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Applied General TopologyUniversidad Polit�ecnica de ValenciaVolume 1, No. 1, 2000pp. 45 - 60

Extension properties and the NiemytzkiplaneHaruto OhtaAbstract. The �rst part of the paper is a brief survey onrecent topics concerning the relationship between C�-embeddingand C-embedding for closed subsets. The second part studiesextension properties of the Niemytzki plane NP . A zero-set, z-,C�-, C-, and P-embedded subsets of NP are determined. Finally,we prove that every C�-embedded subset of NP is a P-embeddedzero-set, which answers a problem raised in the �rst part.2000 AMS Classi�cation: 54C45, 54G20Keywords: C�-embedded, C-embedded, P-embedded, property (U�), uni-formly locally �nite, Niemytzki plane, Tychono� plank1. IntroductionAll spaces are assumed to be completely regular T1-spaces. A subset Y of aspace X is said to be C-embedded in X if every real-valued continuous functionon Y can be continuously extended over X, and Y is said to be C�-embedded inX if every bounded real-valued continuous function on Y can be continuouslyextended over X. Obviously, every C-embedded subset is C�-embedded, butthe converse is not true, in general. In the �rst part of the paper, formed bySections 2, 3 and 4, we discuss several problems concerning the relationshipbetween C�-embedding and C-embedding for closed subsets. For example, thefollowing problem is still open as far as the author knows:Problem 1.1. Does there exist a �rst countable space having a closed C�-em-bedded subset which is not C-embedded?Since a space which answers the above problem positively cannot be normal,the following problem naturally arises:Problem 1.2. Let X be one of the following spaces: The Niemytzki plane (i.e.,the space NP de�ned in Section 4 below); the Sorgenfrey plane ( [3, Example2.3.12]); Michael's product space ([3, Example 5.1.32]). Then, does the spaceX have a closed C�-embedded subset which is not C-embedded?



46 Haruto OhtaIn the second part, formed by Sections 5, 6 and 7, we answer Problem 1.2for the Niemytzki plane NP negatively by determining a zero-set, z-, C�-, C-and P-embedded subsets of NP . The problem, however, remains open for theSorgenfrey plane and Michael's product space.Throughout the paper, let R denote the real line with the Euclidean topology,Q the subspace of rational numbers and N the subspace of positive integers.The cardinality of a set A is denoted by jAj. As usual, a cardinal is the initialordinal and an ordinal is identi�ed with the space of all smaller ordinals withthe order topology. Let ! denote the �rst in�nite ordinal and !1 the �rstuncountable ordinal. All unde�ned terms will be found in [3].2. C�-embedding versus C-embeddingIt is an interesting problem to �nd a closed C�-embedded subset which is notC-embedded. We begin by showing typical examples of such subsets. First, letus consider the subspace � = �R n (�N n N) of �N. The subset N is closed C�-embedded but not C-embedded in �, because � is pseudocompact (cf. [4, 6P,p.97]). More generally, Noble proved in [16] that every space Y can be embeddedin a pseudocompact space pY as a closed C�-embedded subspace. Thus, everynon-pseudocompact space Y embeds in pY as a closed C�-embedded subsetwhich is not C-embedded. Shakhmatov [20] constructed a pseudocompact spaceX with a much stronger property that every countable subset of X is closedand C�-embedded.Now, we give another examples which does not rely on pseudocompactness.For every space X there exist an extremally disconnected space E(X), calledthe absolute of X, and a perfect onto map eX : E(X) ! X (cf. [3, 6.3.20 (b)]).We now call a space X weakly normal if every two disjoint closed sets in X, oneof which is countable discrete, have disjoint neighborhoods.Lemma 2.1. Let X be a space which is not weakly normal. Then E(X) con-tains a closed C�-embedded subset which is not C-embedded.Proof. By the assumption, X has a closed set A and a countable discrete closedset B = fpn : n 2 Ng such that A \ B = ? but they have no disjoint neigh-borhoods. We show that the closed set F = e�1X [B] in E(X) is C�-embeddedbut not C-embedded. Since B is countable discrete closed in X, we can �nda disjoint family U = fUn : n 2 Ng of open-closed sets in E(X) such thate�1X (pn) � Un � E(X) n e�1X [A] for each n 2 N. Let U = SfUn : n 2 Ng;then U is a cozero-set in E(X). Since F and E(X) nU cannot be separated bydisjoint open sets, it follows from Theorem 3.1 below that F is not C-embeddedin E(X). On the other hand, F is C-embedded in U, because each e�1X (pn) iscompact and U is disjoint, and further, U is C�-embedded in E(X) by [4, 1H6,p.23]. Consequently, A is C�-embedded in E(X). �Corollary 2.2. Let X be one of the following spaces: The Niemytzki plane NP ;the Sorgenfrey plane S2; Michael's product space RQ � P ; the Tychono� plankT (see Example 3.3 below). Then E(X) contains a closed C�-embedded subsetwhich is not C-embedded.



Extension properties and the Niemytzki plane 47Proof. It is well known (and easily shown) that the spaces NP , S2 and T arenot weakly normal. Now, we show that Michael's product space RQ � P isnot weakly normal. The space RQ is obtained from R by making each pointof P = R n Q isolated. Enumerate Q as fxn : n 2 Ng and choose yn 2 Pwith jxn � ynj < 1=n for each n 2 N. Let A = fhxn;yni : n 2 Ng andB = fhx;xi : x 2 Pg. Then A and B have no disjoint neighborhoods inRQ � P . Since A is discrete closed in RQ � P , RQ � P is not weakly normal.Hence, the corollary follows from Lemma 2.1. �As another application, we have the following example concerning Problem1.1.Example 2.3. There exists a space X in which every point is a G� and thereexists a closed C�-embedded subset which is not C-embedded. In fact, letR be a maximal almost disjoint family of in�nite subsets of N. Pick a pointpA 2 cl�N A n A for each A 2 R and let R = fpA : A 2 Rg. Then the subspaceX = N [R of �N is extremally disconnected (i.e., E(X) = X) and R is discreteclosed in X. Let E be a countable in�nite subset of R. Then E and RnE haveno disjoint neighborhoods in X by the maximality of R. Hence, by the proofof Lemma 2.1, E is closed C�-embedded in X but not C-embedded. �We change the topology of the space X = N [R in Example 2.3 by declaringthe sets fpAg [ (A n f1;2; � � � ;ng), n 2 N, to be basic neighborhoods of pAfor each A 2 R. The resulting space is �rst countable and is usually called a	-space (see [4, 5I, p.79]). A positive answer to the following problem answersProblem 1.1 positively.Problem 2.4. Does there exist a 	-space having a closed C�-embedded subsetwhich is not C-embedded?For an in�nite cardinal 
, a subset Y of a space X is said to be P
-embeddedin X if for every Banach space B with the weight w(Y ) � �, every continuousmap f : Y ! B can be continuously extended over X. A subset Y of X is saidto be P-embedded in X if Y is P
-embedded in X for every 
. It is known thatY is P
-embedded in X if and only if for every locally �nite cozero-set coverU of Y with jUj � 
, there exists a locally �nite cozero-set cover V of X suchthat fV \ Y : V 2 Vg re�nes U. In particular, Y is C-embedded in X if andonly if Y is P!-embedded in X. For further information about P
-embedding,the reader is referred to [1]. The following problem concerning the relationshipbetween C-embedding and P-embedding is also open:Problem 2.5. Does there exist an example in ZFC of a space X, with jXj =!1, having a closed C-embedded subset which is not P-embedded?Problem 2.6. Does there exist an example in ZFC of a �rst countable spacehaving a closed C-embedded subset which is not P-embedded?It is known that under certain set-theoretic assumption such as MA+:CH,there exists a �rst countable, normal space X which is not collectionwise normal(see [21]). Since a space is collectionwise normal if and only if every closed subset



48 Haruto Ohtais P-embedded, such a space X has a closed C-embedded subset which is notP-embedded (cf. Remark 6.4 in Section 6 below).3. Spaces in which every closed C�-embedded set is C-embeddedWe say that a space X has the property (C� = Q) if every closed C�-embedded subset of X is Q-embedded in X, where Q 2 fC;P
;Pg. A subset Yof a space X is said to be z-embedded in X if every zero-set in Y is the restrictionof a zero-set in X to Y (cf. [2]). Every C�-embedded subset is z-embedded.Two subsets A and B are said to be completely separated in X if there exists areal-valued continuous function f on X such that f[A] = f0g and f[B] = f1g.The following theorem was proved by Blair and Hager in [2, Corollary 3.6.B].Theorem 3.1. [Blair-Hager] A subset Y of a space X is C-embedded in X ifand only if Y is z-embedded in X and Y is completely separated from everyzero-set in X disjoint from Y .Recall from [11] that a space X is �-normally separated if every two disjointclosed sets, one of which is a zero-set, are completely separated in X. Allnormal spaces and all countably compact spaces are �-normally separated. ByTheorem 3.1, we have the following corollary:Corollary 3.2. Every �-normally separated space has the property (C� = C).The converse of Corollary 3.2 does not hold as the next example shows:Example 3.3. The Tychono� plank T = ((!1+1)�(!+1))nfh!1;!ig is not �-normally separated but every closed C�-embedded subset of T is P-embedded,i.e., T has the property (C� = P). To prove these facts, let A = f!1g � !and B = !1 � f!g; then A is closed in T and B is a zero-set in T . Since Aand B cannot be completely separated in T , T is not �-normally separated.Next, let F be a closed C�-embedded subset of T . We have to show that Fis P-embedded in T . Since there is no uncountable discrete closed set in T ,every locally �nite cozero-set cover of F is countable. Hence, it su�ces to showthat F is C-embedded in T . Since F is closed in T , either F includes a closedunbounded subset of B or F \fh�;mi : � < � < !1;n < m � !g = ? for some� < !1 and some n < !. In the former case, every zero-set in T disjoint fromF must be compact. In the latter case, A \ F is �nite since F is C-embedded,which implies that F is compact. In both cases, F is completely separatedform a zero-set disjoint from it. Hence, it follows from Theorem 2.1 that F isC-embedded.The following example shows that the product of a space with the property(C� = P) and a compact space need not have the property (C� = C).Example 3.4. Let T be the Tychono� plank. As we showed in Example 3.3,T has the property (C� = P). We show that T � �E(T) fails to have theproperty (C� = C), where E(T) is the absolute of T . Let eT : E(T) ! T bethe perfect onto map. Then the subspace G = fheT (x);xi : x 2 E(T)g is closedin T � �E(T), because eT is perfect. Since T is not weakly normal, it follows



Extension properties and the Niemytzki plane 49from Lemma 2.1 that E(T) does not have the property (C� = C), and hence, Galso fails to have the property (C� = C), because G is homeomorphic to E(T).Hence, if we prove that G is C�-embedded in T � �E(T), then it would followthat T � �E(T) does not have the property (C� = C). For this end, let f bea bounded real-valued continuous function on G and de�ne g : E(T) ! R byg(x) = f(heT (x);xi) for x 2 E(T). Since g is bounded continuous, g extendsto a continuous function h on �E(T). Then h � � is a continuous extension off over T � �E(T), where � : T � �E(T) ! �E(T) is the projection. Hence, Gis C�-embedded in T � �E(T). �Problem 3.5. Does there exist a space X with the property (C� = C) and ametric space M such that X � M fails to have the property (C� = C)?The positive answer to Problem 1.2 for Michael's product space answersProblem 3.5 positively. We conclude this section by giving a class of spaceshaving the property (C� = P
). Recall from [10, 14] that a family F of subsetsof a space X is uniformly locally �nite in X if there exists a locally �nite cozero-set cover U of X such that every U 2 U intersects only �nitely many membersof F. Let 
 be an in�nite cardinal. A subset Y of a space X is said to beU
-embedded in X if every uniformly locally �nite family F of subsets in Ywith jFj � 
 is uniformly locally �nite in X (cf. [7]). The following theoremwas proved in [15] (see also [7, Proposition 1.6]).Theorem 3.6. [Morita-Hoshina] For every in�nite cardinal 
, a subset Y ofa space X is P
-embedded in X if and only if Y is both z-embedded and U
-embedded in X.Recall from [7] that a space X has the property (U
) (resp. property (U
)�)if every locally �nite (resp. discrete) family F of subsets of X with jFj � 
 isuniformly locally �nite in X. All 
-collectionwise normal and countably para-compact spaces have the property (U
), and all 
-collectionwise normal spaceshave the property (U
)�. Hoshina [7] proved that a space X has the property(U
)� if and only if every closed subset of X is U
-embedded. Combining thiswith Theorem 3.6, we have the following corollary:Corollary 3.7. For every in�nite cardinal 
, every space having the property(U
)� has the property (C� = P
).It will be worth noting that every 
-collectionwise normal Dowker space (see[17]) has the property (U
)� for every 
 but does not have the property (U!).4. ProductsIt is quite interesting to consider the relationship between C�- and C-em-beddings in the realm of product spaces. In spite of extensive studies, thefollowing problem is still unanswered.Problem 4.1. Let A be a closed C-embedded subset of a space X, Y a space,and assume that A�Y is C�-embedded in X �Y . Then, is A�Y C-embeddedin X � Y ?



50 Haruto OhtaIn this section, we summarize partial answers to Problem 4.1 and also discussthe following problem:Problem 4.2. Let X and Y be spaces with the property (C� = C). Under whatconditions on X and Y does X � Y have the property (C� = C)?First, we consider product spaces with a compact factor. Morita-Hoshina[15] proved the following theorem which answers Problem 4.1 positively whenY is a compact space.Theorem 4.3. [Morita-Hoshina] Let A be a subset of a space X, Y an in�nitecompact space, and assume that A�Y is C�-embedded in X �Y . Then A�Yis Pw(Y )-embedded in X � Y , where w(Y ) is the weight of Y .From now on, let 
 denote an in�nite cardinal. The next theorem is ananswer to Problem 4.2.Theorem 4.4. If a space X has the property (U
), then X�Y has the property(C� = P
) for every compact space Y .Proof. If X has the property (U
) and Y is a compact space, then it is easilyproved that X � Y has the property (U
). Hence, X � Y has the property(C� = P
) by Corollary 3.7. �Example 3.4 shows that `property (U
)' in Theorem 4.4 cannot be weakenedto `property (C� = P
)'. The following problem remains open:Problem 4.5. If X�Y has the property (C� = P
) for every compact space Y ,then does X have the property (U
)? More specially, does Theorem 4.4 remaintrue if `property (U
)' is weakened to `property (U
)�'?A space is called �-locally compact if it is the union of countably many closedlocally compact subspaces. Concerning products with a �-locally compact,paracompact factor, the following theorem was proved by Yamazaki in [23] and[25]:Theorem 4.6. [Yamazaki] Let A be a C-embedded subset of a space X, Y a�-locally compact, paracompact space, and assume that A � Y is C�-embeddedin X�Y . Then A�Y is C-embedded in X�Y . Moreover, if A is P
-embeddedin X in addition, then A � Y is also P
-embedded in X � Y .Problem 4.7. Does Theorem 4.4 remain true if `compact' is weakened to `�-locally compact, paracompact'?Next, we consider products with a metric factor. The di�culty of this caseis in the fact that A � Y need not be U!-embedded in X � Y even if A is P-embedded in X (consider Michael's product space). Nevertheless, the followingTheorems 4.8 and 4.9 were proved by Gutev-Ohta [6]:Theorem 4.8. [Gutev-Ohta] Let A be a subset of a space X, Y a non-discretemetric space, and assume that A � Y is C�-embedded in X � Y . Then A � Yis C-embedded in X � Y .



Extension properties and the Niemytzki plane 51Theorem 4.9. [Gutev-Ohta] Let A be a P
-embedded subset of a space X andY a metric space. Then the following conditions are equivalent:(1) A � Y is P
-embedded in X � Y ;(2) A � Y is C�-embedded in X � Y ;(3) A � Y is U!-embedded in X � Y .Corollary 4.10. Let A be a P
-embedded subset of a space X, Y the product ofa �-locally compact, paracompact space K with a metric space M, and assumethat A � Y is C�-embedded in X � Y . Then A � Y is P
-embedded in X � Y .Proof. Since (A � K) � M is C�-embedded in (X � K) � M, A � K is C�-embedded in X � K. Hence, A � K is P
-embedded in X � K by Theorem4.6. Finally, it follows from Theorem 4.9 that (A�K)�M is P
-embedded in(X � K) � M. �Problem 4.11. Does Theorem 4.8 remain true if `metric space' is weakenedto `paracompact M-space' or `La�snev space'?Problem 4.12. Let A be a P
-embedded subset of a space X and Y a paracom-pact M-space. Then, does the condition (2) in Theorem 4.9 imply the condition(1)?Problem 4.13. Let A be a P
-embedded subset of a space X and let Y be oneof the following spaces (i){(iii): (i) a La�snev space; (ii) a strati�able space; (iii)a paracompact �-space. Then, are the conditions (1), (2), (3) in Theorem 4.9equivalent?For the de�nitions of the spaces (i), (ii) and (iii) in Problem 4.13, we referthe reader to [5]. Problems 4.12 and 4.13 were raised in [6].Now, we try to extend Theorems 4.3 and 4.8 to products with a factor spacein wider class of spaces. For this end, we write Y 2 �(Q) if for every space Xand every closed subset A of X, if A�Y is C�-embedded in X �Y , then A�Yis Q-embedded in X �Y , where Q 2 fC;P
g. By Theorem 4.3, Y 2 �(Pw(Y ))for every in�nite compact space Y , and by Theorem 4.8, Y 2 �(C) for everynon-discrete metric space Y . The following results show that the classes �(P
)and �(C) are much wider than we expected.Theorem 4.14. Let Y be a space with Y 2 �(P
). Then Y � Z 2 �(P
) forevery space Z.Proof. Let X be a space with a closed subset A such that A � (Y � Z) isC�-embedded in X � (Y � Z). Then, it is obvious that (A � Z) � Y is C�-embedded in (X � Z) � Y . Since Y 2 �(P
), (A � Z) � Y is P
-embedded in(X � Z) �Y , which means that A � (Y � Z) is P
-embedded in X � (Y � Z).Hence, Y � Z 2 �(P
). �Corollary 4.15. For every space Y , Y � (! + 1) 2 �(C).Proof. Since ! + 1 2 �(C) by Theorem 4.3 (or Theorem 4.8), this followsimmediately from Theorem 4.14. �



52 Haruto OhtaThe next theorem and its corollary were proved by Hoshina and Yamazakiin [9].Theorem 4.16. [Hoshina-Yamazaki] Let Y be a space which is homeomorphicto Y � Y and contains an in�nite compact subset K. Then Y 2 �(Pw(K)).Corollary 4.17. [Hoshina-Yamazaki] For every space Y with jY j � 2, Y 
 2�(P
).Finally, we consider some miscellaneous products. The following theoremwas proved by Yamazaki in [24] and [25]. By a P-space, we mean a P-space inthe sense of Morita [13]. For the de�nition of a �-space, see [5].Theorem 4.18. [Yamazaki] Let A be a closed subset of a normal P-space X,Y a paracompact �-space, and assume that A � Y is C�-embedded in X � Y .Then A � Y is C-embedded in X � Y . Moreover, if A is P
-embedded in X inaddition, then A � Y is P
-embedded in X � Y .Since a P-space is countably paracompact, all normal P-spaces have theproperty (U!) and all 
-collectionwise normal P-spaces have the property (U
).Hence, the following problem naturally arises after Theorem 4.18.Problem 4.19. Let X be a normal P-space and Y a paracompact �-space.Then, does X �Y have the property (C� = C)? Moreover, if X is 
-collection-wise normal in addition, then does X � Y have the property (C� = P
)?Recently, a partial answer to Problem 4.19 was given by Yajima [22].Theorem 4.20. [Yajima] Let X be a collectionwise normal P-space and Ya paracompact �-space. Then every closed C-embedded subset of X � Y isP-embedded in X � Y .5. Zero-sets in the Niemytzki planeIn the remainder of this paper, we consider extension properties of theNiemytzki plane NP , and in the �nal section, we answer Problem 1.2 for NPnegatively. The Niemytzki plane NP is the closed upper half-plane R�[0; +1)with the topology de�ned as follows: For each p = hx;yi 2 NP and " > 0, letS"(p) = (fq 2 NP : d(hx;"i;q) < "g [ fpg for y = 0;fq 2 NP : d(p;q) < "g for y > 0;where d is the Euclidean metric on the plane. The topology of NP is generatedby the family fS"(p) : p 2 NP;" > 0g. Let L = fhx;0i : x 2 Rg � NP .From now on, we always consider a subset of R to be a subspace of R, andconsider a subset of NP to be a subspace of NP unless otherwise stated. Forexample, an interval I is a subspace of R but I � f0g is a subspace of NP .When A � X � NP , we say that A is "-open in X if A is open with respectto the relative topology on X induced from the Euclidean topology. The words"-closed and "-continuous are used similarly.In this section, we determine a zero-set in NP . We �rst state the main resultsin this section, then proceed to the proofs.



Extension properties and the Niemytzki plane 53Theorem 5.1. Let F be a closed subset of NP. Then F is a zero-set in NPif and only if the set fx 2 R : hx;0i 2 Fg is a G�-set in R.Corollary 5.2. If S is a subset of NP with S\L = ?, then clNP S is a zero-setin NP. In particular, every closed subset S of NP with S\L = ? is a zero-setin NP.Proof. This follows from Theorem 5.1 above and Lemma 5.11 below. �The next corollary follows from Corollary 5.2, since F = clNP (F n L) forevery regular-closed set F in NP .Corollary 5.3. Every regular-closed set in NP is a zero-set.Theorem 5.1 also shows that every zero-set in NP is a G�-set with respectto the Euclidean topology. On the other hand, every "-closed set in the closedupper half-plane is a zero-set in NP . Hence, we have the following corollary.Corollary 5.4. For a subset S of NP, S is a Baire set in NP if and only ifS is a Borel set with respect to the Euclidean topology.The �nal theorem of this section describes a zero-set in a subspace of NP .Theorem 5.5. Let Y be a subspace of NP and Y0 = clY (Y n L). Let F be aclosed subset of Y . Then F is a zero-set in Y if and only if A is a G�-set inB, where A = fx 2 R : hx;0i 2 F \ Y0g and B = fx 2 R : hx;0i 2 Y0g.Before proving Theorems 5.1 and 5.5, let us observe some examples of non-trivial zero-sets in NP .Example 5.6. (1) The �rst one is a zero-set E in NP such that E\L = ? butthe set fx 2 R : hx;0i 2 cl" Eg is the Cantor set K, where cl" E is the closure ofE with respect to the Euclidean topology. Let I be the set of all componentsof [0;1] n K. For each open interval I = (a;b) 2 I, de�neEI = fhx;yi : a < x < b; y = minf1 �p1 � (x � a)2;1 �p1 � (x � b)2gg:Then EI is a closed set in NP such that cl" EI n EI = fha;0i;hb;0ig. De�neE = SfEI : I 2 Ig. Then E is a closed set in NP such that E \ L = ? andK = fx 2 R : hx;0i 2 cl" Eg, as required. By Corollary 5.2, E is a zero-set inNP .(2) The second one is a zero-set F of NP such that F = clNP (F nL) and fx 2R : hx;0i 2 Fg = RnQ . Since Q�f0g is countable and discrete closed in NP , wecan �nd a disjoint family S = fS"(x)(hx;0i) : x 2 Qg of basic open sets in NP .De�ne F = NP nSfS : S 2 Sg. Then, fx 2 R : hx;0i 2 Fg = RnQ clearly. Toshow that F = clNP(F n L), consider a point q = hx;0i 2 (R n Q) � f0g. Then,S"(q) \ (F n L) 6= ? for each " > 0, because S is disjoint and the open intervalfy 2 R : hx;yi 2 S"(q) n fqgg cannot be covered by disjoint open intervals Jwith inf J > 0. Hence, q 2 clNP (F n L), which implies that F = clNP (F n L).Finally, F is a zero-set in NP by Corollary 5.2. �



54 Haruto OhtaTo prove Theorems 5.1 and 5.5, we need some de�nitions and lemmas. LetR] = R [ f�1;+1g and consider �1 < x < +1 for each x 2 R. For eacha 2 R], we de�ne a function ha : R ! [0;1] as follows: For a 2 R, de�neha(x) = (1 �p1 � (x � a)2 if jx � aj � 1;1 otherwise,and de�ne h+1(x) = h�1(x) = 1 for x 2 R. By an open interval in R, wemean a set of the form (a;b) = fx 2 R : a < x < bg for a;b 2 R] with a < b.For an open interval J = (a;b) in R, we de�neUJ = fhx;yi : a < x < b;0 � y < minfha(x);hb(x)gg:Lemma 5.7. For every open interval J = (a;b) in R, the following are valid:(1) J � f0g � UJ,(2) J � f0g is a zero-set in NP and UJ is a cozero-set in NP.Proof. (1) is obvious. To prove (2), let H = J � [0;+1). Since UJ is "-open inH, there is an "-continuous function f : H ! [0;1] such that f�1(0) = J �f0gand f�1(1) = H n UJ. We extend f to the function f� : NP ! [0;1] by lettingf�jH = f and f�(p) = 1 for each p 2 NP n H. Then f� is continuous on NPby the de�nition of UJ. Since J � f0g = f�1� (0) and UJ = f�1� [[0;1)], we have(2). �Lemma 5.8. If J is a family of disjoint open intervals in R, then the familyU = fUJ : J 2 Jg is discrete in NP.Proof. Let p = hx;yi 2 NP . If y = 0, then S1(p) meets at most one member ofU. If y > 0, then Sy=2(p) meets at most one member of U. �Let F be a family of subsets of a space X. It is known [14, 18] that F isuniformly locally �nite in X if and only if there exist a locally �nite familyfG(F) : F 2 Fg of cozero-sets in X and a family fZ(F) : F 2 Fg of zero-sets in X such that F � Z(F) � U(F) for each F 2 F. Now, we say thatF is uniformly discrete in X if there exist a discrete family fU(F) : F 2 Fgof cozero-sets in X and a family fZ(F) : F 2 Fg of zero-sets in X such thatF � Z(F) � U(F) for each F 2 F.Lemma 5.9. [15, Lemma 2.3] The union of a uniformly locally �nite family ofzero-sets in a space X is a zero-set in X.Lemma 5.10. If A is a G�-set in R, then A � f0g is a zero-set in NP.Proof. There exist open sets Gn, n 2 N, in R such that A = Tn2N Gn. For eachn 2 N, Gn is the union of a family fJi : i 2 Mg of disjoint open intervals inR. By Lemmas 5.7 and 5.8, fJi � f0g : i 2 Mg is a uniformly discrete familyof zero-sets in NP . Hence, Gn � f0g is a zero-set in NP by Lemma 5.9. Sincethe intersection of countably many zero-sets is a zero-set, A � f0g is a zero-setin NP . �Lemma 5.11. If S is a subset of NP with S \ L = ?, then the set A = fx 2R : hx;0i 2 clNP Sg is a G�-set in R.



Extension properties and the Niemytzki plane 55Proof. For each x 2 RnA, there exists n(x) 2 N such that S1=n(x)(hx;0i)\S =?. For each n 2 N, let Bn = fx 2 R : n(x) = ng. Then it is easily proved thatA \ clR Bn = ?. Since R n A = Sn2N clR Bn, A is a G�-set in R. �Lemma 5.12. Let Y be a subspace of NP such that Y = clY (Y nL) and let Fbe a zero-set in Y . Then A is a G�-set in B, where A = fx 2 R : hx;0i 2 Fgand B = fx 2 R : hx;0i 2 Y g.Proof. Since F is a zero-set in Y , there exist open sets Gn, n 2 N, in Y suchthat F = Tn2N clY Gn. Let H = Tn2N clNP (Gn n L); then F = H \ Y by thecondition of Y . Moreover, the set C = fx 2 R : hx;0i 2 Hg is a G�-set in R byLemma 5.11. Since A = B \ C, A is a G�-set in B. �Lemma 5.13. Let E and F be closed sets in NP such that L � E and E\F =?. Then there exists an open set U in NP such that E � U � clNP U � NPnF.Proof. For each p 2 E, there is n(p) 2 N such that S1=n(p)(p)\F = ?. For eachn 2 N, let En = fp 2 E : n(p) = ng and Un = SfS1=2n(p) : p 2 Eng. Then Unis an open set in NP with En � Un. We show that clNP Un \ F = ? for eachn 2 N. Suppose on the contrary that there is a point q = hx;yi 2 clNP Un \ Ffor some n 2 N. Then y > 0, because F \L = ?. Thus, we can �nd � > 0 suchthat for every x 2 R, if q 62 S1=n(hx;0i), then S�(q) \ S1=2n(hx;0i) = ?. If weput " = minf�;1=2ng, then8p 2 NP (q 62 S1=n(p) ) S"(q) \ S1=2n(p) = ?):Now, since q 2 clNP Un, S"(q) \ S1=2n(p) 6= ? for some p 2 En. By (1), thisimplies that q 2 S1=n(p), which contradicts the fact that S1=n(p) \ F = ?.Hence, clNP Un \F = ? for every n 2 N, and obviously, E � Sn2N Un. On theother hand, since F is Lindel�of, there exists a countable family fVn : n 2 Ng ofopen sets in NP such that F � Sn2N Vn and clNP Vn \ E = ? for each n 2 N.Finally, the set U = Sn2N(Un nTi�n clNP Vi) is a required open set in NP . �We are now ready to prove Theorems 5.1 and 5.5.Proof. (of Theorem 5.1) Let A = fx 2 R : hx;0i 2 Fg. If F is a zero-set in NP ,then A is a G�-set in R by Lemma 5.12. Conversely, assume that A is a G�-setin R, i.e., there exist open sets Gn, n 2 N, in R with A = Tn2N Gn. For eachn 2 N, let Kn = (RnGn)�f0g. Then both A�f0g and Kn are zero-sets in NPby Lemma 5.10. Hence, there exists a continuous function fn : NP ! [0;1]such that fn[A � f0g] = f0g and fn[Kn] = f1g. Let Hn = F \ f�1[[1=2;1]].Then Hn is a closed set in NP with Hn \ L = ?. By using Lemma 5.13 andthe technique used in the proof of Urysohn's lemma, we can de�ne anothercontinuous function gn : NP ! [0;1] such that gn[L] = f0g and gn[Hn] = f1g.De�ne Zn = f�1n [[0;1=2]] [ g�1n [f1g]. Then Zn is a zero-set in NP such thatF � Zn and Zn \ Kn = ?. On the other hand, F [ L is a zero-set in NP ,because it is an "-closed set. Since F = (F [ L) \ Tn2N Zn, F is a zero-set inNP . �Proof. (of Theorem 5.5) If F is a zero-set in Y , then F \ Y0 is a zero-set in Y0.Since Y0 = clY (Y0 nL), it follows from Lemma 5.12 that A is a G�-set in B. To



56 Haruto Ohtaprove the converse, assume that A is a G�-set in B. Since Y n Y0 is a discrete,open and closed subset of Y , F nY0 is a zero-set in Y . Hence, it su�ces to showthat F \ Y0 is a zero-set in Y . To show this, let Z1 = clNP (F n L) \ Y0 andZ2 = (F \ Y0) \ L. Then F \ Y0 = Z1 [ Z2. By Corollary 5.2, Z1 is a zero-setin Y0. On the other hand, by the assumption, there exists a G�-set C in R suchthat A = B \ C. Since Z2 = (C � f0g) \ Y0, Z2 is a zero-set in Y0 by Lemma5.10. Consequently, F \ Y0 is a zero-set in Y0, and hence, in Y , because Y0 isopen and closed in Y . �6. z-embedded subsets in NPA subset A of R is called a Q-set if every subset of A is a G�-set in A. Allcountable sets are Q-sets and the existence of an uncountable Q-set is knownto be independent of the usual axioms of set theory (cf. [12]). It is quite easyto determine a z-embedded set Y in NP such that Y � L. Indeed, the �rsttheorem immediately follows from Theorem 5.1:Theorem 6.1. For a subset A of R, A�f0g is z-embedded in NP if and onlyif A is a Q-set in R.Next, we consider a z-embedded subset in NP which is not necessarily asubset of L.Lemma 6.2. Let Y be a subset of NP such that Y = clY (Y n L). Then Y isz-embedded in NP.Proof. Let F be a zero-set in Y . Let A = fx 2 R : hx;0i 2 Fg and B = fx 2R : hx;0i 2 Y g. Then by Lemma 5.12, A is a G�-set in B, i.e., there is a G�-setC in R with A = B \C. Let Z = (C �f0g)[clNP (F nL). Then Z is a zero-setin NP , because both C � f0g and clNP (F n L) are zero-sets in NP by Lemma5.10 and Corollary 5.2, respectively. Since F = Z \ Y , Y is z-embedded inNP . �Theorem 6.3. Let Y be a subspace of NP and Y0 = clY (Y n L). Then Y isz-embedded in NP if and only if A is a Q-set in R and is a G�-set in B, whereA = fx 2 R : hx;0i 2 Y n Y0g and B = fx 2 R : hx;0i 2 Y g.Proof. First, assume that Y is z-embedded in NP . Then Y nY0 is z-embeddedin NP , because Y0 is open and closed in Y . Hence, it follows from Theorem 6.1that A is a Q-set. Moreover, since Y is z-embedded in NP , there is a zero-setF in NP such that Y n Y0 = F \ Y . By Theorem 5.1, the set C = fx 2 R :hx;0i 2 Fg is a G�-set in R. Since A = B \ C, A is a G�-set in B. Next,we prove the converse. By the assumption, there is a G�-set D in R such thatA = B \ D. Let Z1 = D � f0g and Z2 = clNP (Y n L). Then both Z1 and Z2are zero-sets in NP by Lemma 5.10 and Corollary 5.2, respectively, and theysatisfy that Y n Y0 � Z1, Y0 � Z2, Z1 \ Y0 = ? and Z2 \ (Y n Y0) = ?. Hence,it su�ces to show that both Y n Y0 and Y0 are z-embedded in NP . Since A isa Q-set, Y n Y0 is z-embedded in NP by Theorem 6.1, and Y0 is z-embeddedin NP by Lemma 6.2. �



Extension properties and the Niemytzki plane 57Remark 6.4. It is known that if A � R is a Q-set, then the subspace Y =(A�f0g) [(NP nL) of NP is normal (cf. [21, Example F]). Hence, the closedset A�f0g is then C-embedded in Y . However, this does not mean that A�f0gis C-embedded in NP even if A is countable. In fact, it is known ([8, Example3.14]) that Q �f0g is not C�-embedded in NP ; this also follows from Theorem7.1 below. 7. P-, C- and C�-embedded subsets in NPRecall from [6] that a subset Y of a space X is CU-embedded in X if for everypair of a zero-set E in Y and a zero-set F in X with E \F = ?, E and F \ Yare completely separated in X. The extension properties we have consideredare related as the following diagram, where the arrow `A ! B' means thatevery A-embedded subset is B-embedded:P �! C �! C? �! z# #U! �! CUMoreover, we say that a subset Y � X is uniformly discrete in X if the familyffxg : x 2 Y g is uniformly discrete in X, in other words, there exists a discretefamily fU(x) : x 2 Y g of cozero-sets in X such that x 2 U(x) for each x 2 Y .As is easily shown, every uniformly discrete set in X is P-embedded in X.Finally, we brie
y review scattered sets in R. Let A � R. For every ordinal �,we de�ne the set A(�) inductively as follows: A(0) = A; if � = � + 1, then A(�)is the derived set of A(�); and if � is a limit, then A(�) = TfA(�) : � < �g.A subset A of R is called scattered if A(�) = ? for some �, and then we write�(A) = minf� : A(�) = ?g. It is known that �(A) < !1 for every scattered setA in R.Theorem 7.1. For a subset A of R, the following conditions are equivalent:(1) A is a scattered set in R;(2) A � f0g is uniformly discrete in NP;(3) A � f0g is P-embedded in NP;(4) A � f0g is CU-embedded in NP.Proof. (1) ) (2): We prove this implication by induction on �(A). If �(A) = 0,it is obviously true since A = ?. Now, let � > 0 and assume that the implicationholds for every subset A0 � R with �(A0) < �. Let A � R be a scattered setwith �(A) = �. In case � = � + 1, (A n A(�)) � f0g is uniformly discrete inNP by inductive hypothesis, because �(A n A(�)) < �. Since A(�) is discrete,there is a family fIx : x 2 A(�)g of disjoint open intervals in R such that x 2 Ixfor each x 2 A(�). Hence, it follows from Lemmas 5.7 and 5.8 that A(�) � f0gis also uniformly discrete in NP . Since the union of �nitely many uniformlydiscrete subsets is uniformly discrete, A � f0g is uniformly discrete in NP .In case � is a limit, then U = fA n A(�) : � < �g is an open cover of A.Since every scattered set in R is zero-dimensional, there exists a disjoint openre�nement V of U. By considering order components of each member of V, we



58 Haruto Ohtacan �nd a family J = fJn : n 2 Mg of disjoint open intervals in R such thatJ covers A and fJn \ A : n 2 Mg re�nes V. By Lemmas 5.7 and 5.8 again,the family fJn � f0g : n 2 Mg is uniformly discrete in NP , and hence, so isf(Jn \A) �f0g : n 2 Mg . Moreover, each (Jn \A)�f0g is uniformly discretein NP by inductive hypothesis. Since the union of a uniformly discrete familyof uniformly discrete subsets is also uniformly discrete, A � f0g is uniformlydiscrete in NP .(2) ) (3) ) (4): Obvious.(4) ) (1): Suppose that A is not scattered; then A includes a perfect subsetB which is closed in A. Let K = clR B and take a countable dense subset B0 ofB such that the set B1 = B n B0 is also dense in B, i.e., K = clR B0 = clR B1.Let E = (K n B0) � f0g; then E is a zero-set in NP by Lemma 5.10. Now,B0�f0g is a zero-set in A�f0g, because A�f0g is discrete. On the other hand,B1 �f0g = E \ (A � f0g). Since A� f0g is CU-embedded in NP , there existsa continuous function f : NP ! [0;1] such that f[Bi � f0g] = i for i = 0;1.Let Ci = fx 2 R : f(hx;0i) = ig for each i = 0;1. Then C0 and C1 are disjointG�-sets in R by Theorem 5.1. Hence, we can write K n Ci = Sj2N Di;j, whereeach Di;j is "-closed in K, for each i = 0;1. Since B � C0[C1 and both B0 andB1 are dense in K, Di;j is nowhere dense in K for all i and j. This contradictsthe completeness of K. � �Lemma 7.2. Every CU-embedded subset Y in a �rst countable space X isclosed.Proof. If Y is not closed in X, then there exists a sequence fpn : n 2 Ng � Ywhich converges to a point p 2 X n Y . We may assume that pm 6= pn if m 6= n.Let E = fp2n : n 2 Ng and F = fp2n�1 : n 2 Ng [ fpg. It is easily provedthat F is a compact G�-set in X, and hence, a zero-set in X, because X iscompletely regular. On the other hand, since E [ fpg is also a zero-set in X,E is a zero-set in Y . Since Y is CU-embedded in X, E and F n fpg must becompletely separated in X, which is impossible. �Lemma 7.3. Every scattered subset A of R is a G�-set in R.Proof. This is well-known and also follows from our results. In fact, by Theorem7.1, A�f0g is uniformly discrete in NP , which implies that A�f0g is a zero-setin NP by Lemma 5.9. Hence, A is a G�-set in R by Theorem 5.1. �By Lemma 7.2, we can restrict our attention to closed subsets of NP . The fol-lowing theorem shows that every CU-embedded subset of NP is P-embedded,which answers Problem 1.2 for the Niemytzki plane negatively.Theorem 7.4. Let Y be a closed subspace of NP and let Y0 = clY (Y n L).Then the following conditions are equivalent:(1) The set A = fx 2 R : hx;0i 2 Y n Y0g is a scattered set in R;(2) Y n Y0 is uniformly discrete in NP;(3) Y is P-embedded in NP;(4) Y is CU-embedded in NP.



Extension properties and the Niemytzki plane 59Proof. (1) , (2): This equivalence follows from Theorem 7.1.(1) ) (3): Suppose that (1) is true. Then A is a G�-set in R by Lemma7.3. Hence, the set A � f0g(= Y n Y0) is a zero-set in NP by Lemma 5.10.On the other hand, by the de�nition of Y0, it follows from Corollary 5.2 thatY0 is a zero-set. Consequently, Y0 and Y n Y0 are completely separated in NP .Hence, it su�ces to show that both Y0 and Y n Y0 are P-embedded in NP .By Theorem 6.3 and Corollary 5.2, Y0 is a z-embedded zero-set in NP , whichimplies that Y0 is C-embedded in NP by Theorem 3.1. Since Y0 is separable,Y0 has no uncountable locally �nite cozero-set cover. Hence, Y0 is P-embeddedin NP . On the other hand, Y n Y0 is P-embedded in NP by Theorem 7.1.(3) ) (4): Obvious.(4) ) (1): If Y is CU-embedded in NP , then the set A � f0g(= Y n Y0) isalso CU-embedded in NP , because Y nY0 is open and closed in Y . Hence, thisimplication follows from Theorem 7.1. �By Theorem 7.4, both of the zero-sets E and F de�ned in Example 5.6 areP-embedded in NP .Corollary 7.5. Every CU-embedded subset in NP is a P-embedded zero-set.Proof. Let Y be a CU-embedded set in NP and let Y0 = clY (Y n L). ByTheorem 7.4, Y is P-embedded in NP . Moreover, as I showed in the proof ofTheorem 7.4 (1) ) (3), both Y0 and Y n Y0 are zero-sets in NP . Hence, Y is azero-set in NP . �Recall from [19] that a subset A of a space X is �-embedded in X if A � Yis C�-embedded in X � Y for every space Y . The following problem is open:Problem 7.6. Is every P-embedded subset in NP �-embedded in NP?Acknowledgment. The author would like to thank Ken-iti Tamano for hishelpful suggestions. In particular, Theorem 5.1 was born from the discussionwith him. References[1] R. A. Al�o and H. L. Shapiro, Normal Topological Spaces (Cambridge Univ. Press, Cam-bridge, 1974).[2] L. Blair and A. W. Hager, Extensions of zero-sets and of real-valued functions, Math. Z.136 (1974), 41{52.[3] R. Engelking, General Topology (Heldermann Verlag, Berlin, 1989).[4] L. Gillman and M. Jerison, Rings of Continuous Functions (Van Nostrand, New York,1960).[5] G. Gruenhage, Generalized metric spaces, in: Handbook of Set-theoretical Topology, K.Kunen and J. E. Vaughan, eds. (North-Holland, Amsterdam, 1984), 423{501.[6] V. Gutev and H. Ohta, Does C�-embedding imply C-embedding in the realm of productswith a non-discrete metric factor?, to appear in Fund. Math.[7] T. Hoshina, Spaces with a property related to uniformly local �niteness, Tsukuba J. Math.6 (1982), 51{62.[8] T. Hoshina, Extensions of mappings II, in: Topics in General Topology, K. Morita andJ. Nagata, eds. North-Holland (Amsterdam, 1989), 41{80.



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H. OhtaFaculty of EducationShizuoka UniversityOhya, Shizuoka, 422-8529, JapanE-mail address: echohtaipc.shizuoka.ac.jp