On dense subsets, convergent sequences and projections of Tychonoff products

Topology and its Applications 233 (2018) 52–60

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On dense subsets, convergent sequences and projections of Tychonoff products ✩ A.A. Gryzlov Department of Algebra & Topology, Udmurt State University, Universitetskaya st., 1, Izhevsk, Udmurtiya, 426034, Russia

a r t i c l e

i n f o

Article history: Received 6 October 2016 Received in revised form 24 October 2017 Accepted 29 October 2017 Available online 31 October 2017 MSC: 54A25 54B10 Keywords: Tychonoff product Dense set Convergent sequence Independent matrix Projection

a b s t r a c t It is well know that the Tychonoff product of 2ω many separable spaces is separable [2,3]. We consider for the Tychonoff product of 2ω many separable spaces the problem of the existence of a dense countable subset, which contains no nontrivial convergent in the product sequences. The first result was proved by W.H.  Priestley. He proved [14] that such dense set exists in the Tychonoff product Iα of closed unit intervals. α∈2ω  Zα We prove (Theorem 3.2) that such dense set exists in the Tychonoff product α∈2ω

of 2ω many Hausdorff  separable not single point spaces.  Zα there is a countable dense set Q ⊆ Zα such that for We prove that in α∈2ω α∈2 ω every countable subset S ⊆ Q a set πA (S) is dense in a face Zα for some A, α∈A

|A| = ω.  Iα there is a countable set, that is dense but We prove (Theorem 3.4) that in ω α∈2  sequentially closed in Iα with the Tychonoff topology and is closed and discrete α∈2ω  in Iα with the box topology (Theorem 3.4). α∈2ω

© 2017 Elsevier B.V. All rights reserved.

1. Introduction By Hewitt – Marczewski – Pondiczery theorem (see [2,3]), the Tychonoff product



Xα of 2ω many

α∈2ω

separable spaces is separable. We consider the problem of the existence in the Tychonoff product of 2ω many separable spaces a dense countable subset, which contains no nontrivial convergent in the product sequences. ω In [14] W.H. Priestley proved that such countable dense set exists in I 2 , where I is closed unit interval. ✩

Supported by Ministry of Education and Science of the Russian Federation, project number 1.5211.2017/8.9. E-mail address: [email protected].

https://doi.org/10.1016/j.topol.2017.10.021 0166-8641/© 2017 Elsevier B.V. All rights reserved.

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ω

In [15] P. Simon proved that such countable dense set exists in D2 , where D is a two-point discrete ω space. He proved that in D2 there is a countable dense set such that the closure of every countable subset ω of it has a cardinality 22 . ω In [10] we proved that such countable dense set exists in Z 2 , where Z is a countable discrete space. We ω proved that in Z 2 there is a countable dense set such that every countable subset of Q contains a countable ω discrete in Z 2 subset. Now we in Theorem 3.2 prove that such countable dense set exists in the general case of the product  Zα of not one-point Hausdorff separable spaces. α∈2ω   We prove that in Zα there is a countable dense set Q ⊆ Zα such that for every countable subset α∈2ω α∈2ω  S ⊆ Q a set πA (S) is dense in a face Zα for some A, |A| = ω α∈A

We use the notion of the independent matrix (see Preliminaries), it generalizes the independent family of sets [5,11]. Independent families of sets, in particular, were used by R. Engelking and M. Karlowicz ([4]) for their proof of Hewitt – Marczewski – Pondiczery theorem, by P. Simon ([15]) in his theorem, mentioned above. The notion of independent matrix was defined by J. van Mill [13] as a subfamily of independent linked family, defined by K. Kunen [12]. Families of this type were widely used in the theory of Stone–Chech compactifications of discrete spaces (see in particular [12,13,6,7]). We used independent matrices in [8–10] for an investigation of dense subsets of products. 2. Preliminaries Definitions and notions used in the paper can be found in [1–3]. d(X) denotes the density of a space X, by [A] we denote the closure of a set A, exp A denote the set of all subsets of A and by Exp A we denote the set of all non-empty subsets of a set A. By Y X we denote the set of all mappings from X to Y . We say that X is a countable set if |X| = ω.  Xα on Xα . By πα we denote an α-projection of α∈A

A sequence {xn }∞ n=1 is called trivial if there is n0 ∈ ω such that xn = xn0 for all n ≥ n0 . A subset A ⊆ X is called sequentially closed if A contains limits of all its convergent in X sequences. We will use the notion of the independent matrix of subsets of ω (see [13]). Definition 2.1. ([13]) An indexed family {Aij : i ∈ I, j ∈ J} of subset of ω is called a J by I independent matrix if – whenever j0 , j1 ∈ J are distinct and i ∈ I, then |Aij0 ∩ Aij1 | < ω; – if i1 , . . . , in ∈ I are distinct and j1 , . . . , jn ∈ J, then | ∩ {Aik jk : k = 1, . . . , n}| = ω. For {Aij : i ∈ I, j ∈ J} the family {A∗ij = [Aij ] \ ω : i ∈ I, j ∈ J} is a J by I independent matrix of clopen subsets of ω ∗ . The following construction of the 2ω by 2ω independent matrix of subsets of ω can be found in [13]. Let H  = {< k, u >: k ∈ ω, u ∈ (exp k)exp k }

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and AXY = {< k, u >∈ H  : u(X ∩ k) = Y ∩ k} The family M = {AXY : X, Y ∈ exp ω} is the 2ω by 2ω independent matrix. We will use the following notation (see [8]). Consider the set (Exp k)exp k for k ∈ ω. Elements of this set we will denote by u, v, etc. For k ∈ ω denote Hk = {u ∈ (Exp k)exp k : {n} ∈ u(exp k) for all n < k} H = ∪{Hk : k ∈ ω} For X ∈ exp ω and Y ∈ Exp ω denote Ak (X, Y ) = {u ∈ Hk : u(X ∩ k) = Y ∩ k}, and A(X, Y ) = ∪{Ak (X, Y ) : k ∈ ω} Define the independent matrix M1 = {A(X, Y ) : X ∈ exp ω, Y ∈ Exp ω}. The matrix M1 satisfies the following conditions: (i) if Y1 , Y2 ∈ Exp ω are distinct and X ∈ exp ω, then |A(X, Y1 ) ∩ A(X, Y2 )| < ω; (ii) if X1 , . . . , Xn ∈ exp ω are distinct and Y1 , . . . , Yn ∈ Exp ω, then there is k0 ∈ ω such that (∩{A(Xi , Yi ) : i = 1, ..., n}) ∩ Hk = ∅ for all k > k0 ; (iii) ∪{A(X, Y ) : Y ∈ Exp ω} = H for every X ∈ exp ω. We will use the following properties of M1 , proved in [8]. Lemma 2.1. ([8]) Let u, v ∈ H, u = v. For every B ⊆ exp ω, |B| < 2ω , there is X ∈ exp ω \B and Y ∈ Exp ω such that u ∈ A(X, Y ) and v ∈ / A(X, Y ). Lemma 2.2. ([10]) Let X ∈ exp ω and F ⊆ H be such that |F ∩ Hk | ≤ 1 for all k ∈ ω. Then there is a family T (F, X) ⊆ Exp ω such that (1) |T (F, X)| = ω; (2) |A(X, Y ) ∩ F | < ω for all Y ∈ T (F, X); (3) ∪{A(X, Y ) : Y ∈ T (F, X)} = H. 3. Main results Our proof of main results consists of next parts. At first we will construct, as well as in [10], a matrix M2 , which generates a product Σ of countable discrete spaces.

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 Now the task is to construct a continuous injective mapping from Σ into a product Zα of Hausdorff α∈2ω  separable uncountable spaces and to get a dense subset Q of Zα such that for every countable subset α∈2ω

of Q its image under a projection into one of coordinate spaces is dense. We do this in Theorem 3.1.  Zα there is a countable dense set Q ⊆ By using of Theorem 3.1, we prove (Theorem 3.2) that in α∈2ω   Zα such that for every countable subset S ⊆ Q a set πA (S) is dense in a face Zα for some A, α∈2ω

|A| = ω. Therefore Q contains no convergent sequences.

α∈A

Let us define a matrix M2 , the construction is similar to [10]. Consider M1 = {A(X, Y ) : X ∈ exp ω, Y ∈ Exp ω}. I. Let P be a set of all ordered pairs (u, v) of elements u, v ∈ H. By Lemma 2.1 there is a countable family L = {X(u,v) : (u, v) ∈ P } of sets X(u,v) ∈ exp ω such that for every X(u,v) ∈ L there is Y(u,v) ∈ Exp ω such that u ∈ A(X(u,v) , Y(u,v) ), v∈ / A(X(u,v) , Y(u,v) ) and X(u,v) = X(u ,v ) if (u, v) = (u , v  ). Denote R = exp ω \ L. Let E = {S ⊆ H : |S| = ω} and θ : E → R be a one-to-one mapping from E onto R. Denote θ(S) = XS for S ∈ E. II. Consider S ∈ E and XS ∈ R. There is a countable set TXS ∈ Exp ω such that: – ∪{A(XS , Y ) : Y ∈ TXS } = H; – |{Y ∈ TXS : A(XS , Y ) ∩ S = ∅}| = ω. Indeed, since |S| = ω and |Hk | < ω for all k ∈ ω, we have K = {k ∈ ω : S ∩ Hk = ∅} is countable. Pick a point uk ∈ S ∩ Hk for every k ∈ K. Denote FS = {uk : k ∈ K}. By Lemma 2.2 for FS and XS there is T (FS , XS ) ⊆ Exp ω such that: |T (FS , XS )| = ω, |A(XS , Y ) ∩FS | < ω for all Y ∈ T (FS , XS ) and ∪{A(XS , Y ) : Y ∈ T (FS , XS )} = H. Then |{Y ∈ T (FS , XS ) : Y ∩ F = ∅}| = ω and therefore |{Y ∈ T (F, XS ) : Y ∩ S = ∅}| = ω. Let TXS = T (FS , XS ). III. Consider (u, v) ∈ P , X(u,v) ∈ L and Y(u,v) ∈ Exp ω such that u ∈ A(X(u,v) , Y(u,v) ) and v ∈ / A(X(u,v) , Y(u,v) ). Let

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TX(u,v) ⊆ Exp ω be a countable set such that Y(u,v) ∈ TX(u,v) , |TX(u,v) | = ω and ∪{A(X(u,v) , Y ) : Y ∈ TX(u,v) } = H. Define a family {TX : X ∈ exp ω} of subsets TX ⊆ Exp ω as follows:  TX =

T(u,v) TXS

for X = X(u,v) ∈ L for X = XS ∈ R.

By a similar way as in [10], using the matrix M1 we define a matrix  M2 = {A(X, Y ) : X ∈ exp ω, Y ∈ TX }, which satisfies the following conditions:  (u,v) , Y(u,v) ) (1) for every (u, v) ∈ P there is X = X(u,v) ∈ L ⊆ exp ω and Y ∈ TX(u,v) such that u ∈ A(X  (u,v) , Y(u,v) ); and v ∈ / A(X (2) for every S ∈ E there is X = XS ∈ R = exp ω \ L such that  S , Y ) ∩ S = ∅}| = ω; |{Y ∈ TXS : A(X  (3) ∪{A(X, Y ) : Y ∈ TX } = H for all X ∈ exp ω;   (4) A(X, Y ) ∩ A(X, Y  ) = ∅ for all X ∈ exp ω and Y, Y  ∈ TX , Y = Y  ;  i , Yi ) : i = (5) if X1 , . . . , Xn ∈ exp ω are distinct and Yi ∈ TXi then there is k0 ∈ ω such that (∩{A(X 1, . . . , n}) ∩ Hk = ∅ for every k > k0 . Denote Σ=



TX .

X∈exp ω

We will say that Σ is generated by independent matrix M2 . For ξ ∈ Σ and X ∈ exp ω, let ξ(X) = πX (ξ) ∈ TX . By the properties of the matrix M2 the space Σ satisfies the following conditions: (1) if ξ1 , ξ2 ∈ Σ, ξ1 = ξ2 , then there is X ∈ exp ω such that   ξ2 (X)) = ∅; A(X, ξ1 (X)) ∩ A(X,  (2) | ∩ {A(X, ξ(X)) : X ∈ exp ω}| ≤ 1 for every ξ ∈ Σ; (3) for every u ∈ H there is the only ξu ∈ Σ such that  ∩{A(X, ξu (X)) : X ∈ exp ω} = {u}; (4) if u1 , u2 ∈ H, u1 = u2 , then ξu1 = ξu2 .

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Let μ: H → Σ be a one-to-one mapping from H into Σ, defined by the rule: μ(u) = ξu for every u ∈ H. 

Theorem 3.1. Let

ZX be the Tychonoff product of uncountable separable spaces. Then there is a  countable dense subset Q ⊆ ZX satisfying the following condition: X∈exp ω

X∈exp ω

 of S is dense in (*) for every countable set S ⊆ Q there is X ∈ exp ω such that the X-projection πX (S) ZX . 

Proof. Let

ZX be the Tychonoff product of 2ω many uncountable spaces ZX (X ∈ exp ω). Fix in

X∈exp ω

 every ZX a countable dense subset DX .  Consider the matrix M2 = {A(X, Y ) : X ∈ exp ω, Y ∈ TX }.

Define for every X ∈ exp ω a dense countable set DX ⊆ ZX and a one-to-one mapping ψX : TX → ZX from TX into ZX as follows. We have exp ω = L ∪ R. Let X ∈ L, i.e. X = X(u,v) for some ordered pair (u, v), then TX = T(u,v) . For ψX( u,v) : TX( u,v) → ZX( u,v) we take some one-to-one mapping from TX = TX( u,v) onto DX( u,v) ⊆ ZX( u,v) . Define  DX = DX( u,v) = DX . ( u,v)

Let X ∈ R = exp ω \ L, i.e. X = XS for some S ∈ E, then TX = TXS .   Denote TX = {Y ∈ TXS : A(XS , Y ) ∩ S = ∅}. By the property (2) of M2 , we have |TX | = ω. S S For ψXS : TXS → ZXS   we take some one-to-one mapping from TX = TXS into ZXS such that ψXS (TX ) = DX . This is possible, S S because ZXS is uncountable. Define

DX = DXS = ψXS (TXS ).

Define the mapping Ψ: Σ →

 X∈exp ω

from Σ into

 X∈exp ω

ZX as follows.

ZX

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For ξ ∈ Σ let Ψ(ξ) = z ∈



ZX be such that πX (z) = zX = ψX (ξ(X)).  Ψ is a one-to-one continuous mapping from Σ onto DX . X∈exp ω  Then the mapping Ψ ◦ μ is a one-to-one mapping from H into DX . X∈exp ω

X∈exp ω

Denote for every u ∈ H z u = Ψ(μ(u)) = Ψ(ξu ). Let Ψ(μ(H)) = Q and Ψ(μ(Hk )) = Qk . Let us prove that Q is dense in



ZX .

X∈exp ω d Define the discrete topology on every set DX , we denote this space as DX , and consider the space  d DX . X∈exp ω  d d Let O(X1 , . . . , Xn , a1 , . . . , an ) = {z ∈ DX : zXi = ai , i = 1, . . . , n}, where ai ∈ DX (i = 1, . . . , n), i X∈exp ω  d be a basic open set of the product DX . X∈exp ω

From the definitions of Σ, mappings μ and Ψ, it follows: −1 – z ∈ O(X1 , . . . , Xn , a1 , . . . , an ) if and only if ψX (ai ) = ξ(Xi ) (i = 1, . . . , n) for ξ = Ψ−1 (z); i n   i , ψ −1 (ai )). A(X – z u ∈ O(X1 , . . . , Xn , a1 , . . . , an ) if and only if u ∈ Xi i=1

 i , ψ −1 (ai )) : i = 1, . . . , n}. Consider the set ∩{A(X Xi By the property (5) of the matrix M2 it follows that there is k0 ∈ ω such that  i , ψ −1 (ai )) : i = 1, . . . , n}) ∩ Hk = ∅ for every k > k0 . (∩{A(X Xi  i , ψ −1 (ai )) : i = 1, . . . , n} ∩ Hk we have For u ∈ ∩{A(X Xi z u ∈ O(X1 , . . . , Xn , a1 , . . . , an ) ∩ Qk .  d So O(X1 , . . . , Xn , a1 , . . . , an ) ∩ Q = ∅ and Q is dense in DX . X∈exp ω   Therefore Q is dense in DX and hence Q is dense in ZX . X∈exp ω

X∈exp ω

Let us prove that Q satisfies the condition (*). Let S = {z uk : k ∈ ω} ⊆ Q be a countable set. Then S = {uk : k ∈ ω} is a countable subset of H, i.e. S ∈ E. We have μ(S) = {μ(uk ) : uk ∈ S} = {ξuk : uk ∈ S}, Ψ(μ(S)) = S = {z uk : uk ∈ S} and z uk = Ψ(ξuk ). uk Therefore zX = πX (z uk ) = ψX (ξk (X)).  For S consider corresponding XS ∈ R and ZXS , subsets DX and DXS are dense in ZXS . S  Consider TXS and a countable set TXS = {Y ∈ TXS : A(XS , Y ) ∩ S = ∅}.  We claim that TX = {ξuk (XS ) : uk ∈ S}. S  Indeed, since uk ∈ ξuk (X) for all X ∈ exp ω, we have {ξuk (XS ) : uk ∈ S} ⊆ TX and S S ⊆ ∪{ξuk (XS ) : uk ∈ S}.  By the property (4) of M2 , we have TX = {ξuk (XS ) : uk ∈ S}. S By the definition of the mapping ψXS : TXS → ZXS , we have   ψXS (TX ) = {ψXS (ξuk (XS )) : uk ∈ S} = DX . S S

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uk    Since ψXS (ξuk (XS )) = zX = πXS (z uk ), we have DX = ψXS (TX ) = {πXS (z uk ) : uk ∈ S} = πXS (S). S S S  is dense in ZX . 2 So πXS (S) S



Theorem 3.2. Let

Zα be the Tychonoff product of Hausdorff separable not single point spaces. Then  there is a countable dense set Q ⊆ Xα such that for every countable subset S ⊆ Q a set πA (S) is α∈2ω   dense in a face Zα for some A, |A| = ω, and therefore Q contains no nontrivial convergent in Zα α∈2ω

α∈2ω

α∈A

sequences. 

Proof. Let

Zα be a product of not single point separable spaces. Consider a decomposition σ of the

α∈2ω

set 2ω on countable sets.  For every A ∈ σ let YA = Zα . α∈A     Then YA is naturally homeomorphic to Zα , we will write YA = Zα . YA is an uncountable α∈2ω

A∈σ

A∈σ

α∈2ω

separable Hausdorff space for all A ∈ σ.  YA of uncountable separable Hausdorff spaces. So, we have a product A∈σ   By Theorem 3.1 there is a countable dense subset Q of YA = Xα , satisfying (*). A∈σ

α∈2ω

Suppose there is a nontrivial convergent sequence in Q, and let S = {z uk : k ∈ ω} be its subsequence  = ω. such that z uk = z uk if k = k , we have |S|  is dense in YA , where YA a Hausdorff uncountable space. Then there is a face YA such that πA (S)  But πA (S) is a convergent sequence in YA . Contradiction. 2 By the same way we prove the following theorem. Theorem 3.3. Let



Zα be the Tychonoff product of separable not single point spaces. Then there is a  countable dense subset Q ⊆ Zα satisfying the following condition: α∈2ω

α∈2ω

 of S is dense in Zα . (*) for every countable set S ⊆ Q there is α ∈ 2ω such that the α-projection πα (S) From above we have the following. Theorem 3.4. In the Tychonoff product

 α∈2ω

Iα of intervals Iα = [0, 1] there is a countable subset Q ⊆



Iα

α∈2ω

such that – Q is dense in



Iα ;

α∈2ω

 = Iα ; – for every countable set S ⊆ Q there is α ∈ 2ω such that πα ([S])  Iα sequences; – Q contains no nontrivial convergent in α∈2ω  – Q sequentially closed in Iα ; ω α∈2 – if 0 < aα < bα < 1, then | [aα , bα ] ∩ Q| < ω; α∈2ω  – Q is a closed discrete set in Iα with the box topology on Iα . α∈2ω

α∈2ω

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