Diagonals Separating the Square of a Continuum

A metric continuum X is indecomposable if it cannot be put as the union of two of its proper subcontinua. A subset R of X is said to be continuumwise connected provided that for each pair of points p,q∈R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$p,q\in R$$\end{document}, there exists a subcontinuum M of X such that {p,q}⊂M⊂R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{p,q\}\subset M\subset R$$\end{document}. Let X2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$X^{2}$$\end{document} denote the Cartesian square of X and Δ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta $$\end{document} the diagonal of X2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$X^{2}$$\end{document}. Recently, H. Katsuura asked if for a continuum X, distinct from the arc, X2\Δ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$X^{2}\setminus \Delta $$\end{document} is continuumwise connected if and only if X is decomposable. In this paper, we show that no implication in this question holds. For the proof of the non-necessity, we use the dynamical properties of a suitable homeomorphism of the Cantor set onto itself to construct an appropriate indecomposable continuum X.


Introduction
A continuum is a compact connected non-degenerate metric space.A subcontinuum of a continuum X is a nonempty connected closed subset of X, so singletons are subcontinua.An arc is a continuum homeomorphic to the interval [0, 1].Let X 2 denote the Cartesian square of X and ∆ the diagonal of X 2 .
The authors studied in [6] conditions under which ∆ satisfies some of the properties described in [2, Table 1] and [8,Definition 1.1], for a number of examples and families of continua.
Recently, H. Katsuura [7] proved that the arc is the only continuum for which ∆ does not satisfy that X 2 \ ∆ is connected, and he included the following question [7, p. 4] (Katsuura mentioned that this question was suggested by Wayne Lewis of Texas Tech University in a private conversation): Question 1.1.If X is a continuum other than the arc, is X 2 \ ∆ continuumwise connected if and only if X is decomposable?
As an application of Dynamic Systems to Continuum Theory, we use the dynamic properties of a particular homeomorphism from the Cantor set onto itself to prove that the necessity in Question 2 is not satisfied.In fact, we show that no implication in Question 1.1 is satisfied.

No Sufficiency
A map is a continuous function.Given continua X and Y , and a number ε > 0, an ε-map is an onto map f : X → Y such that for each y ∈ Y , diameter(f −1 (y)) < ε.The continuum X is arc-like provided that for each ε > 0, there exists an ε-map f : X → [0, 1].
Lemma 2.1.Let X be an arc-like continuum and p, q ∈ X such that p = q.Then, for every continuum K ⊂ X 2 containing the points (p, q) and (q, p), K ∩ ∆ = ∅.
Proof.Let d be a metric for X.Let K be a continuum in X 2 such that (p, q), (q, p) ∈ K, for some points p = q ∈ X.
Corollary 2.2.Let X be an arc-like continuum.Then X 2 \∆ is not continuumwise connected.
Since there are decomposable arc-like continua (for example, the sin( 1x )-curve), Corollary 2.2, shows that the sufficiency in Question 1.1 is not satisfied.We also prove in [6,Section 7] that, in fact the sin( 1x )-curve belongs to a family of curves for which ∆ satisfies a stronger property in X 2 .S. B. Nadler, Jr. [10, p. 329] called Elsa continua to the compactifications of the ray [0, ∞) whose remainder is an arc and he proved that they are arc-like [9, Lemma 6] (in fact, it is easy to show that a compactification of [0, ∞) with non-degenerate remainder, is arc-like if and only if the remainder is an arc-like continuum).Then for each Elsa continuum, X is deconposable and X 2 \ ∆ is not continuumwise connected.

No Necessity
In this section we present an indecomposable continuum X such that X 2 \ ∆ is continuumwise connected.Definition 3.1.Let X be a continuum and A a subcontinuum of X with int The construction and proof of the properties of X strongly depends on the dynamic properties of a particular homeomorphism of the Cantor set onto itself.
Let C denote the Cantor ternary set.In this section we will use a homeomorphism f : C → C such that f is minimal (C does not contain any proper nonempty closed subset A such that f (A) = A, or equivalently, every orbit of f is dense) and f is weakly mixing (for any two nonempty subsets U, V of X 2 , there exist We consider the space X obtained by identifying in C × [0, 1], for each p ∈ C, the points (p, 1) and (f (p), 0).Let ϕ : C × [0, 1] → X be the quotient mapping and let σ = ϕ × ϕ : (C × [0, 1]) 2 → X 2 .Let ρ be the metric on C × [0, 1] given by ρ((p, s), (q, t)) = |p − q| + |s − t|).Fix a metric D for the space X.
In the hypothesis of the following theorem, we write the specific properties that we use of the homeomorphism f .Theorem 3.2.Suppose that f : C → C is a homeomorphism such that the orbits of f and f −1 are dense and f × f : C 2 → C 2 has a dense orbit.Then X is an indecomposable continuum such that the diagonal ∆ in X 2 is colocally connected.
This finishes the proof of Claim 1.By the symmetry of the roles of both coordinates in the definition of V 0 , we obtain that the following claim also holds.
Case 3. z ∈ V 3 .This case is similar to Case 2. This completes the proof that X 2 \ σ(K) ⊂ U. Therefore (b) holds.
Finally define V = X 2 \ σ(K).Then V is an open subset of X 2 such that ∆ ⊂ V ⊂ U, and X 2 \ V is connected.This finishes the proof that ∆ is colocally connected in X 2 .