Hypographs of Upper Semi-continuous Maps and Continuous Maps on a Bounded Open Interval
Nada Wu
Department of Mathematics and Statistics, Hanshan Normal University, Chaozhou, Guangdong, China
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To cite this article:
Nada Wu. Hypographs of Upper Semi-continuous Maps and Continuous Maps on a Bounded Open Interval. American Journal of Applied Mathematics. Vol. 4, No. 2, 2016, pp. 75-79. doi: 10.11648/j.ajam.20160402.12
Received: February 3, 2016; Accepted: March 21, 2016; Published: March 25, 2016
Abstract: For any bounded open interval X in the Euclidean space E^{1}, let ↓USC(X) and ↓C(X) be the families of all hypographs of upper semi-continuous maps and continuous maps from X to I=[0,1], respectively. They are endowed with the topology induced by the Hausdorff metric of the metric space Y×I，Y is the closure of X. It was proved in other two papers respectively that ↓USC(X) and ↓C(X) are homeomorphic to s and c_{0 }respectively, where s=(-1,1)^{∞} and c_{0}={(x_{n}): lim_{n→∞}x_{n}=0}. However the topological structure of the pair (↓USC(X), ↓C(X)) was not clear. In the present paper, it is proved in the strongly universal method that the pair of spaces (↓USC(X), ↓C(X)) is pair homeomorphic to which is not homeomorphic to (s, c_{0}). Hence this paper figures out the topological structure of the pair (↓USC(X), ↓C(X)).
Keywords: Hypograph, Upper Semi-continuous Maps, Continuous Maps, Bounded Open Interval, Hausdorff Metric, The Property of Strongly Universal
1. Introduction
For a Tychonoff space X and a subspace L of the real line R with the usual topology, let C(X, L) denote the set of all continuous maps from X to L. C(X, L) can be endowed with different topologies which are interesting for topologists. There are many research results in this field. In 1966 and 1991, two classical results were proved respectively which are listed below.
Theorem 1 [Anderson-Kadec Theorem] [1]. C_{u}(X, L) is homeomorphic to (≈) the Hilbert space l_{2}≈s=(-1,1)^{∞} [2] if X is an infinite compactum (a compactum means a compact metric space) and L=I=[0,1] or L=R, where C_{u}(X, L) is C(X, L) endowed with the uniformly convergence topology.
Theorem 2 [3]. Cp(X, L)≈c_{0} if X is a countable non-discrete metric space and L=R or L=I, where Cp(X, L) denotes C(X, L) endowed with the pointwise convergence topology, and c_{0}={(x_{n}): lim_{n→∞}x_{n}=0}.
In [4] to [7], C(X, I) was endowed with another topology. To introduce this topology, it is necessary to recall the knowledge about hyperspace. For a metric space (M, d), the hyperspace Cld(M) is the set consisting of all non-empty closed subsets of M endowed with Vietoris topology. Hausdorff distance d_{H} defined as follows:
d_{H}(A, B)=Inf{ε: B_{d} (A, ε)⊃B and B_{d}(B, ε)⊃A} (1)
for any A, B∈Cld(M). It is well-known that if M is compact, then d_{H} is a metric and deduces Vietoris topology on Cld (M).
Let X be a space. A (single-valued) function f: X → R is called upper semi-continuous if f^{−1}(−∞, t) is open in X for every t∈R. For a Tychonoff space X and L⊂R, let USC(X, L) denote the family of all upper semi-continuous maps from X to L. For convenience, USC(X, I) and C(X, I) are abbreviated as USC(X) and C(X) respectively.
For every f ∈USC(X), let ↓f be the region below of f, that is, ↓f = {(x, λ)∈X×I: λ ≤ f (x) } then ↓f ∈Cld (X×I).
Hence↓USC(X)={↓f: f∈USC(X)} and ↓C(X)= {↓f: f∈C(X)} can be topologized as subspaces of the hyperspace Cld(X×I). ↓C(X) can be considered as C(X) endowed with another topology which is different with the two former topologies ([4, Corollary 1]).
For two pairs of spaces (X_{1}, Y_{1}) and (X_{2}, Y_{2}) with Y_{1}⊂X_{1} and Y_{2}⊂X_{2}, the symbol (X_{1}, Y_{1})≈(X_{2}, Y_{2}) means that there exists a homeomorphism h: X_{1} →X_{2} such that h(Y_{1})=Y_{2}. For a metric space X, we use X_{0} and cl_{X}(·) to denote the set of all isolated points of X and the closure-operator in X, respectively.
In 2005, it was proved that ↓USC(X)≈Q (Q = [-1, 1]^{∞}is the Hilbert cube) when X is a compactum [8].
In 2006, there was the following result.
Theorem 3 [4]. For a Tychonoff space X, the following
conditions are equivalent:
(a) X is a compactum and cl_{X}(X_{0})≠X;
(b) ↓C(X) ≈ c_{0};
(c) (↓USC(X), ↓C(X)) ≈ (Q, c_{0}).
In 2009, the following theorem was proved which indicated that the topological structure of (↓USC(X), ↓C(X)) was figured out for every compactum X.
Theorem 4 [7]. Let X be a compact metric space, |X| denotes the cardinal number of X, and Σ={(x_{n})∈Q: sup{x_{n}: n∈N^{+}} < 1} be a subspace of Q, then
If X is a non-compact space, ↓USC(X) and ↓C(X) can be endowed with Fell topology. From 2014 to 2016,Yang made a series of research in this area, see [9-11].
If X is a non-compact and topological complete metric space whose completion is compact, then ↓USC(X) and ↓C(X) can be embed in ↓USC(Y) as subspaces [12,13], where Y is the completion of X. In fact, define a map ↓e: ↓USC(X)→↓USC(Y) as following: for any f ∈USC(X),
(2)
It had been proved in [12] that ↓e is an isometric imbedding.
In 2008, there was the following result about ↓USC(X).
Theorem 5 [12]. ↓USC(X) ≈ l_{2} ≈s if (X, ρ) is a non-compact and topological complete metric space whose completion is compact.
However, at that time, we didn’t know the topological structure of ↓C(X) for every X satisfying the condition in Theorem 5. In 2013, we considered the case of cl_{X}(X_{0})≠X, and got the following result.
Theorem 6 [13]. If X is a noncompact, locally compact, totally bounded, separable metric space, then ↓C(X, I) ≈ c_{0} if and only if cl_{X}(X_{0})≠X.
We failed to prove that (↓USC(X,I),↓C(X,I))≈(s,c_{0}) for every noncompact, locally compact, totally bounded, separable metric space X with cl_{X}(X_{0})≠X. In this paper, we take a particular case, that is, consider X as a bounded open interval in the Euclidean space E^{1} and we prove the following main result.
Theorem 7. For any bounded open interval X in the Euclidean space E^{1}, (↓USC(X), ↓C(X))≈.
Remark 1. In fact (Q^{∞}, s^{∞}) ≈ (Q, s) and (Q^{∞}, c_{0}^{∞}) ≈ (Q, c_{0}), however, (s^{∞}, c_{0}^{∞}) is not pair homemomorphic to (s, c_{0}) [14]. Hence the above theorem indicates that (↓USC(X), ↓C(X)) is not homeomorphic to (s, c_{0}).
2. Preliminaries
All spaces under discussion are assumed to be separable metrizable spaces. All definitions on this section can be found in [15] or [16].
Definition 1 A space X is called an absolute retract, abbreviated AR, provided that for every space Y containing X as a closed subspace, X is a retract of Y, that is, there exists a continuous map r: Y → X such that r|_{X} = id_{X}.
Definition 2 A closed subset A of a space X is said to be a Z-set of X if the identity id_{X }can be approximated by continuous maps from X to X \ A. A Zσ-set in a space is a countable union of Z-sets in the space. A space is called a Zσ−space if it is a Zσ−set of itself. We use Z(X) and Zσ(X) to denote the family of all Z-sets and the family of all Zσ-sets in X, respectively. A Z-embedding is an embedding with a Z-set image.
Definition 3 A subset A of a space Y is called homotopy dense in Y if there exists a homotopy h: Y × I → Y such that h_{0} = id_{Y} and h_{t}(Y)⊂A for every t > 0.
Definition 4 Let M_{0} and M_{1} denote the class of compacta and the class of topological complete spaces, respectively. A space is called an absolute Fσ-space if it is an Fσ-set in any space which contains it as a subspace. A space is called an absolute Fσδ-space if it is an Fσδ-set in any space which contains it as a subspace. Let M_{2} denote the class of all absolute Fσδ-spaces. Let (M_{0}, M_{1}, M_{2}) denote the class of all triples of spaces (M, B, A) such that M⊃B⊃A, M∈M_{0}, B∈M_{1} and A∈M_{2}.
Definition 5 Let Fσ denote the class of all absolute Fσ-spaces. Let (X, d) be a copy of Hilbert cube Q and the pair of spaces (X, Y)∈(M_{0}, M_{2})(or res. (M_{0}, Fσ)). We say that (X, Y) is strongly (M_{0}, M_{2}) -universal (or res. strongly (M_{0}, Fσ) universal)provided for each(M, B)∈(M_{0}, M_{2}) (or res. (M_{0}, Fσ)), each continuous map f: M → X, each closed subset K of M such that f|_{K}: K→X is a Z-embedding and each ε > 0, there is a Z-embedding g: M → X such that g|_{K} = f|_{K}, g^{−1}(Y)\K = B\K and d(g(m), f(m)) < ε for each m∈M.
Remark 2. (Q^{∞}, c_{0}^{∞}) is strongly (M_{0}, M_{2}) –universal [14] and (Q,Σ) is strongly (M_{0}, Fσ) universal [15].
Definition 6 Let (X, d) be a copy of Hilbert cube Q and the triple of spaces (X, Y, Z)∈(M_{0}, M_{1}, M_{2}). We say that (X, Y, Z) is strongly (M_{0}, M_{1}, M_{2}) -universal provided for each
(M, B, A)∈(M_{0}, M_{1}, M_{2}), each continuous map f: M → X, each closed subset K of M such that f|_{K}: K→X is a Z-embedding and each ε > 0, there is a Z-embedding
g: M → X such that g|K = f|K, g^{−1}(Y) \ K = B \ K, g^{−1}(Z) \ K = A \ K and d(g(m), f(m)) < ε for each m ∈ M.
Let Ω_{2 }be the absorbing set in R^{∞} for the class M_{2}, constructed in [17].
The following lemma is a statement in [18, page 274].
Lemma 1. Let (R_{C}^{∞}, R^{∞}, D) be a triple of spaces with D≈Ω_{2}, then (R_{C}^{∞}, R^{∞}, D)≈(R_{C}^{∞}, R^{∞}, Ω_{2)} (triple homeomorphism whose definition is similar as pair homeomorphism) if and only if (R_{C}^{∞}, R^{∞}, D) is strongly (M_{0}, M_{1}, M_{2}) –universal, R_{C}= [−∞, +∞].
Corollary 1. Let (A, B, C) be a triple of spaces with (A, B) ≈ (Q, s) and C ≈ c_{0}, then (B,C) ≈ (s^{∞}, c_{0}^{∞}) if (A, B, C) is strongly (M_{0}, M_{1}, M_{2}) -universal.
Proof. Since (A, B) ≈ (Q, s) ≈ (R_{C}^{∞}, R^{∞}), there exists a homeomorphism h: A → R_{C}^{∞}such that h[B] = R^{∞}. Put D = h[C], then (A, B, C) ≈ (R_{C}^{∞}, R^{∞}, D) and D ≈ c_{0}. If (A, B, C) is strongly (M_{0}, M_{1}, M_{2})-universal then (R_{C}^{∞}, R^{∞}, D) is also strongly (M_{0}, M_{1}, M_{2})-universal. Note that Ω_{2 }≈ c_{0}^{17}, (R_{C}^{∞},R^{∞}, D) ≈ (R_{C}^{∞}, R^{∞}, Ω_{2}) by lemma 1. Since (R^{∞}, Ω_{2}) ≈ (s^{∞}, c_{0}^{∞}) [14], (B, C) ≈ (R^{∞}, D) ≈ (R^{∞}, Ω_{2}) ≈ (s^{∞}, c_{0}^{∞}).
3. Proof of the Main Result
In this section, X is always assumed for an bounded open interval in E^{1}, and Y be the closure of X which is compact. We can define a metric d on the product space Y×I as following,
d((x_{1},t_{1}),(x_{2},t_{2}))=Max{|x_{1}-x_{2}|,|t_{1}-t_{2}|}
for any (x_{1}, t_{1}),(x_{2}, t_{2})∈Y×I.
Let d_{H }be the Hausdorff metric on Cld(Y×I).
3.1. Some Lemmas for Proof of Theorem 3
Lemma 2 [13, Corollary 3]. There exists homotopy
H: ↓USC(Y) × I →↓USC(Y) such that
H_{0} = id_{↓USC(Y)}, and H_{t}(↓USC(Y))⊂↓C(Y) for each t > 0, and d_{H}(H(↓f, t), ↓f)≤t for each f∈USC(Y) and each t∈I.
Lemma 3 [5, Lemma 8]. Let A be a metric space and
a, b: A → I two continuous maps with a(y) < b(y) for each y ∈ A. And let M: A × I → I be a map satisfying the following conditions:
(1) for each fixed y0 ∈ A, M(y0, t): I → I is increasing,
(2) for every fixed t0 ∈ I, M(y, t0): A → I is continuous.
Then s: A→ I defined by
is continuous and M(y, a(y)) ≤ s(y) ≤ M(y, b(y)) for every y∈A.
Lemma 4 [19, lemma 2.9]. (Q_{u}, c_{1}) ≈ (Q^{∞}, c_{0}^{∞}), where
Q_{u} =[0,1]^{∞}, c_{1} = {(x_{n}) ∈Q_{u}: lim_{n→∞} x_{n} = 1}.
Lemma 5 [15, Proposition 5.4.6]. (Q,Q\s) ≈ (Q,Σ).
Lemma 6 [4, Lemma 5]. Let F = E∪Z⊂↓USC(Y) be closed. If Z is a Z-set in ↓USC(Y) and for every ↓f∈E there exists a∈Y such that f(a) = 0, then F is a Z-set in ↓USC(Y).
3.2. Proof of Theorem 3
Since (↓USC(Y), ↓USC(X))≈(Q, s) [12], ↓C(X)≈ c_{0} [13], by Corollary 1, it suffices to prove the following lemma.
Lemma 7. The triple of spaces (↓USC(Y), ↓USC(X), ↓C(X)) is strongly (M_{0}, M_{1}, M_{2}) -universal.
Proof. Without loss of generality, assume that X = (−1, 1), Y= [−1, 1]. Let x_{n} = 1/2^{n}, x′_{n} = 1/2^{n} − 1 for every n ∈ N^{+}, x_{0 }= 0 and x_{∞} = −1, then lim_{n→∞} x_{n} = x_{0} and lim_{n→∞} x′_{n} = x_{∞}. Let (D, B, A) be a triple of spaces such that (D, B, A)∈(M_{0}, M_{1}, M_{2}) and K be a closed subset of D. Let Φ: D → USC(Y) be a map such that ↓Φ: D →↓USC(Y) is continuous and ↓Φ| _{K}: K →↓USC(Y) is a Z-embedding. By [7, Lemma 1.1], without loss of generality, we may assume that ↓Φ(K) ∩ ↓Φ(D\K) =. For every ε ∈ (0,1), let δ: Y → [0,1) be a map defined by
δ(y) =(1/5)min{ε,d _{H }(↓Φ(y),↓Φ(K))}.
Then δ is continuous and δ(y) = 0 if and only if y ∈ K. For every k ∈ N^{}_{+}, let
D_{k} = {y ∈ Y: 2^{−k} ≤ δ(y) ≤ 2^{−k+1}}.
Then _{k}_{∈N +} D_{k}= D\K. In what follows, we shall define.
In what follows, we shall define Ψ_{k}:D_{k} → USC(Y) for every k ∈ N_{ +}, and then use these maps and Φ| _{K} to define a map Ψ:D → USC(X) such that ↓Ψ: D →↓USC(Y) is a Z-embedding, Ψ|_{K} = Φ|_{K}, Ψ ^{−1} (USC(X))\K=B\K, Ψ^{−1} (C(X)) \K=A\K and d_{ H }(↓Ψ(y), ↓Φ(y)) < ε for each y∈D. This task will be finished in four steps.
It follows from Lemma 3, that there exists a homotopy H: ↓USC(Y) × I →↓USC(Y) such that
H_{0}= id_{↓USC(Y)}, H_{t}(↓USC(Y))⊂↓C(Y) and
d_{H}(H_{t}(↓f), ↓f) ≤ t for each f ∈USC(Y) and each t ∈ (0,1].
For each y∈D, and t∈I, let
↓h(y) = H(↓Φ(y),δ(y)),
M _{0 }(y,t) = sup{h(y)(x): |x − x_{ 0} | < t},and
M_{∞}(y,t) = sup{h(y)(x): |x − x_{∞}| < t}.
Then h(y)∈C(Y) for each y∈D\K and
↓h|_{ D \K}: D\ K →↓C(Y) is continuous.
Moreover, d_{H}(↓h(y),↓Φ(y)) ≤ δ(y) for every y∈D. It follows from the continuities of δ and H that
M_{0}, M_{∞}: (D\ K) × I → I satisfies the conditions (1) and (2) in Lemma 3.
Thus
is continuous on D\ K and M_{i}(y,δ(y)) ≤ s_{i}(y) ≤ Mi (y,2δ(y)) for every y ∈ D \ K.
By Definition 5 and Lemmas 4 and 5, there exist Z-embeddings α: D → Q_{u} and β: D → Q such that
α^{−1} [c_{ 1}] = A and β^{−1} [Q\Σ] = B. If α(y) = (x_{1}, x_{2}, ···,x_{n}, ···) ∈ Q_{ u} and β(y) = (z_{ 1}, z_{ 2}, ···,z_{ n}, ···) ∈Q, then the symbols α(y)(n) and β(y)(n) denote x_{ n} and z_{ n}, respectively. For each n ∈ N +, define a map V_{ n}: D → I by
V_{n }(y) = max{|β(y)(i)|: i ∈ {1,2,· · ·,n}}
for every y∈D. It is easy to check that V_{n} is continuous and lim_{ n→∞} V_{n}(y) = 1 if and only if y∈B.
Define continuous maps ϕ_{k}: D_{k} → I by
ϕ_{k} (y) = 2 − 2^{k} δ(y). Since y ∈D_{ k}, 2^{−k} ≤ δ(y) ≤2^{−k+1}. Hence 0 ≤ ϕ_{ k} (y) ≤ 1, ϕ_{ k} (y) = 0 if δ(y) = 2^{−k+1} and ϕ_{k} (y) = 1 if δ(y) = 2^{−k}.
Let Y_{i} = {x ∈Y: 2^{−i−1} ≤ |x − x_{ 0} | ≤ 2^{−i}}, and
Y′_{i} = {x ∈ Y: 2^{−i−1} ≤ |x − x_{ ∞} | ≤ 2^{−i}}, then
Y = (1/2, 1]∪∪_{i}_{∈N +} (Y_{ i }∪Y′_{ i}).
Define φ_{i}: Y_{ i} → I by φ_{i} (x) = 2^{i+1} (2^{−i} − |x − x_{ 0} |) for each i∈N_{ +}. Then φ_{ i }(x_{ i}) = 0 and φ_{i }(x_{ i+1}) = 1 for every i ≥ 1.
Define γ_{i}: Y′_{ i} → I by γ_{i} (x) = 2^{i+1} (2^{−i} − |x − x_{ ∞} |) for each i∈N_{+}. Then γ_{i} (x′_{i}) = 0and γ_{i} (x′_{i+1}) = 1 for every i ≥ 1.
Step 2: For every k ∈ N_{+}, define a map Ψ_{k} on D_{k}.
For every k ∈ N_{+}, define a map Ψ_{k}: D_{k} → USC(Y) as follows:
for every y∈D_{ k},
if x∈(1/2, 1]∪∪_{i}_{∈{1,2,…,2k}}(Y_{ i }∪Y′_{ i}),
Ψ_{k} (y)(x) = h(y)(x).
Ψ _{k }(y)(x_{ 2k+1)} = h(y) (x_{ 2k+1)};
Ψ _{k} (y)(x_{ 2k+2}) = (1 − ϕ_{ k} (y))s_{ 0}(y) + ϕ_{ k} (y)h(y) (x_{ 2k+2});
Ψ _{k }(y)(x_{ 2k+3}) = (1 − ϕ_{ k }(y))δ(y) + h(y) (x_{ 2k+3}) ϕ_{ k}(y);
Ψ_{ k}(y)(x_{ 2k+4}) = s_{ 0 }(y);
Ψ_{ k}(y)(x_{ 2k+5}) = δ(y);
Ψ_{k}(y)(x_{2k+6}) = (1 − ϕ_{ k }(y))δ(y)α(y)(1) + ϕ_{ k}(y)s_{0}(y);
if m is odd and m ≥ 7, Ψ_{k}(y)(x_{ 2k+m)} = δ(y);
if m is even and m ≥ 8,
Ψ_{ k }(y)(x_{2k+m}) = δ(y)[(1 − ϕ_{k}(y))α(y)((m+2)/2− 3)
+ ϕ_{k}(y)α(y)(m/2− 3)];
Ψ_{k}(y)(x_{0}) = δ(y);
if x∈Y_{2k+i}, i ∈N_{ +},
Ψ_{k}(y)(x) = (1− φ_{i}(x))Ψ_{k}(y)(x_{2k+i)} + φ_{ i}(x)Ψ_{k}(y)(x_{2k+i+1)}.
Ψ_{k}(y)(x′_{2k+1}) = h(y) (x′_{2k+1});
Ψ_{k}(y)(x′_{2k+2}) = (1− ϕ_{k}(y))s_{∞}(y) + ϕ_{k}(y)h(y)(x′_{2k+2});
Ψ_{k}(y)(x′_{2k+3}) = h(y)(x′_{2k+3})ϕ_{k}(y);
Ψ_{k}(y)(x′_{2k+4}) = s_{∞}(y);
Ψ_{k}(y)(x′_{2k+5}) = 0;
Ψ_{k}(y)(x′_{2k+6}) = (1− ϕ_{k}(y))δ(y)V_{1}(y) + ϕ_{k}(y)s_{∞}(y);
if m is odd and m ≥ 7, Ψ_{k}(y)(x′_{2k+m}) = 0;
if m is even and m ≥ 8,
Ψ_{k} (y)(x′_{2k+m})= δ(y)[(1 − ϕ_{k}(y))V_{ ((m+2)/2−3) }(y)
+ ϕ_{k}(y)V_{ (m/2 −3)} (y)];
Ψ _{k }(y)(x_{∞}) = δ(y);
if x ∈Y′_{2k+i}, i∈N_{+},
Ψ_{k}(y)(x)=(1 −γ_{i} (x)) Ψ_{k}(y) (x′_{2k+i})
+ γ_{i}(x)Ψ_{k}(y)(x′_{2k+i+1}).
Step 3: Checking corresponding properties of maps Ψ_{k}′s to prepare for defining the map Ψ.
Claim 1: For every y∈D_{k}, Ψ_{k}(y) is well-defined on Y, continuous on Y\{x_{0}, x_{∞}} and upper semi-continuous at x_{0} and x_{∞}. Moreover, lim_{ x→x∞}Ψ_{k}(y)(x) = δ(y) =Ψ_{k}(y)(x_{∞}) if and only if lim_{n→∞}V_{n}(y)=1 if and only if y∈B. Therefore, if y ∈ D_{k}, then Ψ_{k}(y)∈USC(X) if and only if y∈B∩D_{k}. Ψ_{k}(y) is continuous at x_{0} if and only if lim_{n→∞}α(y)(n)=1 if and only if y∈A∩D_{k}.
The proof of Claim 1 is trivial.
Claim 2: For each k∈N_{+}, ↓Ψ_{k}:Y_{k} →↓USC(Y) is continuous.
It can be proved like the proof of [4, Proposition 1].
Claim 3: Ψ_{k}(y) = Ψ_{k+1}(y) for every y∈D_{k}∩D_{k+1}.
For every y∈D_{k}∩D _{k+1}, we have δ(y) =1/2^{k}.
Thus, ϕ_{k}(y) = 1 and ϕ_{k+1}(y) = 0.
If x∈(1/2, 1]∪∪_{i}_{∈{1,2,…,2k}}(Y_{ i }∪Y′_{ i}),
Ψ_{k}(y)(x) = h(y)(x)=Ψ_{k+1}(y)(x).
Ψ _{k }(y)(x_{ 2k+1)} = h(y) (x_{ 2k+1)} =Ψ_{k+1}(y)(x_{ 2k+1}).
Ψ _{k} (y)(x_{ 2k+2})= h(y) (x_{ 2k+2})= Ψ_{k+1}(y)(x_{ 2k+2}).
Ψ _{k} (y)(x_{ 2k+3})= h(y) (x_{ 2k+3})= Ψ_{k+1}(y)(x_{ 2k+3}).
Ψ_{ k}(y)(x_{ 2k+4}) = s_{ 0 }(y) = Ψ_{k+1}(y)(x_{ 2k+4}).
Ψ_{ k}(y)(x_{ 2k+5}) = δ(y)= Ψ_{k+1}(y)(x_{ 2k+5}).
Ψ_{k}(y)(x_{2k+6})= s_{ 0 }(y) =Ψ_{k+1}(y)(x_{2k+6}).
If m is odd and m ≥ 7,
Ψ_{k}(y)(x_{ 2k+m)} = δ(y)= Ψ_{k+1}(y)(x_{ 2k+m)}.
If m is even and m ≥ 8,
Ψ_{ k }(y)(x_{2k+m}) = δ(y)α(y)(m/2−3) =Ψ_{ k+1 }(y)(x_{2k+m}).
Ψ_{k}(y)(x_{0}) = δ(y)= Ψ_{k+1}(y)(x_{0}).
If x∈Y_{2k+i} = Y_{2(k+1)+(i-2)},i∈N_{ +},
Ψ_{k}(y)(x) = (1− φ_{i}(x))Ψ_{k}(y)(x_{2k+i)} + φ_{ i}(x)Ψ_{k}(y)(x_{2k+i+1)}
= (1− φ_{i}(x))Ψ_{k+1}(y)(x_{2(k+1)+(i-2))}
+ φ_{ i}(x)Ψ_{k+1}(y)(x_{2(k+1)+(i-2)+1)}
=Ψ_{k+1}(y)(x).
Ψ_{k}(y)(x′_{2k+1}) = h(y) (x′_{2k+1})= Ψ_{k+1}(y)(x′_{2k+1});
Ψ_{k}(y)(x′_{2k+2}) = h(y)(x′_{2k+2}) =Ψ_{k+1}(y)(x′_{2k+2}).
Ψ_{k}(y)(x′_{2k+3}) = h(y)(x′_{2k+3}) =Ψ_{k+1}(y)(x′_{2k+3}).
Ψ_{k}(y)(x′_{2k+4}) = s_{∞}(y) =Ψ_{k+1}(y)(x′_{2k+4}).
Ψ_{k}(y)(x′_{2k+5}) = 0 =Ψ_{k+1}(y)(x′_{2k+5}).
Ψ_{k}(y)(x′_{2k+6}) = (1− ϕ_{k}(y))δ(y)V_{1}(y) + ϕ_{k}(y)s_{∞}(y)
=Ψ_{k+1}(y)(x′_{2k+6}).
If m is odd and m ≥ 7, Ψ_{k}(y)(x′_{2k+m}) = 0=Ψ_{k+1}(y)(x′_{2k+m}).
If m is even and m ≥ 8,
Ψ_{k}(y)(x′_{2k+m})= δ(y)V_{(m/2 −3)} (y) =Ψ_{k+1}(y)(x′_{2k+m}).
Ψ_{k}(y)(x_{∞})=δ(y)=Ψ_{k+1}(y)(x_{∞}).
If x∈Y′_{2k+i}= Y′_{2(k+1)+(i-2)},i∈N_{+},
Ψ_{k}(y)(x)=(1−γ_{i} (x)) Ψ_{k}(y) (x′_{2k+i}) + γ_{i}(x)Ψ_{k}(y)(x′_{2k+i+1})
=(1−γ_{i} (x)) Ψ_{k+1}(y) (x′_{2k+i}) + γ_{i}(x)Ψ_{k+1}(y)(x′_{2k+i+1})
=(1−γ_{i} (x)) Ψ_{k+1}(y) (x′_{2(k+1)+(i-2)})
+ γ_{i}(x)Ψ_{k+1}(y)(x′_{2(k+1)+(i-2)+1})= Ψ_{k+1}(y)(x).
Step 4: Defining the map Ψ and checking its properties
Now we can define a map Ψ: D → USC(Y) as follows:
Then Ψ|_{K} = Φ|_{K}. Therefore, the following claims show that ↓Ψ is as required.
Claim 4: For every y ∈D, d_{H}(↓Ψ(y), ↓Φ(y)) ≤ 4δ(y) < ε.
It can be proved similarly as the proof of [4, Proposition 1].
Claim 5: ↓Ψ: D →↓USC(Y) is a Z-embedding.
Since D is compact, it suffices to prove the following three aspects.
(1). ↓Ψ is continuous.
It follows from Claims 2, 3 and 4.
(2). Ψ is an injection.
For any y1, y2∈D with y1y2, we shall show Ψ(y1) Ψ(y2). By the symmetry, we only consider the following three cases.
Case 1. y1, y2∈K. This fact is trivial.
Case 2. y1∈K and y2∈D\K. Then, by Claim 4,
d_{H}(↓Ψ(y2),↓Φ(y2)) ≤ 4δ(y2).
On the other hand, it follows from the definition of δ that
d_{H}(↓Φ(y1),↓Φ(y2)) ≥ d_{H}(↓Φ(K),↓Φ(y2)) ≥ 5δ(y2) > 0.
We conclude that Ψ(y1) = Φ(y1)Ψ(y2).
Case 3. y1, y2∈D\K. If Ψ(y1) = Ψ(y2), then
δ(y1)=Ψ(y1)(x_{∞})=Ψ(y2)(x_{∞})=δ(y2)0. Thus there exists k such that y1, y2∈D_{k} and ϕ_{k}(y1) = ϕ_{k}(y2).
On the other hand, for every i∈N_{+},
Ψ_{k}(y1)(x_{2k+i}) = Ψ(y1)(x_{2k+i})= Ψ(y2)(x_{2k+i}) = Ψ_{k}(y2)(x_{2k+i}).
Thus, if ϕ_{k}(y1)=ϕ_{k}(y2)=1, then α(y1)(m)=α(y2)(m) for every m∈N_{+ }by δ(y1)=δ(y2) and the definition of Ψ_{k}(y1)(x_{2k+i}) and Ψ_{k}(y2)(x_{2k+i}) for every even number i with i≥8. If ϕ(y1) =ϕ_{k}(y2)1, then α(y1)(m)=α(y2)(m) for every m∈N_{+ }by δ(y1)=δ(y2) and the definition of Ψ_{k}(y1)(x_{2k+i}) and Ψ_{k}(y2)(x_{2k+i}) for every even number i with i≥4. Thus α(y1) = α(y2) in both cases. Since α: D → Q_{u} is injective, we have that y1=y2 which contradicts with the assumption of y1y2.
(3). ↓Ψ(D) is a Z-set of ↓USC(Y).
↓Ψ(D) is compact by (1). Noticing that, for every y∈D_{k}, Ψ(y)(x′_{2k+4}) = Ψ_{k}(y)(x′_{2k+4})=0, it follows from Lemma 6 that ↓Ψ(D) is a Z-set.
Claim 6: Ψ^{−1}(USC(X))\K=B\K and Ψ^{−1}(C(X))\K=A\K.
In fact, for every y∈D\K, there exists k such that y∈D_{k}. It follows form the definition of Ψ and Claim 1. We are done.
4. Conclusion
For any bounded open interval X in the Euclidean space E^{1}, ↓USC(X) is homeomorphic to s [12] and ↓C(X) is homeomorphic to c_{0} [13]. However in this paper, it is proved that (↓USC(X), ↓C(X)) is pair homeomorphic to the pair which is not pair homeomorphic to the pair (s, c_{0}).
Acknowledgements
This work is supported by National Natural Science Foundation of China Youth Science Foundation Project (No. 11505042).
References