WHAT MAKES A SPACE HAVE LARGE WEIGHT?

I. JUHÁSZ, L. SOUKUP, AND Z. SZENTMIKLÓSSY

Abstract. In section 2 of this paper we formulate several conditions (two of them are necessary and sufficient) which imply that a space of small character has large weight. In section 3 we construct a ZFC example of a 0-dimensional space X of size 2^{ω} with w(X) = 2^{ω} and χ(X) = nw(X) = ω, we show that CH implies the existence of a 0-dimensional space Y of size ω_{1} with w(Y) = nw(Y) = ω_{1} and χ(Y) = R(Y) = ω, and we prove that it is consistent that 2^{ω} is as large as you wish and there is a 0dimensional space Z of size 2^{ω} such that w(Z) = nw(Z) = 2^{ω} but χ(Z) = R(Z^{ω}) = ω.

1. Introduction

Since χ(X) ≥ |X| implies w(X) = χ(X), one possible answer to the question in the title is that having large character will make a space have large weight. Thus we arrive at the following more interesting problem: What makes a space have weight larger than its character? Discrete spaces give examples of such spaces but the Sorgenfrey line is first countable, has weight 2^{ω} but it has no uncountable discrete subspace. The reason for the latter space to have weight 2^{ω} is that it is weakly separated, i.e., one can assign to every point x a neighbourhood U_{x} such that x ≠ y implies either x ∉ U_{y} or y ∉ U_{x}. So we may ask now whether every first countable space of “large” weight has a “large” weakly separated subspace? This question was the actual starting point of our investigations, and while we found a negative answer to it we also succeeded in finding successively more and more general conditions that ensure having large weight for spaces of small character.

In section 2 we introduce the notion of irreducible base of a space (see definition 2.3) and investigate its basic properties. This notion is a weakening of weakly separatedness but the existence of such a base still implies that the weight of the space can not be smaller than its

1991 Mathematics Subject Classification. 54E25,03E35.

Key words and phrases. weight, net weight, weakly separated, irreducible base.

The preparation of this paper was supported by the Hungarian National Foundation for Scientific Research grant no. l908 .

The second author was supported by the Magyar Tudományért Foundation.

2 I. JUHÁSZ, L. SOUKUP, AND Z. SZENTMIKLÓSSY

cardinality. The main advantage of this notion, in contrast to weakly separatedness, lies in the fact that, as we will see in section 3, a large space with an irreducible base might have small net weight.

This leads to the formulation of the following problem:

Problem 1. Does every first countable space of uncountable weight contain an uncountable subspace with an irreducible base?

In section 3 we construct examples. First a ZFC example is given of a space Y with |Y | = w(Y) = 2^{ω} and χ(Y) = R(Y) = nw(Y) = ω. After seeing that χ(Y) R(Y) < w(Y) but χ(Y) R(Y) ≥ nw(Y) in the above mentioned example, we asked whether nw(X) ≤ R(X)χ(X) or just nw(X) ≤ R(X^{ω})χ(X) are provable for every T_{2} or regular space X. Using CH a 0-dimensional counterexample is given to the first question and using a c.c.c forcing argument we disprove the second inequality in section 3. However we don’t know ZFC counterexamples.

Problem 2. Is there a ZFC example of a space X satisfying R(X^{ω}) = χ(X) = ω but nw(X) > ω?

We know that under MA the cardinality of such a space must be at least 2^{ω} (see [4]). In [6, p.30] Todorčevič introduced the axiom (W): (W) If X is regular space with R(X)^{ω} = ω then nw(X) = ω. and he claimed that PFA implies (W).

We use standard topological notation and terminology throughout, cf [3].

2. Conditions ensuring large weight

Definition 2.1. Given a topological space ⟨X, τ⟩ and a subspace Y ⊂ X a function f is called a neighbourhood assignment on Y iff f : Y ⟶ τ and y ∈ f(y) for each y ∈ Y.

The notion of weakly separated spaces and the cardinal function R were introduced by Tkačenko in [5].

Definition 2.2. A space Y is weakly separated if we can find a neighbourhood assignment f on Y such that

∀y ≠ z ∈ Y (y ∉ f(z) ∨ z ∉ f(y)),

moreover

R(X) = sup{|Y | : Y ⊂ X is weakly separated}.

Obviously R(X) ≤ nw(X). Tkačenko asked whether R(X) = nw(X) is provable for regular spaces. Hajnal and Juhász, in [2], gave several consistent counterexamples using CH and some c.c.c forcing arguments. However, their spaces were not first countable.

If one wants to construct a first countable space on ω_{1} without uncountable weakly separated subspaces a natural idea is to force with finite approximations of a base of such a space. The space X given by a generic filter satisfies R(X) = ω, but without additional assumptions standard density arguments give w(X) = ω, too. To ensure large weight of the generic space we actually needed that the base should satisfy a certain property. As it turned out this notion proved to be useful not only in the special forcing construction. Its definition is now given below.

Definition 2.3. Let X be a topological space. A base U of X is called irreducible if it has an irreducible decomposition U = ⋃{U_{x} : x ∈ X}, i.e, (i) and (ii) below hold:

(i) U_{x} is a neighbourhood base of x in X for each x ∈ X. (ii) for each x ∈ X the family U_{x}^{−} = ⋃_{y≠x} U_{y} is not a base of X, hence

it doesn’t contain a neighbourhood base of x in X.

Let U be an irreducible base with the irreducible decomposition {U_{x} : x ∈ X}. Then for each x ∈ X, since ⋃_{y≠x} U_{y} does not contain

a neighbourhood base of x in X, we can fix an open neighbourhood U_{x} such that

∀y ∈ X \ {x} ∀V ∈ U_{y} (x ∈ V ⟶ V \ U_{x} ≠ ∅).

Let U_{x}^{∗} = {U ∈ U_{x} : U ⊂ U_{x}}. Then U^{∗} = ⋃{U_{x}^{∗} : x ∈ X} is an irreducible base of X and its irreducible decomposition {U_{x}^{∗} : x ∈ X} has the following property (∗):

∀x ≠ y ∈ X ∀U ∈ U_{x}^{∗} ∀V ∈ U_{y}^{∗} (x ∈ V ∧ y ∈ U) ⟶ V \ U ≠ ∅. (∗)

To simplify our notation we will say that a base U has property (∗) if it has a decomposition U = ⋃{U_{x} : x ∈ X} satisfying (i) and (∗) above. Obviously, any base with property (∗) is irreducible. So we established the following lemma:

Lemma 2.4. A space X has an irreducible base iff it has a base with property (∗).

The next two lemmas establish the basic connection between weakly separatedness, existence of irreducible base and the requirement w(X) ≥ |X|.

Lemma 2.5. If X is weakly separated, then X has an irreducible base.

4 I. JUHÁSZ, L. SOUKUP, AND Z. SZENTMIKLÓSSY

Proof. Let f be a neighbourhood assignment on X witnessing that it is weakly separated. Take U_{x} = {G ∈ τ_{X} : x ∈ G ⊂ f(x)} and U = ⋃{U_{x} : x ∈ X}. Then {U_{x} : x ∈ X} is an irreducible decomposition of the base U.

The converse of this lemma fails as we will see it later (theorem 3.1).

Lemma 2.6. If X has an irreducible base, then w(X) = χ(X)|X|.

Proof. If χ(X) ≥ |X| this is trivial, so assume that λ = χ(X) < |X|. Consider an irreducible base U with irreducible decomposition {U_{x} : x ∈ X}. We can assume that |U| = w(X) and |U_{x}| ≤ λ for each x ∈ X. If W ⊂ U with |W| < |X|, then there is x ∈ X with W ∩ U_{x} = ∅, so W can’t be a base by the irreducibility of U. Thus w(X) = |U| ≥ |X|.

Definition 2.7. Given a topological space X, a subspace Y ⊂ X, a neighbourhood assignment f on Y and a set N ⊂ X let

D_{N}^{f} = {y ∈ Y : y ∈ N ⊂ f(y)}.

The following results show that both weakly separatedness and having an irreducible base may be characterized with the existence of a neighbourhood assignment f such that D_{G}^{f} is “small” in some sense for each open G. For example we have the following easy result whose proof we leave to the reader.

Theorem 2.8. Given a topological space X, a subspace Y ⊂ X is weakly separated iff there is a neighbourhood assignment f on Y such that |D_{G}^{f}| ≤ 1 for each open G ⊂ X.

Lemma 2.9. If a space X has an irreducible base, then there is a neighbourhood assignment f on X such that D_{G}^{f} is closed and discrete in G for all open G ⊂ X.

Proof. Let U be a base of X having a decomposition {U_{x} : x ∈ X} with property (∗) and fix a neighbourhood assignment f with f(x) ∈ U_{x}. Assume on the contrary that x ∈ G is an accumulation point of D_{G}^{f} for some open G ⊂ X. Choose U ∈ U_{x} with x ∈ U ⊂ G. Pick y ∈ D_{G}^{f} ∩ U, y ≠ x. Then y ∈ U ∈ U_{x}, x ∈ G ⊂ f(y) ∈ U_{y}, and U ⊂ f(y), which contradicts property (∗) of U.

Theorem 2.10. The following statements are equivalent for any regular space ⟨X, τ⟩:

(1) X has an irreducible base.

(2) There is a neighbourhood assignment f on X such that D_{G}^{f} is closed and discrete in G for all open G ⊂ X.

(3) There is a neighbourhood assignment f on X such that D_{G}^{f} is a discrete subspace of X for each open G ⊂ X.

Proof.

(1) ⟶ (2). This is just lemma 2.9.

(2) ⟶ (3). Straightforward.

(3) ⟶ (1). Fix a neighbourhood assignment on X witnessing (3). Since X is regular we can assume that f(x) is regular open for each x ∈ X and that f(x) = {x} provided x is isolated. Given an open G ⊂ X set

U(G, x) = (G \ \overline{D_{G}^{f}}) ∪ \{x\}

and put

U_{x} = {U(G, x) : x ∈ D_{G}^{f} ∧ G is regular open}.

Since D_{G}^{f} is discrete and x ∈ D_{G}^{f}, we have that U(G, x) is also open, and U_{x} is a neighbourhood base of x because x ∈ U(G, x) ⊂ G. We claim that U = ⋃{U_{x} : x ∈ X} is an irreducible base because the decomposition {U_{x} : x ∈ X} has property (∗). Assume on the contrary that x ≠ y ∈ X, U(G, x) ∈ U_{x}, U(H, y) ∈ U_{y} with {x, y} ⊂ U(G, x) ∩ U(H, y) and U(G, x) ⊂ U(H, y). Since |G| > 1 and |f(x)| = 1 whenever x is isolated in X, it follows that D_{G}^{f} can not contain isolated points from X. But this set is also discrete, so \overline{D_{G}^{f}} is nowhere dense in X. Since H is regular open, U(G, x) = (G\\overline{D_{G}^{f}})∪\{x\} ⊂ H implies G ⊂ H. Thus y ∈ D_{G}^{f} for y ∈ U(G, x) ⊂ G ⊂ H ⊂ f(y) and so y ∉ U(G, x), which is impossible.

We don’t know if the assumption on the regularity of X is essential in theorem 2.10.

Next we show that the existence of an f with D_{G}^{f} “small” for all open sets G already implies that the weight of our space is large.

Definition 2.11. A topological space Y is pseudo weakly separated if it contains a weakly separated subspace Z with |Z| = |Y |.

Theorem 2.12. Let X be a topological space, Y ⊂ X, f be a neighbourhood assignment on Y and λ ≤ |Y | be a regular cardinal. If D_{G}^{f} is the union of < λ many pseudo weakly separated subspaces for each open G ⊂ X, then w(X) ≥ |Y |.

Proof. Assume on the contrary that B is a base with |B| < |Y | and let κ = |B|^{+} + λ. Since Y = ⋃_{G∈B} D_{G}^{f}, there is a G ∈ B with |D_{G}^{f}| ≥ κ. But D_{G}^{f} is the union of < λ many pseudo weakly separated subspaces, so one of them has cardinality ≥ κ. Thus D_{G}^{f} contains a weakly separated subspace Z with |Z| ≥ κ. Hence w(X) ≥ w(Y) ≥ w(Z) ≥ κ > |B| ≥ w(X), which is impossible.

Since weakly separated spaces have not just large weight but also large net weight, if we assume that D_{N}^{f} is like in theorem 2.12 for all subsets N ⊂ X, the same argument yields that even the net weight of X is large.

Theorem 2.13. Let f be a neighbourhood assignment on a topological space X and λ ≤ |X| be a regular cardinal. If D_{N}^{f} is the union of < λ many pseudo weakly separated subspaces for each N ⊂ X, then nw(X) ≥ |X|.

The following results show that the above type of “smallness” assumptions on D_{G}^{f} can actually be used to characterize spaces of small character and large weight!

Theorem 2.14. Let κ be a cardinal and X a topological space with χ(X) < κ. (If κ is regular then the assumption χ(p, X) < κ for each p ∈ X would suffice.) Then the following are equivalent:

(a) w(X) ≥ κ,

(b) There is a subspace Y ⊂ X of size κ and a neighbourhood assignment f on Y such that D_{G}^{f} is right-separated for each open G ⊂ X.

(c) There is a subspace Y ⊂ X of size κ, a neighbourhood assignment f on Y and a regular cardinal λ ≤ κ such that D_{G}^{f} is the union of < λ many pseudo weakly separated subspaces for each open G ⊂ X.

Proof. (a)⟶(b). For each x ∈ X fix a neighbourhood base B_{x} of x in X with minimal cardinality. Since w(X) ≥ κ we can construct a sequence {y_{η} : η < κ} ⊂ X such that for each η < κ the family ⋃_{ξ<η} B_{y_{ξ}} does not contain a

base of y_{η} in X and we can pick an open set f(η) ∈ τ_{X} which witnesses this, i.e., y_{η} ∈ f(η) and there is no U ∈ ⋃_{ξ<η} B_{y_{ξ}} with y_{η} ∈ U ⊂ f(η).

We claim that the neighbourhood assignment f on Y = {y_{η} : η < κ} has the property that D_{G}^{f} is right separated in its natural order for each open G. Assume on the contrary that there is an open G and ξ < κ such that y_{ξ} ∈ D_{G}^{f} ∩ \overline{\{y_{η} ∈ D_{G}^{f} : η > ξ\}}. Since H ⊂ G implies D_{G}^{f} ∩ H ⊂ D_{H}^{f} we can assume that G ∈ B_{y_{ξ}}. Then there is η > ξ with y_{η} ∈ G ∩ D_{G}^{f}.

Hence y_{η} ∈ G ⊂ f(η) and G ∈ B_{y_{ξ}}, contradicting the choice of y_{η} and f(η). (b)⟶(c). Straightforward. (c)⟶(a) This is immediate from theorem 2.12.

Conditions (b) and (c) in 2.14 have the (perhaps just aesthetic) drawback that the requirements on the subspace Y are external in nature, i.e. they do not only depend on Y. This drawback is eliminated in the following result, which however works only for regular spaces and regular cardinals.

Theorem 2.15. Let X be a regular topological space and κ a regular cardinal with χ(p, X) < κ for each p ∈ X. Then the following are equivalent:

(a) w(X) ≥ κ,

(b) There is a subspace Y ⊂ X of size κ and a neighbourhood assignment f : Y ⟶ τ_{Y} such that |D_{G}^{f}| < κ for each open G ∈ τ_{Y}.

(c) There is a subspace Y ⊂ X of size κ and a neighbourhood assignment f : Y ⟶ τ_{Y} such that D_{G}^{f} is the union of < κ many pseudo weakly separated subspaces for each open G ∈ τ_{Y}.

Proof. (a)⟶(b). If there is a weakly separated subspace Y ⊂ X with |Y | = κ then we are done by theorem 2.8. Otherwise we have h(X) < κ and d(X) < κ, hence we can pick a dense D ⊂ X with |D| < κ. For each x ∈ X fix a neighbourhood base B_{x} of x in X with minimal cardinality containing only regular open sets. Since w(X) ≥ κ we can construct a sequence {y_{η} : η < κ} ⊂ X\D such that for each η < κ writing D_{η} = {y_{ξ} : ξ < η} the family ⋃{B_{y} : y ∈ D∪D_{η}} does not contain a base of y_{η} in X and we can pick a regular open set f′(η) which witnesses this, i.e., y_{η} ∈ f′(η) and there is no U ∈ ⋃{B_{y} : y ∈ D ∪ D_{η}} with y_{η} ∈ U ⊂ f′(η). Let Y = D ∪ {y_{η} : η < κ} and define the neighbourhood assignment f on Y by f(y_{η}) = f′(y_{η}) ∩ Y for η < κ and f(d) = Y for d ∈ D. We claim that the space ⟨Y, τ_{Y} ⟩ and the neighbourhood assignment f have the property that Z_{G} = D_{G}^{f} \ D is right separated in the order inherited from the indexing for each open G. Assume on the contrary that there is an open G and ξ < κ such that y_{ξ} ∈ Z_{G} ∩ \overline{\{y_{η} ∈ Z_{G} : η > ξ\}}. Since H ⊂ G implies D_{G}^{f} ∩ H ⊂ D_{H}^{f} we can assume that there is G′ ∈ B_{y_{ξ}} with G = G′ ∩ Y. Then there is η > ξ with y_{η} ∈ G ∩ Z_{G}. But D ⊂ Y is dense in X and G′ is regular open, so y_{η} ∈ G ⊂ f(η) implies G′ ⊂ f′(η)

8 I. JUHÁSZ, L. SOUKUP, AND Z. SZENTMIKLÓSSY

which contradicts the choice of y_{η} and f′(η). Thus Z_{G} is right separated and so |Z_{G}| < κ. Therefore |D_{G}^{f}| = |Z_{G}| + |D| < κ, which means that ⟨Y, τ_{Y} ⟩ and the neighbourhood assignment f satisfy (b). (b)⟶(c). Straightforward. (c)⟶(a). w(X) ≥ w(Y) ≥ κ by theorem 2.12.

3. Examples of spaces with large weight and small character

Denote by ⟨N , ε⟩ the space of irrational numbers endowed with the Euclidean topology. For x ∈ N and η > 0 write U(x, η) = (x − η, x + η) ∩ N.

Theorem 3.1. There is a set X ⊂ N of size 2^{ω} and a 0-dimensional first countable refinement τ of ε on X such that

(i) X = Y ∪ Z, where τ_{Y} = ε_{Y} and τ_{Z} = ε_{Z}, (ii) ⟨X, τ⟩ has an irreducible base. Thus χ(X) = nw(X) = ω but w(X) = 2^{ω}.

Proof. Let Z = {z_{n} : n ∈ ω} ⊂ N be dense. Fix a nowhere dense closed set Y ⊂ N \ Z of size 2^{ω}. Let X = Y ∪ Z. For each y ∈ Y choose a strictly increasing sequence of pairwise disjoint intervals with rational endpoints, I^{y} = {I_{n}^{y} : n ∈ ω}, such that I^{y} converges to y and J^{y} = ⋃ I^{y} is disjoint from Y. This can be done because Y is nowhere dense. Set J^{z} = ∅ for z ∈ Z. For x ∈ Y and η > 0 let V (x, η) = (U(x, η) \ J^{x}) ∩ X. Let the neighbourhood base of x ∈ Y in τ be

B_{x} = {V (x, η) : η > 0}.

If z = z_{n} ∈ Z then pick η_{n} > 0 such that U(z_{n}, η_{n}) is disjoint from Y ∪ {z_{i} : i < n} and put

B_{z_{n}} = {U(x, η) : η_{n} > η > 0}.

Since

∀η > 0 ∀x ∈ V (y, η) \ {y} ∃δ > 0 U(x, δ) ∩ X ⊂ V (y, η) (†)

it follows that B = ⋃{B_{x} : x ∈ X} is a base of a topology. We claim that ⋃{B_{x} : x ∈ X} is an irreducible decomposition of B because it has property (∗). So let u, v ∈ X, U ∈ B_{u}, V ∈ B_{v} with {u, v} ⊂ U ∩ V. Then u, v ∈ Y because W ∈ B_{z_{n}} implies W ∩ X ⊂ {z_{k} : k ≥ n}. The density of Z in N implies

∀η > 0 ∀δ > 0 U(y, δ) ∩ X ⊄ V (y, η)  (+)

for each y ∈ Y. But v ∈ U implies that there is some η with U(v, η) ⊂ U, so U \ V ≠ ∅. Thus the base B is irreducible. On the other hand, τ_{Y} = ε_{Y} and τ_{Z} = ε_{Z}, because U(y, η) ∩ Y = V (y, η) ∩ Y for y ∈ Y and U(z, η) ∩ Z = V (z, η) ∩ Z for z ∈ Z.

Definition 3.2. Let Y ⊂ N. We say that a topological space ⟨Y, τ⟩ is a standard refinement of ⟨Y, ε⟩ provided that for each y ∈ Y we can choose a sequence of pairwise disjoint intervals with rational end points, I^{y} = {I_{n}^{y} : n ∈ ω}, which converges to y such that taking J^{y} = ⋃ I^{y} the family

B_{y} = {U(y, η) \ J^{y} : η > 0}.

is a neighbourhood base of y in τ.

Theorem 3.3. If CH holds, then there is a 0-dimensional first countable standard refinement τ of ε on N such that

(i) R(⟨N , τ⟩) = ω, (ii) nw(⟨N , τ⟩) = 2^{ω}, (iii) ⟨N , τ⟩ has an irreducible base.

Proof. First observe that for each D ⊂ N the set \{x ∈ \overline{D} : x ∉ \overline{(∞, x) ∩ D}\} is at most countable. Applying CH and this observation for each y ∈ N we can choose a sequence of pairwise disjoint intervals with rational endpoints, I^{y} = {I_{n}^{y} : n ∈ ω}, which is strictly increasing and converges to y, such that taking J^{y} = ⋃ I^{y} the assumptions (A)–(B) below are satisfied:

(A) ∀D ∈ [N ]^{ω} |\{y ∈ \overline{D}^{ε} : y ∉ \overline{D ∩ J^{y}}^{ε}\}| ≤ ω.

To formulate property (B) we need the following notation: for D ∈ [N ]^{ω} and y ∈ N write D[y] = {d ∈ D : y ∉ J^{d}}.

(B) ∀D ∈ [N ]^{ω} |\{y ∈ N : y ∈ \overline{D[y]}^{ε} ∧ D[y] ⊂ J^{y}| ≤ ω. Write V (y, η) = U(y, η) \ J^{y} for η > 0. Let the neighbourhood base of y in τ be

B_{y} = {V (y, η) : η > 0}.

Since

∀η > 0 ∀x ∈ V (y, η) \ {y} ∃δ > 0 U(x, δ) ⊂ V (y, η)  (†)

it follows that B = {B_{y} : y ∈ N } is a base of a topology. Since

∀η > 0 ∀δ > 0 U(y, δ) ⊄ V (y, η)  (+)

it follows that the base B is irreducible.

It is not hard to see that (A) implies that nw(⟨N , τ⟩) > ω. Indeed, assume on the contrary that {M_{m} : m < ω} is a network. Pick countable sets K_{m} ⊂ M_{m} with \overline{K_{m}}^{ε} = \overline{M_{m}}^{ε}. Then, by (A), there is y ∈ N such

10 I. JUHÁSZ, L. SOUKUP, AND Z. SZENTMIKLÓSSY

that for each m ∈ ω either y ∉ \overline{K_{m}}^{ε} = \overline{M_{m}}^{ε} or y ∈ \overline{K_{m} ∩ J^{y}}^{ε} ⊂ \overline{M_{m} ∩ J^{y}}^{ε}. Thus there is no m ∈ ω with y ∈ M_{m} ⊂ N \ J^{y}.

We will show that (B) implies R(⟨N , τ⟩) = ω. Assume on the contrary that X is an uncountable weakly separated subspace of ⟨N , τ⟩. Since ⟨N , ε⟩ has countable weight, we can assume that x ∈ J^{y} or y ∈ J^{x} hold for each x ≠ y ∈ X.

Claim . ∀D ∈ [X]^{ω} |\{x ∈ X : x ∈ \overline{D[x]}^{ε}\}| ≤ ω.

Proof of the claim. If the above defined set is uncountable, then, by (B), there is x ∈ X with D[x] ⊄ J^{x}. Let d ∈ D[x] \ J^{x} be arbitrary. Then d ∉ J^{x} and x ∉ J^{d} which contradicts our assumption on X.

Using this claim, we can find an uncountable subset Y = {y_{μ} : μ < ω_{1}\} of X such that y_{μ} ∉ \overline{Y_{μ}[y_{μ}]}^{ε}, where Y_{μ} = \{y_{ν} : ν < μ\}. So for each μ < ω_{1} we have an interval K_{μ} with rational endpoints such that y_{μ} ∈ K_{μ} and for each ν < μ if y_{ν} ∈ K_{μ} then y_{ν} ∉ Y_{μ}[y_{μ}], that is, y_{μ} ∈ J^{y_{ν}}. Since there are only countable many intervals with rational endpoints, we can assume that K_{ν} = K for each ν < ω_{1}. Since N does not contain uncountable decreasing sequences, there are ν < μ < ω_{1} with y_{ν} < y_{μ}. But J^{y} ⊂ (−∞, y) by the construction, which contradicts y_{μ} ∈ J^{y_{ν}}. So R(⟨N , τ⟩) = ω.

Let us remark that Todorčevič, in [6], proved earlier that CH implies the existence of a 0-dimensional space Y of size ω_{1} with w(Y) = nw(Y) = ω_{1} and χ(Y) = R(Y) = ω.

The next theorem shows that some set-theoretic assumption is necessary to construct a standard refinement having the above described properties. To start with let us recall the Open Coloring Axiom (OCA) (see [6] and [1]).

Open Coloring Axiom . For each second countable T_{3} space X and open H ⊂ [X]^{2} either (i) or (ii) below holds:

(i) X = ⋃_{n∈ω} X_{n} where X_{n} is H-independent,

(ii) X contains an uncountable H-complete subset.

Theorem 3.4. (OCA) If Y ⊂ N and ⟨Y, τ⟩ is a standard refinement of ⟨Y, ε⟩ then either R(Y) > ω or ⟨Y, τ⟩ is σ-second countable.

Proof. For each y ∈ Y choose a sequence of pairwise disjoint intervals with rational end points, I^{y} = {I_{n}^{y} : n ∈ ω}, which witnesses that ⟨Y, τ⟩ is a standard refinement. Let J^{y} = ⋃ I^{y}.

Unfortunately the set E′ = {⟨y, y′⟩ ∈ Y × Y : y ∈ J_{y′} or y′ ∈ J_{y}} is not open in Y × Y, so we need some extra work before applying OCA.

Fix an enumeration {K_{k} : k < ω} of the intervals with rational endpoints. For y ∈ Y let us define the function f_{y} : ω ⟶ 2 by taking f_{y}(k) = 1 iff K_{k} ⊂ J_{y}. Consider the second countable space Z = {⟨y, f_{y}⟩ : y ∈ Y } ⊂ Y × D(2)^{ω} and define the set of edges E on Z as follows:

{⟨y, f_{y}⟩ , ⟨y′, f_{y′}⟩} ∈ E ⟺ (y ∈ J_{y′} or y′ ∈ J_{y}).

It is easy to see that E is open. So OCA implies that either there is an uncountable E-complete Z′ ⊂ Z or Z is the union of countable many E-independent subsets, {Z_{n} : n ∈ ω}.

But by the definition of E, if Z′ is E-complete, then Y′ = {y ∈ Y : ⟨y, f_{y}⟩ ∈ Z′} is weakly separated. On the other hand, if Z_{n} is E-independent, then τ and ε agree on Y_{n} = {y ∈ Y : ⟨y, f_{y}⟩ ∈ Z_{n}}.

Theorem 3.5. For each uncountable cardinal κ there is a c.c.c poset P^{κ} of cardinality κ such that in V^{P^{κ}} there is a 0-dimensional first countable topological space X = ⟨κ, τ⟩ and there are c.c.c posets Q_{0} and Q_{1} satisfying the following conditions:

(a) V^{P^{κ}} ⊧ “X has an irreducible base”,

(b) V^{P^{κ}∗Q_{0}} ⊧ “X is σ-discrete”,

(c) V^{P^{κ}∗Q_{1}} ⊧ “X is σ-second countable”. So, in V^{P^{κ}}, w(X) = κ by (a), nw(X)= κ by (b) and R(X^{ω}) = ω by (c).

Proof. We say that a quadruple ⟨A, n, f, g⟩ is in P_{0}^{κ} provided (1)–(5) below hold:

(1) A ∈ [κ]^{<ω},

(2) n ∈ ω,

(3) f and g are functions,

(4) f : A × A × n ⟶ 2,

(5) g : A × n × A × n ⟶ 3, For p ∈ P_{0}^{κ} we write p = ⟨A^{p}, n^{p}, f^{p}, g^{p}⟩. If p, q ∈ P_{0}^{κ} we set p ≤ q

iff f^{p} ⊇ f^{q} and g^{p} ⊇ g^{q}. If p ∈ P_{0}^{κ}, α ∈ A^{p}, i < n^{p} set U(α, i) = U^{p}(α, i) = {β ∈ A^{p} : f^{p}(β, α, i) = 1}.

A quadruple ⟨A, n, f, g⟩ ∈ P_{0}^{κ} is in P^{κ} iff (i)–(iv) below are also satisfied:

(i) ∀α ∈ A ∀i < n α ∈ U(α, i), (ii) ∀α ∈ A ∀i < j < n U(α, j) ⊂ U(α, i), (iii) ∀α ≠ β ∈ A ∀i, j < n

\begin{matrix}U(α, i) ⊂ U(β, j) & iff & g(α, i, β, j) = 0, \\ U(α, i) ∩ U(β, j) = ∅ & iff & g(α, i, β, j) = 1. \end{matrix}

12 I. JUHÁSZ, L. SOUKUP, AND Z. SZENTMIKLÓSSY

(iv) ∀α ≠ β ∈ A ∀i, j < n

if α ∈ U(β, j) and β ∈ U(α, i) then g(α, i, β, j) = 2. We claim that P^{κ} = ⟨P^{κ}, ≤⟩ satisfies the requirements.

Definition 3.6. Assume that p_{i} = ⟨A^{i}, n^{i}, f^{i}, g^{i}⟩ ∈ P_{0}^{κ} for i ∈ 2. We say that p_{0} and p_{1} are twins iff n_{0} = n_{1}, |A_{0}| = |A_{1}| and taking n = n_{0} and denoting by σ the unique <-preserving bijection between A_{0} and A_{1} we have

(1) σ⌈A_{0} ∩ A_{1} = id_{A_{0}∩A_{1}}.

(2) σ is an isomorphism between p_{0} and p_{1}, i.e. ∀α, β ∈ A_{0}, ∀i, j < n (a) f_{0}(α, β, i) = f_{1}(σ(α), σ(β), i),

(b) g_{0}(α, i, β, j) = g_{1}(σ(α), i, σ(β), j), We say that σ is the twin function of p_{0} and p_{1}. Define the smashing function \overline{σ} of p_{0} and p_{1} as follows: \overline{σ} = σ∪id_{A_{1}}. The function σ^{∗} defined by the formula σ^{∗} = σ ∪ σ^{−1}⌈A_{1} is called the exchange function of p_{0} and p_{1}.

Definition 3.7. Assume that p_{0} and p_{1} are twins and ε : A^{p_{1}} \A^{p_{0}} ⟶ 2. A common extension q ∈ P^{κ} of p_{0} and p_{1} is called an ε-amalgamation of the twins provided

∀α ∈ A^{p_{0}}△A^{p_{1}} f^{q}(α, σ^{∗}(α), i) = ε(\overline{σ}(α)).

Lemma 3.8. If p_{0}, p_{1} ∈ P^{κ} are twins and ε : A^{p_{1}} \ A^{p_{0}} ⟶ 2, then p_{0} and p_{1} have an ε-amalgamation in P^{κ}.

Proof. Write A = A_{0} ∪ A_{1}, f^{−} = f_{0} ∪ f_{1}, g^{−} = g_{0} ∪ g_{1}. Let B and C be disjoint subsets of κ\A of size |A| and let ρ : B ⟶ A and η : C ⟶ A be 1-1. Put q = ⟨A ∪ B ∪ C, n, f, g⟩ where

(1) f^{−} ⊂ f, g^{−} ⊂ g

(2) ∀α ≠ β ∈ A ∀i, j < n

f(α, β, i) = \{ \begin{matrix}f_{1}(\overline{σ}(α), \overline{σ}(β), i) & if σ^{∗}(α) ≠ β, \\ ε(\overline{σ}(α)) & if σ^{∗}(α) = β, \end{matrix}

and

g(α, i, β, j) = \{ \begin{matrix}g_{1}(\overline{σ}(α), i, \overline{σ}(β), j) & if σ^{∗}(α) ≠ β, \\ 2 & if σ^{∗}(α) = β. \end{matrix}

(3) ∀β ∈ B ∪ C ∀i < n U^{q}(β, i) = {β},

(4) ∀α ∈ A ∀i, j < n ∀β ∈ B

β ∈ U^{q}(α, i) iff ∃l < n g(ρ(β), l, α, i) = 0

and

g(β, j, α, i) = \{ \begin{matrix}0 & if ρ(β) ∈ U^{q}(α, i), \\ 1 & if ρ(β) ∉ U^{q}(α, i). \end{matrix}

(5) ∀α ∈ A ∀i, l < n ∀γ ∈ C

γ ∈ U^{q}(α, i) iff \overline{σ}(η(γ)) ∈ U_{1}(\overline{σ}(α), i).

and

g(γ, l, α, i) = \{ \begin{matrix}0 & if η(γ) ∈ U^{q}(α, i), \\ 1 & if η(γ) ∉ U^{q}(α, i). \end{matrix}

Let us remark that (4) and (5) contain no circularity because (1) and (2) define g on A. Obviously q ∈ P_{0}^{κ} and q ≤ p_{0}, p_{1}, so we have to show that q ∈ P^{κ}. (i) is straightforward. Before checking (ii)–(iv) we need some preparation. If α, β ∈ A and i, j < n write ⟨α, i⟩ ◁ ⟨β, j⟩ iff g(α, i, β, j) = 0.

Claim 3.8.1. The relation ◁ is transitive on A × n.

Proof of claim. Assume that ⟨α, i⟩◁⟨β, j⟩◁⟨γ, l⟩. Then, by (2), ⟨\overline{σ}(α), i⟩◁ ⟨\overline{σ}(β), j⟩ ◁ ⟨\overline{σ}(γ), l⟩, so ⟨\overline{σ}(α), i⟩ ◁ ⟨\overline{σ}(γ), l⟩. Thus ⟨α, i⟩ ◁ ⟨γ, l⟩ provided σ^{∗}(α) ≠ γ. Assume that σ^{∗}(α) = γ. Then U_{1}(\overline{σ}(α), i) ⊂ U_{1}(\overline{σ}(β), j) ⊂ U_{1}(\overline{σ}(α), l) and so \overline{σ}(α) ∈ U_{1}(\overline{σ}(β), j) and \overline{σ}(β) ∈ U_{1}(\overline{σ}(α), l). Thus g_{1}(\overline{σ}(β), j, \overline{σ}(α), l) = 2 because p_{1} satisfies (iv). This contradiction proves that σ^{∗}(α) ≠ γ.

Claim 3.8.2. ∀α ≠ β ∈ A ∀i, j < n if g(α, i, β, j) = 0 then U^{q}(α, i) ⊂ U^{q}(β, j).

Proof of claim. We have g_{1}(\overline{σ}(α), i, \overline{σ}(β), j) = 0. Thus

U_{1}(\overline{σ}(α), i) ⊂ U_{1}(\overline{σ}(β), j).  (†)

Let γ ∈ A ∩ U^{q}(α, i). If σ^{∗}(γ) ≠ β then (†) implies γ ∈ U^{q}(β, j) by (2). Assume now that σ^{∗}(γ) = β. Then \overline{σ}(β) = \overline{σ}(γ) ∈ U_{1}(\overline{σ}(α), i) and, on the other hand, \overline{σ}(α) ∈ U_{1}(\overline{σ}(β), j) by (†). Thus g_{1}(\overline{σ}(α), i, \overline{σ}(β), j) = 2 because p_{1} satisfies (iv). Contradiction, σ^{∗}(γ) ≠ β. So we have shown U^{q}(α, i) ∩ A ⊂ U^{q}(β, j) ∩ A. Next we can see that U^{q}(α, i) ∩ B ⊂ U^{q}(β, j)∩B by claim 3.8.1 and by (4). Finally let γ ∈ U^{q}(α, i)∩C. Then \overline{σ}(η(γ)) ∈ U_{1}(\overline{σ}(α), i), so \overline{σ}(η(γ)) ∈ U_{1}(\overline{σ}(β), j). Thus γ ∈ U^{q}(β, j) by (5).

Claim 3.8.3. ∀α ≠ β ∈ A ∀i, j < n if g(α, i, β, j) = 1 then U^{q}(α, i) ∩ U^{q}(β, j) = ∅.

Proof of claim. Since g(α, i, β, j) = 1 we have σ^{∗}(α) ≠ β, so U_{1}(\overline{σ}(α), i)∩ U_{1}(\overline{σ}(β), j) = ∅ implies U^{q}(α, i) ∩ U^{q}(β, j) ∩ A = ∅. Assume now that γ ∈ U^{q}(α, i) ∩ U^{q}(β, j) ∩ B. Then ρ(γ) ∈ U^{q}(α, i) ∩ U^{q}(β, j) ∩ A, contradiction. Finally assume that γ ∈ U^{q}(α, i) ∩ U^{q}(β, j) ∩ C. Then \overline{σ}(η(γ)) ∈ U_{1}(\overline{σ}(α), i) ∩ U_{1}(\overline{σ}(β), j) = ∅, which is impossible.

14 I. JUHÁSZ, L. SOUKUP, AND Z. SZENTMIKLÓSSY

Claim 3.8.4. ∀α ≠ β ∈ A ∪ B ∪ C ∀i, j < n if g(α, i, β, j) = 2 then ∅ ≠ U^{q}(α, i) \ U^{q}(β, j) ≠ U^{q}(α, i).

Proof of claim. The assumption implies α, β ∈ A. If α ≠ σ^{∗}(β) then g(\overline{σ}(α), i, \overline{σ}(β), j) = 2, which implies the statement. So we can assume that α = σ^{∗}(β). Let γ = ρ^{−1}(α). Then γ ∈ U^{q}(α, i) and, on the other hand, γ ∉ U^{q}(β, j), because g(α, l, β, j) = 2 for each l < n by (2). So γ ∈ U^{q}(α, i)\U^{q}(β, j). Let δ = η^{−1}(α). Then \overline{σ}(η(δ)) = \overline{σ}(β) = \overline{σ}(α) ∈ U_{1}(\overline{σ}(α), i) ∩ U_{1}(\overline{σ}(β), j), so δ ∈ U^{q}(α, i) ∩ U^{q}(β, j).

So we have proved (iii) for q, which implies (ii). To check (iv) assume that α ≠ β ∈ A ∪ B ∪ C, i, j < n with α ∈ U^{q}(β, j) and β ∈ U^{q}(α, i). Then α, β ∈ A. If σ^{∗}(α) ≠ β, then \overline{σ}(α) ∈ U_{1}(\overline{σ}(β), j) and \overline{σ}(β) ∈ U_{1}(\overline{σ}(α), i) implies g_{1}(\overline{σ}(α), i, \overline{σ}(β), j) = 2. Hence g(α, i, β, j) = 2. If σ^{∗}(α) = β, then g(α, i, β, j) = 2 by definition. The lemma is proved.

The previous lemma implies that P^{κ} satisfies c.c.c because among uncountably many elements of P^{κ} there are always two twins. Let G be the P^{κ} generic filter and let F = ⋃{f^{p} : p ∈ G}. For each α < κ and n ∈ ω let V (α, i) = {β < κ : F (β, α, i) = 1}. Put B_{α} = {V (α, i) : i < κ} and B = ⋃{B_{α} : α < κ}. By standard density arguments we can see that B is base of a first countable topological space X = ⟨κ, τ⟩. Since P^{κ} satisfies (iv), ⋃{B_{α} : α < κ} is an irreducible decomposition of B.

We are now ready to define the posets Q_{0} and Q_{1} in V^{P^{κ}}.

A triple ⟨B, d, e⟩ is in Q_{0} iff

(a) B ∈ [κ]^{<ω},

(b) d : B ⟶ ω,

(c) e : B ⟶ ω,

(d) ∀α ≠ β ∈ B if d(α) = d(β) then α ∉ V (β, e(β)).

A quadruple ⟨B, m, d, e⟩ is in Q_{1} iff

(A) B ∈ [κ]^{<ω}, (B) d : B ⟶ ω, (C) m ∈ ω, (D) e : B × m ⟶ ω, (E) ∀α, β, γ ∈ B ∀i, j < m if d(α) = d(β) = d(γ) and e(α, i) = e(β, j)

then γ ∈ V (α, i) iff γ ∈ V (β, j). The orderings on Q_{0} and Q_{1} are defined in the straightforward way. If q and r are compatible elements of Q_{i}, then denote by q ∧ r their greatest lower bound in Q_{i}.

Lemma 3.9. P^{κ} ∗ Q_{0} satisfies c.c.c.

Proof. Let ⟨⟨p_{ν}, q_{ν}⟩ : ν < ω_{1}⟩ ⊂ P^{κ}∗Q_{0}. We can assume that p_{ν} decides q_{ν}. Write p_{ν} = ⟨A^{ν}, n^{ν}, f^{ν}, g^{ν}⟩ and q_{ν} = ⟨B^{ν}, d^{ν}, e^{ν}⟩. By standard density arguments we can assume that A^{ν} ⊃ B^{ν}. Applying standard Δ-system and counting arguments we can find ν < μ < ω_{1} such that p_{ν} and p_{μ} are twins and denoting by σ the twin function of p_{ν} and p_{μ} we have d^{ν}(α) = d^{μ}(σ(α)) and e^{ν}(α) = e^{μ}(σ(α)) for each α ∈ B^{ν}.

Define the function ε^{0} : A^{μ} \ A^{ν} ⟶ 2 by the equation ε^{0}(α) = 0. By lemma 3.8 the conditions p^{ν} and p^{μ} have an ε^{0}-amalgamation p. We claim that

p^{∥}—q_{ν} ∧ q_{μ} ∈ Q_{0}.

It is enough to check that if α ∈ B^{ν}, β ∈ B^{μ}, d^{ν}(α) = d^{μ}(β), then p^{∥}—α ∉ V (β, e^{μ}(β)). If σ^{∗}(α) ≠ β, then (2) implies this. If σ^{∗}(α) = β, then p^{∥}—α ∉ V (β, e^{μ}(β)) because p is an ε^{0}-amalgamation.

Lemma 3.10. P^{κ} ∗ Q_{1} satisfies c.c.c.

Proof. Let ⟨⟨p_{ν}, q_{ν}⟩ : ν < ω_{1}⟩ ⊂ P^{κ}∗Q_{1}. We can assume that p_{ν} decides q_{ν}. Write p_{ν} = ⟨A^{ν}, n^{ν}, f^{ν}, g^{ν}⟩ and q_{ν} = ⟨B^{ν}, d^{ν}, m^{ν}, e^{ν}⟩. By standard density arguments we can assume that A^{ν} ⊃ B^{ν}. Applying standard arguments we can find ν < μ < ω_{1} such that p_{ν} and p_{μ} are twins and denoting by σ the twin function of p_{ν} and p_{μ} we have m^{ν} = m^{μ} and d^{ν}(α) = d^{μ}(σ(α)) and e^{ν}(α, i) = e^{μ}(σ(α), i) for each α ∈ A^{ν} and i < m^{μ}.

Define the function ε^{1} : A^{μ} \ A^{ν} ⟶ 2 by the equation ε^{1}(α) = 1. By lemma 3.8 the conditions p^{ν} and p^{μ} have an ε^{1}-amalgamation p. We claim that

p^{∥}—q^{ν} ∧ q^{μ} ∈ Q_{1}.

Set d = d^{ν} ∪ d^{μ} and e = e^{ν} ∪ e^{μ}. Let α, β, γ ∈ B^{ν} ∪ B^{μ} with d(α) = d(β) = d(γ) and e(α, i) = e(β, j). We have to show that

p^{∥}— γ ∈ V (α, i) iff γ ∈ V (β, j).  (∗)

We know that p^{∥}— “\overline{σ}(γ) ∈ V (\overline{σ}(α), i) iff \overline{σ}(γ) ∈ V (\overline{σ}(β), j)”, so we can assume that p^{∥}— “\overline{σ}(γ) ∈ V (\overline{σ}(α), i) and \overline{σ}(γ) ∈ V (\overline{σ}(β), j)”,. So if σ^{∗}(γ) is different from α and β, then we are done. Assume finally that σ^{∗}(γ) = α. Then p^{∥}— “\overline{σ}(γ) ∈ V (\overline{σ}(α), i)”, so p^{∥}— “\overline{σ}(γ) ∈ V (\overline{σ}(β), j)”, thus p^{∥}— “γ ∈ V (β, j)” by (2). But p is an ε^{1}-amalgamation, so p^{∥}— “σ^{∗}(α) ∈ V (α, i)”, i.e. p^{∥}— “γ ∈ V (α, i)”.

Lemma 3.11.

V^{P^{κ}∗Q_{0}} ⊧ “X is σ-discrete”

Proof. Let H be the Q_{0}-generic filter over V^{P^{κ}}. Set d = ⋃{d^{q} : q ∈ H} and e = ⋃{e^{q} : q ∈ H}. By standard density arguments the domains of the functions d and e are κ. We have V (x, e(x)) ∩ d^{−1}{d(x)} = {x} by (d), so d^{−1}(n) is discrete for each n ∈ ω.

Lemma 3.12.

V^{P^{κ}∗Q_{1}} ⊧ “X is σ-second countable”

Proof. Let H be the Q_{1}-generic filter over V^{P^{κ}}. Take d = ⋃{d^{q} : q ∈ H} and e = ⋃{e^{q} : q ∈ H}. By standard density arguments dom(d) = κ and dom(e) = κ × ω. Fix n ∈ ω and let X_{n} = d^{−1}{n}. We claim that w(X_{n}) = ω. Indeed, ⋃{V (α, i) : α ∈ X_{n}, i < ω} is a base of X_{n} and by (E), if α, β ∈ X_{n}, i, j < ω and e(α, i) = e(β, j) then V (α, i) ∩ X_{n} = V (β, j) ∩ X_{n}.

This completes the proof of the theorem.

We have shown in [4] that every first countable T_{2} space satisfying R(X^{ω}) = ω becomes σ-second countable in a suitable c.c.c extension. Thus (c) of theorem 3.5 may be considered as the natural way to insure R(X^{ω}) = ω.

References

[1] U. Abraham, M. Rubin, S. Shelah, On the consistency of some partition theorems for continuous colorings, and the structure of ℵ_{1}-dense real order types, Ann. Pure Appl. Logic 29 (1985), 123-206.

[2] A. Hajnal, I. Juhász, Weakly separated subspaces and networks, Logic ’78, Studies in Logic, 97, 235–245.

[3] I. Juhász, Cardinal Functions - Ten Years Later, Math. Center Tracts 123, Amsterdam, 1980.

[4] I. Juhász, L. Soukup, Z. Szentmiklóssy, Forcing countable networks for spaces satisfying R(X^{ω}) = ω, to appear in Comm. Math. Univ. Carol.,

[5] M. G. Tkačenko, Chains and cardinals, Dokl. Acad. Nauk. SSSR 239 (l978) no 3, 546-549

[6] S. Todorčevič, Partition Problems in Topology, Contemporary Mathematics, vol 84, Providence, 1989

Mathematical Institute of the Hungarian Academy of Sciences

E-mail address: juhasz@math-inst.hu

Mathematical Institute of the Hungarian Academy of Sciences

E-mail address: soukup@math-inst.hu

Eötvös Loránd University, Department of Analysis

E-mail address: szentmiklossy@math-inst.hu