Irreducibility of strong size levels

1 Introduction

A continuum is a compact connected metric space with more than one point, a subcontinuum of X is a compact connected subspace of X, so subcontinua include one-point sets.

In this paper we use the following hyperspaces of X:

All these hyperspaces are considered with the Hausdorff metric [6: Theorem 3.1]. We write C(X) instead of C₁(X). A map is a continuous function.

A continuum X is irreducible provided that there exist points p, q ∈ X such that no proper subcontinuum of X contains both p and q.

A Whitney map for C_n(X) is a map μ : C_n(X) → [0, 1] such that:

1. for each p ∈ X, μ({p}) = 0;

2. μ (X) = 1; and

3. if A, B ∈ C_n(X) and A⊊ B, then μ(A) < μ(B).

Sets of the form μ⁻¹(t) where t ∈ (0, 1) are Whitney levels.

A strong size map for C_n(X) is a map σ : C_n(X) → [0, 1] such that:

1. σ(A) = 0 for each A ∈ F_n(X);

2. σ(X) = 1; and

3. if A, B ∈ C_n(X), A ⊊ B and B ∉ F_n(X), then σ(A) < σ(B).

Sets of the form σ⁻¹(t) where t ∈ [0, 1) are strong size levels.

Notice that the restriction of a strong size map to C(X) is a Whitney map. Whitney levels and strong size levels are continua ([6: Theorem 19.9] and [3: Theorem 2.10]). Hosokawa [3] introduced and studied topological properties of strong size levels. Topological properties of Whitney levels have been studied widely. In [6: Chapter VIII]{in}, there is a very complete account of what was known on Whitney levels up to 1998.

A very important line of research about Whitney and strong size levels is to determine for which topological properties 𝓟 the following implications hold:

1. if X has property 𝓟, then Whitney (strong size) levels have property 𝓟, and

2. if Whitney (strong size) levels have property 𝓟, then X has property 𝓟.

Statements (a) and (b) have been studied for more than 40 years (see [6: Chapter VIII]). The study of strong size levels started with the paper [3] in 2011. Since then statements (a) and (b) have been studied for strong size levels in [1, 2, 3, 4, 5] and [7, 8, 9].

For the case that property 𝓟 is irreducibility and n = 1, in [4], it was shown that if X is a continuum having a Whitney level irreducible with respect to a finite subset, then X is also irreducible with respect to a finite set; and in [9], it was shown an irreducible, hereditarily decomposable continuum such that none of its proper strong size levels is irreducible. In [1], the authors asked whether there exists a continuum with irreducible strong size levels. In this paper we give a complete solution to this question for the case n ≥ 2 with the following theorem.

2 Proof of Theorem 1

Given elements A, B ∈ 2^X such that A ⊊ B, an order arc from A to B is a map α : [0, 1] → 2^X such that α(0) = A, α(1) = B and if 0 ≤ s < t ≤ 1, then α (s) ⊊ α (t). The most important result about order arcs says that if A, B ∈ 2^X, then there exists an order arc from A to B if and only if A ⊊ B and each component of B intersects A [6: Theorem 15.3].

In order to prove that 𝓐 is not irreducible, take A, B ∈ 𝓐. Suppose that A₁, …, A_r are the components of A and B₁, …, B_s are the components of B. Fix points p₁ ∈ A₁, …, p_r ∈ A_r and q₁ ∈ B₁, …, q_s ∈ B_s. We consider two cases.

1. t ₀ = 0. For each i ∈ {1, …, r}, A_i = {p_i} and for each j ∈ {1, …, s}, B_j = {q_j}. Fix two points p₀ ≠ q₀ in X ∖ (A ∪ B). Let 𝓒 = {{p₁} ∪ E : E ∈ F_n−1(X)}. Then 𝓒 is a subcontinuum of 𝓐 such that A, {p₁} ∈ 𝓒 and {p₀, q₀} ∉ 𝓒. Similarly, there exists a subcontinuum 𝓓 of 𝓐 such that B, {q₁} ∈ 𝓓 and {p₀, q₀} ∉ 𝓓. Then 𝓒 ∪ 𝓓 ∪ F₁(X) is a subcontinuum of 𝓐 containing {A, B} and not having the element {p₀, q₀}. Therefore, in this case, 𝓐 is not irreducible.

2. t ₀ > 0. Taking an order arc from a one-point set to X, it is possible to show that there exists a connected element W in 𝓐. Since t₀ < 1, the set X ∖ W is infinite. Fix a point z ∈ X ∖ (W ∪ {p₁, …, p_r} ∪ {q₁, …, q_s}). Using an order arc from {z} to X, it is possible to construct a subcontinuum M₀ of X with the properties: M₀ is non-degenerate, M₀ ⊂ X ∖ (W ∪ {p₁, …, p_r} ∪ {q₁, …, q_s}) and σ(M₀) < t₀. Since σ(W ∪ M₀) > t₀ (n ≥ 2) using again an order arc from a one-point set to W it is possible to obtain a subcontinuum N₀ of W such that the set T = M₀ ∪ N₀ belongs to 𝓐. Notice that M₀ ∩ N₀ = ∅, so T has a component (the continuum M₀) not intersecting {p₁, …, p_r} ∪ {q₁, …, q_s}.

We are going to show that there exist a subcontinuum S of X and a subcontinuum 𝓒 of 𝓐 such that {A, S} ⊂ 𝓒 and T ∉ 𝓒. In the case that A is connected, set S = A and 𝓒 = {A}. Since all the components of A intersect {p₁, …, p_r}, by the previous paragraph, we infer that A ≠ T, so T ∉ 𝓒. Suppose then that A is not connected. Then r > 1.

We apply the Claim 2 to F = A₁ ∪ ⋯ ∪ A_r−1, G = A_r, E = A, L = {p₁, …, p_r−1} and q = p_r. Taking an order arc from F to X, it is possible to find an element S₁ ∈ 𝓐 such that each component of S₁ intersects F. Since each component of F intersects L, it is possible to find an order arc α : [0, 1] → C_n(X) such that α(0) = L, α (1) = S₁ and F ∈ α([0, 1]). Then there exists a subcontinuum 𝓑 of 𝓐 such that {S₁, A} ⊂ 𝓑 and for each K ∈ 𝓑, either each component of K intersects {p₁, …, p_r} or K = K₀ ∪ {p}, where each component of K₀ intersects L and p ∈ X.

Since the component M₀ of T is not degenerate and it does not intersect {p₁, …, p_r}, we have that T ∉ 𝓑. Since each component of S₁ intersects {p₁, …, p_r−1}, we obtain that S₁ has at most r − 1 components. Suppose that D₁, …, D_k are the components of S₁. Then we can choose points p_l1 ∈ D₁, …, p_lk ∈ D_k.

We have shown that A can be connected by a continuum contained in 𝓐 ∖ {T} with an element S₁ ∈ 𝓐 having less components than A. In the case that S₁ is connected, let S = S₁ and 𝓒 = 𝓑. In the case that S₁ is not connected, we can repeat the procedure to obtain an element S₂ ∈ 𝓐 and a continuum 𝓑₁ contained in 𝓐 ∖ {T} such that {S₁, S₂} ⊂ 𝓑₁, S₂ has less components than S₁ and each component of S₂ intersects {p_l1}, …, p_lk}. If S₂ is connected, let S = S₂ and 𝓒 = 𝓑₁ ∪ 𝓑₂. In the case that S₂ is not connected, we repeat this process until we obtain a connected set S_m = S and a continuum 𝓒 = 𝓑₁ ∪ ⋯ ∪ 𝓑_m contained in 𝓐 ∖ {T} such that {A, S} ⊂ 𝓒.

In a similar way, there exist a connected element S₀ ∈ 𝓐 and a continuum 𝓒₀ contained in 𝓐 ∖ {T} such that {S₀, B} ∈ 𝓒₀. Define

Since σ| C(X) is a Whitney map for C(X), we have that σ| C(X))⁻¹(t₀) is a Whitney level for C(X) and then this set is connected. Thus 𝓒^* is a subcontinuum of 𝓐. Since T has two components, T ∉ 𝓒^*. Therefore 𝓒^* is a proper subcontinuum of 𝓐 containing A and B. Hence 𝓐 is not irreducible. This finishes the proof of Theorem 1.