A complete classification of hereditarily equivalent plane continua

TL;DR

This paper proves that the only non-degenerate hereditarily equivalent plane continua are the arc and the pseudo-arc, providing a complete classification of such continua in the plane.

Contribution

It establishes a complete classification of hereditarily equivalent plane continua, showing only the arc and pseudo-arc qualify.

Findings

Arc and pseudo-arc are the only non-degenerate hereditarily equivalent plane continua.

Provides a complete classification of hereditarily equivalent plane continua.

Clarifies the structure of hereditarily equivalent continua in the plane.

Abstract

A continuum is hereditarily equivalent if it is homeomorphic to each of its non-degenerate sub-continua. We show in this paper that the arc and the pseudo-arc are the only non-degenerate hereditarily equivalent plane continua.

Equations

$$C=\{(x,t)\in G\times [0,1]:x\in B_t\}.$$

$$\phi (x,t)=\{\begin{array}{ll}+d(x,B_t)& \text{if }x\in S_t^{+}\\ -d(x,B_t)& \text{if }x\in S_t^{-}\\ 0& \text{otherwise,}\end{array}$$

$$U_1=\{(g,t)\in G\times [0,1]:$$

$$d(g,g^{\prime })<\epsilon \text{ and }|t-t^{\prime }|<\epsilon \}$$

$$U_2=\{(g,t)\in G\times [0,1]:$$

$$d(g,g^{\prime })<\frac{\epsilon }{2}\text{ and }|t-t^{\prime }|<\epsilon \},$$

$$U=\{(x,t)\in G\times (I\setminus \{a,b\}):\Vert x-t\Vert <\epsilon \}$$

$$U=\left\{(x,t)\in G\times (I\setminus \{a,b\}):\Vert x-t\Vert <\frac{\epsilon }{4}\right\}$$

$$\Vert x_1-x_2\Vert$$

$$+\Vert {\pi }_2\circ h(x_2)-{\pi }_1\circ h(x_2)\Vert +\Vert f(x_2)-x_2\Vert <\epsilon .$$

$$C=\{(x,t)\in G\times [0,1]:x\in B_t\}$$

$$U=\left\{(x,t)\in G\times (0,1):\Vert x-f(t)\Vert <\frac{\epsilon }{2}\right\},$$

$$U^{\prime }=\{(x,y)\in G\times f^{\prime }((0,1)):\Vert x-y\Vert <\epsilon \}.$$

Full text

The first named author was partially supported by NSERC grant RGPIN 435518

The second named author was partially supported by NSF-DMS-1807558

1. Introduction

By a continuum, we mean a compact connected metric space. A continuum is non-degenerate if it contains more than one point. We refer to the space $\mathbb{R}^{2}$, with the Euclidean topology, as the plane. The Euclidean distance between two points $x,y$ in $\mathbb{R}^{2}$ (or $\mathbb{R}^{3}$) will be denoted $\|x-y\|$. An arc is a space which is homeomorphic to the interval $[0,1]$. By a map we mean a continuous function.

A continuum $X$ is hereditarily equivalent if it is homeomorphic to each of its non-degenerate subcontinua. This concept was introduced by Mazurkiewicz, who was interested in topological characterizations of the arc. In the second volume of Fundamenta Mathematicae in 1921, Mazurkiewicz [Maz21] asked (Problème 14) whether the arc is the only non-degenerate hereditarily equivalent continuum.

A continuum $X$ is decomposable if it is the union of two proper subcontinua, and indecomposable otherwise. $X$ is hereditarily indecomposable if every subcontinuum of $X$ is indecomposable. $X$ is arc-like (respectively, tree-like) if for every $\varepsilon>0$ there exists an $\varepsilon$-map from $X$ to $[0,1]$ (respectively, to a tree), where $f:X\to Y$ is an $\varepsilon$-map if for each $y\in Y$ the preimage $f^{-1}(y)$ has diameter less than $\varepsilon$. Henderson [Hen60] showed that the arc is the only decomposable hereditarily equivalent continuum. Cook [Coo70] has shown that every hereditarily equivalent continuum is tree-like.

Problème 14 of Mazurkiewicz was formally answered by Moise [Moi48] in 1948, who constructed another hereditarily equivalent plane continuum which he called the “pseudo-arc”, due to this property it has in common with the arc. The pseudo-arc is a one-dimensional fractal-like hereditarily indecomposable arc-like continuum. Such a space was constructed by Knaster [Kna22] in 1922, and another by Bing [Bin48] in 1948 which he proved was topologically homogeneous. Bing [Bin51] proved in 1951 that the pseudo-arc is the only hereditarily indecomposable arc-like continuum. From this characterization it follows that the spaces of Knaster, Moise, and Bing are all homeomorphic, and also it can immediately be seen that the pseudo-arc is hereditarily equivalent.

Since Moise’s article, the question has been: What are all hereditarily equivalent continua? The main result of this paper is:

Theorem 1**.**

If $X$ is a non-degenerate hereditarily equivalent plane continuum, then $X$ is homeomorphic to the arc or to the pseudo-arc.

It remains an open question whether there exists any other hereditarily equivalent continuum in $\mathbb{R}^{3}$.

As part of the sequel we will also give a new characterization (Theorem 7) of the pseudo-arc.

2. Plane strips

If a continuum admits an $\varepsilon$-map to an arc then it can be covered by a chain of open sets whose diameters are less than $\varepsilon$ (i.e. a set that roughly looks like a tube of small diameter). The notion of an $\varepsilon$-strip (see Definition 3 below), introduced in [OT82] in a slightly different form, conveys a similar feeling. However, it was observed in [OT82, Figure 1] that, for arbitrarily small $\varepsilon>0$, there exists an $\varepsilon$-strip which does not admit a $1$-map to an arc. Nevertheless we show in this paper (Theorem 8 below) that if a hereditarily indecomposable plane continuum is contained in an $\varepsilon$-strip for arbitrarily small $\varepsilon>0$, then it must in fact be homeomorphic to the pseudo-arc.

Given two points $x,y$ in the plane $\mathbb{R}^{2}$ we denote by $\overline{xy}$ the straight line segment joining them. Given points $v_{1},\ldots,v_{n}$ in $\mathbb{R}^{2}$, the polygonal arc $A$ with vertices $v_{1},\ldots,v_{n}$ is the union of the straight line segments $\overline{v_{1}v_{2}},\ldots,\overline{v_{n-1}v_{n}}$. Denote the vertex set of $A$ by $V_{A}=\{v_{1},\ldots,v_{n}\}$. If $v_{n}=v_{1}$, then we call $A$ a polygonal closed curve. We will need the following lemma which was proved in [OT82].

Lemma 2** ([OT82], Lemma 2.1).**

Let $T$ be a polygonal closed curve in $\mathbb{R}^{2}$ with vertex set $V_{T}$. Given any $z\in\mathbb{R}^{2}\setminus T$ we say $z$ is odd (respectively, even) with respect to $T$ if there exists a polygonal arc $A$ with vertex set $V_{A}$ from $z$ to a point in the unbounded component of $\mathbb{R}^{2}\setminus T$ so that $A\cap V_{T}=\emptyset=T\cap V_{A}$ and $|T\cap A|$ is odd (respectively, even). Then this notion of odd/even is well-defined, i.e. independent of the choice of $A$.

A component $U$ of $\mathbb{R}^{2}\setminus T$ is called odd (respectively, even) if each point of $U$ is odd (respectively, even) with respect to $T$. Clearly the unbounded complementary domain of $T$ is even.

A map $f:[0,1]\to\mathbb{R}^{2}$ is piecewise linear if there are finitely many points $0=t_{1}<t_{2}<\ldots<t_{n}=1$ such that for each $i=1,\ldots,n-1$, as $t$ runs from $t_{i}$ to $t_{i+1}$, $f(t)$ parameterizes the straight line segment $\overline{f(t_{i})f(t_{i+1})}$. If $f$ is a piecewise linear map, then clearly $f([0,1])$ is a polygonal arc.

Definition 3** ($\varepsilon$-strip).**

Suppose that $f,g:[0,1]\to\mathbb{R}^{2}$ are two piecewise linear maps into the plane such that $f([0,1])\cap g([0,1])=\emptyset$ and for all $t\in[0,1]$, $\|f(t)-g(t)\|<\varepsilon$. Let $B_{t}=\overline{f(t)g(t)}$, and let $T_{t}=B_{0}\cup f([0,t])\cup g([0,t]\cup B_{t}$. We denote the union of all odd (respectively, even) complementary domains of $T_{t}$ by $S^{-}_{t}$ (respectively, $S^{+}_{t}$). If $B_{0}\cap B_{1}=\emptyset$, then we say that $S_{1}^{-}$ is an $\varepsilon$-strip with disjoint ends.

See Figure 1 for an illustration of a simple $\varepsilon$-strip.

We say a continuum $X$ is contained in an $\varepsilon$-strip with disjoint ends if there exist such $f,g$ as in the above definition such that $X\subset S^{-}_{1}$. Observe that in this situation, $X\cap B_{0}=\emptyset=X\cap B_{1}$.

If $X$ is an indecomposable and hereditarily equivalent plane continuum, then it contains uncountably many pairwise disjoint copies of itself. In particular it contains a copy of $X\times C$, where $C$ is the Cantor set [vD93]. This is the key observation behind the following result.

Lemma 4** ([OT84], Theorem 15).**

Suppose that $X$ is a non-degenerate, indecomposable and hereditarily equivalent plane continuum. Then there exists a non-degenerate subcontinuum $Y$ such that for each $\varepsilon>0$, $Y$ is contained in an $\varepsilon$-strip with disjoint ends.

3. Separators

In light of Lemma 4 above, to prove Theorem 1 it suffices to show that any hereditarily indecomposable continuum $X$ contained in arbitrarily small plane strips is homeomorphic to the pseudo-arc (Theorem 8). Our strategy below is to consider a small strip containing the continuum $X$, and to approximate $X$ by a graph $G$ contained in that strip. If we vary $t$ from [math] to $1$, the bridge $B_{t}$ in the strip sweeps across the graph $G$. As it does so, it may wander back and forth in $G$, in a pattern whose essential property is captured in the following result. We will then use the crookedness of the hereditarily indecomposable continuum $X$ to match with that pattern (see Theorems 6 and 7 below) to obtain an $\varepsilon$-map to an arc.

Lemma 5**.**

Suppose that a graph $G$ is contained in an $\varepsilon$-strip $S_{1}^{-}$ with disjoint ends. Let

[TABLE]

Then $C$ separates $G\times\{0\}$ from $G\times\{1\}$ in $G\times[0,1]$.

Proof.

Define the function $\varphi:G\times[0,1]\to\mathbb{R}$ by

[TABLE]

where $d(x,B_{t})=\inf\{\|x-b\|:b\in B_{t}\}$. Then $\varphi$ is a continuous function (see the proof of Lemma 2.3 in [OT82]). Since $S^{-}_{0}=\emptyset=B_{0}\cap G$, $\varphi(x,0)>0$ for each $x\in G$. Similarly, since $G\subset S^{-}_{1}$, $\varphi(x,1)<0$ for each $x\in G$. Hence the set of points $C$ where $\varphi(x,t)=0$ must separate $G\times\{0\}$ from $G\times\{1\}$ in $G\times[0,1]$. ∎

In [HO16, Theorem 20], the authors gave a characterization of hereditarily indecomposable continua in terms of sets as in Lemma 5 which separate $G\times\{0\}$ from $G\times\{1\}$ in the product $G\times[0,1]$ of a graph $G$ with $[0,1]$. Here we give a simplified version of that theorem, which is more broadly applicable.

Theorem 6**.**

A continuum $X$ is hereditarily indecomposable if and only if for any map $f:X\to G$ to a graph $G$, and for any open set $U\subseteq G\times(0,1)$ which separates $G\times\{0\}$ from $G\times\{1\}$ in $G\times[0,1]$, there exists a map $h:X\to U$ such that $f=\pi_{1}\circ h$ (where $\pi_{1}:G\times[0,1]\to G$ is the first coordinate projection).

Proof.

According to [HO16, Theorem 20], a continuum $X$ is hereditarily indecomposable if and only if for any map $f:X\to G$ to a graph $G$ with metric $d$, for any set $M\subseteq G\times(0,1)$ which separates $G\times\{0\}$ from $G\times\{1\}$ in $G\times[0,1]$, for any open set $U\subseteq G\times[0,1]$ with $M\subseteq U$, and for any $\varepsilon>0$, there exists a map $h:X\to U$ such that $d(f(x),\pi_{1}\circ h(x))<\varepsilon$ for all $x\in X$. The condition in the present theorem is clearly stronger than this condition from [HO16, Theorem 20]. Therefore, to prove the present theorem we need only consider the forward implication.

Suppose $X$ is hereditarily indecomposable, let $f:X\to G$ be a map to a graph $G$ with metric $d$, and let $U\subset G\times(0,1)$ be an open set which separates $G\times\{0\}$ from $G\times\{1\}$ in $G\times[0,1]$. It is well-known (see e.g. [Kur68, Theorem §46.VII.3]) that there exists a closed set $M\subset U$ which also separates $G\times\{0\}$ from $G\times\{1\}$ in $G\times[0,1]$. Let $\varepsilon>0$ be small enough so that the open set

[TABLE]

is contained in $U$. Let

[TABLE]

and apply [HO16, Theorem 20] to obtain a map $h^{\prime}:X\to U_{2}$ such that $d(f(x),\pi_{1}\circ h^{\prime}(x))<\frac{\varepsilon}{2}$ for all $x\in X$.

Define $h:X\to U$ by $h(x)=(f(x),\pi_{2}\circ h^{\prime}(x))$, where $\pi_{2}:G\times[0,1]\to[0,1]$ is the second coordinate projection. Clearly this function $h$ is continuous, and $f=\pi_{1}\circ h$. To see that the range of $h$ is really contained in $U$, let $x\in X$, and denote $h^{\prime}(x)=(g,t)$, so that $h(x)=(f(x),t)$. Because $h^{\prime}(x)\in U_{2}$, there exists $(g^{\prime},t^{\prime})\in M$ such that $d(g,g^{\prime})<\frac{\varepsilon}{2}$ and $|t-t^{\prime}|<\varepsilon$. Moreover, by choice of $h^{\prime}$ we have $d(f(x),g)<\frac{\varepsilon}{2}$. So by the triangle inequality, we have $d(f(x),g^{\prime})<\varepsilon$, which means $h(x)\in U_{1}\subseteq U$, as desired. ∎

By Bing’s [Bin51] result a hereditarily indecomposable continuum is homeomorphic to the pseudo-arc if and only if it is arc-like. A new characterization of the pseudo-arc, involving the notion of span zero (see [Lel64]), was obtained in [HO16]. It states that a hereditarily indecomposable continuum is a pseudo-arc if and only if it has span zero. The more technical characterization of the pseudo-arc in Theorem 7 below is useful in cases when (like in the case of hereditarily equivalent plane continua) it is not a priori known that $X$ has span zero.

In the statement below we assume that all spaces (i.e., $X$, $G$, and $I$) are contained in Euclidean space $\mathbb{R}^{3}$. One could just as well use the Hilbert cube $[0,1]^{\mathbb{N}}$, depending on the intended application.

Theorem 7**.**

Suppose that $X\subset\mathbb{R}^{3}$ is a hereditarily indecomposable continuum. Then the following are equivalent:

1. (1)

$X$* is homeomorphic to the pseudo-arc;* 2. (2)

For each $\varepsilon>0$ there exist a graph $G\subset\mathbb{R}^{3}$, a map $f:X\to G$ with $\|(x-f(x)\|<\varepsilon$ for each $x\in X$, and an arc $I\subset\mathbb{R}^{3}$ with endpoints $a$ and $b$, such that the set

[TABLE]

separates $G\times\{a\}$ from $G\times\{b\}$ in $G\times I$.

Proof.

Suppose $X$ is homeomorphic to the pseudo-arc, and fix $\varepsilon>0$. Note $X$ is arc-like and, hence [Lel64], $X$ has span zero. Therefore, according to Theorem 4 of [HO16], there exists $\delta>0$ such that for any graph $G\subset\mathbb{R}^{3}$ and arc $I\subset\mathbb{R}^{3}$ both within Hausdroff distance $\delta$ from $X$, the set $U=\{(x,t)\in G\times I:\|x-y\|<\varepsilon\}$ separates $G\times\{a\}$ from $G\times\{b\}$ in $G\times I$, where $a,b$ are the endpoints of $I$. We may assume that $\delta<\varepsilon$. Since $X$ is arc-like, we may choose an arc $G\subset\mathbb{R}^{3}$ within Hausdorff distance $\delta$ of $X$ and a map $f:X\to G$ such that $\|x-f(x)\|<\varepsilon$ for all $x\in X$. Choose any arc $I$ within Hausdorff distance $\delta$ from $X$. Then $G$, $f$, and $I$ satisfy the conditions of statement (2), as desired.

Conversely, suppose statement (2) holds. To prove that $X$ is homeomorphic to the pseudo-arc, by [Bin51] it suffices to show that for each $\varepsilon>0$ there exists an $\varepsilon$-map from $X$ to an arc. Fix $\varepsilon>0$. Suppose that $G\subset\mathbb{R}^{3}$ is a graph, $f:X\to G$ is a map such that $\|x-f(x)\|<\frac{\varepsilon}{4}$ for all $x\in X$, $I\subset\mathbb{R}^{3}$ is an arc with endpoints $a$ and $b$, and

[TABLE]

separates $G\times\{a\}$ from $G\times\{b\}$ in $G\times I$. Denote by $\pi_{1}:G\times I\to G$ the first coordinate projection and by $\pi_{2}:G\times I\to I$ the second coordinate projection. By Theorem 6 there exists a map $h:X\to U$ such that $f=\pi_{1}\circ h$. We claim that $\pi_{2}\circ h(x):X\to I$ is an $\varepsilon$-map. To see this suppose that $\pi_{2}\circ h(x_{1})=\pi_{2}\circ h(x_{2})$. Then

[TABLE]

∎

4. Proof of main result

We now apply the results established above to prove the following key theorem.

Theorem 8**.**

Let $X\subset\mathbb{R}^{2}$ be a hereditarily indecomposable plane continuum such that for each $\varepsilon>0$, there is an $\varepsilon$-strip with disjoint ends containing $X$. Then $X$ is homeomorphic to the pseudo-arc.

Proof.

Let $\varepsilon>0$, and consider an $\frac{\varepsilon}{2}$-strip with disjoint ends containing $X$. That is, consider piecewise linear maps $f,g:[0,1]\to\mathbb{R}^{2}$ such that $f([0,1])\cap g([0,1])=\emptyset$, $\|f(t)-g(t)\|<\frac{\varepsilon}{2}$ for each $t\in[0,1]$, and $X\subset S^{-}_{1}$. Identify $\mathbb{R}^{2}$ with $\mathbb{R}^{2}\times\{0\}\subset\mathbb{R}^{3}$, and adjust $f$ slightly to obtain a map $f^{\prime}:[0,1]\to\mathbb{R}^{3}$ which is one-to-one (so that $f^{\prime}([0,1])$ is an arc) and $\|f(t)-f^{\prime}(t)\|<\frac{\varepsilon}{2}$ for all $t\in[0,1]$.

Clearly $X$ is $1$-dimensional, so there exists a graph $G\subset S_{1}^{-}$ and a map $h:X\to G$ such that $\|x-h(x)\|<\varepsilon$ for all $x\in X$. By Lemma 5, the set

[TABLE]

separates $G\times\{0\}$ from $G\times\{1\}$ in $G\times[0,1]$. Clearly this set $C$ is contained in

[TABLE]

and the image of this set $U$ under the homeomorphism $\mathrm{id}\times f^{\prime}:G\times[0,1]\to G\times f^{\prime}([0,1])$ is contained in

[TABLE]

Therefore $U^{\prime}$ separates separates $G\times\{f^{\prime}(0)\}$ from $G\times\{f^{\prime}(1)\}$ in $G\times f^{\prime}([0,1])$. Hence, by Theorem 7, $X$ is homeomorphic to the pseudo-arc. ∎

We are now ready to prove our main result, Theorem 1.

Proof of Theorem 1.

Let $X$ be a non-degenerate hereditarily equivalent plane continuum. If $X$ is decomposable, then $X$ is an arc by [Hen60]. Suppose then that $X$ is indecomposable, and hence hereditarily indecomposable. By Lemma 4, we may assume that $X$ is embedded in the plane so that for each $\varepsilon>0$, $X$ is contained in an $\varepsilon$-strip with disjoint ends. It then follows from Theorem 8 that $X$ is homeomorphic to the pseudo-arc. ∎