RaWoAGT.dvi


@
Applied General Topology

c© Universidad Politécnica de Valencia

Volume 7, No. 1, 2006

pp. 73-101

On RG-spaces and the regularity degree

R. Raphael∗ and R. G. Woods†

Abstract. We continue the study of a lattice-ordered ring G(X),
associated with the ring C(X). Following [10], X is called RG when
G(X) = C(Xδ). An RG-space must have a dense set of very weak
P-points. It must have a dense set of almost-P-points if Xδ is Lindelöf,
or if the continuum hypothesis holds and C(X) has small cardinality.
Spaces which are RG must have finite Krull dimension when taken with
respect to the prime z-ideals of C(X). There is a notion of regularity
degree defined via the functions in G(X). Pseudocompact spaces and
metric spaces of finite regularity degree are characterized.

2000 AMS Classification: Primary 54G10; Secondary 46E25, 16E50.

Keywords: P-space, almost-P space, prime z-ideal, RG-space, very weak
P-point.

1. Introduction and elementary results

Throughout X will denote a Tychonoff space and C(X) will be the ring of
continuous real-valued functions on X. The need for examples of epimorphisms
of commutative rings led early on to the consideration of rings of the form C(X).
Later, in connection with the study of the epimorphic hull (cf. [21]), a general
lattice-ordered ring called G(X) was defined. This algebra is an epimorphic
extension of C(X) and has many interesting properties. There is an associated
notion of degree denoted rg(X). It is rarely a ring of continuous functions, but
it is when X is an RG-space. There are persistent open questions concerning
these objects, the most stubborn being whether RG-spaces must have P-points.
These have been previously studied in [21] and [10].

In this article we investigate these topics further. Here are the main results.
An RG-spaces must have a dense set of what we call “very weak P-points”. It

∗Supported by the NSERC of Canada.
†Supported by the NSERC of Canada.



74 R. Raphael and R. G. Woods

must have a dense set of almost-P-points if Xδ is Lindelöf, or if the continuum
hypothesis holds and C(X) has cardinality c.

By studying the spectrum of G(X) we show that RG spaces must have
finite Krull dimension when said dimension is taken with respect to the prime
z-ideals. This allow us to characterize RG spaces when X is scattered and
perfectly normal. This generalizes previous characterizations (see [10]) in the
compact and metric cases.

The regularity degree is studied. The main results characterize (pseudo)
compact spaces and metric spaces which are of finite regularity degree.

For background and notation one should consult [10], [8] and [19]. In par-
ticular βX will denote the Stone-Cech compactification of X, the unique com-
pactification of X in which X is C∗-embedded.

Some of this work was done while the authors were visiting the Mathematics
Department of the University of Auckland. We are grateful to our hosts D.
Gauld and I. Reilly.

Associated with a function f : X → R is the function f∗ : X → R defined
as follows: f∗ vanishes on Z(f) and f∗(x) = 1/f(x) for x ∈ coz(f). In the case
when f ∈ C(X) the function f∗ is rarely X- continuous, but it is continuous
in the Gδ-topology on X. (In ring theory f

∗ is called the quasi-inverse of f in
C(Xδ))). The subalgebra of C(Xδ) generated by C(X) and the quasi-inverses
of the functions in C(X) is denoted G(X). Significantly, G(X) is itself Von
Neumann regular, which is to say that it is closed under taking quasi-inverses.
As well G(X) (like C(X)) is a lattice under point-wise operations.

By definition each function h ∈ G(X) can be written in the form
∑i=n

i=1 aib
∗
i ,

ai, bi ∈ C(X). This representation is not unique but the least n for which such
a representation for h exists is unique, and is called its regularity degree, rg(h).
The regularity degree of X is the supremum of {rg(h)|h ∈ G(X)}. It can be
infinite.

The following result summarizes some useful technical facts about the be-
haviour of the quasi-inverse and the regularity degree.

Proposition 1.1.

(i) (bounded functions) Each function in G(X) has a representation of the
required form using functions from C∗(X). Furthermore, the use of
bounded functions does not change the value of the regularity degree.

(ii) (restrictions) G(X) is well-behaved with respect to restrictions. In par-
ticular if S ⊂ X and h ∈ G(X) then h|S ∈ G(S) and rg(h|S) ≤ rg(h).

(iii) (compression) If rg(X) = ∞, then this can be witnessed by functions
in G(X) whose values lie between 0 and 1.

(iv) If f, g ∈ C(X), (fg∗)∗ = f∗g. In particular if rg(h) = 1, then rg(h∗) =
1.

Proof. Claims (ii) and (iv) are easily verified.

(i) Let h ∈ G(X), say h =
∑1=n

1=1 aib
∗
i , ai, bi ∈ C

∗(X). Then replace aib
∗
i

with (ai/c)(bi/c)
∗ where c = (1 + a2i ) ∨ (1 + b

2
i ).



On RG-spaces and the regularity degree 75

(iii) We claim that for each n ∈ N, there is a function fn ∈ G(X) such that
0 ≤ fn ≤ 1 and rg(fn) > n.

Suppose if possible that rg(X) = ∞ but that there is a positive integer
N that globally bounds the number of terms needed to represent bounded
functions in G(X). Let f ∈ G(X). Since 1 + f2 is not a zero divisor in the
regular ring G(X) its inverse 1/(1 + f2) also lies in G(X). Then f/(1 + f2)
and 1/(1+ f2) can be expressed in terms of N terms. Now the formula for the
quasi-inverse in [20, remark 3.4] shows that the number of terms (of the form
gk∗) needed for the quasi-inverse of a function needing at most N such terms
is a function of N alone. Thus 1 + f2 can be represented by M terms, where
M is an integer that depends only on N. But f is the product of 1 + f2 and
f/(1 + f2) so it requires at most MN terms.

Thus G(X) has bounded functions of arbitrarily high regularity degree.
Since G(X) is a lattice, it also contains non-negative bounded functions of
arbitrarily high regularity degree. Since multiplication of a function by a non-
zero constant does not change its regularity degree, we can obtain our functions
between 0 and 1 as claimed. �

2. The Space of Prime z-ideals of C(X) versus the Stone Space of
Z(G(X))

Let PF(X) denote the set of prime z-filters on C(X). Clearly these are in
1-to-1 correspondence with the set PZ(X) of prime z-ideals on X via the map
j : PZ(X) → PF(X) given by: j(P) = {Z(f) : f ∈ P} (see [8, chapter 2]).
In [16, section 1], there is a discussion of the “patch topology” (although this
term is not used) on PZ(X). Montgomery shows that PZ(X) with this patch
topology is homeomorphic to a compactification of Xδ.

We will present a proof of this result that is clearer than the proof in [16],
and that moreover shows that the patch topology on PZ(X) is “naturally”
homeomorphic to the space of maximal ideals of G(X), or (equivalently) to
the Stone space of the Boolean algebra of zerosets of members of G(X). It
will be technically useful to identify PZ(X), equipped with the patch topology,
with PF(X), equipped with the topology induced by the bijection j mentioned
above.

The Patch Topology.

Let S be a set of prime ideals of C(X) and let f ∈ C(X). In a slight variation
of the notation used in [16, section1], we define: h(f) = {P ∈ S : f ∈ P}
and hc(f) = S\h(f). The patch topology on S is defined to be the topology
for which {h(f) : f ∈ C(X)}

⋃
{hc(g) : g ∈ C(X)} is a subbase. Let S be

PZ(X). Clearly j[h(f)] = {α ∈ PF(X) : Z(f) ∈ α} and j[hc(f)] = {α ∈
PF(X) : Z(f) /∈ α}. Now {α ∈ PF(X) : Z(f) ∈ α} is closed under finite
intersection (as α is a z -filter and hence closed under finite intersection) and
{α ∈ PF(X) : Z(f) /∈ α} is closed under finite intersection (as α is a prime
z-filter). Thus {j[h(f)] ∩ j[hc(g)] : f, g ∈ C(X)} is an open base for the patch



76 R. Raphael and R. G. Woods

topology on PF(X). Clearly this family is just {{F ∈ PF(X) : Z ∈ F and
S /∈ F} : Z, S ∈ Z(X)}.

Stone duality and zero-dimensional compactifications.

We briefly summarize how to construct zero-dimensional compactifications of
zero-dimensional spaces. See [19, 3.2 and 4] for more details. If A is a Boolean
algebra, denote by S(A) the set of all ultrafilters on A. If a ∈ A, denote by
λ(a) the set {α ∈ S(A) : a ∈ α}. Then {λ(a) : a ∈ A} is a clopen base for a
compact Hausdorff zero-dimensional topology on S(A), and thus topologized
S(A) is called the Stone space of A.

Suppose that X is a zero-dimensional Hausdorff space (i.e. the set B(X) of
all clopen sets of X forms an open base of X). Suppose that A is a subalgebra
of B(X) and that A is a base for the open sets of X. In that case the map
i : X → S(A) given by: i(x) = {A ∈ A : x ∈ A} is a homeomorphism
from X onto the subspace i[X] of the Stone space S(A). If we identify X
and i[X], then X is a dense subspace of S(A) so S(A) is a zero-dimensional
compactification of X. Furthermore the map A → λ(A) = clS(A)A is a Boolean
algebra isomorphism from A onto B(S(A)). Finally, every zero-dimensional
compactification of X arises in this way: if K is such a compactification, then
define A to be {A∩X : A ∈ B(K)}; then K and S(A) turn out to be naturally
equivalent compactifications of X.

Theorem 2.1. Let X be a Tychonoff space. Then:

(i) The set Z(G(X)) of zero-sets of functions in G(X) is a Boolean algebra
A of clopen subsets of Xδ that forms an open base for Xδ. Hence S(A)
is a zero-dimensional compactification of Xδ.

(ii) If α ∈ S(A) define α# to be α ∩ Z(X). Then the map k : S(A) →
PF(X) given by: k(α) = α# is a homeomorphism from S(A) onto
PF(X) (with the patch topology) whose restriction to Xδ is a bijection
onto the fixed z-ultrafilters of C(X).

(iii) S(A) is the space of maximal ideals of G(X) (with the hull-kernel topol-
ogy).

Proof. (i) We know that Z(G(X)) is the collection of finite unions of sets of
the form Z ∩ C, where Z ∈ Z(X) and C ∈ coz X (see 5.2 of [21]). This is in
fact the Boolean subalgebra of the power set of X (where sups are unions, infs
are intersections, and Boolean-algebraic complements are set-theoretic comple-
ments) generated by Z(X). Let us denote it by A. Clearly Z(X) ⊆ A so A is
a base for the open sets of Xδ (as Z(X) is). Hence as noted above S(A) is a
compactification of Xδ.

(ii) First we must show that the map k is well-defined, i.e. that if α ∈ S(A)
then α# ∈ PF(X). It is routine to show that if α ∈ S(A), then α# is a z-filter
on X.

If Z, S ∈ Z(X) and Z ∪ S ∈ α# then Z ∪ S ∈ α and as α is prime (being an
ultrafilter on A), either Z ∈ α or S ∈ α. Thus either Z ∈ α# or S ∈ α# and
so α# is prime and hence in PF(X). Thus the map k is well-defined.



On RG-spaces and the regularity degree 77

We show that k is 1-to-1. Suppose that α, β ∈ S(A) and that α 6= β. We
show that α# 6= β#. Without loss of generality suppose that A ∈ α − β. As
A is a union of finitely many sets of the form Z ∩ C, where Z ∈ Z(X) and
C ∈ coz X, and as α is an ultrafilter and hence prime, there exist Z ∈ Z(X)
and C ∈ coz X such that Z∩C ∈ α−β. Thus Z ∈ α and C ∈ α. But Z∩C /∈ β
so either Z /∈ β or else C /∈ β. If Z /∈ β then Z ∈ α# − β# and so α# 6= β#. If
C /∈ β then as β is an ultrafilter on A, it follows that X − C ∈ β ∩ Z(X) = β#.
But C ∈ α so X − C /∈ α. Thus X − C /∈ α# and again α# 6= β#. Thus k is
1-to-1 as claimed.

Next we show that k maps S(A) onto PF(X). Let F ∈ PF(X) and define
α(F) = {A ∈ A} such that there exist Z ∈ F, S ∈ Z(X) − F such that
Z ∩ (X − S) ⊆ A. We will show that α(F) ∈ S(A) and that k(α(F)) = F.

Clearly if A ∈ α(F) and A ⊆ B ∈ A then B ∈ α(F). If A, B ∈ α(F) then
there exist Z, T ∈ F and S, Y ∈ Z(X) − F such that Z ∩ (X − S) ⊆ A and
T ∩ (X − Y ) ⊆ B. Then Z ∩ T ∈ F and S ∪ Y /∈ F as F is prime. Clearly
(Z ∩ T ) ∩ (X − (S ∪ Y )) ⊆ A ∩ B, so A ∩ B ∈ α(F). Thus α(F) is a filter on A.

Next we show that α(F) is an ultrafilter. It suffices to show that if A ∈
A − α(F) then there is an M ∈ α(F) such that M ∩ A = ∅. Now A has the
form A =

⋃
{Z(i)∩C(i) : Z(i), X −C(i) ∈ Z(X), i = 1 . . . n}. As A ∈ A−α(F)

for each i, Z(i) ∩ C(i) /∈ α(F) so either Z(i) /∈ F or X − C(i) ∈ F. Let
I = {i : Z(i) /∈ F} and J = {i : X − C(i) ∈ F}. Then I ∪ J = {1, . . . , n}. Let
G =

⋃
((Z(i) : i ∈ I) and H = X −

⋃
{C(j) : j ∈ J}. Then G ∈ Z(X) − F

because F is prime, and H ∈ F because F is closed under finite intersection.
Let M = H∩(X−G). Clearly M ∈ α(F). We claim that M ∩A = ∅. It suffices
to show that for all i, M ∩Z(i)∩C(i) = ∅. For any such i, as I ∪J = {1, . . . , n}
either i ∈ I or else i ∈ J. In the former case Z(i) ∩ (X − G) = ∅ and in the
latter case, C(i) ∩ H = ∅. Thus M ∩ A = ∅ and so α(F) is an ultrafilter as
claimed, and therefore an element of S(A).

Next we show that k(α(F)) = F. If Z ∈ F it is clear that Z ∈ α(F) ∩
Z(X) = α# = k(α(F)), so F ⊆ k(α(F)). Conversely, suppose that Z ∈
α(F)∩Z(X) = α# = k(α(F)). Thus there exist T ∈ F and S ∈ Z(X)−F such
that T ∩ (X − S) ⊆ Z. Thus T ⊆ Z ∪ S. As T ∈ F it follows that Z ∪ S ∈ F.
But F is prime and S /∈ F so Z ∈ F. Thus k(α(F)) ⊆ F and so k(α(F)) = F.

We have established that k is a well-defined bijection from S(A) onto PF(X).
It remains to show that k is a homeomorphism.

We claim that B = {λ(Z ∩ C) : Z, X − C ∈ Z(X)} is an open base for
S(A). To see this, note that each member of A is the union of finitely many
members of B, so as λ is a Boolean algebra isomorphism (the Stone duality ),
it preserves finite unions and each λ(A) is the union of finitely many members
of B. As each member of B is clopen in S(A), our claim follows.

We must show that the map k is both continuous and open. By the above
claim, to verify “open” it suffices to show that k[λ(Z ∩ C)] is open in PF(X)
whenever Z ∈ Z(X) and C ∈ coz X. But note that k[λ(Z ∩ C)] = {α ∩ Z(X) :
α ∈ λ(Z ∩ C)} = {α ∩ Z(X) : Z ∩ C ∈ α} = {α ∩ Z(X) : Z ∈ α, X − C /∈ α} =
{F ∈ PF(X) : Z ∈ F, X − C /∈ F} (because k is onto).



78 R. Raphael and R. G. Woods

But this latter set is open in the patch topology (see the above discussion),
so k is an open map. But as k is a bijection, it follows that k−1[{F ∈ PF(X) :
Z ∈ F andX −C /∈ F}] = λ(Z ∩C) and because sets of the form {F ∈ PF(X) :
Z ∈ F andX −C /∈ F} form an open base for the patch topology on PF(X) and
λ(Z ∩C) is open in S(A), it follows that k is continuous and a homeomorphism
as claimed.

Finally, if x ∈ Xδ then x corresponds to the ultrafilter α(x) = {A ∈ A : x ∈
A} and so k(x) = k(α(x)) = α(x) ∩ Z(X) = {Z ∈ Z(X) : x ∈ Z}, which is the
fixed z-ultrafilter at x.

(iii) This follows from the fact that the correspondence M → {Z(F) : F ∈
M} is a bijection from the set M of maximal ideals of the ring G(X) onto S(A),
and if M is given the hull-kernel topology then this bijection is a homeomor-
phism. �

The following remark likely appears in [16], but we have not been able to
pinpoint it. We retain the same notation as above.

Proposition 2.2. Let i : Xδ → X be the identity map on the underlying set
of X. Then i can be continuously extended to i∧ : S(A) → βX.

Proof. If α ∈ S(A) let i∧(α) be the unique point in ∩{clβXZ : Z ∈ α∩Z(X) =
α#}. (To see that there is indeed such a unique point, note that as α# is a
prime z-ideal there is a unique y ∈ βX such that Z(Oy) ⊆ α# ⊆ Z(My) (see [8,
7.15]).) To show that i∧ is continuous, it suffices (because {clβXZ : Z ∈ Z(X)})
is a base for the closed sets of βX) to show that (i∧)−1[clβXZ] is closed in S(A)
for each Z ∈ Z(X). But (i∧)−1[clβXZ] = {α ∈ S(A) : i

∧(α) ∈ clβXZ} = {α ∈
S(A) : Z ∈ α} = λ(A) (by Stone duality). But λ(A) is a clopen, hence closed,
subset of S(A). Hence i∧ is continuous. It also maps S(A) onto βX as j maps
Xδ onto the dense subspace X ⊂ βX. �

Note that a portion of Proposition 2.2 has also appeared (with a different
proof) in [4, 3.4, 3.9].

It may be worth noting that if y ∈ βX, then (i∧)−1(y) consists of all prime
z-filters F ∈ PF(X) for which Z(Oy) ⊆ F ⊆ Z(My) (we are still identifying
S(A) and PF(X) via α → α#).

Remark 2.3. If X is an RG-space then G(X) = C(Xδ), Z(G(X)) = Z(Xδ),
and S(Z(G(X))) = β(Xδ).

3. On regularity degree and prime z-ideal length

The following result relies on work in [4] where the Pierce sheaf was used to
study the regularity degree in general commutative rings. Both [4] and [20] are
useful references for the algebraic background needed. The following result is
the key to theorem 3.4 below.



On RG-spaces and the regularity degree 79

Lemma 3.1. Let X be Tychonoff. Suppose that C(X) has a strict ascending
chain of k + 1 prime z-ideals. Then G(X) has a function of regularity degree
at least k + 1. It can be constructed directly from the prime z-ideals in the
ascending chain.

Proof. (Cf. part (3) of [4, theorem 3.1] and the studies of the universal regular
ring by Kennison [11], Olivier [18], and Wiegand [25]). Suppose that P0 ⊂
· · · ⊂ Pk is a strict chain of prime z-ideals in C(X). For i = 0, . . . , k−1, choose
bi ∈ Pi+1 − Pi and let bk = 1. By the proof of [4, theorem 3.1] the element t =
(b0, . . . , bk) ∈ S

′ = Qcl((C(X)/P0)) × Qcl((C(X)/P1)) × · · · × Qcl((C(X)/Pk))
cannot be written as the sum of fewer than k+1 terms of the form cid

∗
i , ci, di ∈

C(X). Now the fields Qcl(C(X)/Pi) are stalks (homomorphic images) of the
universal regular ring T (C(X)) for C(X) whose spectrum is the set of prime
ideals in C(X) under the patch topology. The spectrum of G(X) (cf. 2.1 part
(iii)) is the compact space of prime z-ideals of C(X) under the patch topology.
The universality of T (C(X)) and the fact that C(X) → G(X) is epic show that
G(X) is a homomorphic image of T (C(X))—the spectrum of the former ring is
a subspace of the spectrum of the latter. (See also [4, Proposition 2.5]). Thus
because the Pi are prime z-ideals, the fields Qcl(C(X)/Pi) are homomorphic
images of G(X). Furthermore since G(X) is a regular ring, the finite product
S′ is also a homomorphic image of G(X) (use the regularity of G(X) to obtain
a function ki ∈ ((

⋂
Pj, j 6= i) − Pi) and consider the basic idempotent kik

∗
i ).

So the element t is the image of say h ∈ G(X). Clearly rg(h) ≥ k + 1 because
of the constraint on the regularity degree for t in S′. �

Definition 3.2. By the Krull z-dimension of C(X) we will mean the supremum
of the lengths of chains of prime z-ideals in C(X). The Krull z-dimension of a
maximal ideal will mean the supremum of the lengths of chains of prime z-ideals
lying in it.

Remark 3.3. It is interesting to compare the Krull z-dimension of fixed and
free maximal ideals in C(X) in different cases. When every fixed maximal
ideal is a minimal prime, X is necessarily a P-space, and each free maximal
ideal is also a minimal prime. But in the examples of [9, 4.4, 7.4] each fixed
maximal ideal has Krull z-dimension 1, and there is at least one free maximal
ideal of Krull z-dimension greater than 1. Indeed, we will presently see (in the
remarks after theorem 4.1) that in some models of Ψ, C(Ψ) will have infinite
Krull z-dimension.

Theorem 3.4. If the Krull z-dimension of C(X) is infinite then X is not an
RG space and rg(X) = ∞.

Proof. We will use the fact that if X is RG then SpecG(X), the space of prime
z-ideals of C(X) under the patch topology, is natually homeomorphic to β(Xδ)
(see remark 2.3 and the discussion that precedes it).

Let us first recall some properties of β(Xδ).
1. There is a Boolean isomorphism between the algebra of clopen sets of Xδ

and the algebra of clopen sets of β(Xδ) defined by taking traces in one direction



80 R. Raphael and R. G. Woods

and closures in the other (see the discussion before 2.1). Furthermore, if we
have a countable family of clopen sets {Bn} ⊂ β(Xδ) with traces {An} in Xδ,
then if S =

⋃
Bn, with closure S

′ ⊂ β(Xδ) then S
′ is clopen in β(Xδ) and has

trace equal to
⋃
An. (Thus β(Xδ) is basically disconnected; see [8, 4K(7)]).

2. (All closures are taken in β(Xδ)). If h ∈ G(X) , if B = cl[coz(h)] and
if A is the trace of B in X, then B = cl(A). In particular, if f ∈ C(X) then
coz(f) is clopen in Xδ and cl[coz(f)] is the clopen set of β(Xδ) that consists of
the prime z-ideals of C(X) that do not contain f.

3. Let X be an RG space. Suppose that B is clopen in β(Xδ) with trace A
in Xδ. Let P be a prime z-ideal in B. Suppose that h ∈ G(X), a ∈ C(X) and
h|A = a|A. Then h and a agree at P in the field C(Xδ)/M

P where MP is the
maximal ideal in C(Xδ) corresponding to the point P ∈ β(Xδ). (Reason: h−a
has a clopen zero set in Xδ , it vanishes on A so it vanishes on its closure B).

4. Now assume that C(X) has infinite Krull z-dimension. We could conclude
that rg(X) is infinite directly from Lemma 3.1 but instead our strategy will be
to define a family of pairwise disjoint clopen sets of Xδ and then use them to
define a function on their union that is not in G(X) because it is of ”infinite”
degree. The proof will also show that C(Xδ) has functions of arbitrarily large
regularity degree and thus rg(X) = ∞.

Since the ring C(X) has infinite Krull z-dimension, there are two possibilites.
Case 1. There exists in C(X) a maximal ideal that contains an infinite

ascending sequence {Pn} of prime z-ideals.
Case 2. Case 1 fails, so either there is an infinite descending sequence of

prime z-ideals, or all ascending chains of prime z-ideals are finite, but their
lengths are unbounded globally. One checks easily using [8, 14.8 (a)], if neces-
sary, that in case 2 it is always possible to choose a countable sequence {Ck} of
finite ascending chains {Qk,t} of prime z-ideals with the following properties:

(i) distinct chains are disjoint,
(ii) chain Ck is of length sk, say, and the sk are strictly increasing,
(iii) for each k, Qi,1  Qk,t for all i < k, for all t.
We now contradict the RG property in both cases.
Case 1. Use the ascending sequence {Pn} to choose for each n, bn ∈ Pn+1 −

Pn. Now let B1 = coz(b1) and for n > 1, let Bn = coz(bn) − [
⋃n−1

i=1 coz(bi)].
The {Bn} are non-empty clopen and disjoint in Xδ. Each Bn has Pn but no
other Pi its β(Xδ)-closure. Define h on X by h|Bn = bn|Bn for each n and
h[X −

⋃
Bi] = 0. Then h ∈ C(Xδ) and for each n, h(Pn) = bn(Pn) by part

3. Now, suppose, if possible that h ∈ G(X) say h =
∑i=m

i=1 cid
∗
i . Take the

first m + 2 clopen sets B1, . . . , Bm+2 and their closures cl(B1), . . . , cl(Bm+2)
which are disjoint in β(Xδ) [8, 6.5 III]. Let W = {P1, . . . ., Pm+2}. (Observe the
parallel with the situation of Lemma 3.1). The set W is closed and discrete in
SpecG(X) and h|W can be written as the sum of m terms. But this contradicts
Lemma 3.1 since we have a chain of m + 2 primes, and h has been chosen so
that h|W plays the role of t in 3.1.

Case 2. We inductively define a set of elements {bk,t}, t ∈ {1, . . . , sk − 1}
so that bk,t ∈ Qk,t+1 − Qk,t, and bk,t ∈ Qi,t for all i < k. Start in chain C1 by



On RG-spaces and the regularity degree 81

choosing s1 − 1 elements a1,t ∈ Q1,i+1 − Q1,i. Suppose now that the choices
have been made for the first k − 1 chains, and consider Ck. Since Qk,t+1 is a
prime ideal, condition (iii) shows that there is an element bk,t ∈ Qk,t+1 − Qk,t,
and bk,t ∈ Qi,1 for all i < k.

As in case 1, we want to get disjoint clopen sets and define a function on their
union. Consider cl[coz(bk,t)]. It contains the primes {Qk,j}, j ≤ t, it excludes
the primes Qi,t, t < k, and possibly it contains primes from subsequent chains
{Cr}, r > k.

For each k and t ∈ {1, . . . , sk − 1} let Bk,t = coz(bk,t) − [
⋃j=t−1

j=1 coz(bk,j) ∪

(
⋃

r>k coz(br,t))]. By construction the Bk,t are disjoint clopen subsets of Xδ
and clBk,t contains Qk,t and no other prime from the array. Now define h on⋃
Bk,t to be bk,t on Bk,t and let h vanish off

⋃
Bk,t. Then h ∈ C(Xδ) and

by property 3, h and bk,t agree on Qk,t. Now condition (ii) gives us chains of
arbitrary length so we argue that h /∈ G(X) as in Case 1. �

Lemma 3.5. Suppose that S is a subspace of T that induces an epi C(T ) →
C(S). Then the preimage of a chain of prime z-ideals in C(S) is a chain of
prime z-ideals of C(T ). The lengths of such chains under taking preimages is
either maintained or grows—it cannot decline.

Proof. It is well-known that the preimage of a prime is a prime. It is also well
known (cf. [14]) that taking preimages of primes under an epimorphism is 1−1.
So it suffices to check that the preimage of a z-ideal is a z-ideal. But this is
straightforward: let P be a prime z-ideal in C(S) and let Q be its preimage in
C(T ). Suppose that g ∈ Q, f ∈ C(T ) with Z(f) = Z(g). Now f|S and g|S
have the same zero set. But g|Yn is in the z-ideal P of C(S). So f|S ∈ P , and
therefore f ∈ Q. �

Theorem 3.6. A space X cannot be RG and must be of infinite regularity
degree if it contains an infinite strictly decreasing sequence of epi-inducing (for
example C∗-embedded) subspaces Yn such that for each n, Yn − Yn+1 contains
a cozero set of Yn that is dense in Yn.

Proof. We will show that C(X) contains chains of prime z-ideals of arbitrary
length. Then the conclusion follows from theorem 3.4.

Subspaces that, upon restriction, induce epimorphisms of rings were studied
in [1]. It is immediate that the restriction map C(Yn) → C(Yn+1) is an epi
if the map C(X) → C(Yn+1) is an epi. Suppose that C(Yn+1) has a strictly
ascending chain of k prime z-ideals P1, . . . , Pk. By lemma 3.5 this gives a strict
chain of k prime z-ideals in C(Yn), say Q1, . . . , Qk, where Qi = {f ∈ C(Yn) :
f|Yn+1 ∈ Pi}. Now by assumption, Yn −Yn+1 contains densely coz(h) for some
h ∈ C(Yn). Since h vanishes on Yn+1, it lies in Q1 because Q1 contains the
preimage of any function that is zero on Yn+1. Since coz(h) is dense in Yn , h
is a non-zero divisor in C(Yn). But h ∈ Q1, and it is elementary commutative
algebra that non zero divisors cannot lie in minimal primes (cf. [13]). So Q1
contains properly an additional prime z-ideal, and therefore C(Yn) has a chain
of at least k + 1 prime z-ideals.



82 R. Raphael and R. G. Woods

Now we will see that C(X) has chains of prime z-ideals of arbitrary length.
Given any natural number N, consider C(YN ) which has a chain of prime z-
ideals of length at least 1. By working back through the chain induced by the
inclusions X ⊂ Y1 ⊂ Y2 ⊂ YN one sees that C(X) has a chain of prime z-ideals
of length at least N − 1. �

Remark 3.7. Here are some particular cases of the result or of its method.

(1) X cannot be RG if it has a decreasing sequence Yn of closed epi-inducing
subspaces, with the property that for infinitely many n, Yn is weakly
Lindelöf and Yn+1 is nowhere dense in Yn. Reason: the open set Yn −
Yn+1 is a union of cozero sets so there is a union V of countably many
of them that is dense in Yn − Yn+1. But V is itself a cozero set.

(2) Suppose X is normal, RG, scattered and of infinite CB-index. Then in
the derived sequence of isolated points there cannot be infinitely many
occurances of countable sets. Reason: this is a special case of (i). All
of the remainders are closed and hence C-embedded, and at each stage
the set of isolated points is dense.

In the scattered perfectly normal case we can get a characterization, as
follows:

Theorem 3.8. A scattered perfectly normal space is RG if and only if it is of
finite CB-index.

Proof. By [10, 2.12] finite CB-index implies RG. The converse holds by theorem
3.6 because at each stage the set of isolated points is a dense cozero set. �

4. Relationships that exist between CB(X), rg(X), and Krull
z-dimension

Theorem 4.1.

(i) if X is Tychonoff, then rg(X) = rg(υX),
(ii) The map λ :

∑
aib

∗
i →

∑
aυi (b

υ
i )

∗ is a ring isomorphism from G(X)
onto G(υX). In particular, if X is pseudocompact then G(X) and
G(βX) are naturally isomorphic. Furthermore rg(X) = rg(υX).

(iii) rg(X) ≤ rg(βX).

Proof. (i) and (ii). If f =
∑

aib
∗
i ∈ G(X) let λ(f) =

∑
aυi (b

υ
i )

∗. Clearly
λ(f) ∈ G(υX) and λ(f)|X = f. Since X is Gδ-dense in υX, Xδ is dense
in (υX)δ. Using this denseness one easily verifies that λ is well-defined and
onto. That λ is 1 − 1 and respects ring operations is immediate. Clearly
rg(λ(f)) = rg(f), so rg(X) = rg(υX).

(iii). Let f ∈ G(X), with f =
∑i=n

i=1 aib
∗
i , ai, bi ∈ C

∗(X). Let F =∑i=n
i=1 a

β
i (b

β
i )

∗. Then F ∈ G(βX). If rg(βX) = k, then rg(F) ≤ k. Thus
rg(X) ≤ rg(βX). �



On RG-spaces and the regularity degree 83

Example 4.2. Spaces X with rg(X) = 1 but CB(X) > 1 and even infinite.
The P-space S of [8, 4N] is scattered and of CB-index 2. There is also a

P-space of infinite CB-index as follows : let ωω denote the smallest ordinal
of cardinality ℵω. Remove the ordinals of countable cofinality (i.e. its non P-
points) and call the resulting subspace X. Then X is a scattered P-space of
CB-index ω0.

Example 4.3. Spaces with CB(X) = 2 but with rg(X) varying.

These will all be achieved using the pseudocompact space Ψ of [8] which is
never RG because it is separable and has a zero-set which is an uncountable
discrete set. The first class furnishes examples of the fact that finite regularity
degree does not imply RG (see also example 8.8). The second case shows that
one can have pathology even when all zero sets have discrete boundaries.

First case (finite regularity degree). Take any choice of a maximal almost
disjoint family on N yielding a Ψ which is almost compact (see [8, 6J]). Such
MAD families exist by [17] and [22]. Now βΨ is scattered of CB-index 3 so βΨ is
RG and rg(βΨ) ≤ 7. Let f ∈ G(Ψ). By proposition 1.1 f has a representation
as a finite sum using functions from C∗(Ψ). Each extends to βΨ so there exists
F ∈ G(βΨ) such that F |Ψ = f. By [10, 2.12] rg(F) ≤ 7 so we also have
rg(f) ≤ 7.

Second case (infinite regularity degree). We shall use Teresawa’s theorem
[22, 2.1] choosing [0, 1] as our compact metric space without isolated points,
and work with a Ψ for which βΨ − Ψ is homeomorphic to [0, 1]. Let n ∈ N.
By theorem 10.1 below there exists fn ∈ G[0, 1]such that rg(fn) ≥ n. Since
[0, 1] is compact it is G-embedded in βΨ (cf. [10, 2.1 (c)] ) and there exists
Fn ∈ G(βΨ) such that Fn|[0, 1] = fn. Thus rg(Fn) ≥ n and rg(βΨ) = ∞. Now
by theorem 4.1 that means that rg(Ψ) = ∞.

Remark 4.4.

(1) The previous example reveals an interesting phenomenon in the be-
haviour of rg(X). A space has rg(X) = 1 exactly when it is a P-space
(see [10, 1.4]). So if rg(X) > 1, X has a non P-point x and one easily
verifies that for every neighbourhood U of x, rg(U) > 1. One might
conjecture that in general if rg(X) ≥ n then there must be a point
x ∈ X such that rg(U) ≥ n for every neighbourhood U of x. But
this is false. In the example above rg(Ψ) = ∞, but every point x has
a neighbourhood that is a one point compactification of a countable
discrete set and hence of regularity degree 2.

(2) Note as well that the previous example gives a version of Ψ for which
C(Ψ) has infinite Krull z-dimension because this holds for C[0, 1].



84 R. Raphael and R. G. Woods

5. Almost-P-points and spaces X for which |C(X)| = c

Recall that a point of X is an almost-P-point if it does not belong to a
zero-set with empty interior. We denote the set of almost-P-points of X by
gX. Clearly gX is the intersection of the dense cozero-sets of X. A space
X is called an almost-P-space if every non-empty zero-set has a non-empty
interior. These are precisely the spaces X for which gX = X.

The following facts about gX are useful. The third replies to a question by
M. Tressl (private communication) about powers of R .

Proposition 5.1.

(i) gX is the intersection of the dense Fσ-sets of X.
(ii) Let U = W ∩ S, where W is open in X and S is dense in X . Then

gU = U ∩ gX.
(iii) If X is realcompact, then gX = g(βX). In particular, if gX = ∅ and

X is realcompact, then g(βX) = ∅.

Proof. (i) Let hX denote the intersection of the dense Fσ-sets of X. As each
cozero-set of X is an Fσ-set of X , hX ⊂ gX. Conversely, suppose that p /∈ hX.
Then there is a dense Fσ-set F of X that excludes p. By [8, 3.11(b)] there exists
a zero-set Z of X such that p ∈ Z ⊂ X − F . Thus X − Z is a dense cozero-set
of X that excludes p, and p /∈ gX.

(ii) First suppose that U is open in X. Let p ∈ U. If p /∈ gX, there exists a
dense cozero-set C of X that excludes p. Then U ∩ C is a dense cozero-set of
U that excludes p. Thus p /∈ gU, and so gU ⊂ (gX) ∩ U.

Conversely, suppose that p ∈ U − gU. By (i) there is a dense Fσ-set F of U
that excludes p. Clearly there exists an Fσ-set A ⊂ X such that F = U ∩ A.
As F is dense in U and U is open in X , it follows that (X − U) ∪ A is a dense
Fσ-set of X to which p does not belong. Thus p /∈ hX, and by (i) p /∈ gX.
Thus (gX) ∩ U ⊂ gU.

Next suppose that U is dense in gX. It is routine to show that gU ⊂ U ∩gX.
Conversely, suppose that p ∈ U ∩ gX and let H be a dense Fσ-set of U. Then
there exists an Fσ-set A of X such that H = U ∩ A. As U is dense in X, so
is H and hence A. As p ∈ gX, by (i) p ∈ A. Thus p ∈ U ∩ A = H. Apply (i)
again to conclude that p ∈ gU.

In the general case, observe that W ∩S is dense in W , so by the results above
we have gU = g(W ∩ S) = (W ∩ S) ∩ gW = (W ∩ S) ∩ (W ∩ gX) = U ∩ gX.

(iii) That gX ⊂ g(βX) is a special case of (ii). Conversely let p ∈ g(βX).
As X is realcompact it is the intersection of the (necessarily dense) cozero-sets
of βX that contain X (cf. [19, 5.11(c)]). Thus p ∈ X. If C is a dense cozero-
set of X , then as X is C∗-embedded in βX , there is a cozero-set V of βX
(necessarily dense in X) for which C = V ∩ X. Thus p ∈ V as p ∈ g(βX).
Hence p ∈ V ∩ X = C and p ∈ gX. �



On RG-spaces and the regularity degree 85

Theorem 5.2. Let j : Xδ → X be the identity map on the underlying set and
let jβ : β(Xδ) → βX be its Stone extension. The following are equivalent:

(i) X is an almost-P-space,
(ii) jβ is irreducible.

Proof. (ii) ⇒ (i). Let (i) fail. Then there exists a Z ∈ Z(X) such that Z 6= ∅
and intXZ = ∅. Let A = Xδ − Z. Thus A is a proper clopen subset of Xδ and
therefore clβ(Xδ)A is a proper compact subset of β(Xδ). But j

β[clβ(Xδ)A] =
clβX(j[A]) = clβXA. But intX(X − A) = ∅ so A is dense in X and therefore
clβXA = βX. Thus j

β maps the proper closed subset clβ(Xδ)A onto βX and

therefore jβ is not irreducible, and (ii) fails.
(i) ⇒ (ii). Let (ii) fail. Thus there is a proper closed subset K of β(Xδ) such

that jβ[K] = βX. Since β(Xδ) is 0-dimensional, there exists a proper clopen
subset of β(Xδ) that contains K, and this clopen subset will be of the form
clβ(Xδ)A where A is clopen in Xδ and A 6= Xδ. Since Xδ − A 6= ∅ and is open
in Xδ, there exists an S ∈ Z(X) such that ∅ 6= S ⊂ Xδ − A.

Now βX = jβ[clβ(Xδ)A] = clβXj[A] = clβXA. Therefore A is dense in X,
and intXS = ∅. But S 6= ∅, so X is not an almost-P-space. �

Theorem 5.3. Let X be a non-pseudocompact space with no almost-P-points.
Then X has a non-compact zero-set with empty interior, and hence X has
non-remote points.

Proof. Since X is not pseudocompact choose f ∈ C(X) − C∗(X), without
loss of generality f ≥ 0. Hence there exists a countably infinite subset D =
{d(n) : n ∈ N} of X such that f(d(n + 1)) ≥ f(d(n)) + 1 for each n. Let
Z(n) = f−1[f(d(n)) − 0.25, f(d(n)) + 0.25]. Then d(n) ∈ Z(n) ∈ Z(X) and if
n 6= k then Z(n) ∩ Z(k) = ∅. As X has no almost-P-points for each n ∈ N
there exists a nowhere dense zero-set S(n) of X such that d(n) ∈ S(n). Let
T (n) = Z(n)∩S(n). Then T (n) is a nowhere dense zero-set of X that contains
d(n), and clearly f[T (n)] ⊆ [f(d(n))−0.25, f(d(n))+0.25] (since T (n) ⊆ Z(n)).
Let A(n) = X − f−1[(f(d(n)) − 0.3, f(d(n)) + 0.3)]. Then A(n) ∈ Z(X),
A(n)∩T (n) = ∅, and by the choice of the d(n), if n 6= k then A(n)∪A(k) = X.
Hence for each n there is a gn ∈ C(X) such that A(n) = Z(gn), T (n) = g

−1
n (1),

and 0 ≤ gn ≤ 1. Now define g : X → R by:

g(x) =
∑

n∈N

gn(x) (1)

We check whether this definition makes sense (i.e. whether g(x) ∈ R). If
m ∈ N and if gm(x) 6= 0 then x /∈ A(m) so as noted above if k 6= m then
x ∈ A(k); thus gk(x) = 0 if k 6= m. Thus at most one term in the sum in the
right hand side of (1) is non-zero, and g is well-defined.

Now consider the family

B = {f−1[(∞, f(d(2)))]}
⋃

{f−1[[f(d(n − 1)), f(d(n + 1))]] : n = 2, 3, 4, . . .}



86 R. Raphael and R. G. Woods

Clearly B is a locally finite family of closed subsets of X, and the restriction of
g to any one of them equals the restriction of no more than four different gn to
that set. Thus for each B ∈ B, g|B ∈ C(B) and by [8, 1A(3)], g ∈ C(X).

From our construction it is clear that g−1n (1) =
⋃
{T (n) : n ∈ N}. Thus

D ⊆ g−1(1) and so g−1(1) is a non-compact zero-set of X. Now T (n) = g−1(1)∩
f−1[(f(d(n))−0.3, f(d(n))+0.3)] so each T (n) is open in g−1(1). Hence for each
n ∈ N there exists an open subset W(n) of X such that W(n)∩g−1(1) = T (n).
If intXg

−1(1) 6= ∅ there exists some k ∈ N such that intXg
−1(1) ∩ T (k) 6= ∅.

Then ∅ 6= intXg
−1(1) ∩ W(k) ⊆ T (k), contradicting the construction of T (k).

Thus g−1(1) is a non-compact zero-set of X with empty interior, as required.
As g−1(1) is non-compact there is a point p ∈ clβXg

−1(1) − X, and as g−1(1)
is nowhere dense, such a p is not a remote point of X. �

Remark 5.4. Interestingly, a pseudocompact space with no almost-P-points
and non-measurable cellularity, must also have non-remote points because Ter-
ada [23] has shown that the points of υX − X cannot be remote. Thus any
Tychonoff space of non-measurable cardinality and with no almost-P-points
has non-remote points.

Lemma 5.5. Let V be z-embedded in X, for example a cozero set of X. Then
|C(V )| ≤ |C(X)|.

Proof. In [7, 1.4, 1.5] it is shown that for any space X, |Z(X)| = |Z(X)|ℵ0 and
|C(X)| = |Z(X)|. Now if V is z-embedded, the map Z → Z ∩ V maps Z(X)
onto Z(V ). Thus by the above |C(V )| = |Z(V )| ≤ |Z(X)| = |C(X)|. �

Lemma 5.6. Let X be an RG-space. If X is the union of ℵ1 nowhere dense
zero-sets, then Xδ can be partitioned into ℵ1 non-empty clopen subsets.

Proof. Assume that X =
⋃
{Z(α) : α < ω1}, where each Z(α) is a nowhere

dense zero-set of X. Inductively define a family {A(α) : α < ω} as follows:
A(0) = Z(0). Now let β < ω1 and assume inductively that we have defined
{A(α) : α < β} as follows:

(a) {A(α) : α < β} is a pairwise disjoint countable collection of clopen
subsets of Xδ,

(b)
⋃
{Z(α) : α < β} =

⋃
{A(α) : α < β},

(c) if α < β then A(α) ⊆ Z(α).

Now define A(β) to be Z(β) −
⋃
[A(α) : α < β}. We see that our inductive

hypotheses are satisfied for α < β + 1. Thus {A(α) : α < ω1} is a partition of
X into clopen subsets of Xδ, each of which is a nowhere dense subset of X. Let
S = {α < ω1 : A(α) 6= ∅}. First note that S must be infinite as no topological
space is the union of finitely many nowhere dense subsets. If S were countably
infinite then X =

⋃
{Z(α) : α ∈ S} which would contradict [10, 2.2]. Thus

|S| = ℵ1, and {A(α) : α ∈ S} is the desired partition of Xδ. �

Theorem 5.7. Suppose that X is an RG-space that is the union of ℵ1 nowhere
dense zero-sets. Then |C(X)| ≥ 2ℵ1.



On RG-spaces and the regularity degree 87

Proof. By the preceding lemma there is a partition {A(α) : α ∈ S} of X into
ℵ1 pairwise disjoint non-empty clopen subsets of Xδ, where |S| = ℵ1. If T ⊂ S
define gT : Xδ → R to be the characteristic function of

⋃
{A(α) : α ∈ T },

which is clopen in Xδ. Clearly T → gT is a one-to-one map from the power set
of S into C(Xδ), so |C(Xδ)| ≥ 2

|S| = 2ℵ1. �

Corollary 5.8. If the continuum hypothesis holds then an RG-space X for
which |C(X)| = c must have a dense set of almost-P- points.

Proof. If gX is not dense in X then there is a cozero-set V of X for which V ∩
gX = ∅. Hence gV = ∅ by proposition 5.1 and by the continuum hypothesis V
is the union of ℵ1 nowhere dense zero-sets. As |C(V )| ≤ |C(X)| = c < 2

c = 2ℵ1,
by theorem 5.7 V cannot be an RG-space. Hence by [10, 2.3(f)], X cannot be
an RG-space either. �

Remark 5.9.

(1) “Dense” is the best that we can do in 5.8, as the one-point compactifi-
cation of N satisfies the hypotheses of 5.8 but the unique non-isolated
point is not an almost-P-point.

(2) Another way to prove 5.8 is to note that |C(V )| = c (by the argument
above) and that V is an RG-space (by [10, 2.3 (f)]) so by 5.7, V must
have an almost-P-point, which will be an almost P-point of X as V is
open in X.

6. Results and examples involving Lindelöf spaces

We begin with a result that does not require any assumptions on Xδ.

Theorem 6.1. Let X = S ∪ T where S and T are complementary dense
realcompact z-embedded (for example, Lindelöf) suspaces of X. Then X is not
RG.

Proof. By [3, 2.6 (b)] S and T are closed and hence clopen in Xδ so χS ∈ C(Xδ).
But since S and T are both dense in X, χS is not X-continuous at any point of
X. Therefore χS /∈ G(X), as functions in G(X) are continuous on dense open
subsets of X ([10, 2.1 (a)] ). Thus X is not RG. �

The following result in the positive direction is an improvement to [10, 2.3
(d)].

Theorem 6.2. Suppose that X is RG, and that Xδ is normal. Let Y ⊂ X and
let Y be z-embedded and realcompact. Then Y is RG. In particular Y is RG if
it is Lindelöf.

Proof. By [3, 2.6] Yδ is closed in Xδ. Now apply [10, 2.3 (g)]. �

The following result is negative and considers disjoint dense subspaces of a
space X. But we again need a constraint on Xδ.

Theorem 6.3. Let S and T be disjoint dense realcompact z-embedded (say
Lindelöf) subspaces of X. If Xδ is normal then X is not RG.



88 R. Raphael and R. G. Woods

Proof. Again S and T are disjoint and closed in Xδ and by the normality
of Xδ there is a function f ∈ C(Xδ), 0 ≤ f ≤ 1 such that f[S] = 0 and
f[T ] = 1. Now A = f−1[0, 1/2] ∈ Z(Xδ) and hence it is clopen in Xδ. Again
χA[S] = 1, χA[T ] = 0. But S and T are dense in X so χA is not continuous at
any point of X. Hence as in 6.1 χA /∈ G(X), and X is not RG. �

Corollary 6.4. [CH] Let |X| = c. Then if X contains disjoint dense realcom-
pact z-embedded subspaces then X is not RG.

Proof. By [15, 4] Xδ is paracompact and hence normal. Now apply the theorem.
�

We can relax the demand on Xδ if we know that Sδ is Lindelöf.

Theorem 6.5. Let S and T be disjoint dense subspaces of X with Sδ Lindelöf
and T realcompact and z-embedded. Then X is not RG.

Proof. As above, S and T are disjoint closed sets of Xδ. Now for all x ∈ S,
there exists Vx ∈ coz(Xδ) such that x ∈ Vx and Vx∩T = ∅ (because the point x
does not belong to the closed subset T of Xδ). Therefore Sδ ⊂ {

⋃
Vx : x ∈ S}.

As S is a Lindelöf subset of Xδ, we can take a countable subcover and then
take its union to get W ∈ coz(Xδ) such that S ⊂ W and W ∩ T = ∅. Thus W
is clopen in Xδ and again χW ∈ C(Xδ) − G(X) so X is not RG. �

One way to ensure that Sδ is Lindelöf is to know that S is a union of
countably many scattered Lindelöf spaces (cf. [15, 5.2]). So we have:

Corollary 6.6. Let S and T be disjoint dense Lindelöf subspaces of X with S
the union of countably many Lindelöf scattered subspaces. Then X is not RG.

Remark 6.7. We do not know if this corollary holds if S is the union of
uncountably many dense Lindelöf subspaces. Note that if X satisfies the con-
ditions of the preceding corollary then X is weakly Lindelöf and that there
are weakly Lindelöf spaces whose Lindelöf subspaces are nowhere dense—for
example see [2, Ex. 2.2].

Theorem 6.8. If X is an RG-space and if X has a dense subspace A such
that Aδ is Lindelöf then gX is dense in X.

Proof. Case 1. Suppose, if possible, that gX = ∅. Then X =
⋃
{Zα : α ∈ I},

where each Zα ∈ Z(X) and intX[Zα] = ∅. Therefore Aδ =
⋃
{Aδ ∩Zα : α ∈ I},

so as Aδ is Lindelöf there exist {αi.i ∈ N}, such that Aδ =
⋃
{Aδ∩Zαi : i ∈ N}.

Clearly A = ∪{A ∩ Zαi : i ∈ N}.
We claim that for each i ∈ N, intA(A ∩ Zαi) = ∅. For if not, there exists a

V open in X such that ∅ 6= V ∩ A ⊂ A ∩ Zαi. As Zαi is closed nowhere dense
in X, V − Zαi is a non-empty open set of X so (V − Zαi) ∩ A 6= ∅ as A is
dense in X, contradicting the choice of V . Thus the claim holds.

Now by the claim, and the fact that A ∩ Zαi ∈ Z(A) it follows by [10,
2.2] that A is not an RG-space, and hence by [10, 2.3 (b)] that X is not an
RG-space.



On RG-spaces and the regularity degree 89

Case 2. Suppose that gX is not dense in X. Then there exists V ∈ coz(X)
such that (gX) ∩ V = gV = ∅. Now A ∩ V is dense in V because A is dense
in X, and (A ∩ V )δ = Aδ ∩ Vδ. Since Vδ is clopen in Xδ and Aδ is Lindelöf,
(A∩V )δ is a closed subspace of a Lindelöf space and hence Lindelöf itself. Thus
by case 1 (with V replacing X and A ∩ V replacing A), we have that V is not
RG and hence that X is not RG. �

Corollary 6.9. If X is RG and has a dense subspace that is a countable union
of scattered Lindelöf spaces, then gX is dense in X.

Theorem 6.10. If a space X is the union of countably many scattered Lindelöf
subspaces, then Xδ is Lindelöf and |C(Xδ)| = |C(X)|.

Proof. Let X =
⋃
{L(n) : n ∈ N}, where each L(n) is a scattered Lindelöf

space. By [15, 5.2] each (L(n))δ is Lindelöf, so as the subspace topology that
L(n) inherits from Xδ is the same as that of (L(n))δ , it follows that Xδ is the
union of countably many Lindelöf subspaces and hence is Lindelöf.

Let w(Y ) and wL(Y ) denote respectively the weight and the weak Lindelöf
number of the space Y . In [5] it is proved that for any space Y, |C(Y )| ≤
w(Y )wL(Y ). We apply this to Xδ. Since Xδ is Lindelöf we have wL(Xδ) = ℵ0.
Furthermore we have w(Xδ) ≤ |Z(X)| = |C(X)| (the latter equality is part
of [7, 1.4]). Hence we have |C(Xδ)| ≤ |C(X)

ℵ0|. But it is well-known that
|C(X)|ℵ0| = |C(X)| (see [7]). Thus |C(Xδ)| ≤ |C(X)|. But the opposite
inequality obviously holds as Xδ is just X with a stronger topology. �

Remark 6.11. Since the Comfort/ Hager inequality in the proof of 6.10 uses
the weak Lindelöf number, one might be tempted to try to strengthen 6.10
by looking for conditions on X that would imply that Xδ is weakly Lindelöf
(rather than Lindelöf). However by [21, 5.14] a weakly Lindelöf P-space is
Lindelöf, so no additional generality can be gained this way.

7. Some properties incompatible with being RG

Let i : Xδ → X be the identity map on the underlying set of X.

Theorem 7.1. X is not an RG-space if it satisfies the condition: there exist
disjoint subsets C and D of β(Xδ) so that

(i) iβ(C) and iβ(D) are both dense in βX,
(ii) for some h ∈ C(β(Xδ)), h[C] = {0} and D ⊂ coz(h).

Proof. Suppose that we have such a function h ∈ C(β(Xδ)). We will show that
h|X /∈ G(X), a contradiction. Throughout the superscript α will denote the
Stone extension of a bounded function on X and the superscript γ will denote
the Stone extension of a bounded function on Xδ.

Suppose, if possible, that h|X =
∑n

i=1 aib
∗
i , ai, bi ∈ C

∗(X) (using part (i)
of 1.1). Let b = Πbi. We will show that b = 0. Clearly b

γ = bα ◦ iβ . By
computation, (h|X)b2 ∈ C∗(X) and it determines [(h|X)b2]α ∈ C(βX). Now
the functions [(h|X)b2]α◦iβ , (h|X)b2]γ and h(bγ)2 are all in C(β(Xδ)) and they
agree on Xδ so they are equal. But h vanishes on C and therefore [(h|X)b

2]α



90 R. Raphael and R. G. Woods

vanishes on iβ[C] which is dense in βX. Thus [(h|X)b2]α is the zero function
and therefore h(bγ)2 is the zero function on β(Xδ). But h is non-zero on all
elements of D, so bγ vanishes on D which implies that bα vanishes on iβ[D], so
again by denseness, bα = 0 and b = 0.

The above argument shows the product of all the bi equals 0. In fact, the
techniques just used also show that the bi are pairwise orthogonal. For example,
if di is the product of the (n − 1) bj obtained by ignoring bi, then in (h|X)d

2
i

the term aibi(di)
2 vanishes and since (h|X)d2i ∈ C

∗(X) we get di = 0 by again
using the fact that iβ[D] is dense in βX and D ⊂ coz(h). Thus all (n − 1)-
fold products vanish. Now continue by finite descent to get that the {bi} are
orthogonal.

Now let t = b21 + · · · + b
2
n, and let s = (b

∗
1)

2 + · · · + (b∗n)
2. Since the {bi}

are orthogonal, so are the b∗i and by computation (h|X)t ∈ C
∗(X). Again the

properties of h show that ht = 0. But by computation, (h|X) = (h|X)ts, so
h|X = 0, which implies that h = 0 and this is false since h is non-zero at the
elements of D. �

Remark 7.2.

(1) If gX = ∅ and X is RG then there is a compact set in β(Xδ) disjoint
from X that contains the maximal ideals of C(β(Xδ)) that correspond
to the minimal prime ideals in C(X). (For each x ∈ X, there is a non-
zero divisor fx ∈ C(X) that lies in Mx). Let Lx be the (patch-open)
subset of β(Xδ) consisting of all prime z-ideals of C(X) that contain
fx. No minimal prime lies in any Lx and X ∪ Lx is open in β(Xδ) so
one takes its complement.

(2) It is easy to derive the non-vanishing of gX in 6.8 from theorem 7.1 us-
ing [8, Exercise 3B.1] and remark (i). Moreover the methods show that
if in addition to the hypotheses of 6.8, X is also cozero-complemented
and an RG-space then X must have a P-point (if not the minimal
primes of C(X) form a compact subset of β(Xδ) disjoint from Aδ and
7.1 gives a contradiction).

(3) It is interesting to compare the condition in the Theorem 7.1 with
Smirnov’s theorem that says that a space is Lindelöf if and only if it
is normally placed in its Stone-Cech compactification. If gX = ∅ we
can apply 7.1 if the compact subset of β(Xδ) from part (i) is a subset
of a zero set disjoint from a dense subset of X. One suspects that
there are spaces that satisfy the condition in 7.1 but do not satisfy the
hypotheses of 6.8.

Our next result uses P-spaces that are functionally countable. These were
characterized by A.W. Hager (cf. [15, 3.1,3.2]) and include (properly) all Lin-
delöf P-spaces.

Theorem 7.3. Let X be an RG space for which Xδ is functionally countable.
Then no family of at most ℵ1 nowhere dense zero-sets of X can have X as its
union.



On RG-spaces and the regularity degree 91

Proof. Suppose that X is an RG-space that is a union of ℵ1 nowhere dense
zero-sets. By lemma 5.6 there is a partition {A(α) : α < ω1} of X into ℵ1
non-empty clopen subsets of Xδ. Let H = {rα : α < ω1} be a subset of R of
cardinality ℵ1 and define f : X → R by f[A(α)] = rα. Then f ∈ C(Xδ) and
|f[Xδ]| = ℵ1, so Xδ is not functionally countable. �

8. The structure of G(X). G-embedded subspaces

Proposition 8.1. (G(X) as a functor). Let k : X → Y be continuous. Then
under composition k induces a (natural) ring homomorphism G(k) : G(Y ) →
G(X). Also rg(G(k))(f) ≤ rg(f)

Proof. By a straightforward computation (b ◦ k)∗ = b∗ ◦ k for any b ∈ C(Y ).
Furthermore if f ∈ G(Y ), then f ◦ k ∈ G(X) and rg(f ◦ k) ≤ rg(f) as follows.

Let rg(f) = n, f =
∑i=n

i=1 aib
∗
i , ai, bi ∈ C(X). So f ◦ k = (

∑i=n
i=1 aib

∗
i ) ◦ k =∑i=n

i=1 (ai ◦ k)(b
∗
i ◦ k) =

∑i=n
i=1 (ai ◦ k)(bi ◦ k)

∗. Now ai ◦ k, bi ◦ k ∈ C(X) so
f ◦ k ∈ G(X) and rg(f ◦ k) ≤ n = rg(f). �

Example 8.2. Note that it is possible that k be a surjection and that rg(f ◦
k) < rg(f). The perfect irreducible surjection βN → N∗ is a simple example.
More generally, suppose that X is an almost-P-space with no P-points, and let
i : Xδ → X be the identity map on the underlying set. We know by 5.2 that
k = βi : β(Xδ) → βX is a perfect irreducbile continuous surjection, a very
well-behaved map. Since X has no P-points, rg(X) ≥ 2 so let f ∈ G∗(X) with
rg(f) ≥ 2. Now fβ ∈ G(βX) and rg(fβ) ≥ 2 since rg(f) = rg(fβ|X) ≤ rg(fβ).
But f ◦ i ∈ C∗(Xδ) as G

∗(X) ⊂ C∗(Xδ). So (f ◦ i)
β ∈ C∗((βX)δ) and so

rg((f ◦ i)β = 1. But fβ ◦ βi|Xδ = (f ◦ i)
β|Xδ = f ◦ i, so f

β ◦ βi = (f ◦ i)β and
therefore rg(fβ◦βi) = 1 while rg(fβ) = 2. Note that since βXδ is not a P-space,
(its regularity degree exceeds 1), yet for all f ∈ G(X), fβ ◦ βi ∈ G(β(Xδ)) and
rg(fβ ◦ βi) = 1.

Remark 8.3. If k : X → Y is a continuous surjection and S ⊂ Y , then one
easily has χS ◦ k = χk−1[S]. Thus if S ∈ Z(Y ) is not open, rg(χS) = 2, while

rg(χS ◦k) = rg(χk−1[S]). Now if S ∈ Z(Y ) then k
−1[S] ∈ Z(X) so rg(χk−1[S]) =

1 if and only if k−1[S] is clopen in X. Thus if S ∈ Z(Y ), rg(χS) = rg(χS ◦ k) if
and only if either S and k−1[S] are both non-clopen in Y and X respectively,
or else S and k−1[S] are both clopen respectively in Y and X.

Lemma 8.4. Let V be a cozero-set of X and let f : V → R be a function.
Define f∧ to coincide with f on V and to vanish on X − V . Then:

(i) (f∧)∗ = (f∗)∧,
(ii) if f ∈ G(V ), then f∧ ∈ G(X) and rg(f) = rg(f∧). (In particular V is

G-embedded in X).

Proof. Result (i) is readily checked. (ii). First assume that f ∈ C(V ), and
let V = coz(g), g ∈ C∗(X). Let k = f[1 + |f|]−1. Then k ∈ C∗(V ) so by
[3, 1.1] there exists h ∈ C(X) such that h[X − V ] = 0 and h|V = kg. Now
[1+|f|]−1 ∈ C∗(V ) so for the same reason there is a t ∈ C(X) with t[X−V ] = 0



92 R. Raphael and R. G. Woods

and t|V = g[1 + |f|]−1. Thus ht∗ ∈ G(X) and rg(ht∗) = 1. One checks that
ht∗ = f∧ so rg(f∧) = 1 = rg(f) establishing (b) when f ∈ C(V ).

Now assume that f ∈ G(V ) and that rg(f) = n. There exist ai, bi ∈ C(V )

so that f =
∑i=n

i=1 aib
∗
i . Using (i) one easily verifies that

f∧ =
∑

a∧i (b
∧
i )

∗ (∗∗)

By part (iv) of 1.1 and the previous paragraph each (b∧i )
∗ is in G(X) and has

regularity degree 1. Similarly each a∧i is in G(X) and has regularity degree 1.
Thus it follows from (∗∗) that f∧ ∈ G(X) and rg(f∧) ≤ n = rg(f). So by part
(ii) of proposition 1.1 rg(f∧) = rg(f). �

Lemma 8.5. Let V ∈ coz(X) and V ⊂ T ⊂ X. Then if T − V is G-embedded
in X, then T is G-embedded in X.

Proof. Let S = T − V . Let f ∈ G(T ). Now f|S ∈ G(S) so there exists
k ∈ G(X) such that k|S = f|S. Let V = coz(g). By lemma 8.4 there exists
f∧ ∈ G(X) such that f∧|V = f and rg(f∧) = 1. Let h = f∧gg∗ + k(1 − gg∗).
Then h ∈ G(X) and one easily verifies that h|T = f and T is G-embedded in
X. �

Theorem 8.6. Let X be a cozero-complemented space. Then:

(i) If f ∈ G(X) there is a dense cozero-set V of X such that f|V ∈ C(V ).
(ii) If X is also normal and the nowhere dense zero-sets of X are RG-

spaces then G(X) is the set of f ∈ C(Xδ) for which there exists a
dense cozero-set V of X such that f|V ∈ C(V ).

(iii) If X is normal, and there is an integer k such that rg(Z) ≤ k for each
nowhere dense zero-set Z ∈ Z(X) then rg(X) ≤ 2k + 1.

Proof. (i) Let f ∈ G(X). By the proof [10, 2.1(a)] there is a finite family
{Z(i) : i = 1, . . . , n} of zero-sets of X such that f is continuous on

⋂
{(X −

Z(i)) ∪ intXZ(i)}. As X is cozero-complemented, for each i there exists a
cozero-set V (i) of X that is dense in intXZ(i). (If intXZ(i) is empty then so
is V (i), but that will cause no problems.) Let W(i) = (X − Z(i)) ∪ V (i) and
let V =

⋂
{W(i)}. Then V is a dense cozero-set of X contained in

⋂
{(X −

Z(i)) ∪ intXZ(i)} and f|V ∈ C(W) as required.
(ii) Suppose that f ∈ C(Xδ) and there exists g ∈ C

∗(X) such that coz g
is dense in X and f|coz g ∈ C(coz g). By the proof of lemma 8.4 there exists
f∧ ∈ G(X) such that f∧|coz g = f|coz g and rg(f∧) = 1. By hypothesis Z(g)
is an RG-space, and f|Z(g) ∈ C((Z(g))δ), so f|Z(g) ∈ G(Z(g)). Hence there
exist n ∈ N, ai, bi ∈ C(Z(g)) such that f|Z(g) =

∑n
1 aib

∗
i . As X is normal,

Z(g) is C-embedded in X and so for each i there exist ji, ki ∈ C(X) such that
ji|Z(g) = ai and ki|Z(g) = bi. Let m = j1k1

∗ · · · + jnkn
∗. Then m ∈ G(X)

and m|Z(g) = f|Z(g). We now claim that

f = f∧gg∗ + (1 − gg∗)m, (♯)

(where f∧ is as defined in 8.4).



On RG-spaces and the regularity degree 93

For if x ∈ coz g then (1−gg∗)(x) = 0 and f∧gg∗(x) = f(x)g(x)/g(x) = f(x),
so the right hand side of equation ♯ evaluated at x gives f(x) + 0 = f(x). If
x ∈ Z(g) then g∗(x) = 0 and the right hand side of equation ♯ evaluated at x
gives = 0 + (1)m(x) = f(x). Our claim is verified and so f ∈ G(X).

(iii) Let f ∈ G(X). By (i) there exists g ∈ C∗(X) such that Z(g) is nowhere
dense in X and f|X − Z(g) ∈ C(X − Z(g)). As f|Z ∈ G(Z), rg(f|Z) ≤ k

so there are ai, bi ∈ C(Z) so that f|Z =
∑k

1 aib
∗
i . As X is normal Z is

C-embedded in X so there exist A1, . . . , Ak, B1, . . . , Bk ∈ C(X) such that
Ai|Z = ai and Bi|Z = bi. By part (ii) of 8.4 (f|X − Z(g))

∧ ∈ G(X) and
rg(f|X − Z(g))∧ = 1. Let m = A1B

∗
1 + · · · + AkB

∗
k; then by part(ii) of 8.4

m ∈ G(X) and rg(m) ≤ k. It is routine to show that f = (f|X − Z(g))∧gg∗ +
(1 − gg∗)m, from which it follows that rg(f) ≤ 2k + 1. �

Remark 8.7.

(1) Example [10, 2.10] (call it X) satisfies all the hypotheses of 8.6(ii)
except normality, but it does not satisfy its conclusion. (As usual,
let I(X) denote the set of isolated points of X. Observe that every
nowhere dense set of X is discrete and hence an RG-space, and that
if V ∈ coz X then I(X) − V is the complementary cozero-set.) As
f|I(X) is continuous for each f ∈ C(Xδ), the set occuring in part (b)
of the statement of 8.6 is just C(Xδ ), but this does not equal G(X)
because as noted in [10, 2.10] X is not an RG-space. This shows that
the hypothesis of normality cannot be dropped from 8.6(ii).

(2) As zero-sets of RG-spaces are RG-spaces, one consequence of 8.6 part
(a) is that if X is a normal cozero-complemented RG-space, and if
f ∈ C(Xδ), then there exists a dense cozero-set V of Xsuch that f|V ∈
C(V ). Thus if X is a normal basically disconnected RG-space and if
f ∈ C(Xδ), then there exists a dense cozero-set V of X such that
f|V ∈ C(V ).

(3) In [10, 5.6] it is shown that nowhere dense z-embedded zero-sets of
quasi-P spaces are P-spaces. Thus a normal cozero-complemented
quasi-P space X will have rg(X) finite (see the argument in the re-
marks below). In particular:

Example 8.8. A countable nodec space of finite regularity degree.

The space X presented in [9, 5.10] is a countable nowhere locally compact
nodec (i.e. closed nowhere dense subspaces are discrete) extremally discon-
nected quasi-P space without P-points. Thus it satisfies the hypotheses of 8.6
and because its nowhere dense zero-sets are RG-spaces, the function m con-
structed in the proof of 8.6 can be taken to be in C(X) (as the function f|Z(g)
will be in C(Z(g))). From equation (♯) in 8.6 it follows that if f ∈ G∗(X) then
rg(f) ≤ 3. As remarked in 8.6 part (ii) above, that means that there is an
integer k such that rg(f) ≤ k for all f ∈ G(X). But X is not an RG-space by
[10, 2.2]. This is yet another example of a space X with rg(X) finite without
X being an RG-space.



94 R. Raphael and R. G. Woods

9. Very weak P-points and RG spaces

Definition 9.1. A point p ∈ X is called a very weak P-point if p is not a
limit point of any countable discrete subset of X. We denote the set of very
weak P-points of X by vwP(X).

Recall that p is a weak P-point of X if p is not a limit point of any countable
subset of X. Weak P-points and very weak P-points were introduced (with
different names) by Kunen [12] who showed that the implications P-point ⇒
weak P-point ⇒ very weak P-point are strict. (See also [24]). Note that each
point of the space of example 8.8 is a very weak P-point, but no point is a weak
P-point. The goal of this section is to show that if vwP(X) is not dense in X
then X is not an RG space.

Theorem 9.2. Let X be any Tychonoff space and let S be any countable scat-
tered subspace of X such that CB(S) is finite. Then S is G-embedded in X.

Proof. We induct on CB(S). If CB(S) = 1 then S is discrete. If S is finite
it is C-embedded and we are done. So suppose S = {s(i) : i ∈ N}. As X is
Tychonoff there is a pairwise disjoint family {V (i) : i ∈ N} of cozero-sets of X
such that s(i) ∈ V (i) for each i. Let f ∈ C(S). Let V =

⋃
{V (i) : i ∈ N} and

extend f to F ∈ C(V ) by letting F [V (i)] = {f(s(i))}. As V ∈ coz X by 8.4
(b) F can be extended to a member of G(X). Hence S is G-embedded in X if
CB(S) = 1.

Now suppose that T is G-embedded in X if T is any countable scattered
subspace of X and if CB(T ) ≤ n. Suppose that S is a countable scattered
subspace of X and that CB(S) = n + 1. Let L = S − I(S). Then L is
countable, scattered, and CB(L) = n. As I(S) is a countable open subset of
S there is a cozero-set coz g of X for which S ∩ coz g = I(S). Let f ∈ C(S).
By the preceding paragraph there is an F ∈ G(X) such that F |S = f|S.
By the induction hypothesis there is a K ∈ G(X) such that K|L = f|L.
A straightworward computation shows that Fgg∗ + K(1 − gg∗)|S = f. As
Fgg∗ + K(1 − gg∗) ∈ G(X) we are done. �

The following result generalizes [10, 3.3] and also gives a shorter proof.

Lemma 9.3. Let X be a zero-dimensional Tychonoff space. Let D = {d(i) :
i ∈ N} be a countable discrete subset of X. Let q be a limit point of D. Suppose
that f ∈ G(X) and that f satisfies these conditions:

(i) if i ∈ N and d(i) ∈ A and A is a clopen subset of X then rg(f|A) ≥ n,
(ii) there exists r > 0 such that if i ∈ N then f(d(i)) = r,
(iii) there exists s > 0 such that s 6= r and f(q) = s.

Then rg(f) ≥ n + 1.

Proof. Suppose to the contrary that f = h1k
∗
1 + · · · + hnk

∗
n where the hi, ki ∈

C(X). Let J = {j ∈ {1, . . . , n} : kj(q) 6= 0}. Observe that J 6= ∅ or else
k∗j (q) = 0 for each j ∈ {1, . . . , n} and so f(q) = 0, contradicting (c). As k

∗
j

is continuous on the open set X − Z(kj), if W =
⋂
{X − Z(kj) : j ∈ J} then



On RG-spaces and the regularity degree 95

q ∈ W and if t is defined by t =
∑

{hjk
∗
j : j ∈ J} then t is continuous on W

and hence at q.
Let E be a clopen subset of X contained in W and containing q. Let

I = {1, . . . , n} − J. Then |I| < n and

f|E = t|E +
∑

{hjk
∗
j : j ∈ I} (1)

Claim 1 : f|(D ∩ E) ∪ {q} is not continuous at q.
To prove claim 1, note that since E is open and contains q and since q ∈ clXD

it follows that q ∈ clX(D ∩E). But f(q) = s and f[D ∩E] = r 6= s so f cannot
be continuous at q.

Claim 2: There exists m ∈ N such that d(m) ∈ D ∩ E and f(d(m)) 6=
t(d(m)).

To prove claim 2, note that by the definition of J and I that kj(q) = 0 (
and hence k∗j (q) = 0) for every j ∈ I. Thus f(q) = t(q). Hence if our claim

failed then f|(D ∩ E) ∪ {q} = t|(D ∩ E) ∪ {q}, which contradicts claim 1 since
t|(D ∩ E) ∪ {q} is continuous at q by our choice of E.

So, let m be as in claim 2. By (i) there exists i ∈ I such that k(i)∗(d(m)) 6= 0.
As above there is a clopen subset T of E containing d(m) and such that k(i)∗

is continuous on T . By (1) this means that

f|T = (t + hik
∗
i )|T +

∑
{hjk

∗
j : j ∈ I − {i}} (2)

But |I−{i}| ≤ n−2 and (t+hik
∗
i )|T ∈ C(T ) so rg(f|T ) ≤ n−1, contradicting

assumption 1. Hence rg(f) ≥ n + 1 as claimed. �

Theorem 9.4. Let X be a countable scattered space for which CB(X) = n.
Using the notation of [10, 3.2], define fn : X → R by: fn[Ii(X)] = {i + 1}, i =
0, . . . , n − 1). Then:

(i) fn ∈ G(X).
(ii) If x ∈ Dn−1(X) and A is a clopen set containing x then rg(fn|A) ≥ n.

In particular rg(X) ≥ n.

Proof. Induct on n. If n = 1 the result is trivial. Suppose that fk satisfies
(i) and (ii) for each countable scattered space X for which CB(X) ≤ k. Let
Y be a countable scattered space for which CB(Y ) = k + 1. Then Dk(Y ) 6=
∅ and Dk+1(Y ) = ∅ (see [10, 3.2]). Evidently CB(Y − Dk(Y )) = k and
fk+1|Y − Dk(Y ) = fk. Hence by the induction hypotheses fk+1|Y − Dk(Y ) ∈
G(Y −Dk(Y )). Also fk+1|Dk(Y ) ∈ G(Dk(Y )) as Dk(Y ) is discrete. Thus by 9.2
there exist functions F, G ∈ G(Y ) such that F |Y − Dk(Y ) = fk+1|X − Dk(Y )
and G|Dk(Y ) = fk+1|Dk(Y ). As Dk(Y ) is closed in the countable space Y
there is a cozero-set coz g of Y for which Y − coz g = Dk(Y ). Then fk+1 =
Fgg∗ + G(1 − gg∗) and so fk+1 ∈ G(Y ). This, together with our induction
hypotheses, shows that fk+1 satisfies the hypotheses of 9.3. Hence by 9.3
rg(fk+1) ≥ k + 1. It remains to show that if p ∈ Dk(Y ) and if A is any
clopen subset of X that contains p, then rg(fk+1|A) ≥ k + 1. But for such an



96 R. Raphael and R. G. Woods

A, CB(A) = k + 1 and fk+1|A has exactly the same properties (re A) as fk+1
has. So the proof that rg(fk+1) ≥ k+1 also shows that rg(fk+1|A) ≥ k+1. �

Lemma 9.5. Let S be a space for which vwP(S) = ∅. Then for each n, S has
a countable scattered subspace S(n) with CB(S(n)) = n.

Proof. The argument is by induction. Suppose that we have such a space
S(k) for the integer k. Let I(S(k)) = {dn : n ∈ N}. Since the set {dn} is
discrete, there exist pairwise disjoint cozero-sets {Vn} in X with dn = Vn∩S(k).
By hypothesis dn is the limit of a countable discrete set T (n) and if we let
H(n) = Vn ∩ T (n) then dn is a limit point of the countable discrete set H(n).
Let S(k + 1) = S(k) ∪ (

⋃
H(n), n ∈ N). One verifies that I(S(k + 1)) =

S(k + 1) − S(k) and S(k + 1) is countable scattered of CB-index k + 1. �

Corollary 9.6. Suppose that X is a space whose countable subspaces are scat-
tered, and suppose that vwP(X) is empty. Then for each n ∈ N there is a
countable scattered subspace S(n) such that CB(S(n)) = n.

Theorem 9.7. Let X be a space with a subspace S such that vwP(S) is empty.
Then :

(i) rg(X) = ∞,
(ii) X is not an RG-space.

Proof. (i) As S has no isolated points it has a countable discrete subset {d(n) :
n ∈ N}. As X is Tychonoff there exists a pairwise disjoint family {V (n) : n ∈
N} of cozero-sets of X with d(n) ∈ V (n). Let n ∈ N. Now vwP(V (n) ∩ S) =
V (n) ∩ vwP(S) = ∅ so by 9.5 V (n) ∩ S has a countable scattered subspace
S(n) for which CB(S(n)) = n. By 9.4 there exists f(n) ∈ G(S(n)) such that
rg(f(n)) ≥ n. By 9.2 there exists k(n) ∈ G(X) such that k(n)|S(n) = f(n).
Thus rg(k(n)) ≥ rg(f(n)). It follows that rg(X) = ∞.

(ii) Define H : X → R by: H|V (n) = k(n)|V (n) and H[X −
⋃
{V (n) : n ∈

N}] = 0, each set in question being clopen in Xδ. Since kn|V (n) ∈ G(V (n))
the restriction of H to each set in the partition is continuous so H ∈ C(Xδ).
But rg(H|V (n)) = rg(kn|V (n)) ≥ n for each n so H /∈ G(X) and X is not an
RG-space. �

Corollary 9.8. If the set of very weak P-points of X is not dense in X then
rg(X) = ∞ and X is not an RG-space.

Proof. If vwP(X) is not dense there exists a cozero-set V of X disjoint from
it. It follows that vwP(V ) = ∅. By 9.7 rg(V ) = ∞ and V is not an RG space.
Since V is G-embedded in X, rg(X) = ∞, and X is not an RG-space since V
is not one by [10, 2.3(f)]. �

Corollary 9.9. If X has a first countable subspace without isolated points then
rg(X) = ∞ and X is not an RG-space.

Proof. If S is first countable without isolated points then vwP(S) = ∅. Now
apply the theorem. �



On RG-spaces and the regularity degree 97

10. Regularity degree for compact and related spaces

We now apply the above methods to some other spaces.
We now use 9.4 to characterize compact (and related) spaces of infinite

regularity degree. (Our original proof was different and used the fact that a
compact space without P-points contains a compact zero set with an infinite
boundary).

Theorem 10.1. Let K be a compact space that is not scattered. Then for each
n, there exists a countable scattered subspace S(n) ⊂ K such that CB(S(n)) =
n. Consequently rg(K) = ∞ and the Krull z-dimension of C(K) is infinite.

Proof. Since K is not scattered it has a compact subspace A without isolated
points. As in [15, 3.1], there is a compact subspace L ⊂ A without isolated
points that maps continuously onto the Cantor set C. If rg(L) = ∞ then
rg(K) = ∞ because L is C∗-embedded in K. Thus it suffices to assume that
there exists a continuous surjection f : K → C.

Although the following is likely folklore, we will now in detail construct
a family F of countably many non-empty clopen subsets of C which, when
partially ordered by inclusion, forms a tree of height n with the property

(*) for all F ∈ F, |{G ∈ F : G ⊂ F}| = ℵ0.
Let s = (s1, . . . , sk) ∈ N

k be an ordered k-tuple of positive integers. Let |s|
denote the number of components in s. If s = (s1, . . . , sk) ∈ N

k and if j ∈ N
we will denote the element (s1, . . . , sk, j) ∈ N

k+1 by s + j. If s is an initial
sequence of t (i.e. t = (s1, . . . , sn, a, b, . . .)) we will write s ≤ t. (By convention
s ≤ t if s = t). Now we construct F.

Let {Cs : |s| ∈ N
1} be a pairwise disjoint family of clopen non-empty subsets

of C. If s ∈ N1 let {Cs+j : j ∈ N}, be a pairwise disjoint family of clopen
non-empty subsets of Cs.

Now let k < n and suppose that we have defined {Cs : s ∈ N
m, 1 ≤ m ≤ k}

with the following properties:

(1) for all m ∈ {1, . . . , k − 1}, for all s ∈ Nm, {Cs+j : j ∈ N} is a pairwise
disjoint family of clopen subsets of C.

(2) If |s| < m ≤ k and j ∈ N, then Cs+j ⊆ Cs.

Now define {Cs+j : |s| = k, j ∈ N} as follows. The set {Cs+j : j ∈ ω} is
a pairwise disjoint family of non-empty clopen subsets of Cs. Then (1) and
(2) are satisfied when k is replaced by k + 1 and F = {Cs : 1 ≤ s ≤ n} is the
required family.

We have a continuous surjection f : K → C. Then {f−1[F ] : F ∈ F} =

{C
′

s : s ∈
⋃m=n

m=1 N
m} (where C

′

s = f
−1[Cs]) satisfies (1) and (2) when Cs is

replaced by C
′

s and C is replaced by K.

Now we define S(n) inductively from the “bottom up”. (1) If |s| = n choose
d(s) ∈ Cs. If |s| = n − 1 , let d(s) be a limit point of {d(s + j), j ∈ N}. (There
will be such a limit point as K is compact and {d(s) : |s| = n} is discrete by
(1)).



98 R. Raphael and R. G. Woods

If we have chosen {d(s) : s ∈
⋃n

i=m+1 N
i} and if s ∈ Nm, choose d(s) to be

a limit point of {d(s + j) : j ∈ N}. Let S(n) = {d(s) : s ∈
⋃n

i=1 N
i}. Then

I(S(n)) = {d(s) : |s| = n}, I(S(n) − I(S(n))) = {d(s) : |s| = n − 1} and so on.
We see that CB(S(n)) = n and that S(n) is countable and scattered.

The fact that rg(K) is infinite now follows from 9.2 and 9.4. The fact that
K is of infinite Krull z-dimension also follows from Theorem 3.6. �

Corollary 10.2. Let K be a compact space. Then the following are equivalent:

(i) K is scattered and CB(K) < ∞,
(ii) rg(K) < ∞,
(iii) K is an RG space.

Proof. The equivalence (i) ⇔ (ii) is in [10, 3.4] as is the implication (iii) ⇒ (ii).
Suppose (ii). Then K is scattered by the above theorem, and CB(K) is finite
by Theorem 9.4 so K is RG by [10, 2.12]. Thus (ii) ⇒ (iii). �

Corollary 10.3. Any space containing a compact space that is not RG is of
infinite regularity degree and of infinite Krull z-dimension. In particular this
applies to a locally compact space that is not scattered.

Proof. The first assertion is immediate. For the second, suppose that X is
locally compact and not scattered. Then its Cantor-Bendixon kernel A is non-
empty, locally compact, and has no isolated points. Local compactness shows
that A has a compact subset K that has no isolated points. Thus K is not RG,
and the conclusion follows. �

The following is immediate from our theorem and Theorem 4.1.

Corollary 10.4. Any pseudocompact space without isolated points is of infinite
regularity degree and of infinite Krull z-dimension.

Corollary 10.5. Let K be a metric space. Then the following are equivalent:

(i) K is scattered and CB(K) < ∞,
(ii) rg(K) < ∞,
(iii) K is an RG space.

Proof. Theorem [10, 3.4] shows that (i) ⇔ (iii) and that (iii) ⇒ (ii). To see
that (ii) ⇒ (i) observe that, with (ii), K is scattered by part (i) of theorem
9.7, and CB(K) is finite by Theorem [10, 3.3]. �

Remark 10.6. The example before Theorem 9.2 and [10, 2.4] show that we
cannot generalize Theorem 10.1 to Lindelöf spaces. The space Ψ is always
scattered locally compact, of finite CB-index, and never RG so there is no
possibility of a generalizing 10.2 from compact to locally compact spaces.

11. Does RG imply finite regularity degree?

By [10] and our work above we already have a positive answer to the above
question in the pseudocompact, metric, scattered Lindelöf, and scattered per-
fectly normal cases. We now give two other instances where the answer is



On RG-spaces and the regularity degree 99

positive. Since we don’t know whether perfectly normal RG-spaces are scat-
tered, the first result is pertinent.

Theorem 11.1. Let X be perfectly normal and RG. Then rg(X) is finite.

Proof. Let f ∈ G(X). By [10, 2.1 (a)] there is a dense open subset of X on
which f is continuous. Let V0 be the union of the open subsets of X = X0 on
which f is continuous. Then V0 is a dense cozero set of X0 whose complement
is the nowhere dense zero-set Z1 = X0 − V0. Since Z1 is perfectly normal and
also RG (by [10, 2.3(f)]) the process can be repeated— f|Z1 is continuous on
a dense cozero set V1 of Z1 with complement Z2 nowhere dense in Z1. The
process can be continued to get subsets Zn with dense cozero sets Vn. Since
the Zn are C-embedded in X, by theorem 3.6 the process must stop with an N
for which ZN+1 = ∅. Furthermore by theorem 3.4 the number N is global—it
works for all f ∈ G(X).

Now we need a representation for f. We know that X is the union of the
Vn and that f is continuous on each of them. Suppose that Vn = coz(gn). We
can assume that each gn ∈ C(X) because each Zn is C-embedded in X. One

readily verifies that f = fg0g
∗
0 +

∑j=N
j=1 [Π

k=j−1
k=0 (1 − gkg

∗
k)]gjg

∗
j f. Since it uses

a number of terms that depends only on N, rg(X) is finite. �

Proposition 11.2. Suppose that X is an RG-space. Then X must be of finite
regularity degree, if it contains countably many pairwise disjoint subspaces Yn
each homeomorphic to X with the property that each Yn is clopen in Xδ.

Proof. Assume if possible that rg(X) = ∞. As each Yn is homeomorphic to
X, there exists for each n, a function fn ∈ G(Yn) such that rg(fn) ≥ n. Now
define F : Xδ → R by F |Yn = fn, F |[X − ∪Yn] = 0. Since each Yn is clopen
in Xδ (as are countable unions of clopen sets), F ∈ C(Xδ) by [8, 1A]. Since

X is RG, we have a representation F =
∑i=k

i=1 aib
∗
i , ai, bi ∈ C(X). But if one

restricts this representation of F to Yn one sees that the constant k globally
bounds the regularity degrees of the {fn}, and this is not possible. �

Corollary 11.3. If X is RG and rg(X) = ∞, then the free union of ω copies
of X is not RG.

Corollary 11.4. Suppose that X is an RG-space of countable pseudocharacter.
Then X must be of finite regularity degree, if it contains countably many disjoint
subspaces Yn each homeomorphic to X.

Remark 11.5. It is natural to question the restrictiveness of the hypothesis
of proposition 11.2. If such subspaces Yn exist then there is a constraint on
the P-space Xδ—it must have countably many pairwise disjoint clopen subsets
each a copy of itself. This demand holds in some but certainly not all P-spaces.
For example, under Martin’s Axiom βN − N will have a dense set of P-points,
and if one takes a maximal family of pairwise disjoint clopen subsets {Aα} in
βN − N, then the set of P-points in their union has this property since it is
the free union of c pairwise disjoint clopen copies of itself. There are many



100 R. Raphael and R. G. Woods

similar P-spaces (cf. [6]) where the constraint holds. It clearly fails if Xδ is the
one-point Lindelöfization of an uncountable discrete set.

12. Miscellaneous

Theorem 12.1.

(i) TFAE for a subspace Y of a space X:
(a) If f ∈ G(X) then f|Y ∈ C(Y )
(b) If V ∈ coz X then V ∩ Y is clopen in Y.

(ii) If Y satisfies the (equivalent) conditions of (1) and Y is z-embedded in
X then Y is a P-space.

Proof. To show that (a) implies (b) let V ∈ coz X. Then the characteristic
function χV of V belongs to G(X). Now χV |Y = χV ∩Y so by (a) χV ∩Y ∈ C(Y ).
This immediately implies (b).

To show that (b) implies (a), let f ∈ G(X).Then f = g1h
∗
1 + · · · + gnh

∗
n,

gi, hi ∈ C(X). To show that f|Y ∈ C(Y ) it clearly suffices to show that h
∗
i |Y ∈

C(Y ) for each i. But h∗i |coz(hi) ∈ C(coz(hi)) so h
∗
i |Y ∩coz(hi) ∈ C(Y )∩coz(hi).

Similarly h∗i |Z(hi) ∈ C(Z(hi)) (as h
∗
i [Z(hi)] = 0) so h

∗
i |Y ∩ Z(hi) ∈ C(Y ∩

Z(hi)). Thus (by (b)) h
∗
i |Y is continuous on each of complementary clopen

subsets of Y . Hence h∗i |Y ∈ C(Y ) and we are done.
If Y is z-embedded in X then by (b) every cozero-set of Y is clopen and so

Y is a P-space. �

13. Open questions

(1) If X is RG, is rg(X) finite?
(2) (cf. 1.1 and the proof of proposition 11.2) If X has infinite regularity

degree, can this be witnessed by a family of functions in G(X) that are
pairwise orthogonal?

(3) Must an RG-space have a dense set of P-points?

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On RG-spaces and the regularity degree 101

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Received July 2004

Accepted March 2005

R. Raphael (raphael@alcor.concordia.ca)
Mathematics and Statistics, Concordia University, Montréal, Canada, H4B
1R6.

R. G. Woods (rgwoods@cc.umanitoba.ca)
Mathematics, University of Manitoba, Winnipeg, Canada, R3T 2N2.