Georgian Mathematical Journal 1(1994), No. 5, 479-486

GENERALIZED SIERPINSKI SETS

A. KHARAZISHVILI

Abstract. The notion of a Sierpinski topological space is introduced and some properties of such spaces connected with Borel measures are considered.

We assume that all topological spaces E to be considered below possess the following property: any singleton in E is a Borel subset of E. In particular, all Hausdorff topological spaces possess this property.

We say that a topological space E is a Luzin space if each σ-finite continuous, i.e., difffused, measure defined on the Borel σ-algebra of E is identically zero.

We say that a topological space E is a Sierpinski space if E contains none of Luzin spaces with the cardinality equal to card(E).

The classical Luzin set on the real line R gives us a nontrivial example of an uncountable Luzin topological space (see, for example, [1]). Similarly, the classical Sierpinski set on R gives us a nontrivial example of an uncountable Sierpinski topological space (see [1]). Therefore Luzin topological spaces (accordingly, Sierpinski topological spaces) may be considered as generalized Luzin sets (accordingly, as generalized Sierpinski sets). Some properties of Luzin and Sierpinski topological spaces are investigated in [2] and [3]. In this paper we investigate some other properties of Sierpinski topological spaces.

It is obvious that the cardinality of any Luzin topological space is strictly majorized by the first measurable (in the broad sense) cardinal number. The following simple proposition gives a characterization of measurable (in the broad sense) cardinal numbers in the terms of Sierpinski topological spaces.

Proposition 1. Let E be the main base set. Then the next two relations are equivalent:

1) card(E) is a measurable (in the broad sense) cardinal number;

2) the topological space (E, T) is a Sierpinski space for every topology T on the set E.

479 Proof. Indeed, if card(E) is a measurable (in the broad sense) cardinal number, then for each set X ⊂ E with card(X) = card(E) there exists a probability continuous measure defined on the family of all subsets of X. Therefore for any topology T on the set E the space (E, T) contains none of Luzin spaces with the cardinality equal to card(E). Thus we see that in this situation the topological space (E, T) is a Sierpinski space. Conversely, let us assume that the topological space (E, T) is a Sierpinski space for every topology T on the set E. If we set T = a discrete topology on E, then we immediately find that card(E) is a measurable (in the broad sense) cardinal number.

1991 Mathematics Subject Classification. 54H05.

1072-947X/1994/0900-0479$07.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1994 Plenum Publishing Corporation

It is not difficult to verify that if E is a Sierpinski topological space and X is a subspace of E with card(X) = card(E), then X too is a Sierpinski topological space.

Proposition 2. Let E be a topological space and let

E = ⋃_{i∈I} E_{i},

where (E_{i})_{i∈I} is a finite family of Sierpinski subspaces of E. Then the topological space E is a Sierpinski space. In particular, the topological sum of any finite family of Sierpinski spaces is also a Sierpinski space.

Proof. Let X be an arbitrary subspace of the space E with card(X) = card(E). Assume that X is a Luzin subspace of the space E. Since the set of indices I is finite, there exists an index i ∈ I such that the equality card(E_{i} ∩ X) = card(E_{i}) = card(E) is fulfilled.

Let us consider the set E_{i} ∩ X. This set, being a subset of the Luzin topological space X, is also the Luzin space. At the same time, this set is a subspace of the topological space E_{i}. Therefore we see that the Siepinski topological space E_{i} contains the Luzin topological space E_{i} ∩ X with the same cardinality, which is impossible. The obtained contradiction proves the proposition.

We shall ascertain below that the result of Proposition 2, generally speaking, does not hold for topological sums of infinite families of Sierpinski spaces.

Let E be the main base set whose cardinality is not cofinal with the least infinite cardinal number ω = ω_{0}. We set T (E) = {X ⊂ E : card(E\X) < card(E)}∪{∅}. It is not difficult to verify that T (E) is a topology in E and the topological space (E, T (E)) is a Sierpinski space. It is further easy to ascertain that if card(E) = ω_{ξ+1}, then any subset of (E, T (E)), having the cardinality ω_{ξ}, is discrete. Hence it follows that if card(E) = ω_{ξ+1} and the cardinal number ω_{ξ} is not measurable (in the broad sense), then any subset of the space (E, T (E)), having the cardinality ω_{ξ}, is a Luzin topological space.

Let us now consider a countable disjoint familiy of sets (E_{n})_{n∈ω} where (∀n)(n ∈ ω ⇒ card(E_{n}) = ω_{n+1}).

Provide each set E_{n} with the topology T (E_{n}) defined above and denote by E the space which is the topological sum of the family of spaces (E_{n}, T (E_{n}))_{n∈ω}. We assert that E is not a Sierpinski space. Indeed, each space (E_{n}, T (E_{n})) contains a discrete subspace L_{n} with the cardinality equal to ω_{n}. By the well-known result of Ulam the cardinal number ω_{n} is not measurable (in the broad sense). Therefore L_{n} is a Luzin space. Thus the space E contains the topological sum L of the countable family (L_{n})_{n∈ω} of Luzin spaces. Obviously, L itself is a Lusin space (we have the general fact by virtue of which the topological sum of the family of Luzin spaces is a Luzin space if the cardinality of the set of indices of this family is not measurable in the broad sense). Finally, it is absolutely clear that card(L) = card(E) = ω_{ω} and thus the topological space E is not a Sierpinski space.

Similar arguments are used to prove

Proposition 3. Let us assume that all cardinal numbers are not measurable in the broad sense (this assumption does not contradict the standard axioms of the modern set theory). Then for any infinite cardinal number a there exists a family (E_{i})_{i∈I} of topological spaces such that

1) card(I) = a;

2) each E_{i} is a Sierpinski topological space;

3) the topological sum of the family (E_{i})_{i∈I} is not a Sierpinski space.

On the other hand, we would like to note here that by Proposition 1 if (E_{i})_{i∈I} is a family of Sierpinski spaces and card(I) is a measurable (in the broad sense) cardinal number, then the topological sum (E_{i})_{i∈I} too is a Sierpinski space.

Example 1. Let E be the main base set and Φ be a filter in E. Consider the topology T_{Φ} on E associated with the filter Φ. This topology is defined by the equality T_{Φ} = {X : X ∈ Φ} ∪ {∅}. It is clear that the topology T (E) considered above is the particular case of the topology associated with the filter. Assume that for the filter Φ the following conditions hold:

1) for each element x ∈ E the set E\{x} belongs to the filter Φ;

2) the filter Φ is countably complete (ω_{1}-complete), i.e., Φ is closed with respect to intersections of arbitrary countable families of its elements.

Now we can see that the topological space (E, T_{Φ}) is not a Luzin space, since in this situation there exists a two-valued probability continuous Borel measure on E. Assume also that the cardinality of E is not measurable in the broad sense. Then the said topological space (E, T_{Φ}) is a Sierpinski space if and only if the relation X ∩ Y ≠ ∅ is fulfilled for each set X ⊂ E with card(X) = card(E) and for each set Y ∈ Φ.

Let now E be a topological space. Assume that E is not a Luzin space. Then there naturally arises the question: does the space E contain at least one Sierpinski subspace? It will be shown below that the answer to this question is negative even in the case when the cardinality of the space E is equal to the first uncountable cardinal number ω_{1}. For this we need one auxiliary assertion.

Lemma. Let ω_{α} be an arbitrary uncountable regular initial order number provided with the standard order topology and X be an arbitrary unbounded subset of ω_{α}. Then there exists an unbounded nonstationary (in ω_{α}) set Y ⊂ X.

Proof. The required set Y ⊂ X can be readily constructed by the method of transfinite recursion. Indeed, using the regularity of the ordinal number ω_{α}, it is possible to define by transfinite recursion the family of sets (Z_{ξ})_{ξ<ω_{α}} satisfying the following conditions:

1) for any index ξ < ω_{α} the set Z_{ξ} is an open bounded interval in the ordered set ω_{α};

2) the intersection Z_{ξ} ∩ X is not empty for any index ξ < ω_{α};

3) for any indices ξ < ω_{α} and ζ < ω_{α} such that ξ < ζ, the right end-point of the interval Z_{ξ} is strictly less that the left end-point of the interval Z_{ζ}.

Having constructed the family (Z_{ξ})_{ξ<ω_{α}} we choose for each ordinal number ξ < ω_{α}, an element y_{ξ} from the nonempty intersection Z_{ξ} ∩ X and set

Y = ⋃_{ξ<ω_{α}}{y_{ξ}}.

Now it is not difficult to verify that the set Y is the required one, since it is unbounded and nonstationary in ω_{α} and is certainly entirely contained in the original set X.

Example 2. Let us consider the first uncountable ordinal number ω_{1} provided with its order topology. Let Φ be the filter in the topological space ω_{1} generated by the family of all unbounded closed subsets of this space. Further, let T_{Φ} be the topology in the set ω_{1} associated with the filter Φ. Then the topological space (ω_{1}, T_{Φ}) is not a Luzin space. Indeed, if μ is the standard two-valued probability continuous Borel measure in the space ω_{1}, then the domain of definition of the measure μ̄ (of the usual completion of μ) coincides with the Borel σ-algebra of the topological space (ω_{1}, T_{Φ}). Thus we see that there exist nontrivial σ-finite Borel measures on the space (ω_{1}, T_{Φ}). Let us ascertain that at the same time the space (ω_{1}, T_{Φ}) does not contain any Sierpinski subspace. Let X be an arbitrary subspace of the space (ω_{1}, T_{Φ}). Without loss of generality it can be assumed that the cardinality of X is equal to ω_{1}. Then, according to the above lemma, there exists a nonstationary (in ω_{1}) set Y ⊂ X with the cardinality also equal to ω_{1}. Note now that the subspace Y of the space X is discrete. Taking into consideration the fact that by the result of Ulam the cardinal number ω_{1} is not measurable in the broad sense, we see that Y is a Luzin subspace of the space X. Finally, we ascertain that the topological space X is not a Sierpinski space. Therefore our topological space (ω_{1}, T_{Φ}) does not contain Seirpinski subspaces.

A similar example can evidently be constructed for an arbitrary uncountable regular nonmeasurable (in the broad sense) cardinal number ω_{α}.

For our further consideration we need two simple notions from the general set theory.

Let E be a base set and D be a family of subsets of E covering E. We put cov(D) = inf{card(D′) : D′ ⊂ D and D′ covers E}.

It is easy to see that the cardinal number cov(D) is an invariant with respect to bijective mappings. Therefore cov(D) can be regarded as some cardinal-valued characteristic of the given family D.

Let E be a base set again and D be an arbitrary family of subsets of E. We put

cof(D) = inf{card(D′) : D′ ⊂ D and for any set X ∈ D there exists

a set Z ∈ D′ such that X ⊂ Z}.

The cardinal number cof(D) is also an invariant with respect to bijective mappings and therefore can be regarded as another cardinal-valued characteristic of the given family D. If D is an ideal of subsets of E and D′ is a subfamily of D such that (∀X)(X ∈ D ⇒ (∃Z)(Z ∈ D′&X ⊂ Z)), then we say that D′ is a base of the ideal D. Clearly, this notion is dual to the well-known notion of a filter base.

We have the following proposition which gives us some sufficient conditions for the existence, in the topological space E, of a Sierpinski subspace with the cardinality equal to card(E).

Proposition 4. Let E be a topological space, μ be a nonzero σ-finite continuous Borel measure in E and D(μ) be the countably additive ideal of subsets of E generated by all μ-measure-zero sets. It is also assumed that the next two conditions are fulfilled for the ideal D(μ):

1) if card(I) < card(E), then the union of any family (X_{i})_{i∈I} ⊂ D(μ) does not coincide with the space E (in other words, the equality cov(D(μ)) = card(E) is true);

2) there exists a base of the ideal D(μ) with the cardinality less than or equal to the cardinality of E (in other words, the inequality cof(D(μ)) ≤ card(E) is true).

Then the topological space E contains some Sierpinski subspace S with card(S) = card(E).

Proof. Let ω_{α} be the initial ordinal number corresponding to the cardinality of the given topological space E. Fix any base of the ideal D(μ) with the cardinality less than or equal to ω_{α}. It is obvious that we can represent this base as a familiy of sets (Z_{ξ})_{ξ<ω_{α}}. Now, using the method of transfinite recursion, let us define the ω_{α}-sequence (s_{ξ})_{ξ<ω_{α}} of elements of the given space E. Assume that for an ordinal number ξ < ω_{α} we have already defined the partial ξ-sequence (s_{ζ})_{ζ<ξ} of elements of E. Consider the set

( ⋃_{ζ<ξ} Z_{ζ}) ⋃ ( ⋃_{ζ<ξ}{s_{ζ}}).

Since the condition 1) holds, this set does not coincide with the space E. Therefore there exists an element s_{ξ} ∈ E which does not belong to the mentioned set. In this manner we shall construct the ω_{α}-sequence (s_{ξ})_{ξ<ω_{α}} and, having done so, set

S = ⋃_{ξ<ω_{α}}{s_{ξ}}.

It remains for us to verify that the set S is a Sierpinski subspace of the space E. Indeed, let X be an arbitrary subset of S such that card(X) = card(S) = card(E) = ω_{α}. Then the procedure of the construction of the set S immediately implies that the set X is not contained in any of the sets Z_{ξ} (ξ < ω_{α}). Thus we have the inequality μ^{∗}(X) > 0 where μ^{∗} denotes an outer measure associated with the given measure μ. The latter fact leads to a conclusion that the topological space X is not a Luzin subspace. Therefore the constructed topological space S is the Sierpinski space.

One may easily note here that if card(E) = ω_{1}, then the condition 1) in the formulation of Proposition 4 becomes superfluous. The next simple example, on the other hand, shows that if card(E) = ω_{2}, then this condition plays an essential role.

Example 3. Assume that the relation 2^{ω_{1}} = ω_{2} is fulfilled. Take two disjoint sets E_{1} and E_{2} such that card(E_{1}) = ω_{1}, card(E_{2}) = ω_{2}. Identify the set E_{1} with the ordinal number ω_{1} and equip E_{1} with the topology T_{Φ} (see Example 2). Further, equip the set E_{2} with a discrete topology. Denote by E the topological sum of two spaces E_{1} and E_{2}. Now it is not difficult to verify that the topological space E satisfies the following conditions:

a) card(E) = ω_{2};

b) there exists a probability continuous Borel measure μ in the space E such that cof(D(μ)) ≤ ω_{2};

c) the space E does not contain any Sierpinski subspace.

Example 4. Let ω_{α} be an arbitrary uncountable regular initial ordinal number not measurable in the broad sense. Equip ω_{α} with its order topology and denote by Φ the filter in ω_{α} generated by the family of all closed unbounded subsets of ω_{α}. Further, set E = ω_{α} and equip E with the topology T_{Φ}. Then the topological space E satisfies the following conditions:

a) card(E) = ω_{α};

b) there exists a probability continuous Borel measure μ in the space E such that cov(D(μ)) = ω_{α};

c) the space E does not contain any Sierpinski subspace.

Thus Example 4 shows us that the condition 2) in the formulation of Proposition 4 also plays an essential role and cannot be omitted.

Example 5. Let E be a metric space with the cardinality continuum 2^{ω}. It is clear that if the cardinality continuum is measurable in the broad sense, then any subspace X of E with card(X) = card(E) is a Sierpinski topological space (in particular, E is not a Luzin topological space). Now let us consider some situation when the cardinality continuum is not measurable in the broad sense. Assume that Martin’s axiom holds and that the given metric space E is not a Luzin space. Then one can establish that there always exists a subset S of E satisfying the next two conditions:

a) card(S) = card(E) = 2^{ω};

b) S is a Sierpinski subset of the space E.

The existence of such a space S ⊂ E is proved in [3]. Note that the proof is essentially based on the following well-known fact: every σ-finite Borel measure defined in the space E is concentrated on some separable subspace of E (in other words, for every σ-finite Borel measure given in the space E there exists a separable support in E).

Example 6. Let E be again a metric space with the cardinality continuum 2^{ω}. Assume that the continuum hypothesis holds and that E is not a first category space. Then there always exists a subset L of E satisfying the next two conditions:

a) card(L) = card(E) = 2^{ω};

b) L is a Luzin subspace of the space E.

The existence of such a set L ⊂ E is also proved in [3]. To establish this result we need to consider two cases. We begin by assuming that the space E is not separable. Then there exists a discrete set L ⊂ E such that card(L) = ω_{1} = 2^{ω}. As we know, the least uncountable cardinal number ω_{1} is not measurable in the broad sense. Therefore the subspace L of E is a Luzin topological space with the cardinality continuum. Now assume E to be a separable metric space. In that case every σ-finite continuous Borel measure defined in the space E is concentrated on a first category subset of E. Using the classical transfinite construction due to Luzin (see, for instance, [1]), we can define a set L ⊂ E with the cardinality continuum such that we shall have the inequality card(L∩X) ≤ ω for each first category set X ⊂ E.

Now it is not difficult to verify that such a set L is a Luzin subspace of our space E.

From the foregoing arguments it also follows that any nonseparable metric space E contains an uncountable Luzin subspace.

Finally, let us formulate one unsolved problem concerning Sierpinski topological spaces.

Problem. Give a characterization of Sierpinski spaces in purely topological terms.

Remark. In the literature the term ”Luzin topological space” is used in other senses as well. For example, Luzin topological spaces in the sense of N. Bourbaki coincide with Borel subsets of complete separable metric spaces while in the modern set-theoretical topology Luzin topological spaces are uncountable Hausdorff spaces which contain neither isolated points nor uncountable first category subsets.

References

1. J. Oxtoby, Measure and Category, Springer-Verlag, New York–Heidelberg–Berlin, 1971.

2. A.B. Kharazishvili, Luzin spaces and Bernstein sets. (Russian) Bull. Acad. Sci. Georgian SSR, 138(1990), No.1, 25-28.

3. —–, Remark on Luzin sets and Sierpinski sets. (Russian) Bull. Acad. Sci. Georgia, 143(1991), No.3, 237-240.

(Received 03.06.1993)

Author’s address:

I. Vekua Institute of Applied Mathematics

of Tbilisi State University

2, University St., Tbilisi 380043

Republic of Georgia