Abstract. We construct a family of spaces with “nice” structure which is univer- sal in the class of all compact metrizable spaces of large transfinite dimension ω

145 (1994)

Universal spaces in the theory of transfinite dimension, II

by

Wojciech O l s z e w s k i (Warszawa)

Abstract. We construct a family of spaces with “nice” structure which is univer- sal in the class of all compact metrizable spaces of large transfinite dimension ω

, or, equivalently, of small transfinite dimension ω

; that is, the family consists of compact metrizable spaces whose transfinite dimension is ω

, and every compact metrizable space with transfinite dimension ω

is embeddable in a space of the family. We show that the least possible cardinality of such a universal family is equal to the least possible cardinality of a dominating sequence of irrational numbers.

1. Introduction. In Part I of the paper we have proved that there is no universal space in the class of all compact metrizable spaces X with Ind X = ω

, or equivalently, with ind X = ω

(see [3], Proposition 4.11, or Lemma 2.3 of this paper). That class will be denoted by D. We have also shown that there is no universal space in the class of all separable metrizable spaces X with Ind X = ω

, to be denoted by C. In this part we introduce the notion of a universal family which is a generalization of the notion of a universal space, and we study universal families for C and D.

1.1. Definition. Let C be a class of topological spaces. A family A of spaces belonging to C is said to be a universal family in C if every space in C is embeddable in a space belonging to A.

Universal families can play a role similar to that played by universal spaces. Universal families of small cardinality consisting of spaces with

“nice” structure are of particular interest.

In Sections 4 and 5 we construct a universal family A in D consisting of spaces with “nice” structure. Since every separable metrizable space X has a compactification Z such that Ind Z = Ind X (see [5], and [6] for the proof), the family A is also universal in C. In Section 6 we estimate the least possible cardinality of a universal family in D; by the above compactification theorem, it is equal to the least possible cardinality of a universal family in C.

1991 Mathematics Subject Classification: Primary 54F45, 54A25.

[121]

In order to formulate our result, we have to recall a few notions.

Irrational numbers can be viewed as sequences of natural numbers. We denote by N

the set of irrational numbers, i.e., the set of all sequences σ = (σ(k))

of natural numbers. On N

we consider the relation ≤

defined by letting

σ ≤

τ if σ(k) ≤ τ (k) for all but a finite number of k ∈ N . A subset D ⊆ N

cofinal in N

is said to be dominating, i.e., D is domi- nating if for every σ ∈ N

, there exists a τ ∈ D such that σ ≤

τ . We set d = min{|D| : D ⊆ N

is a dominating sequence}.

One can prove that

ℵ

≤ d ≤ c ;

one can also prove that each of the following formulae is consistent with the axioms of set theory:

ℵ

= d = c , ℵ

= d < c , ℵ

< d < c , ℵ

< d = c .

For a deeper discussion and the proofs of the above statements we refer the reader to E. K. van Douwen’s survey [1].

Section 6 contains the proof of the equality

min{|A| : A is a universal family in D} = d , which, in particular, gives

min{|A| : A is a universal family in C} = d .

Acknowledgements. The paper contains some of the results of my Ph.D. thesis supervised by Professor R. Engelking whom I would like to thank for his comments and improvements. I also wish to express my thanks to Professors J. Chaber, J. Krasinkiewicz and W. Marciszewski whose com- ments have helped me to improve the presentation significantly.

2. Three lemmas. For a topological space and its closed subset A, we denote by X/A the quotient space obtained by identifying A to a point (see [4], Example 2.4.12); we denote this point by a, and the natural quotient mapping by q.

2.1. Lemma. Let X be a compact metrizable space, and A its closed sub- set. Let ε be a positive real number. If f : X → I

has the property that f |A is an ε-mapping, then there exist m > n and g : X → I

such that the diagonal f M g is an ε-mapping and

(f M g)

(I

× {(0, . . . , 0)}) = A .

P r o o f. Since X/A is a compact metrizable space, there exists an em- bedding

h = (h

, h

, . . .) : X/A → I

= I × I × . . .

Let h

: X/A → I be a function such that (h

)

(0) = {a}; set φ = (h

, h

· h

, h

· h

, . . .). It is easy to check that φ : X/A → I × I

is also an embedding, and therefore f M (φ ◦ q) : X → I

× I × I

is an ε-mapping. By compactness of X, so is f M (p

◦ φ ◦ q) for suffi- ciently large m, where p

: I × I

→ I × I

denotes the projec- tion, i.e., p

((x

, x

, x

, . . .)) = (x

, x

, x

, . . . , x

) for x

∈ I and (x

, x

, . . .) ∈ I

. Consider an m with this property.

Set g = p

◦ φ ◦ q. Then (f M g)

(I

× {(0, . . . , 0)}) = g

((0, . . . , 0))

= (h

◦ q)

(0) = q

(a) = A.

2.2. Lemma. Let m and n be natural numbers such that n ≥ 2m + 2.

Let X be a compact metrizable space, and A and B its closed subspaces with Ind B ≤ m. Then there exists f : X → I

such that f |B−A is an embedding, and f

((0, . . . , 0)) = A.

P r o o f. Consider the quotient mapping q : X → X/A. Since we have Ind({a} ∪ q(B)) ≤ m and n − 1 ≥ 2m + 1, there exists an embedding h : {a} ∪ q(B) → I

. Let h

be an extension of h onto X/A, and g : X/A → I a function such that g

(0) = {a}. Since I

is embeddable in the geometrical boundary bd I

of I

, and bd I

is homogeneous, we can assume that bd I

is the range of h

, and h

(a) = (0, . . . , 0). Let φ : (bd I

)×I → I

be an embedding with φ(x, 0) = x for every x ∈ bd I

.

It is easy to check that f = φ ◦ (h

M g) ◦ q has the required properties.

Note that the assumption n ≥ 2m + 2 in Lemma 2.2 can be replaced by n ≥ 2m + 1. The proof under this weaker assumption is similar to that of Corollaries 2.5 and 2.7 in [7]. However, we will only need the lemma in the form given above.

2.3. Lemma. A compact metrizable space satisfies Ind X ≤ ω

if and only if for every pair of distinct points x, y ∈ X there exists a finite-dimensional partition L between x and y.

P r o o f. The necessity is obvious. To show the sufficiency, we first prove that for any x ∈ X and any neighbourhood U ⊆ X of x, there exists an open set V ⊆ X such that Ind bd V < ω

, i.e., ind X ≤ ω

.

For every y ∈ X − U , consider a finite-dimensional partition L

between x and y; let U

and V

be disjoint open subsets of X such that x ∈ U

, y ∈ V

, and L

= X − (U

∪ V

). Then X − U ⊆ S

{V

: y ∈ X − U }; by

compactness of X − U , there exists a finite family V ⊆ {V

: y ∈ X − U }

such that X − U ⊆ S

V. It follows immediately that V = X − [

{cl V

: V

∈ V}

is an open subset of X with x ∈ V ⊆ U , and bd V ⊆ [

{bd V

: V

∈ V} ⊆ [

{L

: V

∈ V} ; since V is finite,

Ind bd V ≤ max{Ind L

: V

∈ V} < ω

.

Now, let A ⊆ X be an arbitrary closed set, and U ⊆ X an open set containing A. For every x ∈ A, consider an open set V

⊆ X such that x ∈ V

⊆ U and Ind bd V

< ω

. By compactness of A, there exists a finite family V ⊆ {V

: x ∈ A} such that A ⊆ S

V. Then V = S

V is an open subset of X such that A ⊆ V ⊆ U and bd V ⊆ S

{bd V

: V

∈ V}; since V is finite,

Ind bd V ≤ max{Ind bd V

: V

∈ V} < ω

.

3. The structure of spaces X with Ind X ≤ ω

. In this section we shall prove that compact metrizable spaces of a certain structure have large transfinite dimension not greater than ω

(see Theorem 3.2); actually, it turns out (see Theorem 5.1) that each compact metrizable space X with Ind X ≤ ω

has that structure.

Let {M

, r

} be an inverse sequence; then M denotes the inverse limit of {M

, r

}, and r

: M → M

, for k ∈ N, denotes the projection. Each family {M

, r

}, where r

: M

→ M

, determines an inverse sequence {M

, r

}; to wit, it suffices to set r

= r

◦ . . . ◦ r

for k > j and r

equal to the identity mapping of M

. For simplicity, we shall also call each such family {M

, r

} an inverse sequence.

The next lemma is a technical one and will only be used in the proofs of Theorem 3.2 and Lemma 6.1.

3.1. Lemma. Let n be a fixed natural number , and let x, y ∈ M be distinct.

Suppose there is a j ∈ N such that for every k ≥ j, there exist pairwise disjoint subsets U

, V

and L

of M

, where U

, V

are open and L

is closed, which satisfy the following conditions:

(3.1) M

= U

∪ V

∪ L

, (3.2) Ind L

≤ n,

(3.3) r

(x) ∈ U

and r

(y) ∈ V

,

(3.4) (r

)

(U

) ⊆ U

and (r

)

(V

) ⊆ V

.

Then there exists an at most n-dimensional partition in M between x and y.

P r o o f. Define L =

\

r

(L

) , U = [

r

(U

) , and V = [

r

(V

) .

From (3.1) it follows directly that M = U ∪ V ∪ L. Clearly, U, V are open and L is closed in M . Since U

∩ V

= ∅, we have U ∩ V = ∅ by (3.4), and since U

∩ L

= ∅ = L

∩ V

, we also have U ∩ L = ∅ = L ∩ V . By (3.3), x ∈ U and y ∈ V , and thus L is a partition in M between x and y.

Since L

∩ U

= ∅ = V

∩ L

, by (3.1) and (3.4), r

(L

) ⊆ L

, so we can regard r

|L

as a mapping to L

; thus L coincides with lim ←−{L

, r

|L

}, and so Ind L ≤ n by the theorem on the dimension of the limit of an inverse sequence (see [2], Theorem 1.13.4) and (3.2).

3.2. Theorem. Let {M

, r

} be an inverse sequence of finite-dimen- sional compact metrizable spaces M

in which all bonding mappings r

are retractions. Suppose that for every k ∈ N, there exist a covering A

of M

and a metric %

on M

with the following properties:

(3.5) the sets A

− M

, where A

∈ A

, are open in M

and pairwise disjoint,

(3.6) for every A

∈ A

, there exists an A

∈ A

such that r

(A

)

⊆ A

,

(3.7) for every i ∈ N, the sequence of real numbers (sup{diam

r

(A

) : A

∈ A

})

converges to 0.

Then Ind M ≤ ω

.

P r o o f. First for every A

∈ A

, we define A

∈ A

for j ≤ k in such a way that

(3.8) r

(A

) ⊆ A

,

(3.9) (A

)

= A

whenever j ≤ i ≤ k.

For instance, one can define A

to be an arbitrary member A

of A

such that r

(A

) ⊆ A

for A

∈ A

(its existence is guaranteed by (3.6)), and then set by induction A

= (A

)

. Of course, we put A

= A

.

We can now begin the proof of Ind M ≤ ω

. Since M , as the inverse

limit of a sequence of compact spaces, is compact, it suffices to find a

finite-dimensional partition between any two distinct points x, y ∈ M (see Lemma 2.3). To this end, we shall apply Lemma 3.1.

Take the smallest i ∈ N such that r

(x) 6= r

(y) and define ε =

%

(r

(x), r

(y)). By (3.7), there exists a j > i such that (3.10) diam

r

(A

) < ε/3 for each A

∈ A

.

Let

U

= {z ∈ M

: %

(z, r

(x)) < ε/3} , V

= {z ∈ M

: %

(z, r

(x)) > 2ε/3} , L

= {z ∈ M

: ε/3 ≤ %

(z, r

(x)) ≤ 2ε/3} , and U

= (r

)

(U

), V

= (r

)

(V

), L

= (r

)

(L

).

Put

U

= {A

∈ A

: A

∩ V

= ∅} , V

= {A

∈ A

: A

∩ V

6= ∅}

for k ≥ j, and U

=  [

U



− L

, V

=  [ V



− L

, L

= L

for k > j; since the bonding mappings are retractions, L

is a subset of M

for k > j.

We first check that

(3.11) U

, V

and L

are pairwise disjoint.

Obviously, U

∩ V

= ∅. Suppose, on the contrary, that U

∩ V

6= ∅ for a k > j. Then there exists a z ∈ (A

∩ B

) − L

for some A

∈ U

and B

∈ V

. By (3.5), we have z ∈ M

and since r

is a retraction, we conclude, using (3.8), that

z ∈ A

∩ B

∩ M

⊆ A

∩ B

.

Observe that A

∈ U

and B

∈ V

; indeed, as A

∈ U

(B

∈ V

) we have A

∩ V

= ∅ (B

∩ V

6= ∅); hence by (3.9), (A

)

∩ V

= A

∩ V

= ∅ ((B

)

∩ V

= B

∩ V

6= ∅), and so A

∈ U

(B

∈ V

). In the same manner we can show by induction that z ∈ A

∩ B

and A

∈ U

, B

∈ V

.

Consequently, A

∩ V

= ∅ and z 6∈ V

; however, z 6∈ L

, and so, by the definition of U

, V

, L

, we conclude that z ∈ U

; thus B

∩ U

6= ∅. On the other hand, B

∩ V

6= ∅ (since B

∈ V

).

This shows that

%

(r

(z

), r

(x)) < ε/3 and %

(r

(z

), r

(x)) > 2ε/3

for some z

, z

∈ B

, contrary to (3.10). Thus U

∩ V

= ∅. That U

∩ L

=

∅ = L

∩ V

follows directly from the definition of U

, V

, L

.

We now show that for n = Ind M

,

(3.12) U

, V

and L

satisfy conditions (3.1)–(3.4) of Lemma 3.1.

Condition (3.1) is obvious for k = j. Take k > j. Since M

⊇ U

∪ V

∪ L

= h [

U



− L

i

∪ h [ V



− L

i

∪ L

⊇  [ U



∪  [ V



= [

A

= M

, (3.1) also holds for k > j.

Conditions (3.2) and (3.3) follow immediately from the definitions of U

, V

and L

.

Let z ∈ M

and r

(z) ∈ U

. Take an A

∈ A

such that z ∈ A

and suppose that A

∈ V

.

Then A

∩ V

6= ∅, so (A

)

∩ V

6= ∅ (see (3.9)) and A

∈ V

. If k = j, then, by (3.8), r

(z) ∈ A

, therefore also A

∩U

6= ∅, contrary to (3.10). If k > j, then A

⊆ V

∪L

and, by (3.8), r

(z) ∈ V

∪L

; but we know (see (3.11)) that (V

∪ L

) ∩ U

= ∅, and therefore r

(z) 6∈ U

, a contradiction. Thus A

∈ U

.

This clearly forces A

⊆ U

∪ L

. We have z 6∈ L

, because otherwise z ∈ L

= L

, and hence r

(z) = z ∈ L

∩ U

= ∅ (see (3.11)). Thus z ∈ A

− L

⊆ U

, and the first part of (3.4) is proved.

The second part of (3.4) can be shown similarly, and thus the proof of (3.12) is complete.

We see at once that

(3.13) L

is a closed subset of M

.

In order to check that the assumptions of Lemma 3.1 are satisfied, it remains to show that

(3.14) U

and V

are open subsets of M

.

We prove this for U

by induction on k; the same argument works for V

. For k = j, (3.14) is evident. Assume that (3.14) holds for numbers less than some k > j.

First, we verify that

(3.15) U

∩ M

= U

.

Indeed,

U

⊆ M

∩ (r

)

(U

) ⊆ M

∩ U

(see (3.12) and (3.4); recall that r

is a retraction). On the other hand, if

there were z ∈ (U

∩ M

) − U

, we would have

z ∈ (M

− U

) ∩ U

⊆ (L

∪ V

) − (L

∪ V

)

⊆ (L

∪ V

) − (L

∪ V

) = ∅ (recall that L

= L

and V

⊆ V

; see (3.12) and (3.4)).

We now return to the inductive proof. Take a z ∈ U

. If z ∈ M

, then z ∈ U

by (3.15); then (r

)

(U

) is a neighbourhood of z by the inductive assumption, and it follows from (3.12) and (3.4) that (r

)

(U

) ⊆ U

. If z 6∈ M

, then z ∈ A

− M

for some A

∈ U

; by (3.5), A

− M

is a neighbourhood of z, and since L

⊆ M

, this neighbourhood is contained in U

.

We have thus verified that the assumptions of Lemma 3.1 are satisfied.

Consequently, there exists a finite-dimensional partition (more precisely, a partition of dimension not greater than n = Ind M

) between x and y in the space M .

4. The spaces I

. In this section, for any increasing sequence σ = (σ(k))

of positive integers, and any sequence a = (a(k))

of real num- bers such that 1/2 < a(k) < 1 for every k and Q

k=1

a(k) = 0 we construct a compact metrizable space I

with Ind I

= ω

. For our purposes it suf- fices to restrict attention to any fixed sequence a = (a(k))

which has the above properties, except for Section 6, where we will additionally need the condition

(∗)

Y

2a(k) < ∞ .

Thus the reader can assume that a = (1/2 + 1/2

)

; it is easy to see that this sequence has all the required properties.

From now on, σ stands for an increasing sequence of positive integers, and a for a sequence of real numbers such that 1/2 < a(k) < 1 and Q

k=1

a(k) = 0; for a given σ, we denote by σ|k the sequence σ(0), σ(1), . . . , σ(k − 1);

in particular, σ|0 is the empty sequence. We denote by S

the set of all sequences

γ : {1, 2, . . . , σ(k)} → {0, 1 − a(k + 1)} ,

i.e., the set of all sequences of σ(k) elements equal to either 0 or 1 − a(k + 1);

S

consists of the empty sequence. Let

S

= S

, S

= S

× S

× . . . × S

, S

= [

S

.

For every sequence s = (γ

, γ

, . . . , γ

) ∈ S

and m < k, let s|m stand

for the sequence (γ

, γ

, . . . , γ

) ∈ S

.

Fix σ and a. First, we construct by induction an inverse sequence {I

, r

}, where r

: I

→ I

; simultaneously, we define a covering {I

: s ∈ S

} of I

by σ(k)-dimensional cubes for k = 0, 1, . . .

Let I

be the σ(0)-dimensional cube I

and I

= I

for the unique s ∈ S

. Assume that we have already defined I

and its covering {I

: s ∈ S

}.

For every s ∈ S

, we identify I

and the standard σ(k)-dimensional cube I

. For t = (s, γ) ∈ S

, set I

= {(x

, . . . , x

) ∈ I

: γ(i) ≤ x

≤ γ(i) + a(k + 1) for i ≤ σ(k)}; that is, I

is a smaller cube placed in a corner of I

such that the length ratio of their edges is a(k + 1).

Roughly speaking, we glue a σ(k + 1)-dimensional cube, denoted here by I

, along its σ(k)-dimensional face, which is identified with I

, to every cube I

⊆ I

, where t ∈ S

, in such a way that the sets I

− I

are pairwise disjoint. Our I

is the space so obtained.

Precisely, the space I

can be defined as follows. Let Q

, where t ∈ S

, be a copy of the (σ(k + 1) − σ(k))-dimensional cube I

. Set

I

= {(y, {y

: s ∈ S

}) ∈ I

× P{Q

: s ∈ S

} :

y ∈ I

and y

= (0, . . . , 0) for s 6= t}

for t ∈ S

, and

I

= [

{I

: t ∈ S

} .

It is easily seen that the covering {I

: s ∈ S

} consists of σ(k + 1)- dimensional cubes.

The orthogonal projections of the σ(k + 1)-dimensional cubes I

onto their σ(k)-dimensional faces I

, where s ∈ S

, determine a retraction of I

onto I

; denote it by r

. More precisely,

r

((y, {y

: s ∈ S

})) = (y, {z

: s ∈ S

}) , where z

= (0, . . . , 0) for s ∈ S

, for every (y, {y

: s ∈ S

}).

Thus, the inductive construction of {I

, r

} is complete.

In Fig. 4.1 the first steps in constructing {I

, r

}, where σ(k) = k+1 for k = 0, 1, . . . , and a(k) = 1/2+1/2

for k = 1, 2, . . . , are exhibited.

Let

I

= lim ←−{I

, r

} .

We list several properties of the space I

.

I

I

z }| {

| {z }

I

Fig. 4.1

It is easy to check that for k = 0, 1, . . . , I

is a closed subspace of I

× P{Q

: s ∈ S

}; thus the spaces I

are compact and metrizable, and so is I

. Since a(k + 1) > 1/2, we have I

= S

{I

: t ∈ S

and t|k = s} for every s ∈ S

; in particular, I

= S

{I

: t ∈ S

}. Hence I

⊆ [

{I

: t ∈ S

and t|k = s} for every s ∈ S

, (4.1)

I

⊆ I

for k = 0, 1, . . . (4.2)

Since a(k + 1) > 1/2, we also have

(4.3) \

{I

: t ∈ S

and t|k = s} 6= ∅ for every s ∈ S

. It follows immediately from the definition that the bonding mappings r

are retractions; thus we can assume that

(4.4) I

⊆ I

for k = 0, 1, . . . From the definition it also follows that

(4.5) r

(I

) = I

⊆ I

for every s ∈ S

.

For k = 0, 1, . . . , there exists a metric %

on I

with the property that (4.6) diam

r

◦ r

◦ . . . ◦ r

(I

)

≤

 Y

l=k+1

a(l)



diam

I

for any m > k and s ∈ S

. The metrics %

can be defined by induction in the following way:

%

is the standard Euclidean metric on I

= I

, and %

is the restriction to I

of the Cartesian product metric of I

× P{Q

: s ∈ S

}, where I

is equipped with the metric %

and every Q

is equipped with the standard Euclidean metric.

We now prove that for each increasing sequence σ of natural numbers, Ind I

= ω

.

Since I

, and hence I

, contains a σ(k)-dimensional cube and σ(k) →

∞ as k → ∞, we have Ind I

≥ ω

.

The opposite inequality follows from Theorem 3.2 applied to M

= I

for k ∈ N; we take {I

: s ∈ S

} for A

. It is clear that (3.5) is satis- fied; (3.6) follows from (4.5), and (3.7) follows from (4.6) and the equality Q

k=1

a(k) = 0.

5. Every space X with Ind X ≤ ω

is embeddable in some I

. Let a = (a(k))

be an arbitrary sequence of real numbers such that 1/2 <

a(k) < 1 for each k ∈ N and Q

k=1

a(k) = 0.

5.1. Theorem. Every compact metrizable space X with Ind X ≤ ω

is embeddable in I

for some increasing sequence σ of natural numbers.

P r o o f. Fix a metric on X. We shall define inductively an increasing sequence σ = (σ(k))

of natural numbers and a sequence (h

)

of map- pings, where h

: X → I

, such that

(5.1)

r

◦ h

= h

for every k > 0 and

(5.2)

h

is a 1/(k + 1)-mapping.

Simultaneously, we shall define families {G

: s ∈ S

} of pairwise disjoint open subsets of X (for the definition of S

, see Section 4) satisfying the following conditions:

h

(G

) ⊆ I

for every s ∈ S

, (5.3)

h

X − [

{G

: s ∈ S

} is an embedding.

(5.4)

By Lemma 2.1, there exist an m ∈ N and a 1-mapping g

: X → I

. Since Ind X ≤ ω

, there exists a finite-dimensional partition L

in X between

E

= g

({(z

, . . . , z

) ∈ I

: z

≤ 1 − a(1)}) and

E

= g

({(z

, . . . , z

) ∈ I

: z

≥ a(1)}),

for i = 1, . . . , m; let G

and G

be disjoint open subsets of X such that E

⊆ G

, E

⊆ G

, and X − L

= G

∪ G

. Let

σ(0) = max{2(Ind L

) + 1 : i = 1, . . . , m} + m . Consider a mapping f

: X → I

such that f

| S

i=1

L

is an embedding and

f

(X) ⊆ {(z

, z

, . . . , z

) ∈ I

:

z

≤ a(1) for i = m + 1, m + 2, . . . , σ(0)} , and set h

= g

M f

; obviously, h

: X → I

= I

. The family {G

: s ∈ S

} is defined by letting

G

= G

∩ G

∩ . . . ∩ G

for s = γ = (γ(i))

∈ S

such that γ(i) = 0 for i > m, and G

= ∅ for other s ∈ S

; observe that this family consists of pairwise disjoint open subsets of X.

It is a simple matter to verify that conditions (5.2)

–(5.4)

are satisfied.

Assume that the numbers σ(0) < σ(1) < . . . < σ(k), the mappings h

, and the families {G

: s ∈ S

} of pairwise disjoint subsets of X are defined and satisfy (5.1)

–(5.4)

.

Fix an s ∈ S

. The set I

⊆ I

is closed, and so h

(cl G

) ⊆ I

by (5.3)

. Since the sets G

are open and pairwise disjoint, bd G

⊆ X − S

{G

: t ∈ S

}. Hence, by (5.4)

, h

|bd G

is an embedding of bd G

in I

.

By Lemma 2.1 applied to the space cl G

and to its closed subset bd G

, there exist a natural number m > Ind I

= σ(k) and a mapping g

: cl G

→ I

such that

(h

|cl G

) M g

is a 1/(k + 2)-mapping, (5.5)

((h

|cl G

) M g

)

(I

× {(0, . . . , 0)}) = bd G

. (5.6)

Since S

is finite, one can assume that m does not depend on s.

Let

E

= ((h

|cl G

) M g

)

({(z

, . . . , z

) ∈ I

×I

: z

≤ 1−a(k+2)}) , E

= ((h

|cl G

) M g

)

({(z

, . . . , z

) ∈ I

× I

: z

≥ a(k + 2)}) . Since Ind cl G

≤ Ind X ≤ ω

, there exists a finite-dimensional partition L

in the space cl G

between E

and E

; let G

and G

be disjoint open subsets of cl G

such that E

⊆ G

, E

⊆ G

, and cl G

− L

= G

∪ G

.

Set σ(k + 1) = max{2(Ind L

) + 2 : i = 1, . . . , m and s ∈ S

} + m.

By Lemma 2.2 applied to the space cl G

and its closed subsets A = bd G

and B = S

i=1

L

, there exists a mapping f

: cl G

→ I

such that f

[

L

− bd G

is an embedding, (5.7)

f

((0, . . . , 0)) = bd G

; (5.8)

obviously, one can additionally assume that

(5.9) f

(cl G

) ⊆ {(z

, z

, . . . , z

) ∈ I

:

z

≤ a(k + 2) for i = m + 1, m + 2, . . . , σ(k + 1)} . Let h

= (h

|cl G

) M g

M f

; then

h

: cl G

→ I

× I

× I

= I

× I

.

Identifying in the natural way I

× I

and I

, we can assume that I

is the range of h

.

By (5.6) and (5.8), we have h

(x) = h

(x) for every x ∈ bd G

and s ∈ S

; thus the mappings h

, s ∈ S

, and h

|X− S

{G

: s ∈ S

} are

compatible. Denote by h

the combination (∇{h

: s ∈ S

})∇(h

|X − S {G

: s ∈ S

}); obviously, h

: X → I

.

Since the sets I

− I

are pairwise disjoint, we have (5.10) h

(I

− I

) = G

for every s ∈ S

(see (5.6) and (5.8)).

We now show that h

satisfies (5.1)

and (5.2)

.

From the definition of h

it follows immediately that h

(x) = h

(x) for x ∈ X− S

{G

: s ∈ S

}; since r

is a retraction, r

(h

(x))

= h

(x) for these points. On the other hand, if x ∈ G

for some s ∈ S

, then r

(h

(x)) = r

(h

(x)) = r

(((h

|cl G

) M g

M f

)(x))

= (h

|cl G

)(x) = h

(x). Thus we have shown that (5.1)

is satisfied.

In order to show (5.2)

take an arbitrary point z ∈ I

. If z ∈ I

⊆ I

, then h

(z) ⊆ X − S

{G

: s ∈ S

} by (5.10); since h

(x) = h

(x) for every x ∈ X − S

{G

: s ∈ S

}, by (5.2)

, h

(z) is either empty or a one-point set. If z 6∈ I

, then z ∈ I

− I

for some s ∈ S

, and so h

(z) ⊆ G

by (5.10); thus

h

(z) = h

(z) = ((h

|cl G

) M g

M f

)

(z) ,

and so diam h

(z) ≤ 1/(k + 2) by (5.5). We have shown that (5.2)

is satisfied.

Define the family {G

: t ∈ S

} by letting

G

= G

∩ G

∩ G

∩ . . . ∩ G

for t = (s, γ) ∈ S

such that γ(i) = 0 for i = m + 1, m + 2, . . . , σ(k + 1), and G

= ∅ for other t ∈ S

.

It is easily seen that the sets G

are open, pairwise disjoint, and X − [

{G

: t ∈ S

} =



X − [

{G

: s ∈ S

}

 (5.11)

∪ [

{L

: i = 1, . . . , m and s ∈ S

} . We now show (5.3)

–(5.4)

. Take a t = (s, γ) ∈ S

. If γ(i) = 1 − a(k + 2) for some i > m, then G

= ∅, and so (5.3)

is satisfied.

Assume therefore that γ(i) = 0 for i = m + 1, m + 2, . . . , σ(k + 1). Since G

∩ E

= ∅ for i = 1, . . . , m, we have

((h

|cl G

) M g

)(G

) ⊆ {(z

, . . . , z

) ∈ I

× I

:

γ(i) ≤ z

≤ γ(i) + a(k + 2) for i = 1, . . . , m} ;

hence, by (5.9),

h

(G

) = ((h

|cl G

) M g

M f

)(G

) ⊆ I

, and therefore (5.3)

is satisfied for every s ∈ S

.

In order to prove (5.4)

it suffices to show that h

|X − S {G

: s ∈ S

} is 1-1. Put

C = X − [

{G

: s ∈ S

} , D =  [

{L

: s ∈ S

, i = 1, . . . , m}



∩  [

{G

: s ∈ S

}



; observe that X − S

{G

: t ∈ S

} = C ∪ D (see (5.11)).

Consider a pair of distinct points x, y ∈ X − S

{G

: t ∈ S

}. If one of them, say x, belongs to C, and the other to D, then h

(x) = h

(x) ∈ I

, whereas h

(y) 6∈ I

(see (5.10)); if x, y ∈ C, then h

(x) = h

(x) 6=

h

(y) = h

(y) by (5.4)

. If x, y ∈ ( S

i=1

L

) ∩ G

for some s ∈ S

, then h

(x) = h

(x) 6= h

(y) = h

(y) by (5.7). If x ∈ G

and y ∈ G

for some distinct s, t ∈ S

, then h

(x) ∈ I

− I

and h

(y) ∈ I

− I

(see (5.10)); since (I

− I

) ∩ (I

− I

) = ∅, we have h

(x) 6= h

(y). Thus (5.4)

is satisfied.

The inductive construction of the mappings h

, h

, . . . is complete.

Since the sequence (h

)

satisfies (5.1)

for k = 1, 2, . . . , it determines a mapping h : X → I

; from (5.2)

, k = 0, 1, . . . , it follows that h is 1-1, and so it is a homeomorphic embedding by compactness of X.

5.2. Corollary. For every sequence a = (a(k))

such that 1/2 <

a(k) < 1 for each k ∈ N and Q

k=1

a(k) = 0 the family {I

: σ is an increas- ing sequence of natural numbers} is universal in the class of all compact metrizable spaces X with Ind X = ω

.

6. Cardinalities of universal families. Denote by D the class of all compact metrizable spaces X with ind X = ω

; let

m = min{|A| : A is a universal family in D} . In this section we show that

(6.1) m = d .

The proof will be preceded by two lemmas. We denote by st(Y ) the star of a set Y ⊆ I

with respect to the covering {I

: t ∈ S

}, that is,

st(Y ) = [

{I

: t ∈ S

and Y ∩ I

6= ∅} .

6.1. Lemma. Let j be a natural number , and let A be a subspace of I

.

If the image of A under the projection of I

onto I

is not contained in

any st(I

), where s ∈ S

, then for some x, y ∈ A, there exists an at most σ(j)-dimensional partition in A between x and y.

P r o o f. Denote by A

the image of A under the projection of I

onto I

. Since A 6= ∅ and I

= S

{I

: s ∈ S

}, it follows that A

∩ I

6= ∅ for some s ∈ S

; take an x ∈ A whose image under the above projection—to be denoted by x

—belongs to I

. By assumption, there exists a y ∈ A whose image, say y

, belongs to I

− st(I

). Let

U

= I

− [

{I

: t ∈ S

and I

∩ I

= ∅}, V

= I

− st(I

) and L

= I

− (U

∪ V

) .

The sets U

and V

are open, and x

∈ U

, y

∈ V

; hence L

is a partition in I

between x

and y

. Of course,

Ind L

≤ Ind I

≤ σ(j) .

From the definitions it follows immediately that none of the sets I

, where t ∈ S

, intersects U

and V

simultaneously.

Set

U

= {I

: t ∈ S

and I

∩ I

6= ∅} ,

V

= {I

: t ∈ S

and I

∩ I

= ∅} for k ≥ j , and

U

=  [ U



− L

, V

=  [ V



− L

, and L

= L

for k > j . A reasoning similar to that in the proof of Theorem 3.2 shows that the sets U

, V

, L

satisfy the assumptions of Lemma 3.1 for n = Ind L

≤ σ(j);

therefore there exists an at most σ(j)-dimensional partition L in I

between x and y.

The set A ∩ L is a partition in A between x and y, and Ind(A ∩ L) ≤ Ind L ≤ σ(j).

From now on, we will only be concerned with sequences a = (a(k))

which additionally satisfy the following condition:

(∗)

Y

2a(k) < ∞

(see the beginning of Section 4).

6.2. Lemma. Let σ and τ be increasing sequences of natural numbers; let

n

= min{i = 0, 1, . . . : τ (k) ≤ σ(k + i)} for k ∈ N. Then I

is embeddable

in I

if and only if the sequence (n

)

is bounded.