mathbb{R}$. Hence both $X$ and \textbf{\@text@daccent{\nabla}}X are completely metrizable.

This concludes Example 1.

Example 2**.**

There is a rim-discrete widely-connected $G_{\delta}$-subset of the plane.

Let $X$ be as defined in Example 1. In the bucket-handle continuum $K$ there is a Cantor set $\Delta$ such that $K\setminus\Delta$ is the union of $\omega$-many pairwise disjoint open sets $K_{i}\simeq C\times(0,1)$. For each $i<\omega$, let $\varphi_{i}\mathrel{\mathop{\mathchar 58\relax}}C\times(0,1)\to K_{i}$ be a homeomorphism, and put $X_{\omega}=\bigcup\{\varphi_{i}[X]\mathrel{\mathop{\mathchar 58\relax}}i<\omega\}$. Essentially, $X_{\omega}\cup\Delta$ is the set $W[X]$ from [11]. By the particular construction in [11], we may assume $(K_{i})$ is a null-sequence; this will be important later in Example 4.

Recall $X$ is a dense hereditarily disconnected subset of $C\times(0,1)$, and \textbf{\@text@daccent{\nabla}}X is connected. By [11, Proposition 1], $X_{\omega}\cup\Delta$ widely-connected. By the closing remark in Example 1 and [11, Proposition 6], $X_{\omega}\cup\Delta$ is completely metrizable. And by treating each point of $\Delta$ like $\langle\frac{1}{2},0\rangle$ in Claim 4, we see that $X_{\omega}\cup\Delta$ is rim-discrete.

4. Suslinian compactifications

Here we define Suslinian upper semi-continuous decompositions of the Cantor fan and Knaster bucket-handle. The respective continua will homeomorphically contain ̣\nabla$$X and $X_{\omega}\cup\Delta$.

Example 3**.**

There is a Suslinian continuum \mathcal{F}\supseteq\textnormal{{\@text@daccent{\nabla}}}X.

By Claim 3 of Example 1, there is a null-sequence $D_{0},D_{1},...\subseteq\nabla(C\times(0,1))$ of disjoint copies of $D$ such that $\nabla(C\times(0,1))\setminus\bigcup\{D_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ is zero-dimensional, and $\nabla X\cap\bigcup\{D_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ is a countable point set (the set of all midpoints of arcs in the $D_{n}$’s).

Let $f$ be the Dębski function from the beginning of Example 1. Observe that

[TABLE]

is the $C\times[0,1]$-closure of the graph of $f$; $E=\overline{\{\langle c,f(c)\rangle\mathrel{\mathop{\mathchar 58\relax}}c\in C\}}$. Thus, $E$ is compact.

Let $F$ denote the the Cantor fan $\nabla(C\times[0,1])$.

For each $n<\omega$ let $E_{n}\subseteq D_{n}$ be the natural copy of $E$ in $D_{n}$, so that E_{n}\cap\textnormal{{\@text@daccent{\nabla}}}X=\varnothing for each $n<\omega$. Identify each $E_{n}$ with the singleton $e_{n}\mathrel{\mathop{\mathchar 58\relax}}=\{E_{n}\}$, and put $\mathcal{F}=(F\setminus\bigcup\{E_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\})\cup\{e_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$. Since $(E_{n})$ is a null-sequence of compact sets, $\mathcal{F}$ is an upper semi-continuous decomposition of $F$. Therefore $\mathcal{F}$ is a continuum. Additionally, $\mathcal{F}$ contains ̣\nabla$$X. It remains to show $\mathcal{F}$ is Suslinian.

For each $n<\omega$, let $\mathcal{D}_{n}=(D_{n}\setminus E_{n})\cup\{e_{n}\}\subseteq\mathcal{F}$ denote the quotient of $D_{n}$ obtained by shrinking $E_{n}$ to $e_{n}$. We note that $\mathcal{D}_{n}$ is (homeomorphic to) a Hawaiian earring.

Claim 5**.**

Each subcontinuum of $\mathcal{F}\setminus\{e_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ is an arc in some $\mathcal{D}_{m}$.

Proof of Claim 5.

Let $\mathcal{S}\subseteq\mathcal{F}\setminus\{e_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ be a continuum.

Since $\mathcal{F}\setminus\bigcup\{\mathcal{D}_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ is zero-dimensional, there exists $m<\omega$ such that $S\cap\mathcal{D}_{m}\neq\varnothing$. This alternatively follows from the fact that $X$ is punctiform.

Let $s\in\mathcal{S}\cap\mathcal{D}_{m}$. Then there is a simple closed curve $\mathcal{J}\subseteq\mathcal{D}_{m}$ such that $s\in\mathcal{J}\setminus\{e_{m}\}$. Note that $\mathcal{J}$ corresponds to a vertical jump $J\subseteq D_{m}$.

Let $k<\omega$ be the unique integer such that $(\nabla\circ\Xi)^{-1}(D_{m})$ spans a rectangle from $\mathcal{R}_{k}$. For each $n<\omega$ let $i(n),j(n)<|\mathcal{R}_{k+1+n}|$ be such that:

[TABLE]

Let $U_{n}$ be the open region of $F$ between $D^{0}_{n}\mathrel{\mathop{\mathchar 58\relax}}=(\nabla\circ\Xi)(T^{k+1+n}_{i(n)}[D])$ and $D_{m}$.

Let $V_{n}$ be the open region of $F$ between $D^{1}_{n}\mathrel{\mathop{\mathchar 58\relax}}=(\nabla\circ\Xi)(T^{k+1+n}_{j(n)}[D])$ and $D_{m}$.

Let $a$ and $b$ be the top and bottom points of $J$. The set

[TABLE]

is open in $F$. By construction $G_{n}$ is also a union of elements of $\mathcal{F}$, and

[TABLE]

Let $\mathcal{W}$ be an open subset of $\mathcal{F}$ such that $e_{m}\in\mathcal{W}$ and $\mathcal{W}\cap\mathcal{S}=\varnothing$. Then $E_{m}\subseteq W$, $D^{0}_{n}\to a\in W$, $D^{1}_{n}\to b\in W$, and $\mathcal{W}$ contains all but finitely many simple closed curves in $\mathcal{D}_{m}$. So there is an integer $N<\omega$ such that $\partial G_{n}\subseteq W$ for all $n\geq N$.

Note that $H_{n}\mathrel{\mathop{\mathchar 58\relax}}=\overline{G_{n}}\cup E_{m}$ is a closed subset of $F$, and $H_{n}$ is a union of elements of $\mathcal{F}$. For any $n\geq N$ we have

[TABLE]

so $\mathcal{S}$ misses the boundary of $\mathcal{G}_{n}$. Since $\mathcal{S}$ is connected and $s\in\mathcal{G}_{n}$, it follows that $\mathcal{S}\subseteq\mathcal{G}_{n}$. In conclusion,

[TABLE]

Therefore $\mathcal{S}$ is an arc in $\mathcal{D}_{m}$. ∎

By Claim 5, $\mathcal{F}$ is the union of a punctiform set $\mathcal{P}$ and a countable set $\mathcal{Q}$. Indeed, let $\mathcal{Q}\supseteq\{e_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ be a countable set which contains a dense subset of each $\mathcal{D}_{n}$, and put $\mathcal{P}=\mathcal{F}\setminus\mathcal{Q}$. Every subcontinuum of $\mathcal{F}$ must intersect $\mathcal{Q}$, so $\mathcal{F}$ is Suslinian.

This concludes Example 3.

Example 4**.**

There is a Suslinian continuum $\mathcal{K}\supseteq X_{\omega}\cup\Delta$.

According to the construction in Example 2, $X_{\omega}$ is the union of a null-sequence of disjoint open sets homeomorphic to $X$. Thus, there is a null-sequence $D_{0},D_{1},...\subseteq K\setminus\Delta$ of disjoint copies of $D$ such that $K\setminus\bigcup\{D_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ is zero-dimensional and $X_{\omega}\cap\bigcup\{D_{n}\mathrel{\mathop{\mathchar 58\relax}}n<\omega\}$ is countable. As in Example 3, it is now possible to define a Suslinian upper semi-continuous decomposition $\mathcal{K}$ of $K$, such that $X_{\omega}\cup\Delta\subseteq\mathcal{K}$.

We remark that each example in this section is the union of a punctiform set and a countable set. This is materially stronger than the Suslinian property by Piotr Minc’s example in [13].

5. Conclusion

The continua in Examples 3 and 4 are not rational. Arguments in Claim 2 of Example 1 show that, in fact, every countable separator of $\mathcal{F}$ (Example 3) contains the vertex $\langle\frac{1}{2},0\rangle$. Likewise, every co-countable subset of $\mathcal{K}$ (Example 4) is connected. Rational compactifications of ̣\nabla$$X and $X_{\omega}\cup\Delta$ exist nevertheless by [15, Theorem 1], or the stronger [6, Theorem 8].

A point $e$ in a connected space $X$ is called an explosion point if for every two points $x$ and $y$ in $X\setminus\{e\}$, there is a relatively clopen subset of $X\setminus\{e\}$ which contains $x$ but not $y$. The dispersion point $\langle\frac{1}{2},0\rangle$ in Example 1 is not an explosion point because \textnormal{{\@text@daccent{\nabla}}}X\setminus\{\langle\frac{1}{2},0\rangle\} has non-degenerate quasi-components. Thus, the following variation of [1, Problem 79] is still open.

Question 1**.**

Is it true that no biconnected set with an explosion point can be embedded into a rational continuum?

Added September 2019.

Question 1 was recently answered by Jan J. Dijkstra and the author in [3]. Essentially, we use the Axiom of Choice to show \textnormal{{\@text@daccent{\nabla}}}X (Example 1) contains a connected set with an explosion point.