The product of two chainable Kelley continua is also a Kelley continuum

Topology and its Applications 269 (2020) 106924

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The product of two chainable Kelley continua is also a Kelley continuum Jimmy A. Naranjo-Murillo Instituto de Matemáticas, Universidad Nacional Autónoma de México, Circuito Exterior, Cd. Universitaria, Ciudad de México, 04510, México

a r t i c l e

i n f o

Article history: Received 12 June 2019 Received in revised form 3 October 2019 Accepted 24 October 2019 Available online 28 October 2019 MSC: primary 54F15 secondary 54B10

a b s t r a c t For many years, it was an open question whether the product of two Kelley continua is also a Kelley continuum. In 1977, Roger W. Wardle presented a Kelley continuum X such that X × X is not a Kelley continuum. Many years later, in 2004, Janusz J. Charatonik and Wlodzimierz J. Charatonik presented a Kelley continuum X such that X × [0, 1] is not a Kelley continuum. In this paper we prove that the product of two chainable Kelley continua is also a Kelley continuum. © 2019 Elsevier B.V. All rights reserved.

Keywords: Chainable Continuum Kelley continuum Product

1. Introduction A continuum is a nonempty compact connected metric space. Given a continuum X, the hyperspace of closed nonempty subsets of X and the hyperspace of subcontinua of X are denoted by 2X and C(X), respectively, considered with the Hausdorff metric HX . A chain for a continuum X is an open cover U = {U1 , U2 , ..., Un } of X such that for each i, j ∈ {1, ..., n}, Ui ∩ Uj = ∅ if and only if |i − j| ≤ 1. A continuum X is chainable provided that for each ε > 0, there exists a chain U for X such that for each U ∈ U, diameter(U ) < ε. Given a continuum X, a point p ∈ X and ε > 0, the ball in X centered in p with radius ε will be denoted by B(p, ε). A continuum X is a Kelley continuum provided that for each ε > 0, there exists δ > 0 such that if A ∈ C(X), p ∈ A and q ∈ B(p, δ), there exists B ∈ C(X) such that q ∈ B and HX (A, B) < ε. Kelley continua are also known as continua with the property of Kelley, precisely because it was J. L. Kelley who defined this property. He introduced it in [4, p. 26], where he called it property 3.2 and used it E-mail address: [email protected]. https://doi.org/10.1016/j.topol.2019.106924 0166-8641/© 2019 Elsevier B.V. All rights reserved.

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to study the contractibility of hyperspaces. However, the property of Kelley has been considered interesting by itself because of its relation with other properties in continua. For many years, it was an open question whether the product of two Kelley continua is a Kelley continuum. This was answered in 1977, when Roger W. Wardle presented a Kelley continuum X such that X × X is not a Kelley continuum [7, Example 4.7, p. 297]. Despite this result, it seemed natural that the product of a Kelley continuum with the interval [0, 1] was a Kelley continuum. However, in 2004, Janusz J. Charatonik and Wlodzimierz J. Charatonik presented a Kelley continuum X such that X × [0, 1] is not a Kelley continuum [1, Theorem 1, p. 3]. On the other hand, it is known that if a product of continua is a Kelley continuum, then so are the factor spaces [7, Corollary 4.6, p. 297]. In this direction, the following problem seems to be interesting. Problem 1.1. Find classes C of continua with the following property: if X and Y are Kelley continua belonging to C, then X × Y is a Kelley continuum. Recall that a dendroid is an arcwise connected continuum X such that the intersection of each pair of its subcontinua is connected. The following particular case of Problem 1.1 is an open question. Problem 1.2. Is the product of two Kelley dendroids a Kelley continuum? This paper is devoted to prove the product of two chainable Kelley continua is also a Kelley continuum. In order to prove this result, we use and adapt the technique developed by Illanes, Martínez-Montejano and Villarreal in Theorem 5.4 of [2]. This technique strongly depends on the Mountain Climbing Theorem (which is the base to prove Lemma 2.2 below). A similar use of this technique is used to prove the main theorem in [6]. 2. Chainable Kelley continua A mapping is a continuous function. Given a mapping between continua f : X → Y and ε > 0, f is an ε-mapping if it is onto and diameter(f −1 (y)) < ε for each y ∈ Y . A continuum X is arc-like if for each ε > 0, there exists an ε-mapping f : X → [0, 1]. A taut chain U = {U1 , U2 , ..., Un } for a chainable continuum X is a chain such that U1  ∪{clX (Ui ) : i ≥ 2} = ∅, Un  ∪{clX (Ui ) : i ≤ n − 1} = ∅ and for each i, j ∈ {1, ..., n}, clX (Ui ) ∩ clX (Uj ) = ∅ if and only if |i − j| ≤ 1. It is known that if X is a chainable continuum, then for each ε > 0, there exists a taut chain U = {U1 , U2 , ..., Un } for X such that diameter(Ui ) < ε for each i ∈ {1, ..., n} [5, Lemma 12.10, p. 235]. It is also known that a continuum is chainable if and only if it is arc-like [5, Theorem 12.11, p. 235]. Given a continuum X, an order arc in C(X) is an arc α in C(X) such that if A, B ∈ α then A ⊂ B or B ⊂ A [3, Definition 14.1, p. 110]. It is known that if X is a continuum and A, B ∈ C(X), A  B, then there is an order arc in C(X) from A to B [3, Theorem 14.6, p. 112]. In order to prove the main result of the paper we need the following two lemmas. Lemma 2.1. Let X be a chainable continuum and ϕ : X → [0, 1] be an onto mapping. Then for each η > 0 there exist r ∈ N, an η-mapping ψ : X → [0, r] and a piecewise linear mapping μ : [0, r] → [0, 1] such that |(μ ◦ ψ)(x) − ϕ(x)| < η for each x ∈ X. Proof. Let d be a metric for X. Let θ > 0 be such that if x, y ∈ X and d(x, y) < θ, then |ϕ(x) − ϕ(y)| < η2 . Let γ > 0 be such that γ < min{η, θ}. Take a taut chain of open subsets U = {U1 , ..., Ur } of X such that for each i ∈ {1, ..., r}, diameter(Ui ) < γ2 and U covers X. Fix points u0 ∈ U1 ∪{clX (Ui ) : i ≥ 2} and ur ∈ Ur ∪{clX (Ui ) : i ≤ r −1}. Using Uryhson’s lemma it is possible to define an onto mapping ψ : X → [0, r] such that ψ(u0 ) = 0, ψ(ur ) = r and for each i ∈ {1, ..., r}, ψX (clX (Ui )) ⊂ [i − 1, i].

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Notice that ψ is an η-mapping. For each i ∈ {1, ..., r − 1} fix a point ui ∈ Ui ∩ Ui+1 . For each i ∈ {1, ..., r}, consider the simplest linear mapping μ defined in the interval [i − 1, i] which satisfies μ(i − 1) = ϕ(ui−1 ) and μ(i) = ϕ(ui ). We denote by μ : [0, r] → [0, 1] the common extension of all these mappings. Take x ∈ X. Let i ∈ {1, ..., r} be such that x ∈ Ui . Then max{d(ui , ui−1 ), d(x, ui )} < diameter(Ui ) < θ. By the definition of ψ, ψ(x) ∈ [i − 1, i]. By the definition of μ, we have that ϕ(ui−1 ) = μ(i − 1) ≤ μ(ψ(x)) ≤ μ(i) = ϕ(ui ) or ϕ(ui ) = μ(i) ≤ μ(ψ(x)) ≤ μ(i − 1) = ϕ(ui−1 ). By the choice of θ, |ϕ(ui−1 ) − ϕ(ui )| <

η 2

and |ϕ(x) − ϕ(ui )| < η2 . Then |μ(ψ(x)) − ϕ(x)| ≤

|μ(ψ(x)) − ϕ(ui )| + |ϕ(ui ) − ϕ(x)| ≤ |ϕ(ui−1 ) − ϕ(ui )| + |ϕ(ui ) − ϕ(x)| < η η + = η. 2 2 2 Lemma 2.2. [2, Lemma 5.3, p. 178] Suppose that n ∈ N, an ≤ an−1 ≤ ... ≤ a1 = b1 ≤ ... ≤ bn−1 ≤ bn , cn ≤ cn−1 ≤ ... ≤ c1 = e1 ≤ ... ≤ en−1 ≤ en , f, g : [0, n] → R are piecewise linear mappings and for each i ∈ {1, ..., n}, f (i) = bi , g(i) = ei , f ([i − 1, i]) = [ai , bi ] and g([i − 1, i]) = [ci , ei ] (then f |[0,1] and g|[0,1] are constant). Let λ = max({max{|ai − ci | , |bi − ei |} : i ∈ {1, ..., n}}∪ {max{ai−1 − ai , bi − bi−1 , ci−1 − ci , ei − ei−1 } : i ∈ {2, ..., n}}). Then there exist piecewise linear mappings α, β : [0, n] → [0, n] such that α(0) = 0 = β(0), α(n) = n = β(n), and for each t ∈ [0, n], |(f ◦ α)(t) − (g ◦ β)(t)| ≤ 2λ. Now we proceed to prove the main result of the paper. Theorem 2.3. If X and Y are chainable Kelley continua, then X × Y is also a Kelley continuum. Proof. Let d and ρ be metrics for X and Y , respectively. We may suppose that the metric D for X × Y is given by D((x, y), (p, q)) = max{d(x, p), ρ(y, q)} for each (x, y), (p, q) ∈ X × Y . Let ε > 0.

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Since X and Y are chainable continua, by [5, Theorem 12.11, p. 235], there exist 2ε -mappings ϕX : X → [0, 1] and ϕY : Y → [0, 1]. Then, there exists η > 0 such that η < ε and for each subinterval J of [0, 1] with diameter(J) < η, we have that diameter(ϕ−1 X (J)) <

ε ε and diameter(ϕ−1 Y (J)) < . 2 2

By the uniform continuity of ϕX and ϕY , there exists ζ > 0 such that: if x, u ∈ X and d(x, u) < ζ, then |ϕX (x) − ϕX (u)| <

η 48

and if v, y ∈ Y and ρ(v, y) < ζ, then |ϕY (v) − ϕY (y)| <

η . 48

Since X and Y are Kelley continua, there exists δ > 0 such that δ < min{ 2ε , ζ} and i) if A ∈ C(X), p ∈ A and q ∈ B(p, δ), then there exists B ∈ C(X) such that q ∈ B and HX (A, B) < ζ; ii) if A ∈ C(Y ), p ∈ A and q ∈ B(p, δ), then there exists B ∈ C(Y ) such that q ∈ B and HY (A, B) < ζ. Let A ∈ C(X × Y ), (p0 , q0 ) ∈ A and (x0 , y0 ) ∈ B((p0 , q0 ), δ). We will find B ∈ C(X × Y ) such that (x0 , y0 ) ∈ B and HX×Y (A, B) < ε. By [3, Theorem 14.6, p. 112], there exists an order arc from {(p0 , q0 )} to A. Then, there exists an injective mapping σ : [0, 1] → C(A) such that σ(0) = {(p0 , q0 )}, σ(1) = A and: if 0 ≤ s ≤ t ≤ 1, then σ(s) ⊂ σ(t). By the uniform continuity of σ, there is n ∈ N such that if s, t ∈ [0, 1] and |s − t| ≤ n1 , then HX×Y (σ(s), σ(t)) < ζ. For each j ∈ {0, 1, ..., n}, set tj = nj , Aj = πX (σ(tj )) and Cj = πY (σ(tj )). For each j ∈ {1, ..., n}, by the choice of δ and (x0 , y0 ), there exist subcontinua Kj and Lj of X and Y , respectively, such that x0 ∈ Kj , HX (Kj , Aj ) < ζ, y0 ∈ Lj and HY (Lj , Cj ) < ζ. Set K0 = {x0 } = B0 , L0 = {y0 } = D0 and for each j ≥ 1, Bj = K0 ∪K1 ∪...∪Kj and Dj = L0 ∪L1 ∪...∪Lj . Then HX (A0 , B0 ) = d(p0 , x0 ) < ζ, HY (C0 , D0 ) = ρ(q0 , y0 ) < ζ and for each j ≥ 1, Bj and Dj are subcontinua of X and of Y , respectively, such that HX (Aj , Bj ) < ζ and HY (Cj , Dj ) < ζ. Let j ∈ {0, 1, ..., n}. Since HX (Aj , Bj ) < ζ and HY (Cj , Dj ) < ζ, the choice of ζ implies that H[0,1] (ϕX (Aj ), ϕX (Bj )) <

η η < 48 16

H[0,1] (ϕY (Cj ), ϕY (Dj )) <

η η < . 48 16

and

Similarly, for each j ∈ {1, ..., n}, H[0,1] (ϕX (Aj−1 ), ϕX (Aj )) <

η η < 48 16

H[0,1] (ϕY (Cj−1 ), ϕY (Cj )) <

η η < . 48 16

and

For each j ∈ {1, ..., n}, we have H[0,1] (ϕX (Bj−1 ), ϕX (Bj )) ≤

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H[0,1] (ϕX (Bj−1 ), ϕX (Aj−1 )) + H[0,1] (ϕX (Aj−1 ), ϕX (Aj )) + H[0,1] (ϕX (Aj ), ϕX (Bj )) < η η η η + + = . 48 48 48 16 Similarly, H[0,1] (ϕY (Dj−1 ), ϕY (Dj )) <

η . 16

The required continuum B will be constructed as the limit in 2X×Y , of a sequence of finite sets (m) (m) {Em }∞ , yi ) : m=1 . We will construct, for each m ∈ N, a number km ∈ N and a finite set Em = {(xi (m) (m) ε i ∈ {0, 1, ..., km }} ⊂ X × Y such that HX×Y (Em , A) < 2 , (x0 , y0 ) = (x0 , y0 ), and for each i ∈ {0, 1, ..., km − 1}, (m)

d(xi

(m)

, xi+1 ) <

1 1 (m) (m) and ρ(yi , yi+1 ) < . m m

η 1 Let m ∈ N and let θ > 0 be such that θ < min{ 2ε , 16 , m }.

Claim 1. There exist rX , rY ∈ N, two θ-mappings ψX : X → [0, rX ], ψY : Y → [0, rY ], ω > 0 (ω < θ) and two piecewise linear mappings μX : [0, rX ] → [0, 1] and μY : [0, rY ] → [0, 1] satisfying: if x, u ∈ X and |ψX (x) − ψX (u)| < ω, then d(x, u) < θ; if v, y ∈ Y and |ψY (v) − ψY (y)| < ω, then ρ(v, y) < θ; η for each x ∈ X, |(μX ◦ ψX )(x) − ϕX (x)| < θ < 16 ; η for each y ∈ Y , |(μY ◦ ψY )(y) − ϕY (y)| < θ < 16 ; for each i ∈ {1, ..., n}, H[0,1] (μX (ψX (Ai−1 )), μX (ψX (Ai ))) < H[0,1] (μY (ψY (Ci−1 )), μY (ψY (Ci ))) < η4 ; (6) for each i ∈ {0, 1, ..., n}, H[0,1] (μX (ψX (Ai )), μX (ψX (Bi ))) < H[0,1] (μY (ψY (Ci )), μY (ψY (Di ))) < η4 ; (7) for each i ∈ {1, ..., n}, H[0,1] (μX (ψX (Bi−1 )), μX (ψX (Bi ))) < H[0,1] (μY (ψY (Di−1 )), μY (ψY (Di ))) < η4 .

(1) (2) (3) (4) (5)

η 4

and

η 4

and

η 4

and

In order to prove Claim 1, we apply Lemma 2.1. First we apply it using the continuum X, the mapping ϕX and the number θ. So, we obtain rX ∈ N, a θ-mapping ψX : X → [0, rX ] and a piecewise linear mapping μX : [0, rX ] → [0, 1] satisfying (3). Now, we apply Lemma 2.1 using the continuum Y , the mapping ϕY and the number θ. This way, we obtain rY ∈ N, a θ-mapping ψY : Y → [0, rY ] and a piecewise linear mapping μY : [0, rY ] → [0, 1] satisfying (4). Since ψX is a θ-mapping, there exists ωX > 0, ωX < θ, such that if x, u ∈ X and |ψX (x) − ψX (u)| < ωX , then d(x, u) < θ. Since ψY is a θ-mapping, there exists ωY > 0, ωY < θ, such that if v, y ∈ Y and |ψY (v) − ψY (y)| < ωY , then ρ(v, y) < θ. Put ω = min{ωX , ωY }. Then the functions ψX and ψY are θ-mappings satisfying (1), (2), (3) and (4). In order to prove (5), take i ∈ {1, ..., n}. By (3), we have H[0,1] (μX (ψX (Ai−1 )), μX (ψX (Ai ))) ≤ H[0,1] (μX (ψX (Ai−1 )), ϕX (Ai−1 )) + H[0,1] (ϕX (Ai−1 ), ϕX (Ai )) + H[0,1] (ϕX (Ai ), μX (ψX (Ai ))) <

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η η η η + + < . 16 16 16 4 Similarly, by (4), H[0,1] (μY (ψY (Ci−1 )), μY (ψY (Ci ))) <

η . 4

In order to prove (6), take i ∈ {0, 1, ..., n}. By (3), we have H[0,1] (μX (ψX (Ai )), μX (ψX (Bi ))) ≤ H[0,1] (μX (ψX (Ai )), ϕX (Ai )) + H[0,1] (ϕX (Ai ), ϕX (Bi )) + H[0,1] (ϕX (Bi ), μX (ψX (Bi ))) < η η η η + + < . 16 16 16 4 Similarly, by (4), H[0,1] (μY (ψY (Ci )), μY (ψY (Di ))) <

η . 4

In order to prove (7), take i ∈ {1, ..., n}. By (3), we have H[0,1] (μX (ψX (Bi−1 )), μX (ψX (Bi ))) ≤ H[0,1] (μX (ψX (Bi−1 )), ϕX (Bi−1 )) + H[0,1] (ϕX (Bi−1 ), ϕX (Bi )) + H[0,1] (ϕX (Bi ), μX (ψX (Bi ))) < η η η η + + < . 16 16 16 4 Similarly, by (4), H[0,1] (μY (ψY (Di−1 )), μY (ψY (Di ))) <

η . 4

Therefore, the proof of Claim 1 is finished. For each i ∈ {0, 1, ..., n}, let X X X [aX i , bi ] = μX (ψX (Ai )), [ci , ei ] = μX (ψX (Bi )),

[aYi , bYi ] = μY (ψY (Ci )) and [cYi , eYi ] = μY (ψY (Di )). Then X [aX 0 , b0 ] = {μX (ψX (p0 ))}, X aX 0 = b0 , X [cX 0 , e0 ] = {μX (ψX (x0 ))}, X cX 0 = e0 ,

[aY0 , bY0 ] = {μY (ψY (q0 ))}, aY0 = bY0 , [cY0 , eY0 ] = {μY (ψY (y0 ))}, cY0 = eY0 .

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  X η X X X X  for each i ∈ {0, 1, ..., n}, H[0,1] ([aX i , bi ], [ci , ei ]) < 4 , which implies that max{ ai − ci ,  XBy (6),  b − eX } < η . By (5), for each i ∈ {1, ..., n}, H[0,1] ([aX , bX ], [aX , bX ]) < η , which implies that i i i−1 i−1 i i 4 4 η η X X X X X X X max{aX i−1 − ai , bi − bi−1 } < 4 . By (7), for each i ∈ {1, ..., n}, H[0,1] ([ci−1 , ei−1 ], [ci , ei ]) < 4 , which η X X X implies that max{cX i−1 − ci , ei − ei−1 } < 4 . Therefore,     X  X X max({max{aX i − ci , bi − ei } : i ∈ {0, 1, ..., n}}∪ X X X X X X X {max{aX i−1 − ai , bi − bi−1 , ci−1 − ci , ei − ei−1 } : i ∈ {1, ..., n}}) η < . 4

Similarly, using (6), (5) and (7), it can be shown that     max({max{aYi − cYi  , bYi − eYi } : i ∈ {0, 1, ..., n}} ∪ {max{aYi−1 − aYi , bYi − bYi−1 , cYi−1 − cYi , eYi − eYi−1 } : i ∈ {1, ..., n}}) η < . 4 For each k ∈ {1, 2}, let πk : R2 → R be the respective projection on the kth-coordinate. Given i ∈ {0, 1, ..., n}, since Ai = πX (σ(ti )) and Ci = πY (σ(ti )), we have π1 ((ψX × ψY )(σ(ti ))) = ψX (Ai ) ⊂ [0, rX ] and π2 ((ψX × ψY )(σ(ti ))) = ψY (Ci ) ⊂ [0, rY ]. Then (ψX × ψY )(σ(ti )) is a subcontinuum of the set ψX (Ai ) × ψY (Ci ) and ψX (Ai ) × ψY (Ci ) is an arc or a 2-cell. Thus, there exists a mapping fi : [i, i + 1] → ψX (Ai ) × ψY (Ci ) such that: each coordinate of fi is piecewise linear, Im fi is as close as we want from (ψX × ψY )(σ(ti )) and Im fi contains any previous chosen point in (ψX × ψY )(σ(ti )). We choose fi satisfying the following conditions. Since (ψX × ψY )(σ(t0 )) = {(ψX (p0 ), ψY (q0 ))}, we can choose f0 to be the constant mapping that only takes the value (ψX (p0 ), ψY (q0 )). Since μX × μY is uniformly continuous, fi can be chosen in such a way that H[0,1]2 ((μX × μY )(Im fi ), (μX × μY )((ψX × ψY )(σ(ti )))) <

η , and 8

H[0,rX ]×[0,rY ] (Im fi , (ψX × ψY )(σ(ti )))) < ω.

(8)

Note that X π1 ((μX × μY )(Im fi )) = μX (π1 (Im fi )) ⊂ μX (ψX (Ai )) = [aX i , bi ],

π1 ((μX × μY )((ψX × ψY )(σ(ti )))) = X μX (π1 ((ψX × ψY )(σ(ti )))) = μX (ψX (Ai )) = [aX i , bi ],

and also π2 ((μX × μY )(Im fi )) = μY (π2 (Im fi )) ⊂ μY (ψY (Ci )) = [aYi , bYi ], and π2 ((μX × μY )((ψX × ψY )(σ(ti )))) = μY (π2 ((ψX × ψY )(σ(ti )))) = μY (ψY (Ci )) = [aYi , bYi ]. X We choose points wi , xi , yi , zi in (ψX × ψY )(σ(ti )) such that μX (π1 (wi )) = aX i , μX (π1 (xi )) = bi , Y Y μY (π2 (yi )) = ai and μY (π2 (zi )) = bi . Then we can assume that wi , xi , yi , zi belong to Im fi . Then X Y Y aX i , bi ∈ μX (π1 (Im fi )) and ai , bi ∈ μY (π2 (Im fi )). Therefore,

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X π1 ((μX × μY )(Im fi )) = [aX i , bi ] and

π2 ((μX × μY )(Im fi )) = [aYi , bYi ]. In fact, retracing a part of the path fi , if necessary, we can ask that fi ends at xi . So, fi (i + 1) = xi and π1 ((μX × μY )(fi (i + 1))) = bX i . If i ≥ 1, since σ(ti−1 ) ⊂ σ(ti ), we can also ask that Im fi starts at the point xi−1 . So, fi (i) = xi−1 and π1 ((μX × μY )(fi (i))) = bX i−1 . Define fX : [0, n + 1] → [0, 1] as the common extension of the piecewise linear mappings π1 ◦ (μX × μY ) ◦ f0 , ..., π1 ◦ (μX × μY ) ◦ fn . We consider the common extension f ∗ : [0, n + 1] → [0 , rX ] × [0, rY ] of the piecewise linear mappings f0 , ..., fn . Note that fX = π1 ◦ (μX × μY ) ◦ f ∗ = μX ◦ π1 ◦ f ∗ . For each i ∈ {0, 1, ..., n}, fix a point pi ∈ ψX (Bi ) such that μX (pi ) = eX i . In the case that i = 0,  pi = ψX (x0 ). If i ≥ 1, since ψX (Bi ) is a subinterval of [0, rX ] and pi−1 ∈ ψX (Bi−1 ) ⊂ ψX (Bi ), there exists a piecewise linear mapping gi : [i, i + 1] → ψX (Bi ) such that gi (i) = pi−1 , gi (i + 1) = pi and X gi ([i, i + 1]) = ψX (Bi ). Thus μX (gi (i)) = eX i−1 , μX (gi (i + 1)) = ei and X μX (gi ([i, i + 1])) = μX (ψX (Bi )) = [cX i , ei ].

Consider the constant mapping g0 : [0, 1] → {ψX (x0 )}. Let g : [0, n + 1] → [0, 1] be the common extension of the piecewise linear mappings μX ◦ g0 , ..., μX ◦ gn , and let g ∗ : [0, n + 1] → [0, rX ] be the common extension of the mappings g0 , ..., gn . Note that g = μX ◦ g ∗ . By Lemma 2.2, there exist piecewise linear mappings αX , βX : [0, n + 1] → [0, n + 1] such that αX (0) = 0 = βX (0), αX (n + 1) = n + 1 = βX (n + 1), and for each t ∈ [0, n + 1], |(fX ◦ αX )(t) − (g ◦ βX )(t)| ≤

η . 2

(9)

Recall that, for each i ∈ {0, 1, ..., n}, the points yi and zi were chosen in Im fi satisfying μY (π2 (yi )) = aYi and μY (π2 (zi )) = bYi . Claim 2. There exists a piecewise linear onto mapping λ : [0, n + 1] → [0, n + 1] such that λ(0) = 0 and for each i ∈ {0, 1, ..., n}, (f ∗ ◦ αX ◦ λ)(i + 1) = zi and π2 ((μX × μY )((f ∗ ◦ αX ◦ λ)([i, i + 1]))) = [aYi , bYi ]. In order to prove Claim 2, define r0 = 0 and for each i ∈ {1, ..., n + 1}, ri = sup{t ∈ [0, n + 1] : αX ([0, t]) ⊂ [0, i]}. Notice that αX (ri ) = i, rn+1 = n + 1 and, for each t ∈ [r0 , r1 ], it holds that αX (t) ∈ [0, 1] and f ∗ (αX (t)) = (ψX (p0 ), ψY (q0 )). Let i ∈ {1, ..., n + 1}. Then f ∗ (αX (ri )) = f ∗ (i) = xi−1 and [i − 1, i] ⊂ αX ([ri−1 , ri ]). Since the domain of fi−1 is [i − 1, i] and zi−1 ∈ Im fi−1 , there exists si ∈ [ri−1 , ri ] such that fi−1 (αX (si )) = zi−1 . Put s0 = 0 and define λi : [i − 1, i] → [si−1 , ri ] as the simplest piecewise linear mapping satisfying λi (i − 1) = si−1 , λi (i − 1 + 12 ) = ri and λi (i) = si . For each i ∈ {1, ..., n}, λi (i) = si = λi+1 (i), so that the common extension λ : [0, n + 1] → [0, n + 1] of all the mappings λ1 , ..., λn+1 is well defined. The function λ is continuous because λ1 , ..., λn+1 are continuous. The mapping λ is onto because λ(0) = λ1 (0) = s0 = 0 and λ(n + 12 ) = λn+1 (n + 12 ) = rn+1 = n + 1. Let i ∈ {0, 1, ..., n}. Then λ(i + 1) = λi+1 (i + 1) = si+1 and (f ∗ ◦ αX ◦ λ)(i + 1) = fi (αX (si+1 )) = zi . It remains to show that π2 ((μX × μY )((f ∗ ◦ αX ◦ λ)([i, i + 1]))) = [aYi , bYi ].

J.A. Naranjo-Murillo / Topology and its Applications 269 (2020) 106924

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First we take t ∈ [i, i + 1]. Then λ(t) = λi+1 (t) ∈ [si , ri+1 ] and, by the definition of ri+1 , αX (λ(t)) ∈ [0, i + 1]. Let j ∈ {0, 1, ..., i} be such that αX (λ(t)) ∈ [j, j + 1]. Then f ∗ (αX (λ(t))) = fj (αX (λ(t))). Since π2 ((μX × μY )(Im fj )) = [aYj , bYj ] ⊂ [aYi , bYi ], we have that π2 ((μX × μY )(fj (αX (λ(t))))) ∈ [aYi , bYi ]. Now, we take u ∈ [aYi , bYi ]. Since π2 ((μX × μY )(Im fi )) = [aYi , bYi ], there exists v ∈ [i, i + 1] such that π2 ((μX × μY )(fi (v))) = u. Since [i, i + 1] ⊂ αX ([ri , ri+1 ]), there exists s ∈ [ri , ri+1 ] such that αX (s) = v. Since [ri , ri+1 ] ⊂ [si , ri+1 ] and λi+1 : [i, i + 1] → [si , ri+1 ] is onto, there exists t ∈ [i, i + 1] such that λi+1 (t) = s. Thus, π2 ((μX × μY )((f ∗ ◦ αX ◦ λ)(t))) = u. Therefore, the proof of Claim 2 is finished. Let fY = π2 ◦ (μX × μY ) ◦ (f ∗ ◦ αX ◦ λ) : [0, n + 1] → [0, 1]. Notice that fY = μY ◦ π2 ◦ (f ∗ ◦ αX ◦ λ) and fY is piecewise linear. For each i ∈ {0, 1, ..., n}, fix a point qi ∈ ψY (Di ) such that μY (qi ) = eYi . In the case that i = 0,  qi = ψY (y0 ). If i ≥ 1, since ψY (Di ) is a subinterval of [0, rY ] and qi−1 ∈ ψY (Di−1 ) ⊂ ψY (Di ), there  is a piecewise linear mapping hi : [i, i + 1] → ψY (Di ) that satisfies hi (i) = qi−1 , hi (i + 1) = qi and hi ([i, i + 1]) = ψY (Di ). Thus, μY (hi (i)) = eYi−1 , μY (hi (i + 1)) = eYi and μY (hi ([i, i + 1])) = μY (ψY (Di )) = [cYi , eYi ]. Consider the constant mapping h0 : [0, 1] → {ψY (y0 )}. Let h : [0, n +1] → [0, 1] be the common extension of all the piecewise linear mappings μY ◦h0 , ..., μY ◦hn , and let h∗ : [0, n + 1] → [0, rY ] be the common extension of all the mappings h0 , ..., hn . By Lemma 2.2, there exist piecewise linear mappings αY , βY : [0, n + 1] → [0, n + 1] such that αY (0) = 0 = βY (0), αY (n + 1) = n + 1 = βY (n + 1), and for each t ∈ [0, n + 1], |(fY ◦ αY )(t) − (h ◦ βY )(t)| ≤

η . 2

(10)

By the uniform continuity of g ∗ ◦ βX ◦ λ ◦ αY , h∗ ◦ βY , g ◦ βX ◦ λ ◦ αY and h ◦ βY there exists ϑ > 0 such that if s, t ∈ [0, n + 1] and |s − t| < ϑ then |(g ∗ ◦ βX ◦ λ ◦ αY )(s) − (g ∗ ◦ βX ◦ λ ◦ αY )(t)| < ω, |(h∗ ◦ βY )(s) − (h∗ ◦ βY )(t)| < ω, |(g ◦ βX ◦ λ ◦ αY )(s) − (g ◦ βX ◦ λ ◦ αY )(t)| < |(h ◦ βY )(s) − (h ◦ βY )(t)| <

η and 4

η . 4

(11)

Fix a partition 0 = τ0 < τ1 < ... < τk = n + 1 of the interval [0, n + 1] such that for each i ∈ {1, ..., k}, (m) (m) τi − τi−1 < ϑ. For each i ∈ {0, 1, ..., k}, we choose points xi ∈ X and yi ∈ Y satisfying (m)

ψX (xi

) = (g ∗ ◦ βX ◦ λ ◦ αY )(τi ) and ψY (yi

(m)

) = (h∗ ◦ βY )(τi ).

Since (g ∗ ◦βX ◦λ ◦αY )(τ0 ) = g ∗ (0) = ψX (x0 ), we choose x0 = x0 . Since (h∗ ◦βY )(τ0 ) = h∗ (0) = ψY (y0 ), (m) (m) (m) we choose y0 = y0 . Put km = k and Em = {(xi , yi ) : i ∈ {0, 1, ..., km }}. (m)

Claim 3. For each i ∈ {0, 1, ..., km − 1}, 1 , m 1 (m) (m) ρ(yi , yi+1 ) < , m (m)

d(xi

(m)

, xi+1 ) <

J.A. Naranjo-Murillo / Topology and its Applications 269 (2020) 106924

10

and HX×Y (Em , A) <

ε . 2

In order to prove the first part of the claim, take i ∈ {0, 1, ..., km − 1}. We have that τi − τi−1 < ϑ, the choice of ϑ implies that    (m) (m)  ψX (xi ) − ψX (xi+1 ) = |(g ∗ ◦ βX ◦ λ ◦ αY )(τi ) − (g ∗ ◦ βX ◦ λ ◦ αY )(τi+1 )| < ω. (m)

Then, the choice of ω guarantees that d(xi

(m)

, xi+1 ) < θ <

1 m.

(m)

Similarly, ρ(yi

(m)

1 m. (m) (m) (xi , yi )

, yi+1 ) <

In order to prove the rest of the claim, take i ∈ {0, 1, ..., km } we will see that ∈ N (A, 2ε ). Let j ∈ {0, 1, ..., n} be such that (αX ◦ λ ◦ αY )(τi ) ∈ [j, j + 1]. Then, f ∗ ((αX ◦ λ ◦ αY )(τi )) = fj ((αX ◦ λ ◦ αY )(τi )). By (8), there exists (m1 , m2 ) ∈ σ(tj ) ⊂ A such that (μX × μY )(fj ((αX ◦ λ ◦ αY )(τi ))) − (μX × μY )(ψX × ψY )(m1 , m2 ) <

η . 8

This implies that η and 8 η |μY (π2 (fj ((αX ◦ λ ◦ αY )(τi ))) − μY (ψY (m2 ))| < . 8

|μX (π1 (fj ((αX ◦ λ ◦ αY )(τi ))) − μX (ψX (m1 ))| <

Then |fX ((αX ◦ λ ◦ αY )(τi )) − μX (ψX (m1 ))| < |fY (αY (τi )) − μY (ψY (m2 ))| <

η and 8

η . 8

By (3) and (4), |μX (ψX (m1 )) − ϕX (m1 )| <

η η and |μY (ψY (m2 )) − ϕY (m2 )| < . 8 8

Thus, |fX ((αX ◦ λ ◦ αY )(τi )) − ϕX (m1 )| < |fY (αY (τi )) − ϕY (m2 )| <

η and 4

η . 4

(12)

By the choice of αX and βX , |(fX ◦ αX )(λ ◦ αY )(τi )) − (g ◦ βX )((λ ◦ αY )(τi ))| ≤

η . 2

By definition, (m)

μX (ψX (xi Thus, by (3) and (12),

)) = μX ((g ∗ ◦ βX ◦ λ ◦ αY )(τi )) = g((βX ◦ λ ◦ αY )(τi )).

J.A. Naranjo-Murillo / Topology and its Applications 269 (2020) 106924

11

    (m) ϕX (xi ) − ϕX (m1 ) ≤    (m) (m)  ϕX (xi ) − μX (ψX (xi )) +

    (m) μX (ψX (xi )) − fX ((αX ◦ λ ◦ αY )(τi )) + |fX ((αX ◦ λ ◦ αY )(τi )) − ϕX (m1 )| < η η η 7η + + = . 8 2 4 8 (m)

By the choice of η, d(xi , m1 ) < 2ε . (m) By the choice of αY and βY , |(fY ◦ αY )(τi ) − (h ◦ βY )(τi )| ≤ η2 . By definition, μY (ψY (yi )) = μY ((h∗ ◦ βY )(τi )) = h(βY (τi )). Thus, by (4) and (12),     (m) ϕY (yi ) − ϕY (m2 ) ≤    (m) (m)  ϕY (yi ) − μY (ψY (yi )) +     (m) μY (ψY (yi )) − fY (αY (τi )) + |fY (αY (τi )) − ϕY (m2 )| < η η η 7η + + = . 8 2 4 8 (m)

By the choice of η, ρ(yi , m2 ) < 2ε . (m) (m) Therefore, (xi , yi ) ∈ B((m1 , m2 ), 2ε ) ⊂ N (A, 2ε ). This proves that Em ⊂ N (A, 2ε ). Now, take (m1 , m2 ) ∈ A. Since A = σ(tn ), by (8), there exists t ∈ [n, n + 1] such that (μX × μY )(fn (t)) − (μX × μY )((ψX × ψY )(m1 , m2 )) <

η . 8

This implies that η and 8 η |μY (π2 (fn (t))) − μY (ψY (m2 ))| < . 8

|μX (π1 (fn (t)) − μX (ψX (m1 ))| <

Since αX ◦ λ ◦ αY : [0, n + 1] → [0, n + 1] is onto, there exists s ∈ [0, n + 1] such that (αX ◦ λ ◦ αY )(s) = t. Because fX = μX ◦ π1 ◦ f ∗ and fY = μY ◦ π2 ◦ (f ∗ ◦ αX ◦ λ), the previous inequalities can be rewritten as |(fX ◦ αX ◦ λ ◦ αY )(s) − μX (ψX (m1 ))| < |(fY ◦ αY )(s) − μY (ψY (m2 ))| <

η and 8

η . 8

Take i ∈ {0, 1, ...., km − 1} such that s ∈ [τi , τi+1 ]. (m) Since μX (ψX (xi )) = (g ◦ βX ◦ λ ◦ αY )(τi ) and |s − τi | ≤ τi+1 − τi < ϑ, by (11) and (9) we have that     (m) μX (ψX (xi )) − μX (ψX (m1 )) ≤ |(g ◦ βX ◦ λ ◦ αY )(τi ) − (g ◦ βX ◦ λ ◦ αY )(s)| + |(g ◦ βX ◦ λ ◦ αY )(s) − (fX ◦ αX ◦ λ ◦ αY )(s)| +

12

J.A. Naranjo-Murillo / Topology and its Applications 269 (2020) 106924

|(fX ◦ αX ◦ λ ◦ αY )(s) − μX (ψX (m1 ))| < 7η η η η + + = . 4 2 8 8 η 16 ,

Thus, using (3) and the inequality θ <

we have that

    (m) ϕX (xi ) − ϕX (m1 ) ≤    (m) (m)  ϕX (xi ) − μX (ψX (xi )) +     (m) μX (ψX (xi )) − μX (ψX (m1 )) + |μX (ψX (m1 )) − ϕX (m1 )| < η η + |μX (ψX (xm < η. i )) − μX (ψX (m1 ))| + 16 16 (m)

By the choice of η, we have that d(xi , m1 ) < 2ε . (m) Since μY (ψY (yi )) = (h ◦ βY )(τi ) and |s − τi | ≤ τi+1 − τi < ϑ, by (11) and (10), we have that     (m) μY (ψY (yi )) − μY (ψY (m2 )) ≤ |(h ◦ βY )(τi ) − (h ◦ βY )(s)| + |(h ◦ βY )(s) − (fY ◦ αY )(s)| + |(fY ◦ αY )(s) − μY (ψY (m2 ))| < 7η η η η + + = . 4 2 8 8 Thus, using (4) and the inequality θ <

η 16 ,

we have that

    (m) ϕY (yi ) − ϕY (m2 ) ≤    (m) (m)  ϕY (yi ) − μY (ψY (yi )) +     (m) μY (ψY (yi )) − μY (ψY (m2 )) + |μY (ψY (m2 )) − ϕY (m2 )| <   η η   (m) + μY (ψY (yi )) − μY (ψY (m2 )) + < η. 16 16 (m)

By the choice of η, we have that ρ(yi , m2 ) < 2ε . (m) (m) Therefore, (m1 , m2 ) ∈ B((xi , yi ), 2ε ) ⊂ N (Em , 2ε ). This proves that A ⊂ N (Em , 2ε ). We have shown that HX×Y (Em , A) < 2ε , which ends the proof of Claim 3. X×Y Without loss of generality, we may assume that {Em }∞ . Since, m=1 converges to some element B ∈ 2 ε for each m ∈ N, (x0 , y0 ) ∈ Em , (x0 , y0 ) ∈ B. Since, for each m ∈ N, HX×Y (Em , A) < 2 , we have that HX×Y (B, A) ≤ 2ε < ε. Claim 4. B is connected.

J.A. Naranjo-Murillo / Topology and its Applications 269 (2020) 106924

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Suppose that there are disjoint closed nonempty subsets K and L of X × Y such that B = K ∪ L. Put dist(K, L) = min{D( a, b) :  a ∈ K and b ∈ L}. Since K and L are compact and disjoint, dist(K, L) > 0. Then, there exists r > 0 such that dist(K, L) > 3r. 1 We may suppose that (x0 , y0 ) ∈ K. Since {Em }∞ m=1 converges to B, there is M ∈ N such that M < r and for each m ≥ M , HX×Y (B, Em ) < r. Then, for each m ≥ M , Em ⊂ N (B, r) = N (K, r) ∪ N (L, r). Let (m) (m) m ≥ M . We have that (x0 , y0 ) = (x0 , y0 ) ∈ K, so that (m)

l0 = max{l ∈ {0, 1, ..., km } : for each j ∈ {0, ..., l}, (xj (m)

(m)

, yj

) ∈ N (K, r)}

(m)

is well defined and l0 ≥ 0. If l0 < km , then (xl0 +1 , yl0 +1 ) ∈ N (L, r). Thus, there exist p ∈ K and q ∈ L such that

(m) (m) D( p, (xl0 , yl0 ))

< r and

(m) (m) D( q , (xl0 +1 , yl0 +1 ))

< r. Therefore,

dist(K, L) ≤ D( p, q) ≤ (m) (m) D( p, (xl0 , yl0 ))

+

(m) (m) (m) (m) D((xl0 , yl0 ), (xl0 +1 , yl0 +1 ))

r+

(m)

(m)

+ D((xl0 +1 , yl0 +1 ), q) <

1 + r < 3r. m

This contradicts the choice of r, so l0 = km . Then, Em ⊂ N (K, r). Since this holds for each m ≥ M , we have that B ⊂clX×Y (N (K, r)) and L = ∅. This contradicts that L = ∅. Therefore, we conclude that B is connected. We have shown that B is a subcontinuum of X × Y , (x0 , y0 ) ∈ B and HX×Y (A, B) < ε. This concludes the proof that X × Y is a Kelley continuum. 2 Acknowledgements I am very grateful to my advisor Alejandro Illanes, for his support, guidance and permanent availability. To Verónica Martínez-de-la-Vega and Jorge M. Martínez-Montejano for their support during my graduate studies. I also wish to thank the anonymous referee for his/her suggestions and careful reading of the paper. This paper was supported by the scholarship number 295666 of Consejo Nacional de Ciencia y Tecnología (CONACyT). This paper was also partially supported by the project “Teoría de Continuos, Hiperespacios y Sistemas Dinámicos III” (IN106319) of PAPIIT, DGAPA, UNAM. References [1] J.J. Charatonik, W.J. Charatonik, Property of Kelley for the Cartesian products and hyperspaces, Proc. Am. Math. Soc. 136 (1) (2008) 341–346. [2] A. Illanes, J.M. Martínez-Montejano, K. Villarreal, Connected neighborhoods in products, Topol. Appl. 241 (2018) 172–184. [3] A. Illanes, S.B. Nadler Jr., Hyperspaces: Fundamentals and Recent Advances, Monographs and Textbooks in Pure and Applied Mathematics, vol. 216, Marcel Dekker, Inc., New York, 1999. [4] J.L. Kelley, Hyperspaces of a continuum, Trans. Am. Math. Soc. 52 (1942) 22–36. [5] S.B. Nadler Jr., Continuum Theory. An Introduction, Monographs and Textbooks in Pure and Applied Mathematics, vol. 158, Marcel Dekker, Inc., New York, 1992. [6] J.A. Naranjo-Murillo, The product of two chainable Kelley continua has the fupcon property, preprint. [7] R.W. Wardle, On a property of J. L. Kelley, Houst. J. Math. 3 (2) (1977) 291–299.