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When do the Banach lattices C([0,α],X) determine the ordinal intervals [0,α]?
J. Math. Anal. Appl. 443 (2016) 1362–1369
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
When do the Banach lattices C([0, α], X) determine the ordinal intervals [0, α]? Elói Medina Galego ∗ , Michael A. Rincón-Villamizar University of São Paulo, Department of Mathematics, IME, Rua do Matão 1010, São Paulo, Brazil
a r t i c l e
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Article history: Received 15 April 2016 Available online 15 June 2016 Submitted by Richard M. Aron Keywords: C(K, X) Banach lattices Kaplansky’s theorem Lattices without copies of c0 Cancellation law
a b s t r a c t Suppose that X is a Banach lattice containing no Banach sublattice isomorphic to c0 and consider the Banach lattices C([0, α], X) of X-valued continuous functions defined on the ordinal intervals [0, α], provided with the supremum norm. We prove that if the finite sums of X satisfy a certain geometric condition, then for all ordinals α and β the following assertions are equivalent: (1) The Banach lattices C([0, α], X) and C([0, β], X) are isomorphic. (2) The intervals of ordinals [0, α] and [0, β] are homeomorphic. As application of this cancellation law we obtain the complete classification, up to Banach lattices isomorphism, of certain C(K ⊕ [0, α], X) spaces, where K is an arbitrary perfect compact Hausdorff space. © 2016 Elsevier Inc. All rights reserved.
1. Introduction and the main theorem We will use the standard terminology and notation of Banach lattice and Banach space theory. For unexplained definitions and notation we refer to [1–3,19]. Let K be a compact Hausdorff space and let X be a Banach lattice. We denote by C(K, X) the Banach lattice of X-valued continuous functions on K, provided with the supremum norm. In the case where X = R we write C(K) instead of C(K, R). All Banach lattices considered in this paper are assumed to be real. Let α be an ordinal, by [0, α] denote the interval of ordinals {ξ : 0 ≤ ξ ≤ α} endowed with the order topology. When n is a finite ordinal we will denote C([0, n], X) by X n . In 1947 Kaplansky [14] stated that as a lattice alone C(K) determines the topology of the space K. More precisely, if T : C(K) → C(S) is a lattice isomorphism, that is T (f ∨ g) = T f ∨ T g for each f, g ∈ C(K), * Corresponding author. E-mail addresses: [email protected] (E. Medina Galego), [email protected] (M.A. Rincón-Villamizar). http://dx.doi.org/10.1016/j.jmaa.2016.06.022 0022-247X/© 2016 Elsevier Inc. All rights reserved.
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then K and S are homeomorphic. Then, in 1995 Behrends and Pelant obtained a vector-valued version of Kaplansky theorem. More specifically, in [5] they showed that there exists a compact Hausdorff space L such that the existence of a homeomorphism between the topological products K × L and S × L implies that K and S are homeomorphic for arbitrary compact Hausdorff spaces K and S. So, if we suppose that C(K, C(L)) is Banach lattice isomorphic to C(S, C(L)) for some compact Hausdorff spaces K and S then the following Banach lattices isomorphisms hold C(K × L) ∼ = C(K, C(L)) ∼ = C(S, C(L)) ∼ = C(S × L). Thus by the Kaplansky theorem K × L is homeomorphic to S × L and consequently K is isomorphic to S. But in general the Banach lattice C(K, X) do not determine the compact Hausdorff spaces K even when X is a two dimensional Banach space. Indeed, in 2006 Yamamoto and Yamashita [20] proved that there are non homeomorphic uncountable compact metric spaces K and S such that their topological sums K ⊕ K and S ⊕ S are homeomorphic [20, Theorem 1.1]. Then, we have the following Banach lattice isometric isomorphisms 2 2 C(K, l∞ )∼ ), = C(K ⊕ K) ∼ = C(S ⊕ S) ∼ = C(S, l∞ 2 where l∞ denotes the real space R2 endowed with the maximum norm. 2 ) determine the ordinal intervals [0, α]. Indeed, assume On the other hand, the Banach lattices C([0, α], l∞ 2 2 that C([0, α], l∞ ) is Banach lattice isomorphic to C([0, β], l∞ ) for some ordinals α and β. This means that the Banach lattices C([0, α2]) and C([0, β2]) are Banach lattice isomorphic. Hence the Kaplansky theorem implies that [0, α2] and [0, β2] are homeomorphic. Consequently, [0, α] and [0, β] are homeomorphic. In contrast with this, the Banach lattices C([0, α], X) do not determine the ordinal intervals [0, α] when X is the classical Banach lattice lp with 1 ≤ p < ∞. Indeed, denote by ω the first infinite ordinal. Then, we have the following Banach lattice isomorphisms.
C([0, ω], X) ∼ = C([0, ω], X 2 ) ∼ = C([0, ω], X) ⊕ C([0, ω], X) ∼ = C([0, ω2], X),
(1.1)
but of course [0, ω] is not homeomorphic to [0, ω2]. These results lead naturally to the following problem. Problem 1.1. When do the Banach lattices C([0, α], X) determine the ordinal intervals [0, α]? In other words, we are led to look for an extension of Kaplansky theorem for the family of C([0, α], X) spaces, where α is an arbitrary ordinal. In the affirmative case, notice that according to (1.1) X must somehow be very different from the classical Banach lattices. In fact, our answer to this question is as follows. Theorem 1.2. Suppose that X is a Banach lattice containing no Banach sublattice isomorphic to c0 . If for any positive integer n, there is no Banach lattice isomorphism from X n+1 into X n , then for all infinite ordinals α and β, the following assertions are equivalent: (a) C([0, α], X) and C([0, β], X) are Banach lattice isomorphic. (b) [0, α] and [0, β] are homeomorphic. Remark 1.3. Although among the “natural” Banach lattices there is no infinite dimensional Banach lattice that satisfies the hypotheses of Theorem 1.2, there are 2ℵ0 isomorphically different separable infinite dimensional Banach lattices X satisfying these hypotheses. Indeed, let (pi )∞ i=1 be a fixed strictly decreasing
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sequence of real numbers greater than 2. Let p ∈ (1, limi pi ] and a > 0 be fixed. For a real number x, let [x] denote the greatest integer less than or equal to x. By [9, Main result], there is a sequence (ni)∞ i=1 such that the uniformly convex Banach lattice
Xa =
∞ i=1
i] lp[an i
p
satisfies the hypotheses of Theorem 1.2. Moreover, if 0 < b < a then Xa is not Banach isomorphic to a subspace of Xb [8, Proposition 2]. The homeomorphic classification of the ordinal intervals [0, α] can be found in [4, Corollary 3]. On the other hand, by the classical theorem of Mazurkiewicz and Sierpińki [15] we know that an infinite countable compact metric space is homeomorphic to an ordinal interval [0, α], where ω ≤ α < ω1 . Here ω1 the first uncountable ordinal. Thus, Theorem 1.2 provides the Banach lattice classification of the C(K, X) spaces for countable compact metric spaces K, whenever X contains no Banach lattice copy of c0 and X n contains no Banach lattice copy of X n+1 , for every positive integers n. As a consequence of Theorem 1.2 this classification can be extended for another family of compact metric spaces. Indeed, let us recall that a subset of a topological space is a perfect set if is closed set and contains no isolated points. Moreover, a well-known Cantor–Bendixson theorem states that every compact metric space can be decomposed uniquely into the union of two disjoint sets, one perfect and other countable [19, Theorem 8.5.2]. So, the following corollary which is at same times an extension of Theorem 1.2 (the case K = ∅) covers the cases where the countable sets of the above-mentioned decomposition are closed. Corollary 1.4. Suppose that X is a Banach lattice containing no Banach sublattice isomorphic to c0 and such that for any positive integer n, there is no Banach lattice isomorphism from X n+1 into X n . Then for all perfect compact space K and ordinals α and β, the following assertions are equivalent: (a) C(K ⊕ [0, α], X) and C(K ⊕ [0, β], X) are Banach lattice isomorphic. (b) [0, α] and [0, β] are homeomorphic. Remark 1.5. The hypothesis of Banach lattice isomorphism can not be omitted in the statement of Corollary 1.4 even for the C(2ℵ0 ⊕[0, α]) spaces, where 2ℵ0 is the Cantor cube and α is countable. Indeed, according to the classical Milutin’s theorem (see [16] or [17]), C(2ℵ0 ⊕ [0, ω]) is Banach isomorphic to C(2ℵ0 ⊕ [0, ω2]). However [0, ω] is not homeomorphic to [0, ω2]. We refer to [10] for a Banach space isomorphic classification of the C(K ⊕ [0, α]) spaces, where m is a cardinal number and K = 2m is the product of m copies of the two-point space 2, endowed with the product topology. Remark 1.6. The statement of Corollary 1.4 does not remain true if we change the topological sums K ⊕[0, α] and K ⊕[0, β] by the topological products K ×[0, α] and K ×[0, β] respectively. Indeed, a theorem of Brouwer [6] characterizes 2ℵ0 as the unique totally disconnected, compact metric space without isolated points. In particular, 2ℵ0 × [0, ω] and 2ℵ0 × [0, ω2] are homeomorphic. Consequently, C(2ℵ0 × [0, ω]) is Banach lattice isomorphic to C(2ℵ0 × [0, ω2]) and once again it is enough to observe that [0, ω] is not homeomorphic to [0, ω2]. A Banach space isomorphic classification of the C(2m × [0, α]) spaces can be found in [11]. That the statement (b) of Theorem 1.2 implies the statement (a) of Theorem 1.4 is trivial. Thus, from now on it is our task to prove that the statement (a) of Theorem 1.2 implies the statement (b) of Theorem 1.2, see section 3. The proof of Corollary 1.4 will be done in the section 4.
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2. Some auxiliary results In this section we will prove some auxiliary statements in order to prove Theorem 1.2. For a Banach space X, we denote by SX the unit sphere of X. Recall that an operator T between Banach lattices is called positive if T x ≥ 0 when x ≥ 0 [19, p. 60]. If S is a topological space, we denote the closure of a subset A of S by A. The derivative is the space S (1) obtained by deleting from S its isolated points. The αth derivatives S (α) are defined recursively setting S (0) = S, S (α+1) = (S (α) )(1) and S (α) = β<α S (β) when α is an ordinal limit. Recall that a compact space S is scattered if S (α) = ∅ for some ordinal α. In this case, the height of S, denoted by ht(S), is the minimal α satisfying S (α) = ∅. Lemma 2.1. Let X be a Banach lattice containing no copy of c0 and K and S locally compact Hausdorff spaces. Suppose that T : C0 (K, X) → C0 (S, X) is a positive into isomorphism satisfying T −1 ≤ 1. Let h ∈ C0 (K, X) be such that h ≥ 0 and h ≤ 1. Fix 0 < r < 1. If α is an ordinal such that h|K (α) > r then {y ∈ S : T g(y) ≥ r} ∩ S (α) = ∅. h≤g, g≤1
Proof. We use transfinite induction on α. Let α = 0 and set A= {y ∈ S : T f (y) ≥ r}. h≤f ≤1
Positiveness of T implies that A ⊃ {y ∈ S : T h(y) ≥ r} = ∅, because T h ≥ h > r. This proves the case α = 0. The remainder of the proof follows as in [12, Proposition 3.1]. 2 Corollary 2.2. Let X be a Banach lattice containing no copy of c0 and K and S locally compact Hausdorff spaces. Suppose that T is a Banach lattice isomorphism from C0 (K, X) into C0 (S, X) satisfying T −1 = 1. Fix 0 < r < 1. If g ∈ C0 (K, X) and for some ordinal α we have g = g|K (α) then T g|S (α) ≥ r g . Proof. Let h = |g| and suppose that g = g|K (α) = 1. Note that g = h|K (α) = 1 > r. By Lemma 2.1 there is a y ∈ S (α) such that T h(y) ≥ r. Since T is a Banach lattice isomorphism, we have T h = |T g|. Thus T h(y) = |T g|(y) = |T g(y)|. Moreover, |T g(y)| = T g(y) . Then, we deduce that T g|S (α) ≥ r.
2
According to [7] if R is a set-valued map from K to S then for any set F ⊂ S the image inverse of F by R is defined as R−1 (F ) = {x ∈ K : R(x) ∩ F = ∅}. If F = {y} is a singleton, we write R−1 (y) instead of R−1 ({y}).
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Proposition 2.3. Let K and S be locally compact Hausdorff spaces and X a Banach lattice containing no copy of c0 . If T is a positive isomorphism from C0 (K, X) into C0 (S, X) then for each ordinal α we have (1) if S (α) is finite then K (α) is also finite. (2) if S (α) is infinite then |K (α) | ≤ |S (α) |. Moreover, in the case where K and S are compact Hausdorff, if S is scattered then so is K and ht(K) ≤ ht(S). Proof. We assume that T −1 = 1. Take 0 < r < 1 and e ∈ SX with e ≥ 0. Let x ∈ K be given and consider the set-valued map Ψ(x) defined by Ψ(x) = {y ∈ S : T (f · e)(y) ≥ r for all f ∈ Fx }, where Fx = {f ∈ C0 (K) : 0 ≤ f ≤ 1 and f (x) > r}. Claim 2.4. For each ordinal α such that K (α) = ∅, we have Ψ(x) ∩ S (α) = ∅ whenever x ∈ K (α) . Indeed, let α be an ordinal such that K (α) = ∅ and pick x ∈ K (α) . For f ∈ Fx , let Ψf = {y ∈ S : T (f · e)(y) ≥ r}. Consider the collection {Ψf ∩ S (α) : f ∈ Fx }. By compactness of Ψf , it suffices to prove that this collection has the finite intersection property. If f1 , . . . , fn ∈ Fx , setting h(t) = (min1≤j≤n fj (t)) · e for every t ∈ K, we see that h ∈ C0 (K, X), h ≥ 0, h ≤ 1 and h|K (α) > r. Therefore, it follows from Theorem 2.1 that n j=1
(Ψfj ∩ S (α) ) ⊃
{y ∈ S : T g(y) ≥ r} ∩ S (α) = ∅.
h≤g,g≤1
So, Ψ(x) ∩ S (α) =
(Ψf ∩ S (α) ) = ∅,
f ∈Fx
and the claim is proved. Next, for each y ∈ S consider the sets Ψ−1 (y). Then by Claim 2.4 K (α) ⊂
Ψ−1 (y)
(2.1)
y∈S (α)
To end the proof, we show that Ψ−1 (y) is finite for all y ∈ S. Indeed, in this case equation (2.1) proves that K (α) is finite in the case where S (α) is finite. Now, if S (α) is infinite we obtain |K (α) | ≤ Ψ−1 (y) ≤ |S (α) |. y∈S (α) Suppose that there is a y ∈ S for which Ψ−1 (y) is infinite. Then there exists a sequence of disjoint open sets {Un : n ∈ N} such that
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Un ∩ Ψ−1 (y) = ∅, for all n ∈ N. Take xn ∈ Un ∩ Ψ−1 (y) for every n ∈ N. By Urysohn lemma [18, p. 39], exists fn ∈ C0 (K) satisfying 0 ≤ fn ≤ 1, fn (xn ) = 1 and fn (K \ Un ) = 0. Note that fn ∈ Fxn for all n ∈ N. Thus T (fn · e)(y) ≥ r for every n ∈ N, that is, inf{ T (fn · e)(y) : n ∈ N} > 0.
(2.2)
∞ It is not difficult to show that n=1 T (fn · e)(y) is a weakly unconditionally convergent series. Moreover, by ∞ (2.2), n=1 T (fn ·e)(y) is not unconditionally convergent. Then, according to [2, Theorem 2.4.11] X contains a linear subspace which is Banach isomorphic to c0 . Therefore Ψ−1 (y) is finite for each y ∈ S. Finally observe that if K and S are compact Hausdorff spaces with S scattered then equation (2.1) implies that K is also scattered and ht(K) ≤ ht(S). 2 3. On Banach lattice isomorphisms between C([0, α], X) spaces In this section we prove that the statement (a) of Theorem 1.2 implies the statement (b) of Theorem 1.2. Firstly, notice that by a famous result [3, Theorem 14.12] a Banach lattice contains a sublattice isomorphic to c0 if and only if it has a Banach subspace isomorphic to c0 . So, it suffices to prove the statements of Theorem 1.2 for Banach lattices without copy of c0 . Theorem 3.1. Let X be a Banach lattice containing no copy of c0 such that for any positive integer n, there is no Banach lattice isomorphism from X n+1 into X n . Let K and S be ordinal intervals. If C(K, X) is Banach lattice isomorphic to C(S, X) then K and S are homeomorphic. Proof. Let T : C(K, X) → C(S, X) be a Banach lattice isomorphism. If K and S are finite, we easily infer from our hypotheses that K and S are homeomorphic. Then, suppose that K and S are infinite. Without loss of generality assume that T −1 = 1. Note that K and S are scattered [19, Proposition 8.6.7]. By applying Proposition 2.3 twice, we obtain ht(K) = ht(S). On the other hand, by [19, Proposition 8.6.5] we may assume that K = [1, ω α n] and S = [1, ω β m] for some ordinals α, β, m and n different from zero, where m, n are finite numbers. Thus α + 1 = ht(K) = ht(S) = β + 1 [19, Proposition 8.6.6], and so α = β. Now, we have to show that m = n. Fix 0 < r < 1 and assume that m < n. Then K (α) = {ω α , . . . , ω α n}. Let K0 = {ω α , . . . , ω α (m + 1)}. Take a sequence {U1 , . . . , Um+1 } of disjoint open subsets of K such that ω α j ∈ Uj for each 1 ≤ j ≤ m + 1. By Urysohn lemma, for each 1 ≤ j ≤ m + 1, there is hj ∈ C(K) satisfying 0 ≤ hj ≤ 1, hj (K \ Uj ) = 0 and hj (ω α j) = 1. Let L : C(K0 , X) → C(K, X) be defined by L(f )(x) =
m+1
hi (x)f (ω α i),
i=1
for every f ∈ C(K0 , X). Note that L is a Banach lattice isometry, that is L(f ) = f and L(f ∨ g) = L(f ) ∨ L(g) for all f, g ∈ C(K0 , X). Let R : C(S, X) → C(S (α) , X) be the natural restriction operator. For all f ∈ C(K0 , X) we have f = L(f ) = L(f )|K (α) . Hence, it follows from Corollary 2.2 that R ◦ T ◦ L(f ) ≥ r f .
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Thus R ◦ T ◦ L is a Banach lattice isomorphism from X m+1 into X m which contradicts our assumption. Therefore m < n is impossible. The same argument shows that n < m is false. So, m = n and we are done. 2 4. On Banach lattice isomorphisms between C(K ⊕ [0, α], X) spaces The aim of this section is to prove the nontrivial part of Corollary 1.4, that is Corollary 4.2. We start with a vector-valued extension of [13, Lemma 3.2]. To state it, recall that the perfect kernel of a topological space S, denoted by P (S), is defined by P (S) = α≥1 S (α) . Lemma 4.1. Let K and S be locally compact Hausdorff spaces and X a Banach lattice containing no copy of c0 . If T is a Banach lattice isomorphism from C0 (K, X) onto C0 (S, X) then there exists a Banach lattice isomorphism Tˆ from C0 (S \ P (S), X) into C0 (K \ P (K), X). Proof. Let ψ : C0 (S \ P (S), X) → C0 (S, X) be defined by
ψ(f )(y) =
f (y),
if y ∈ S \ P (S)
0,
if y ∈ P (S),
for every f ∈ C0 (S \ P (S), X). Then ψ is an isometry and ψ(f ∨ g) = ψ(f ) ∨ ψ(g) for each f, g ∈ C0 (S \ P (S), X). Next, assume that T −1 = 1 and fix 0 < r < 1. Let f ∈ C0 (S \ P (S), X), f = 0 and consider the function g=
T −1 (ψ(f )) . T −1 (ψ(f ))
We claim that g|P (K) < 1. Indeed, if g|P (K) ≥ 1 then g|K (α) = g = 1 for each ordinal α. According to Lemma 2.1 we know that {y ∈ S : T g(y) ≥ r} ∩ S (α) = ∅, for every ordinal α. Therefore, by compactness we have {y ∈ S : T g(y) ≥ r} ∩ P (S) = ∅, which is impossible. Thereby if R : C0 (K, X) → C0 (K \ P (K), X) denotes the natural restriction operator then Rg = 1, that is R(T −1 (ψ(f ))) = T −1 (ψ(f )) . Define Tˆ : C0 (S \ P (S), X) → C0 (K \ P (K), X) by Tˆ (f ) = (R ◦ T −1 ◦ ψ)(f ). Then, for all f, g ∈ C0 (S \ P (S), X) we have Tˆ(f ∨ g) = Tˆ (f ) ∨ Tˆ (g) and 1 f ≤ Tˆ (f ) ≤ f . T Thus, Tˆ is a Banach lattice isomorphism. 2
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Corollary 4.2. Let X be a Banach lattice containing no copy of c0 such that for any positive integer n, there is no Banach lattice isomorphism from X n+1 into X n . Let K and S be ordinal intervals and L a perfect compact Hausdorff space. If C(K ⊕ L, X) is Banach lattice isomorphic to C(S ⊕ L, X) then K and S are homeomorphic. Proof. Since K and S are scattered it follows by [19, Proposition 8.6.7] that P (K⊕L) = L and P (S⊕L) = L. So, by Lemma 4.1 there exist into Banach lattice isomorphisms T1 : C(K, X) → C(S, X) and T2 : C(S, X) → C(K, X). If K and S are finite, we are done. If not, once again by [19, Proposition 8.6.5] we may assume that K = [1, ω α n] and S = [1, ω β m] for some ordinals α, β, m and n different from zero, where m, n are finite numbers. By applying Proposition 2.3 twice, we have ht(K) = ht(S). According to [19, Proposition 8.6.6] we see that α + 1 = ht(K) = ht(S) = β + 1. Thus α = β. Now, proceeding as in the proof of Theorem 3.1 we conclude that m = n. 2 References [1] Y.A. Abramovich, C.D. Aliprantis, Positive operators, in: Handbook of the Geometry of Banach Spaces, Vol. I, NorthHolland, Amsterdam, 2001, pp. 85–122. [2] F. Albiac, N.J. Kalton, Topics in Banach Space Theory, Grad. Texts in Math., vol. 233, Springer, New York, 2006. [3] C.D. Aliprantis, O. Burkinshaw, Positive Operators, Springer, Dordrecht, 2006. [4] J.W. Baker, Compact spaces homeomorphic to a ray of ordinals, Fund. Math. 76 (1) (1972) 19–27. [5] E. Behrends, J. Pelant, The cancellation law for compact Hausdorff spaces and vector-valued Banach–Stone theorems, Arch. Math. (Basel) 64 (4) (1995) 341–343. [6] L. Brouwer, On the structure of perfect sets of points, Proc. Akad. Amst. 12 (1910) 785–794. [7] R.S. Burachik, A.N. Iusem, Set-Valued Mappings and Enlargements of Monotone Operators, Optim. Appl., vol. 8, Springer, New York, 2008. [8] P.G. Casazza, C.A. Kottman, B.L. Lin, Remarks on Figiel’s reflexive Banach spaces not isomorphic to their square, J. Math. Anal. Appl. 63 (3) (1978) 750–752. [9] T. Figiel, An example of infinite dimensional reflexive Banach space non-isomorphic to its Cartesian square, Studia Math. 42 (1972) 295–306. [10] E.M. Galego, On isomorphism classes of C(2m ⊕ [0, α]) spaces, Fund. Math. 204 (1) (2009) 87–95. [11] E.M. Galego, An isomorphic classification of C(2m × [0, α]) spaces, Bull. Pol. Acad. Sci. Math. 57 (3–4) (2009) 279–287. [12] E.M. Galego, M.A. Rincón-Villamizar, Weak forms of Banach–Stone theorem for C0 (K, X) spaces via the αth derivatives of K, Bull. Sci. Math. 139 (2015) 880–891. [13] Y. Gordon, On the distance coefficient between isomorphic function spaces, Israel J. Math. 8 (1970) 391–397. [14] I. Kaplansky, Lattices of continuous functions, Bull. Amer. Math. Soc. (N.S.) 53 (1947) 617–623. [15] S. Mazurkiewicz, W. Sierpińki, Contributions à la topologie des ensembles dénombrables, Fund. Math. 1 (1920) 513–522. [16] A.A. Milutin, Isomorphisms of spaces of continuous functions on compacts of power continuum, Tieoria Func. (Kharkov) 2 (1966) 150–156 (in Russian). [17] H.P. Rosenthal, The Banach space C(K), in: Handbook of the Geometry of Banach Spaces, North-Holland Publishing Co., Amsterdam, 2001, pp. 1547–1602. [18] W. Rudin, Real and Complex Analysis, third edition, McGraw-Hill Book Co., New York, 1987. [19] Z. Semadeni, Banach Spaces of Continuous Functions. Vol. 1, Monogr. Mat., vol. 55, PWN-Polish Sci. Publ., Warszawa, 1971. [20] S. Yamamoto, A. Yamashita, A counterexample related to topological sums, Proc. Amer. Math. Soc. 134 (12) (2006) 3715–3719.
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
When do the Banach lattices C([0, α], X) determine the ordinal intervals [0, α]? Elói Medina Galego ∗ , Michael A. Rincón-Villamizar University of São Paulo, Department of Mathematics, IME, Rua do Matão 1010, São Paulo, Brazil
a r t i c l e
i n f o
Article history: Received 15 April 2016 Available online 15 June 2016 Submitted by Richard M. Aron Keywords: C(K, X) Banach lattices Kaplansky’s theorem Lattices without copies of c0 Cancellation law
a b s t r a c t Suppose that X is a Banach lattice containing no Banach sublattice isomorphic to c0 and consider the Banach lattices C([0, α], X) of X-valued continuous functions defined on the ordinal intervals [0, α], provided with the supremum norm. We prove that if the finite sums of X satisfy a certain geometric condition, then for all ordinals α and β the following assertions are equivalent: (1) The Banach lattices C([0, α], X) and C([0, β], X) are isomorphic. (2) The intervals of ordinals [0, α] and [0, β] are homeomorphic. As application of this cancellation law we obtain the complete classification, up to Banach lattices isomorphism, of certain C(K ⊕ [0, α], X) spaces, where K is an arbitrary perfect compact Hausdorff space. © 2016 Elsevier Inc. All rights reserved.
1. Introduction and the main theorem We will use the standard terminology and notation of Banach lattice and Banach space theory. For unexplained definitions and notation we refer to [1–3,19]. Let K be a compact Hausdorff space and let X be a Banach lattice. We denote by C(K, X) the Banach lattice of X-valued continuous functions on K, provided with the supremum norm. In the case where X = R we write C(K) instead of C(K, R). All Banach lattices considered in this paper are assumed to be real. Let α be an ordinal, by [0, α] denote the interval of ordinals {ξ : 0 ≤ ξ ≤ α} endowed with the order topology. When n is a finite ordinal we will denote C([0, n], X) by X n . In 1947 Kaplansky [14] stated that as a lattice alone C(K) determines the topology of the space K. More precisely, if T : C(K) → C(S) is a lattice isomorphism, that is T (f ∨ g) = T f ∨ T g for each f, g ∈ C(K), * Corresponding author. E-mail addresses: [email protected] (E. Medina Galego), [email protected] (M.A. Rincón-Villamizar). http://dx.doi.org/10.1016/j.jmaa.2016.06.022 0022-247X/© 2016 Elsevier Inc. All rights reserved.
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then K and S are homeomorphic. Then, in 1995 Behrends and Pelant obtained a vector-valued version of Kaplansky theorem. More specifically, in [5] they showed that there exists a compact Hausdorff space L such that the existence of a homeomorphism between the topological products K × L and S × L implies that K and S are homeomorphic for arbitrary compact Hausdorff spaces K and S. So, if we suppose that C(K, C(L)) is Banach lattice isomorphic to C(S, C(L)) for some compact Hausdorff spaces K and S then the following Banach lattices isomorphisms hold C(K × L) ∼ = C(K, C(L)) ∼ = C(S, C(L)) ∼ = C(S × L). Thus by the Kaplansky theorem K × L is homeomorphic to S × L and consequently K is isomorphic to S. But in general the Banach lattice C(K, X) do not determine the compact Hausdorff spaces K even when X is a two dimensional Banach space. Indeed, in 2006 Yamamoto and Yamashita [20] proved that there are non homeomorphic uncountable compact metric spaces K and S such that their topological sums K ⊕ K and S ⊕ S are homeomorphic [20, Theorem 1.1]. Then, we have the following Banach lattice isometric isomorphisms 2 2 C(K, l∞ )∼ ), = C(K ⊕ K) ∼ = C(S ⊕ S) ∼ = C(S, l∞ 2 where l∞ denotes the real space R2 endowed with the maximum norm. 2 ) determine the ordinal intervals [0, α]. Indeed, assume On the other hand, the Banach lattices C([0, α], l∞ 2 2 that C([0, α], l∞ ) is Banach lattice isomorphic to C([0, β], l∞ ) for some ordinals α and β. This means that the Banach lattices C([0, α2]) and C([0, β2]) are Banach lattice isomorphic. Hence the Kaplansky theorem implies that [0, α2] and [0, β2] are homeomorphic. Consequently, [0, α] and [0, β] are homeomorphic. In contrast with this, the Banach lattices C([0, α], X) do not determine the ordinal intervals [0, α] when X is the classical Banach lattice lp with 1 ≤ p < ∞. Indeed, denote by ω the first infinite ordinal. Then, we have the following Banach lattice isomorphisms.
C([0, ω], X) ∼ = C([0, ω], X 2 ) ∼ = C([0, ω], X) ⊕ C([0, ω], X) ∼ = C([0, ω2], X),
(1.1)
but of course [0, ω] is not homeomorphic to [0, ω2]. These results lead naturally to the following problem. Problem 1.1. When do the Banach lattices C([0, α], X) determine the ordinal intervals [0, α]? In other words, we are led to look for an extension of Kaplansky theorem for the family of C([0, α], X) spaces, where α is an arbitrary ordinal. In the affirmative case, notice that according to (1.1) X must somehow be very different from the classical Banach lattices. In fact, our answer to this question is as follows. Theorem 1.2. Suppose that X is a Banach lattice containing no Banach sublattice isomorphic to c0 . If for any positive integer n, there is no Banach lattice isomorphism from X n+1 into X n , then for all infinite ordinals α and β, the following assertions are equivalent: (a) C([0, α], X) and C([0, β], X) are Banach lattice isomorphic. (b) [0, α] and [0, β] are homeomorphic. Remark 1.3. Although among the “natural” Banach lattices there is no infinite dimensional Banach lattice that satisfies the hypotheses of Theorem 1.2, there are 2ℵ0 isomorphically different separable infinite dimensional Banach lattices X satisfying these hypotheses. Indeed, let (pi )∞ i=1 be a fixed strictly decreasing
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sequence of real numbers greater than 2. Let p ∈ (1, limi pi ] and a > 0 be fixed. For a real number x, let [x] denote the greatest integer less than or equal to x. By [9, Main result], there is a sequence (ni)∞ i=1 such that the uniformly convex Banach lattice
Xa =
∞ i=1
i] lp[an i
p
satisfies the hypotheses of Theorem 1.2. Moreover, if 0 < b < a then Xa is not Banach isomorphic to a subspace of Xb [8, Proposition 2]. The homeomorphic classification of the ordinal intervals [0, α] can be found in [4, Corollary 3]. On the other hand, by the classical theorem of Mazurkiewicz and Sierpińki [15] we know that an infinite countable compact metric space is homeomorphic to an ordinal interval [0, α], where ω ≤ α < ω1 . Here ω1 the first uncountable ordinal. Thus, Theorem 1.2 provides the Banach lattice classification of the C(K, X) spaces for countable compact metric spaces K, whenever X contains no Banach lattice copy of c0 and X n contains no Banach lattice copy of X n+1 , for every positive integers n. As a consequence of Theorem 1.2 this classification can be extended for another family of compact metric spaces. Indeed, let us recall that a subset of a topological space is a perfect set if is closed set and contains no isolated points. Moreover, a well-known Cantor–Bendixson theorem states that every compact metric space can be decomposed uniquely into the union of two disjoint sets, one perfect and other countable [19, Theorem 8.5.2]. So, the following corollary which is at same times an extension of Theorem 1.2 (the case K = ∅) covers the cases where the countable sets of the above-mentioned decomposition are closed. Corollary 1.4. Suppose that X is a Banach lattice containing no Banach sublattice isomorphic to c0 and such that for any positive integer n, there is no Banach lattice isomorphism from X n+1 into X n . Then for all perfect compact space K and ordinals α and β, the following assertions are equivalent: (a) C(K ⊕ [0, α], X) and C(K ⊕ [0, β], X) are Banach lattice isomorphic. (b) [0, α] and [0, β] are homeomorphic. Remark 1.5. The hypothesis of Banach lattice isomorphism can not be omitted in the statement of Corollary 1.4 even for the C(2ℵ0 ⊕[0, α]) spaces, where 2ℵ0 is the Cantor cube and α is countable. Indeed, according to the classical Milutin’s theorem (see [16] or [17]), C(2ℵ0 ⊕ [0, ω]) is Banach isomorphic to C(2ℵ0 ⊕ [0, ω2]). However [0, ω] is not homeomorphic to [0, ω2]. We refer to [10] for a Banach space isomorphic classification of the C(K ⊕ [0, α]) spaces, where m is a cardinal number and K = 2m is the product of m copies of the two-point space 2, endowed with the product topology. Remark 1.6. The statement of Corollary 1.4 does not remain true if we change the topological sums K ⊕[0, α] and K ⊕[0, β] by the topological products K ×[0, α] and K ×[0, β] respectively. Indeed, a theorem of Brouwer [6] characterizes 2ℵ0 as the unique totally disconnected, compact metric space without isolated points. In particular, 2ℵ0 × [0, ω] and 2ℵ0 × [0, ω2] are homeomorphic. Consequently, C(2ℵ0 × [0, ω]) is Banach lattice isomorphic to C(2ℵ0 × [0, ω2]) and once again it is enough to observe that [0, ω] is not homeomorphic to [0, ω2]. A Banach space isomorphic classification of the C(2m × [0, α]) spaces can be found in [11]. That the statement (b) of Theorem 1.2 implies the statement (a) of Theorem 1.4 is trivial. Thus, from now on it is our task to prove that the statement (a) of Theorem 1.2 implies the statement (b) of Theorem 1.2, see section 3. The proof of Corollary 1.4 will be done in the section 4.
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2. Some auxiliary results In this section we will prove some auxiliary statements in order to prove Theorem 1.2. For a Banach space X, we denote by SX the unit sphere of X. Recall that an operator T between Banach lattices is called positive if T x ≥ 0 when x ≥ 0 [19, p. 60]. If S is a topological space, we denote the closure of a subset A of S by A. The derivative is the space S (1) obtained by deleting from S its isolated points. The αth derivatives S (α) are defined recursively setting S (0) = S, S (α+1) = (S (α) )(1) and S (α) = β<α S (β) when α is an ordinal limit. Recall that a compact space S is scattered if S (α) = ∅ for some ordinal α. In this case, the height of S, denoted by ht(S), is the minimal α satisfying S (α) = ∅. Lemma 2.1. Let X be a Banach lattice containing no copy of c0 and K and S locally compact Hausdorff spaces. Suppose that T : C0 (K, X) → C0 (S, X) is a positive into isomorphism satisfying T −1 ≤ 1. Let h ∈ C0 (K, X) be such that h ≥ 0 and h ≤ 1. Fix 0 < r < 1. If α is an ordinal such that h|K (α) > r then {y ∈ S : T g(y) ≥ r} ∩ S (α) = ∅. h≤g, g≤1
Proof. We use transfinite induction on α. Let α = 0 and set A= {y ∈ S : T f (y) ≥ r}. h≤f ≤1
Positiveness of T implies that A ⊃ {y ∈ S : T h(y) ≥ r} = ∅, because T h ≥ h > r. This proves the case α = 0. The remainder of the proof follows as in [12, Proposition 3.1]. 2 Corollary 2.2. Let X be a Banach lattice containing no copy of c0 and K and S locally compact Hausdorff spaces. Suppose that T is a Banach lattice isomorphism from C0 (K, X) into C0 (S, X) satisfying T −1 = 1. Fix 0 < r < 1. If g ∈ C0 (K, X) and for some ordinal α we have g = g|K (α) then T g|S (α) ≥ r g . Proof. Let h = |g| and suppose that g = g|K (α) = 1. Note that g = h|K (α) = 1 > r. By Lemma 2.1 there is a y ∈ S (α) such that T h(y) ≥ r. Since T is a Banach lattice isomorphism, we have T h = |T g|. Thus T h(y) = |T g|(y) = |T g(y)|. Moreover, |T g(y)| = T g(y) . Then, we deduce that T g|S (α) ≥ r.
2
According to [7] if R is a set-valued map from K to S then for any set F ⊂ S the image inverse of F by R is defined as R−1 (F ) = {x ∈ K : R(x) ∩ F = ∅}. If F = {y} is a singleton, we write R−1 (y) instead of R−1 ({y}).
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Proposition 2.3. Let K and S be locally compact Hausdorff spaces and X a Banach lattice containing no copy of c0 . If T is a positive isomorphism from C0 (K, X) into C0 (S, X) then for each ordinal α we have (1) if S (α) is finite then K (α) is also finite. (2) if S (α) is infinite then |K (α) | ≤ |S (α) |. Moreover, in the case where K and S are compact Hausdorff, if S is scattered then so is K and ht(K) ≤ ht(S). Proof. We assume that T −1 = 1. Take 0 < r < 1 and e ∈ SX with e ≥ 0. Let x ∈ K be given and consider the set-valued map Ψ(x) defined by Ψ(x) = {y ∈ S : T (f · e)(y) ≥ r for all f ∈ Fx }, where Fx = {f ∈ C0 (K) : 0 ≤ f ≤ 1 and f (x) > r}. Claim 2.4. For each ordinal α such that K (α) = ∅, we have Ψ(x) ∩ S (α) = ∅ whenever x ∈ K (α) . Indeed, let α be an ordinal such that K (α) = ∅ and pick x ∈ K (α) . For f ∈ Fx , let Ψf = {y ∈ S : T (f · e)(y) ≥ r}. Consider the collection {Ψf ∩ S (α) : f ∈ Fx }. By compactness of Ψf , it suffices to prove that this collection has the finite intersection property. If f1 , . . . , fn ∈ Fx , setting h(t) = (min1≤j≤n fj (t)) · e for every t ∈ K, we see that h ∈ C0 (K, X), h ≥ 0, h ≤ 1 and h|K (α) > r. Therefore, it follows from Theorem 2.1 that n j=1
(Ψfj ∩ S (α) ) ⊃
{y ∈ S : T g(y) ≥ r} ∩ S (α) = ∅.
h≤g,g≤1
So, Ψ(x) ∩ S (α) =
(Ψf ∩ S (α) ) = ∅,
f ∈Fx
and the claim is proved. Next, for each y ∈ S consider the sets Ψ−1 (y). Then by Claim 2.4 K (α) ⊂
Ψ−1 (y)
(2.1)
y∈S (α)
To end the proof, we show that Ψ−1 (y) is finite for all y ∈ S. Indeed, in this case equation (2.1) proves that K (α) is finite in the case where S (α) is finite. Now, if S (α) is infinite we obtain |K (α) | ≤ Ψ−1 (y) ≤ |S (α) |. y∈S (α) Suppose that there is a y ∈ S for which Ψ−1 (y) is infinite. Then there exists a sequence of disjoint open sets {Un : n ∈ N} such that
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Un ∩ Ψ−1 (y) = ∅, for all n ∈ N. Take xn ∈ Un ∩ Ψ−1 (y) for every n ∈ N. By Urysohn lemma [18, p. 39], exists fn ∈ C0 (K) satisfying 0 ≤ fn ≤ 1, fn (xn ) = 1 and fn (K \ Un ) = 0. Note that fn ∈ Fxn for all n ∈ N. Thus T (fn · e)(y) ≥ r for every n ∈ N, that is, inf{ T (fn · e)(y) : n ∈ N} > 0.
(2.2)
∞ It is not difficult to show that n=1 T (fn · e)(y) is a weakly unconditionally convergent series. Moreover, by ∞ (2.2), n=1 T (fn ·e)(y) is not unconditionally convergent. Then, according to [2, Theorem 2.4.11] X contains a linear subspace which is Banach isomorphic to c0 . Therefore Ψ−1 (y) is finite for each y ∈ S. Finally observe that if K and S are compact Hausdorff spaces with S scattered then equation (2.1) implies that K is also scattered and ht(K) ≤ ht(S). 2 3. On Banach lattice isomorphisms between C([0, α], X) spaces In this section we prove that the statement (a) of Theorem 1.2 implies the statement (b) of Theorem 1.2. Firstly, notice that by a famous result [3, Theorem 14.12] a Banach lattice contains a sublattice isomorphic to c0 if and only if it has a Banach subspace isomorphic to c0 . So, it suffices to prove the statements of Theorem 1.2 for Banach lattices without copy of c0 . Theorem 3.1. Let X be a Banach lattice containing no copy of c0 such that for any positive integer n, there is no Banach lattice isomorphism from X n+1 into X n . Let K and S be ordinal intervals. If C(K, X) is Banach lattice isomorphic to C(S, X) then K and S are homeomorphic. Proof. Let T : C(K, X) → C(S, X) be a Banach lattice isomorphism. If K and S are finite, we easily infer from our hypotheses that K and S are homeomorphic. Then, suppose that K and S are infinite. Without loss of generality assume that T −1 = 1. Note that K and S are scattered [19, Proposition 8.6.7]. By applying Proposition 2.3 twice, we obtain ht(K) = ht(S). On the other hand, by [19, Proposition 8.6.5] we may assume that K = [1, ω α n] and S = [1, ω β m] for some ordinals α, β, m and n different from zero, where m, n are finite numbers. Thus α + 1 = ht(K) = ht(S) = β + 1 [19, Proposition 8.6.6], and so α = β. Now, we have to show that m = n. Fix 0 < r < 1 and assume that m < n. Then K (α) = {ω α , . . . , ω α n}. Let K0 = {ω α , . . . , ω α (m + 1)}. Take a sequence {U1 , . . . , Um+1 } of disjoint open subsets of K such that ω α j ∈ Uj for each 1 ≤ j ≤ m + 1. By Urysohn lemma, for each 1 ≤ j ≤ m + 1, there is hj ∈ C(K) satisfying 0 ≤ hj ≤ 1, hj (K \ Uj ) = 0 and hj (ω α j) = 1. Let L : C(K0 , X) → C(K, X) be defined by L(f )(x) =
m+1
hi (x)f (ω α i),
i=1
for every f ∈ C(K0 , X). Note that L is a Banach lattice isometry, that is L(f ) = f and L(f ∨ g) = L(f ) ∨ L(g) for all f, g ∈ C(K0 , X). Let R : C(S, X) → C(S (α) , X) be the natural restriction operator. For all f ∈ C(K0 , X) we have f = L(f ) = L(f )|K (α) . Hence, it follows from Corollary 2.2 that R ◦ T ◦ L(f ) ≥ r f .
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Thus R ◦ T ◦ L is a Banach lattice isomorphism from X m+1 into X m which contradicts our assumption. Therefore m < n is impossible. The same argument shows that n < m is false. So, m = n and we are done. 2 4. On Banach lattice isomorphisms between C(K ⊕ [0, α], X) spaces The aim of this section is to prove the nontrivial part of Corollary 1.4, that is Corollary 4.2. We start with a vector-valued extension of [13, Lemma 3.2]. To state it, recall that the perfect kernel of a topological space S, denoted by P (S), is defined by P (S) = α≥1 S (α) . Lemma 4.1. Let K and S be locally compact Hausdorff spaces and X a Banach lattice containing no copy of c0 . If T is a Banach lattice isomorphism from C0 (K, X) onto C0 (S, X) then there exists a Banach lattice isomorphism Tˆ from C0 (S \ P (S), X) into C0 (K \ P (K), X). Proof. Let ψ : C0 (S \ P (S), X) → C0 (S, X) be defined by
ψ(f )(y) =
f (y),
if y ∈ S \ P (S)
0,
if y ∈ P (S),
for every f ∈ C0 (S \ P (S), X). Then ψ is an isometry and ψ(f ∨ g) = ψ(f ) ∨ ψ(g) for each f, g ∈ C0 (S \ P (S), X). Next, assume that T −1 = 1 and fix 0 < r < 1. Let f ∈ C0 (S \ P (S), X), f = 0 and consider the function g=
T −1 (ψ(f )) . T −1 (ψ(f ))
We claim that g|P (K) < 1. Indeed, if g|P (K) ≥ 1 then g|K (α) = g = 1 for each ordinal α. According to Lemma 2.1 we know that {y ∈ S : T g(y) ≥ r} ∩ S (α) = ∅, for every ordinal α. Therefore, by compactness we have {y ∈ S : T g(y) ≥ r} ∩ P (S) = ∅, which is impossible. Thereby if R : C0 (K, X) → C0 (K \ P (K), X) denotes the natural restriction operator then Rg = 1, that is R(T −1 (ψ(f ))) = T −1 (ψ(f )) . Define Tˆ : C0 (S \ P (S), X) → C0 (K \ P (K), X) by Tˆ (f ) = (R ◦ T −1 ◦ ψ)(f ). Then, for all f, g ∈ C0 (S \ P (S), X) we have Tˆ(f ∨ g) = Tˆ (f ) ∨ Tˆ (g) and 1 f ≤ Tˆ (f ) ≤ f . T Thus, Tˆ is a Banach lattice isomorphism. 2
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Corollary 4.2. Let X be a Banach lattice containing no copy of c0 such that for any positive integer n, there is no Banach lattice isomorphism from X n+1 into X n . Let K and S be ordinal intervals and L a perfect compact Hausdorff space. If C(K ⊕ L, X) is Banach lattice isomorphic to C(S ⊕ L, X) then K and S are homeomorphic. Proof. Since K and S are scattered it follows by [19, Proposition 8.6.7] that P (K⊕L) = L and P (S⊕L) = L. So, by Lemma 4.1 there exist into Banach lattice isomorphisms T1 : C(K, X) → C(S, X) and T2 : C(S, X) → C(K, X). If K and S are finite, we are done. If not, once again by [19, Proposition 8.6.5] we may assume that K = [1, ω α n] and S = [1, ω β m] for some ordinals α, β, m and n different from zero, where m, n are finite numbers. By applying Proposition 2.3 twice, we have ht(K) = ht(S). According to [19, Proposition 8.6.6] we see that α + 1 = ht(K) = ht(S) = β + 1. Thus α = β. Now, proceeding as in the proof of Theorem 3.1 we conclude that m = n. 2 References [1] Y.A. Abramovich, C.D. Aliprantis, Positive operators, in: Handbook of the Geometry of Banach Spaces, Vol. I, NorthHolland, Amsterdam, 2001, pp. 85–122. [2] F. Albiac, N.J. Kalton, Topics in Banach Space Theory, Grad. Texts in Math., vol. 233, Springer, New York, 2006. [3] C.D. Aliprantis, O. Burkinshaw, Positive Operators, Springer, Dordrecht, 2006. [4] J.W. Baker, Compact spaces homeomorphic to a ray of ordinals, Fund. Math. 76 (1) (1972) 19–27. [5] E. Behrends, J. Pelant, The cancellation law for compact Hausdorff spaces and vector-valued Banach–Stone theorems, Arch. Math. (Basel) 64 (4) (1995) 341–343. [6] L. Brouwer, On the structure of perfect sets of points, Proc. Akad. Amst. 12 (1910) 785–794. [7] R.S. Burachik, A.N. Iusem, Set-Valued Mappings and Enlargements of Monotone Operators, Optim. Appl., vol. 8, Springer, New York, 2008. [8] P.G. Casazza, C.A. Kottman, B.L. Lin, Remarks on Figiel’s reflexive Banach spaces not isomorphic to their square, J. Math. Anal. Appl. 63 (3) (1978) 750–752. [9] T. Figiel, An example of infinite dimensional reflexive Banach space non-isomorphic to its Cartesian square, Studia Math. 42 (1972) 295–306. [10] E.M. Galego, On isomorphism classes of C(2m ⊕ [0, α]) spaces, Fund. Math. 204 (1) (2009) 87–95. [11] E.M. Galego, An isomorphic classification of C(2m × [0, α]) spaces, Bull. Pol. Acad. Sci. Math. 57 (3–4) (2009) 279–287. [12] E.M. Galego, M.A. Rincón-Villamizar, Weak forms of Banach–Stone theorem for C0 (K, X) spaces via the αth derivatives of K, Bull. Sci. Math. 139 (2015) 880–891. [13] Y. Gordon, On the distance coefficient between isomorphic function spaces, Israel J. Math. 8 (1970) 391–397. [14] I. Kaplansky, Lattices of continuous functions, Bull. Amer. Math. Soc. (N.S.) 53 (1947) 617–623. [15] S. Mazurkiewicz, W. Sierpińki, Contributions à la topologie des ensembles dénombrables, Fund. Math. 1 (1920) 513–522. [16] A.A. Milutin, Isomorphisms of spaces of continuous functions on compacts of power continuum, Tieoria Func. (Kharkov) 2 (1966) 150–156 (in Russian). [17] H.P. Rosenthal, The Banach space C(K), in: Handbook of the Geometry of Banach Spaces, North-Holland Publishing Co., Amsterdam, 2001, pp. 1547–1602. [18] W. Rudin, Real and Complex Analysis, third edition, McGraw-Hill Book Co., New York, 1987. [19] Z. Semadeni, Banach Spaces of Continuous Functions. Vol. 1, Monogr. Mat., vol. 55, PWN-Polish Sci. Publ., Warszawa, 1971. [20] S. Yamamoto, A. Yamashita, A counterexample related to topological sums, Proc. Amer. Math. Soc. 134 (12) (2006) 3715–3719.