EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 2, 2022, 774-783
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Results about P -Normality

Lutfi Kalantan1, Mai Mansouri1,∗

1 King Abdulaziz University, Department of Mathematics, P.O.Box 80203, Jeddah 21589,
Saudi Arabia

Abstract. A topological spaceX is called P -normal if there exist a normal space Y and a bijective
function f : X −→ Y such that the restriction f|A : A −→ f(A) is a homeomorphism for each
paracompact subspace A ⊆ X. In this paper we present some new results on P -normality. We
study the invariance and inverse invariance of P -normality as a topological property. We also
investigate the Alexandroff Duplicate of a P -normal space, the closed extension of a P -normal
space, the discrete extension of a P -normal space and the Dowker topological space. Furthermore,
we introduce a new property related to P -normality which we call strong P -normality.

2020 Mathematics Subject Classifications: 54D15, 54C10

Key Words and Phrases: Normal, P -normal, L-normal, C-normal, Strong P -noramlity, Alexan-
droff Duplicate, Invariance, Closed extension, Discrete extension, Paracompact, Product

1. Introduction

We introduced P -normality in our previous paper [10]. The purpose of this paper is
to study some new results about P -normality. We investigate some types of invariance.
We also discuss the Alexandroff Duplicate, the closed extension space and the discrete
extension space of a P -normal space. We examine whether P -normality is preserved
in these spaces or not. Finally, we define a new topological property called strong P -
normality. Throughout this paper, we denote an ordered pair by ⟨x, y⟩, the set of positive
integers by N and the set of real numbers by R. A T4 space is a T1 normal space and a
Tychonoff space is a T1 completely regular space. We do not assume T2 in the definition of
compactness, paracompactness and countable compactness. We do not assume regularity
in the definition of Lindelöfness. For a subset A of a space X, intA and A denote the
interior and the closure of A, respectively. An ordinal γ is the set of all ordinal α such
that α < γ. The first infinite ordinal is ω0, the first uncountable ordinal is ω1, and the
successor cardinal of ω1 is ω2. We begin by recalling the following definitions:

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i2.4387

Email addresses: LNKalantan@hotmail.com (L. Kalantan), mfmansouri1@kau.edu.sa (M. Mansouri)

https://www.ejpam.com 774 © 2022 EJPAM All rights reserved.



L. Kalantan, M. Mansouri / Eur. J. Pure Appl. Math, 15 (2) (2022), 774-783 775

Recall that a topological space (X , τ ) is paracompact if any open cover has a locally
finite open refinement. For a subspace A of X, A is paracompact if (A , τA ) is paracom-
pact, i.e., any open (open in the subspace) cover of A has a locally finite open (open in
the subspace) refinement. We do not assume T2 in the definition.

Definition 1. A topological space X is called P -normal if there exist a normal space Y
and a bijective function f : X −→ Y such that the restriction f|A : A −→ f(A) is a
homeomorphism for each paracompact subspace A ⊆ X [10].

2. New Results on P -normality

In [10], we proved P -normality is a topological property, studied its independence of
other topological properties, and investigated whether or not P -normality was an additive
property, a multiplicative property and a hereditary property. Until now, we still don’t
know if P -normality is hereditary with respect to closed subspaces. But under some con-
ditions it is hereditary with respect to compact subspaces. Before we state such conditions
we introduce some results:

Proposition 1. If X is a T1 space, f : X −→ Y is a one to one and onto map, and the
restriction of f on any finite subset of X is a homeomorphism. Then Y is also T1.

Proof. SinceX is T1 and f is a bijection then Y has more than one element. Let a, b ∈ Y
be arbitrary such that a ̸= b. Then there exist unique c, d ∈ X such that f(c) = a and
f(d) = b and c ̸= d. Now, {c, d} ⊆ X is a finite subset of X. So f |{c,d} : {c, d} −→ {a, b} is
a homeomorphism. Now, a = f(c) and c is isolated in {c, d} because {c, d} ⊆ X is discrete.
Also, b = f(d) and d is isolated in {c, d}. So there exist Y -open subset U containing a
such that U ∩ {a, b} = {a}, and there exists Y -open subset V containing b such that
V ∩ {a, b} = {b}. Then b /∈ U ∋ a and a /∈ V ∋ b. Which implies that Y is T1.

Corollary 1. If X is a T1, P -normal space then the witness Y is T4.

Using the previous proposition we can state the following theorems:

Corollary 2. If X is T1 and the only paracompact subspaces of X are the finite subspaces,
then X is P -normal.

Theorem 1. Let X be a T1, Fréchet P -normal space. Then, any compact subspace of X
is P -normal.

Proof. Let Y and f be a witness space and function respectively of the P -normality
of X. Since X is T1, then Y is T4 by the above corollary. Now, f is continuous since X
is Fréchet by [10, Theorem 5]. Let A ⊆ X be any compact subset of X. The continuous
image of a compact subset is compact so f(A) ⊆ Y is compact in Y . Moreover, since Y is
T2 that means f(A) is closed in Y . Y is normal and normality is hereditary with respect



L. Kalantan, M. Mansouri / Eur. J. Pure Appl. Math, 15 (2) (2022), 774-783 776

to closed sets so f(A) is normal and hence will be a witness for the P -normality of A. Let
g = f |A : A −→ f(A). Let C ⊆ A be any paracompact subspace of A. Since a subspace
of a subspace is a subspace, then C is paracompact in X. Therefore, g|C = f |A|C

= f |C is
a homeomorphism. Which means A, the arbitrary compact subset of X, is P -normal and
we are done.

In the same way we proved the previous theorem, we can deduce the following corollary
about P -normality being hereditary with respect to countably compact subspaces with
additional conditions. recall that a C-closed space Y is a T2 space where every countably
compact subset A ⊆ Y is closed [9].

Corollary 3. Let X be a T1, Fréchet P -normal space. Let Y a witness of the P -normality
of X be a C-closed space. Then, any countably compact subspace of X is P -normal.

Recall that a topological space (X , τ ) is called epinormal if there is a coarser topology
τ ′ on X such that (X , τ ′ ) is T4 [3].

Theorem 2. Let X be a T1, Fréchet P -normal space, then X is epinormal.

By the above discussion we have seen that if X is T1 and P -normal then Y , the
witness of P -normality, is T4. Now, since X is Fréchet that means the witness function f
is continuous by [10, Theorem 5]. This allows us to consider Y as a coarser space of X [7,
2.4].

Note that since in this case Y is coarser than X and Y is T4 hence T2 then X has to be
T2. Which means, if a space X is not T2, but T1 and Fréchet then it cannot be P -normal.

3. Invariance

We begin by studying the invariant properties of P -normality. P -normality is not
invariant in general.

Example 1. In [10] we showed that the Dieudonné plank (X,τ ) is not P -normal. Now,
consider (X,τ ′), where τ ′ is generated by making any element on the right side of the
plank A isolated. Consider idX : (X,τ ′) −→ (X,τ ). Since τ is coarser than τ ′ then
idX : (X,τ ′) −→ (X,τ ) is continious, one to one and onto. (X,τ ′) is P -normal being
T2-paracompact i.e normal but the Dieudonné plank (X,τ ) is not.

By a theorem of Ponomarev [7, 4.2.D] which says:“ a T0 spaceX is first countable if and
only if X is a continuous image of a metrizable space under an open mapping”. Consider
(R,RS), which we have shown is not P -normal in [10], but it is Tychonoff and first
countable. So there exists a metrizable space X and an open function g : X −→ (R,RS).
Since any metrizable space is P -normal, then this example shows that P -normality is not
open invariant and hence cannot be quotient invariant either.



L. Kalantan, M. Mansouri / Eur. J. Pure Appl. Math, 15 (2) (2022), 774-783 777

Example 2.

The function f : (R,RS) −→ ( {0, 1} , D ) defined by

f(x) =

{
0 ; if x < 0
1 ; if x ≥ 0

is both closed and open. Where D is the discrete topology. Now, as we previously mentioned
(R,RS) is not P -normal but ( {0, 1} , D ) is P -normal since it is normal. This example
shows that P -normality is not inverse invariant in general. Moreover, it is not inverse
open invariant nor inverse closed invariant.

4. Generating Spaces and P -normality

In this section, we start with the study of the Alexandroff Duplicate space of a P -
normal space. Let us first recall the definition of the Alexandroff Duplicate topological
space. Let X be an infinite topological space. We denote the family of all finite subsets
of X by [X ]<ω0 , i.e.,

[X ]<ω0 = {E ⊂ X : E is finite }.

Put X ′ = X × {1}. So, X ′ is just a copy of X and we have X ∩ X ′ = ∅. The ground
set of the Alexandroff duplicate space A(X) of X is A(X) = X ∪ X ′. To simplify the
symbols, we do the following: For an element x ∈ X, we denote the element ⟨x, 1⟩ in X ′

by x′ and for any subset B ⊆ X, put B′ = {x′ : x ∈ B} = B × {1} ⊆ X ′. For each
x′ ∈ X ′, put B(x′) = {{x′}}, so any element in X ′ will be isolated in A(X). For each
x ∈ X, put B(x) = {U ∪ (U ′ \ E′ ) : U is open in X with x ∈ U and E ∈ [X ]<ω0 }.
Then B = {B(y) : y ∈ A(X) } generates a unique topology on A(X) such that B is its
neighborhood system. A(X) with this topology is called the Alexandroff Duplicate of X,
see [2] and [6]. Observe that for any open set U in X, we have that U ∪ U ′ is open in
A(X) and for any x ∈ U , we have U ∪ (U ′ \ {x′ } ) is a basic open neighborhood of x.

Theorem 3. If X is P -normal, then so is its Alexandroff Duplicate A(X).

Proof. Let X be any P -normal space. Pick a normal space Y and a bijective function
f : X −→ Y such that f|C : C −→ f(C) is a homeomorphism for each paracompact
subspace C ⊆ X. Consider the Alexandroff Duplicate spaces A(X) and A(Y ) of X and Y
respectively. Y is normal then so is A(Y ) [2].

Let us define g : A(X) −→ A(Y ) by g(a) = f(a) if a ∈ X. If a ∈ X ′, let b be the unique
element in X such that b′ = a, then define g(a) = (f(b))′. Then g is a bijective function.
Now, a subspace C ⊆ A(X) is paracompact in A(X) if and only if C ∩X is paracompact
in X. To prove this, assume that C is paracompact in A(X). Let U = {Uα ⊆ C∩X : Uα is
open in C ∩X for each α ∈ λ} be any open cover (open in C ∩X) for C ∩X. That means
for each α ∈ λ there exists Vα ⊆ X open in X such that Uα = Vα∩(C∩X). Now, for every
α ∈ λ, Vα is open in X and therefore Vα∪V ′

α is open in A(X). So (Vα∪V ′
α)∩C is open in C.

This implies (Vα∪V ′
α)∩C = (Vα∩C)

⋃
(V ′

α∩C) is open in C. Take the unions of these sets



L. Kalantan, M. Mansouri / Eur. J. Pure Appl. Math, 15 (2) (2022), 774-783 778

: (∪α∈λ(Vα∩C))
⋃
(∪α∈λ(V

′
α∩C)). For each x′ ∈ C \(∪α∈λ(V

′
α∩C)) consider the singleton

{x′}. Consider W = {(Vα ∩ C)
⋃
(V ′

α ∩ C), {x′} : α ∈ λ, and x′ ∈ C \ (∪α∈λ(V
′
α ∩ C))}.

This is an open cover of C in A(X) (open in C). Since C is assumed to be paracompact in
A(X) then this cover W has a locally finite open refinement (in C). That is, there exists
{Gs : s ∈ S} such that for each s ∈ S, Gs is open in C and for each s ∈ S there exists α ∈ λ
such that either Gs ⊆ (Vα∩C)

⋃
(V ′

α∩C) or Gs = {x′} for some x′ ∈ C \ (∪α∈λ(V
′
α∩C))}.

Let S ′ = {s ∈ S : Gs ⊆ Vα ∩ C}. Then {Gs : s ∈ S ′} is a sub-family of {Gs : s ∈ S}
which is locally finite, hence {Gs : s ∈ S ′} is locally finite as well. So {Gs : s ∈ S ′} is a
locally finite open (open in C ∩ X) refinement of U . On the other hand, assume C ∩ X
is a paracompact subset in X. Let G = {Gα ∩ C : α ∈ λ;Gα ⊆ A(X) open in A(X) for
each α ∈ λ} be an arbitrary open (open in C) cover of C in A(X). So C ⊆

⋃
G. Consider

GX = {(Gα ∩ C) ∩X : α ∈ λ} = {Gα ∩ (C ∩X) : α ∈ λ}.Then C ∩X ⊆
⋃
GX , i.e, this

is an open (open in C ∩ X) cover for C ∩ X. By assumption there exists {Hs : s ∈ S}
locally finite open (open in C ∩X) refinement of GX . That is, for each s ∈ S there exists
αs ∈ λ such that Hs ⊆ Gαs . For every s ∈ S there exists Ks ⊆ X open in X such that
Ks ∩ (X ∩ C) = Hs. Consider {(Ks ∪K ′

s) ∪ C : s ∈ S}= {(Ks ∩ C)
⋃
(K ′

s ∩ C) : s ∈ S}.
Since {Ks ∩ C : s ∈ S} is locally finite then so is {(Ks ∩ C)

⋃
(K ′

s ∩ C) : s ∈ S}. For
every x′ ∈ (C ∩X ′) \ (

⋂
s∈S(K

′
s ∩ C)) there exists αx′ ∈ λ such that x′ ∈ Gα′

x
. Consider

K = {(Ks ∩ C) ∪ (K ′
s ∩ C), {x′} : s ∈ S;x′ ∈ (C ∩X ′) \ (

⋂
s∈S(K

′
s ∩ C))}. K is a locally

finite open refinement (open in C) of G.
Now, let C ⊆ A(X) be any paracompact subspace. We show g|C : C −→ g(C) is a

homeomorphism. Let a ∈ C be arbitrary. If a ∈ C ∩X ′, let b ∈ X be the unique element
such that b′ = a. For the smallest basic open neighborhood {(f(b))′} of the point g(a) we
have that {a} is open in C ∩X

′
and g({a}) ⊆ {(f(b))′}. If a ∈ C ∩X. Let W be any open

set in Y such that g(a) = f(a) ∈ W . Consider H = (W ∪ (W ′ \ {f(a)′})) ∩ g(C) which
is a basic open neighborhood of f(a) in g(C). Since f|C∩X

: C ∩ X −→ f(C ∩ X) is a
homeomorphism, then there exists an open set U in X with a ∈ U and f|C∩X

(U ∩C) ⊆ W .
Now, (U∪(U ′\{a′}))∩C = G is open in C∩X such that a ∈ G and g|C (G) ⊆ H. Therefore,
g|C is continuous. Now, we show that g|C is open. Let K ∪ (K ′ \ {k′}), where k ∈ K and
K is open in X, be any basic open set in A(X), then (K ∩C)∪ ((K ′ ∩C) \ {k′}) is a basic
open set in C. Since X ∩C is compact in X, then g|C (K ∩ (X ∩C)) = f|X∩C

(K ∩ (X ∩C))
is open in Y ∩f(C∩X) as f|X∩C

is a homeomorphism. Thus K∩C is open in Y ∩f(X∩C).
Also, g((K ′ ∩C) \ {k′}) is open in Y ′ ∩ g(C) being a set of isolated points. Thus g|C is an
open function. Therefore, g|C is a homeomorphism.

Next, we present a result about Dowker topological spaces. This result may seem
repeated as it was mentioned about L-normality in [11] but it is so interesting that we
mention it again with regards to P -normality. Recall that a Dowker space is a T4 space
whose product with I, I = [0, 1] with its usual metric, is not normal. M. E. Rudin used
the existence of a Suslin line to obtain a Dowker space which is hereditarily separable and
first countable [12]. Using CH, I. Juhász, K. Kunen, and M. E. Rudin constructed a first
countable hereditarily separable real compact Dowker space [8]. Weiss constructed a first
countable separable locally compact Dowker space whose existence is consistent with MA



L. Kalantan, M. Mansouri / Eur. J. Pure Appl. Math, 15 (2) (2022), 774-783 779

+ ¬ CH [14]. We already know that such spaces are consistent examples of Dowker spaces
whose product with I are not L-normal [11]. This means that they cannot be P -normal
either; since any regular P -normal space is L-normal [10].

We move on to studying the P -normality of the closed extension. Let (X , τ ) be a
topological space and let p be an object not in X, i.e., p ̸∈ X. Put Xp = X ∪{ p }. Define
a topology τ ⋆ on Xp by τ ⋆ = { ∅ } ∪ {U ∪ { p } : U ∈ τ }. The space (Xp , τ ⋆ ) is called
the closed extension space of (X , τ ), [13, Example 12].

Since characterizing all paracompact subspaces [10] is a core subject in the notion of
P -normality, we will start with characterizing all paracompact subspaces of the closed
extension space (Xp , τ ⋆ ) of a given space (X , τ ).

Proposition 2. Let (X , τ ) be a topological space. Consider the closed extension space
(Xp , τ ⋆ ) of (X , τ ). Let A ⊆ Xp.

If p ̸∈ A, then A is a paracompact subset in (Xp , τ ⋆ ) if and only if A is a paracompact
subset in (X , τ ).

If p ∈ A, then A is a paracompact subset in (Xp , τ ⋆ ) if and only if A \ {p} is a
compact subset in (X , τ ).

The proof of this proposition can be found in [5]. A space is called ultra-connected if
any two non-empty closed sets intersect [13]. Since any normal space is P -normal (just by
taking in Definition 1, Y = X and f to be the identity function) then by [5, Theorem 1.4]
, we get the following theorem: If (X , τ ) is ultra-connected, then its closed extension
(Xp , τ ⋆ ) is P -normal [5]. Recall that a topological space X is called C-normal if there
exist a normal space Y and a bijective function f : X −→ Y such that the restriction
f|C : C −→ f(C) is a homeomorphism for each compact subspace C ⊆ X [4]. Now, in [5]
it was proved that the closed extension space (Xp , τ ⋆ ) is not C-normal if (X , τ ) is not
ultra-connected. In [10], we showed that P -normality implies C-normality. Combining all
the information above together we get:

Theorem 4. If (X , τ ) is not ultra-connected, then the closed extension space (Xp , τ ⋆ )
is not P -normal .

Now, we discuss a new result about P -normality and whether it’s preserved or not in
the discrete extension space. To do this let us recall the definition of the discrete extension
space: Let M be a non-empty proper subset of a topological space (X , τ ). Define a new
topology τ (M) on X as follows: τ (M) = {U ∪K : U ∈ τ and K ⊆ X \M }. (X , τ (M) )
is called a discrete extension of (X , τ ) and we denote it by XM [13], see also [7, 5.1.22].
We will now show that P -normality is not preserved by a discrete extension. That is, the
discrete extension of a P -normal space need not be P -normal.

Example 3. We know that (R,RS) where RS is the rational sequence topology on R,
is a Tychonoff locally compact non compact space [13, Example 65]. Thus R with the



L. Kalantan, M. Mansouri / Eur. J. Pure Appl. Math, 15 (2) (2022), 774-783 780

rational sequence topology has a one-point compactification. Let X = R ∪ {p}, where
p ̸∈ R, be a one-point compactification of R. Since X is T2-compact, then it is T4, hence
P -normal [10]. Now, take the discrete extension of X denoted by XR. Observe that in XR,
the singleton {p} is closed-and-open. XR is first countable and Tychonoff because R with
the rational sequence topology is, thus XR is of countable tightness. XR is also separable
because (R,RS) is separable and Q ∪ {p} is a countable dense subset of XR [1]. Now, R
with the rational sequence topology is not normal. Since R is closed in XR, we conclude
that XR is not normal. Using the theorem: “If Y is T3, separable, P -normal and of
countable tightness, then Y is normal.” [10], we conclude that XR cannot be P -normal.

This example shows that [1, Theorem 12] is not true for P -normality. That is if Y is a
Tychonoff space, then a discrete extension XM of any compactification X of Y need not
be P -normal.

5. Strong P -Normality

Definition 2. A topological space X is called strongly P -normal if there exists a bijective
function f : X −→ I, where I = [ 0 , 1 ] the closed unit interval considered with its usual
metric topology, such that the restriction f|A : A −→ f(A) is a homeomorphism for each
paracompact subspace A ⊆ X.

It is clear from the definition that any strongly P -normal space is P -normal. The
converse is not always true.

Example 4. ω2+1 with its usual ordered topology is P -normal because it is normal being
T2 compact. But ω2 + 1 cannot be strongly P -normal because |[ 0 , 1 ]| = |R| = c < ω2 =
|ω2 + 1|.

Example 5. (R,U) is not strongly P -normal. (R,U) is homeomorphic to the open interval
(0, 1) with the usual topology. So, if (R,U) is strongly P -normal, then there will be a
bijection f : (0, 1) −→ [0, 1] such that f |A is a homeomorphism for every paracompact
subset A ⊆ (0, 1). Since (0, 1) is Fréchet, f is continuous by [10, Theorem 5]. Since f is
bijection, there is unique a, b ∈ (0, 1) such that f(a) = 0 and f(b) = 1. Assume without
loss of generality that a < b. Then, (0, 1) \ [a, b] ̸= ∅ and clearly f is continuous on
[a, b]. Now using the Intermediate Value, for every y ∈ (f(a), f(b)) = (0, 1), there exists
x ∈ (a, b) such that f(x) = y. Hence, f−1([0, 1]) ⊆ [a, b] ⊂ (0, 1). This implies that for
every x ∈ (0, 1) \ [a, b] (which is non-empty), x has no image in [0, 1] which contradicts
that f is a function. Hence there is no continuous bijection between R and I. Therefore,
(R,U) is not strongly P -normal.

Theorem 5. Strong P -normality is a topological property.



L. Kalantan, M. Mansouri / Eur. J. Pure Appl. Math, 15 (2) (2022), 774-783 781

Proof. Let X be any strongly P -normal space. Assume that X ∼= Z, so there exists a
homeomorphism k : Z −→ X. Since X is strongly P -normal then there exists a witness
function h : X −→ I which is a bijection with the restriction h|C : C −→ f(C) is a
homeomorphism for any paracompact subspace C of X. . Then h ◦ k : Z −→ I satisfies
the requirements.

Theorem 6. If X is Fréchet and strongly P -normal, then any function witnessing the
strong P -normality of X is continuous.

Proof. Assume that X is strongly P -normal and Fréchet. Let f : X −→ I be a witness
of the strong P -normality of X. Let A ⊆ X and pick y ∈ f(A). Pick the unique x ∈ X
such that f(x) = y. Thus x ∈ A. Since X is Fréchet, there exist a sequence (an) ⊆ A such
that an −→ x. The subspace B = {x, an : n ∈ N} of X is paracompact being compact ,
thus f|B : B −→ f(B) is a homeomorphism. Now, let W ⊆ I be any open neighborhood
of y, then W ∩ f(B) is open in the subspace f(B) containing y. By continuity of the
homeomorphism f|B , f

−1(W ∩ f(B)) = f−1(W ) ∩ B is an open neighborhood of x in B.
Then,(f−1(W ) ∩ B) ∩ {an : n ∈ N} ≠ ∅. So (f−1(W ) ∩ B) ∩ A ̸= ∅. Therefore we have,
∅ ̸= f((f−1(W ) ∩ B) ∩ A) ⊆ f(f−1(W ) ∩ A) = W ∩ f(A) then W ∩ f(A) ̸= ∅. Hence
y ∈ f(A), thus f(A) ⊆ f(A). Therefore, f is continuous.

Example 6. It is clear that I with its usual metric topology is strongly P -normal. We
show that the product I × I is not strongly P -normal.

Proof. Suppose to the contrary that I × I is strongly P -normal. Pick a bijection
f : I × I −→ I such that f |A : A −→ f(A) is a homeomorphism for each paracompact
subspace A ⊆ I × I. Now I × I is first countable and hence Fréchet. This implies that f
is continuous, see Theorem 6, which contradicts the fact that there exists no continuous
bijection f : I × I −→ I. Because if there were, then f would be a homeomorphism since
I × I is compact and I is T2. This is a contradiction since I × I is connected with no cut
points (I × I) \ {⟨x, y⟩} is connected for every point ⟨x, y⟩ ∈ I × I, while I is connected
with cut points (Take any x ∈ ( 0 , 1 ) ⊂ I, then I \ {x} = [0, x) ∪ (x, 1] where both [0, x)
and (x, 1] are non-empty disjoint open subsets of I).

Therefore, I × I is not strongly P -normal.

So, a product of two strongly P -normal spaces may not be strongly P -normal. But,
a product of two strongly P -normal spaces is P -normal. To show this, we start with a
lemma.

Lemma 1. If A is a paracompact subset of the product X ×Z, then p1(A) and p2(A) are
both paracompact in X and Z respectively. Where p1 and p2 are the natural projection
functions.

Proof. Let A be a paracompact subset of the product X×Z. Suppose that p1(A) is not
paracompact subset in X, i.e., p1(A) as a subspace of X is not paracompact. Then there



REFERENCES 782

exist an open cover U = {Uα ⊆ p1(A) : Uα is open in p1(A) for each α ∈ Λ } for p1(A) such
that any open (open in p1(A)) refinement of U is not locally finite. Now, let x ∈ p1(A)
and fix an αx ∈ Λ such that x ∈ Uαx . For each z ∈ Z such that there exists x ∈ p1(A)
with ⟨x, z⟩ ∈ A, let Wz be an open neighborhood of z in Z. Note that A ⊆ p1(A)× p2(A).
Consider the family K = { (U)αx ×Wz) ∩ A : ⟨x, z⟩ ∈ A } which is an open (open in A)
cover for A.

Claim: K has no locally finite open refinement.
Proof of Claim: Suppose that K has a locally finite open refinement, say {Gs × Hs :
s ∈ S } (we can assume that this refinement is of the basic open set form in the product
X × Z), then the family {Gs ∩ pa(A) : s ∈ S } would be a locally finite open refinement
of U which is a contradiction and Claim is proved.

So, K is an open (open in A) cover for A which has no locally finite open refinement
and this contradicts that A is a paracompact subset in X × Z. Therefore, p1(A) is a
paracompact subset in X.

Similarly, p2(A) is a paracompact subset of Z.

Theorem 7. If X and Z are both strongly P -normal, then X × Z is P -normal.

Proof. Assume that X and Z are both strongly P -normal. Pick two bijection func-
tions f : X −→ I and g : Z −→ I such f|A : A −→ f(A) is a homeomorphism for
each paracompact subspace A ⊆ X and g|A : A −→ f(A) is a homeomorphism for each
paracompact subspace A ⊆ Z. I × I is normal being T2 compact. Put h = f × g, i.e.,
h : X×Z −→ I×I is defined by h(⟨x, z⟩) = ⟨f(x), g(z)⟩ for each ⟨x, z⟩ ∈ X×Z. It is clear
that h is a bijection function. Let A be any paracompact subset of X × Z. By Lemma 1,
we have that p1(A) is a paracompact subset of X and p2(A) is a paracompact subset of Z.
Thus f|p1(A)

: p1(A) −→ f(p1(A)) is a homeomorphism and g|p2(A)
: p2(A) −→ g(p2(A)) is

a homeomorphism. Since a product of two homeomorphisms is a homeomorphism [7], we
get that hp1(A)×p2(A) = (f|p1(A)

)× (g|p2(A)
) : p1(A)× p2(A) −→ (f(p1(A)))× (g(p2(A))) =

h(p1(A) × p2(A)) is a homeomorphism. Since A ⊆ p1(A) × p2(A) and a restriction of a
homeomorphism is a homeomorphism, we conclude that hA : A −→ h(A) is a homeomor-
phism. Therefore, X × Z is P -normal.

The following problems are still open:

1. Is P -normality hereditary with respect to closed sets?

2. Is there a Tychonoff P -normal space which is not normal?

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