Countable Tightness and the Grothendieck Property in C p -Theory
aa r X i v : . [ m a t h . GN ] J u l COUNTABLE TIGHTNESS AND THE GROTHENDIECKPROPERTY IN C p -THEORY FRANKLIN D. TALL Abstract.
The Grothendieck property has become important in re-search on the definability of pathological Banach spaces [CI], [HT], andespecially [HT20]. We here answer a question of Arhangel’ski˘ı by prov-ing it undecidable whether countably tight spaces with Lindel¨of finitepowers are Grothendieck. We answer another of his questions by prov-ing that PFA implies Lindel¨of countably tight spaces are Grothendieck.We also prove that various other consequences of MA ω and PFA con-sidered by Arhangel’ski˘ı, Okunev, and Reznichenko are not theorems ofZFC. Introduction
For a topological space X , C p ( X ) is the set of continuous real-valued func-tions on X , given the pointwise topology inherited from R X . The classictheorem of Grothendieck [Gro52] states: Proposition 1.
Let X be countably compact and let A ⊆ C p ( X ) be suchthat every infinite subset of A has a limit point in C p ( X ) . Then the closureof A in C p ( X ) is compact. This theorem has many applications in Analysis. We became interested in itdue to its applications in Model Theory (see [CI], [HT], and [HT20]). Theseinvolve questions of definability, especially of pathological Banach spaces.The upshot is that if certain topological spaces ( type spaces ) associated witha logic have the closure property X has in the above theorem, then theseBanach spaces are not definable in that logic. We are therefore interestedin what classes of topological spaces other than the countably compact onessatisfy the conclusion of Proposition 1. We restrict ourselves to only considerinfinite completely regular spaces. Date : July 20, 2020. Research supported by NSERC Grant A-7354. . 54C35, 54A35, 54G20, 54A20, 54A25.
Key words and phrases . Grothendieck property, countable tightness, C p ( X ), Lindel¨of,Fr´echet-Urysohn, PFA, surlindel¨of. Definition 1 [Arh98] . A ⊆ X is countably compact in X if every infinitesubset of A has a limit point in X . X is a g -space if each A ⊆ X whichis countably compact in X has compact closure. X is a Grothendieck space (resp. weakly Grothendieck space ) if C p ( X ) is a hereditary g -space (resp. a g -space). Definition 2. X is countably tight if whenever A ⊆ X and x ∈ A , there is acountable B ⊆ A such that x ∈ B . X is realcompact if X can be embeddedas a closed subspace of a product of copies of the real line. Theorem 2 [Arh98] . If X is countably tight, then X is weakly Grothendieck. This is stated as “clear” in [Arh98]. Here is a proof:Clearly,
Lemma 3.
A closed subspace of a realcompact space is realcompact.
Lemma 4 [Eng89] . A completely regular space is compact if and only if itis realcompact and countably compact.
Definition 3.
A space is wD if whenever { d n : n < ω } is a closed discretesubspace, there is an infinite S ⊆ ω and a discrete collection of open sets { U n : n ∈ S } with d n ∈ U n for all n ∈ S . Lemma 5 [Dou84], [Vau78] . Every realcompact space is wD.
Lemma 6 (folklore) . Let X be wD. Let Y be countably compact in X . Then Y is countably compact.Proof. Suppose not. Let { d n : n < ω } be a closed discrete subspace of Y . Let { U n : n ∈ S } be a discrete collection of open subsets of X , with d n ∈ U n for every n ∈ S , where S ⊆ ω is infinite. Pick e n ∈ U n ∩ Y . Then { e n : n ∈ S } is a closed discrete subspace of Y , contradiction. (cid:3) Lemma 7 [Arh92] . If X is countably tight, then C p ( X ) is realcompact.Proof of Theorem 2. Let X be countably tight. Then C p ( X ) is realcompactand hence wD. Let Y be countably compact in C p ( X ). Then Y is countablycompact. But Y is realcompact, so Y is compact. (cid:3) Applications of the Proper Forcing Axiom
In [Arh98], Arhangel’ski˘ı proved:
Proposition 8. MA + ¬ CH implies that if X is countably tight and X n isLindel¨of for all n < ω , then X is Grothendieck. OUNTABLE TIGHTNESS & GROTHENDIECK 3
In fact, MA ω suffices.A dramatic strengthening of Proposition 8 is Theorem 9.
PFA implies Lindel¨of countably tight spaces are Grothendieck.Proof.
This actually follows easily from known results. First, a definition:
Definition 4.
A space is surlindel¨of if it is a subspace of C p ( X ) for someLindel¨of X .Arhangel’ski˘ı [Arh92] proved: Lemma 10.
PFA implies that every surlindel¨of compact space is countablytight.
Okunev and Reznichenko [OR07] proved:
Lemma 11. MA ω implies that every separable surlindel¨of compact count-ably tight space is metrizable. Definition 5.
A space X is Fr´echet-Urysohn if whenever x is a limit pointof Z ⊆ X , there is a sequence in Z converging to x .It follows quickly that: Theorem 12.
PFA implies that every surlindel¨of compact space is Fr´echet-Urysohn.Proof.
Metrizable spaces are clearly Fr´echet-Urysohn. By countable tight-ness, if K is compact and L ⊆ K and p ∈ L , then there is a countable M ⊆ L such that p ∈ M . But M is separable compact and so metrizable. (cid:3) Arhangel’ski˘ı proved:
Lemma 13 [Arh98] . X is Grothendieck if and only if it is weakly Grothen-dieck and compact subspaces of C p ( X ) are Fr´echet-Urysohn. This proves Theorem 9. (cid:3)
Okunev and Reznichenko [OR07] point out that the conclusions of Lem-mas 10 and 11 can be simultaneously consistently achieved without largecardinals, so we have:
Theorem 14. If ZFC is consistent, so is
ZFC plus “every Lindel¨of countablytight space is Grothendieck”.
FRANKLIN D. TALL
Lemmas 10 and 11 actually consistently solve several other problems ofArhangel’ski˘ı:
Problem 1 [Arh98] . If X is separable and compact and Y ⊆ C p ( X ) is Lin-del¨of, does Y have a countable network? Problem 2 [Arh92] . If X is separable and compact and C p ( X ) is Lindel¨of,must X be hereditarily separable? Notice that a positive answer to the first of these yields a positive answerto the second, since a space with a countable network is clearly hereditarilyseparable.
Lemma 15 [Arh92, I.1.3] . X has a countable network if and only if C p ( X ) does. Okunev [Oku95] considers versions of Problem 1 with the additional hypoth-esis that finite powers of Y are Lindel¨of. He proves: Proposition 16.
MA + ¬ CH implies that if Y is a space with all finitepowers Lindel¨of and X is a separable compact subspace of C p ( Y ) , then X ismetrizable. He states that this is a reformulation of
Proposition 17.
MA + ¬ CH implies that if X is a separable compact spaceand Y ⊆ C p ( X ) has all finite powers Lindel¨of, then Y has a countablenetwork. Okunev and Reznichenko note that actually MA ω suffices for these insteadof MA + ¬ CH. Okunev and Reznichenko also prove:
Proposition 18 [OR07, 1.8] . PFA implies that every surlindel¨of compactseparable space is metrizable.
Proposition 19 [OR07, 1.9] . PFA implies every surlindel¨of compact spaceis ℵ -monolithic, where a space is ℵ -monolithic if the closure of every count-able set has countable network weight. We can use Lemmas 10 and 11 to prove:
Theorem 20.
PFA implies that if X is a separable compact space and Y ⊆ C p ( X ) is Lindel¨of, then Y has a countable network.Proof. We closely follow part of the argument in [Oku95] for Proposition 17.He starts by recalling some material from [Arh92] (or see [Tka15]). Given
OUNTABLE TIGHTNESS & GROTHENDIECK 5 a continuous map p : X → Y , the dual map p ∗ : C p ( Y ) → C p ( X ) is definedby p ∗ ( f ) = f ◦ p , for all f ∈ C p ( Y ). The dual map is always continuous; itis an embedding if and only if p is onto. If Y ⊆ C p ( X ), then the reflectionmap ϕ XY : X → C p ( Y ) is defined by ϕ XY ( x )( y ) = y ( x ), for all x ∈ X and y ∈ Y . The reflection map is continuous.Suppose X is a separable compact space and Y is a Lindel¨of subspace of C p ( X ) which does not have a countable network. We consider the reflectionmap ϕ XY : X → C p ( Y ) and let X = ϕ XY ( X ). Then X is separable andcompact. Next, consider the dual map ϕ ∗ XY : C p ( X ) → C p ( X ). It’s anembedding, so Y = ( ϕ ∗ XY ) − ( Y ) is a subspace of C p ( X ) homeomorphicto Y . Since Y does not have a countable network, neither does Y . Thenneither does C p ( X ), so neither does X . But by Lemmas 10 and 11, X ismetrizable. This is a contradiction, since compact metrizable spaces have acountable network. (cid:3) Let us mention some more open problems.
Problem 3.
Are Lindel¨of first countable spaces Grothendieck?
Although we can’t fully answer Problem 3, we can weaken the hypothesis ofTheorem 9 in the first countable case:
Theorem 21. MA ω implies that every Lindel¨of first countable space isGrothendieck. Before proving this, we need to mention some more general facts about C p ,taken from [Oku95]. Lemma 22.
Let Y ⊆ C p ( X ) . Let x , x ∈ X . Let x ∼ Y x if y ( x ) = y ( x ) for all y ∈ Y . Let X be the set of equivalence classes and π : X → X thenatural map. For any y ∈ Y , there is a y ′ : X → R such that y = y ′ ◦ π .Give X the weakest topology that makes all of the y ′ ’s continuous. With thistopology, X is homeomorphic to ϕ XY ( X ) . Then ( ϕ ∗ XY ) − ( Y ) is a subspaceof C p ( X ) homeomorphic to Y . In particular, this tells us that if K is a separable compact subspace of C p ( Y ), then Y is homeomorphic to a subspace of C p ( K ), where K is acontinuous image of K and hence is separable and compact. Proof of Theorem 21.
Let Y be Lindel¨of and first countable. Let K be acompact separable subspace of Y . Let K be a continuous image of K , and Y be a homeomorphic copy of Y included in C p ( K ). We now invoke twoapplications of MA ω : Lemma 23 [OR07] . MA ω implies that if K is a compact separable space,then every Lindel¨of subspace of C p ( K ) is hereditarily Lindel¨of. FRANKLIN D. TALL
Lemma 24 [Sze80] . MA ω implies that every first countable hereditarilyLindel¨of space is hereditarily separable. But,
Lemma 25 [Arh98, 5.26] . Every hereditarily separable space is Grothen-dieck. (cid:3)
A corollary of what we just proved is of interest.
Corollary 26. MA ω implies that if K is a compact subspace of C p ( Y ) ,where Y is Lindel¨of and first countable, then K is metrizable.Proof. In the previous proof, we showed Y was hereditarily separable. Arhan-gel’ski˘ı proved: Lemma 27 [Arh97, 3.13] . If Y is separable and K is a compact subspace of C p ( Y ) , then K is metrizable. (cid:3) In the spirit of Problem 3, one can ask:
Problem 4. If X is a separable compact space and Y is a Lindel¨of firstcountable subspace of C p ( X ) , does Y have a countable network? We have a partial answer:
Theorem 28. MA ω implies that if X is a separable compact space and Y is a Lindel¨of first countable subspace of C p ( X ) , then Y has a countablenetwork. This follows from what we have just done by the same argument as forTheorem 20.Note that:
Theorem 29. If Y is a hereditary g -space, then countably compact subspacesof Y are compact.Proof. Let Z ⊆ Y be countably compact. Then it is countably compact initself and its closure in itself is compact. (cid:3) Problem 5.
If countably compact subspaces of C p ( X ) are compact, is X Grothendieck?
There is a necessary and sufficient condition on X so that C p ( X ) is countablytight (see Lemma 34 below), and there is even a necessary and sufficientcondition on X that ensures C p ( X ) is Fr´echet-Urysohn [GN82], but theseconditions are too onerous and entail more than we need. OUNTABLE TIGHTNESS & GROTHENDIECK 7
Problem 6.
Find a necessary and sufficient condition on X such that com-pact subspaces of C p ( X ) are countably tight. Definition 6.
A sequence { x α : α < κ } is free if for all β < κ , { x α : α < β } ∩ { x α : α ≥ β } = ∅ . It is well-known that:
Lemma 30. If X is Lindel¨of and countably tight, then X does not includean uncountable free sequence.Proof. Suppose F = { x α : α < ω } is free. Let F β = { x α : α < β } . Then { F β : β < ω } is a decreasing family of closed subspaces of X . Since X isLindel¨of, there is an x ∈ T { F β : β < ω } . Then x ∈ F but x / ∈ A for anycountable A ⊆ F , contradicting countable tightness. (cid:3) Todorcevic proved:
Theorem 31 [Tod93] . PFA implies: if X includes no uncountable free se-quences, then every countably compact subspace of C p ( X ) is compact. The hypothesis is weaker than that of Theorem 9, but so is the conclusion.
Problem 7.
Does
PFA imply that if X includes no uncountable free se-quences, then X is (weakly) Grothendieck? Todorcevic also proved:
Lemma 32 [Tod93] . Suppose every countably compact subspace of C p ( X ) iscompact. Then every compact subspace of C p ( X ) is countably tight.Proof. Suppose there is a Z ⊆ C p ( X ) such that there is a y ∈ Z − S { Z : Z ⊆ Z is countable } . But S { Z : Z ⊆ Z is countable } is countablycompact, hence compact, hence closed, a contradiction. (cid:3) Corollary 33 [Tod93] . PFA implies that if X does not include any uncount-able free sequences, then compact subspaces of C p ( X ) are countably tight. Counterexamples
In [Arh98], Arhangel’ski˘ı asked whether the conclusion of Proposition 8 istrue in ZFC. It is not:
Example 1.
Assuming ♦ plus Kurepa’s Hypothesis, Ivanov [Iva78] con-structs a compact space Y of cardinality 2 c such that Y n is hereditarilyseparable for all n < ω . C p ( Y ) is the required counterexample. FRANKLIN D. TALL
To see this, we require several results from the literature.
Lemma 34 [Arh92] . X n is Lindel¨of for every n < ω if and only if C p ( X ) is countably tight. Lemma 35 [Arh92] . X embeds into C p ( C p ( X )) . Clearly, separable Fr´echet-Urysohn spaces have cardinality ≤ c . Ivanov’sspace Y is too big to be Fr´echet-Urysohn, yet it embeds in C p ( C p ( Y )),so C p ( Y ) cannot be Grothendieck, although it is weakly Grothendieck.( C p ( Y )) n is, however, (hereditarily) Lindel¨of for all n < ω by the Velichko-Zenor theorem: Lemma 36 [Vel81], [Zen80] . If X n is hereditarily separable for all n < ω ,then ( C p ( X )) n is hereditarily Lindel¨of for all n < ω . Ivanov’s space also provides counterexamples for various other propositionsproved by Arhangel’ski˘ı, Okunev, and Reznichenko under MA ω or PFA. Y is surlindel¨of, compact, countably tight, separable, but not metrizable. Thisviolates the conclusion of Lemma 11. Definition 7. An S -space is a hereditarily separable space that is not hered-itarily Lindel¨of. A strong S -space is an S -space with all finite powers hered-itarily separable. Lemma 37 [Tod89] . b = ℵ implies there is a compact strong S -space. b = ℵ is weaker than CH, which is weaker than ♦ . Todorcevic’s spacewill work for the purpose of violating the conclusion of Lemma 11 as wellas Y . (To be more precise, Todorcevic constructs a locally compact, locallycountable strong S -space T , but then its one-point compactification T ∗ = T ∪ {∗} is a compact strong S -space).Both Y and T ∗ , embedded in C p ( C p ( Y )) and C p ( C p ( T ∗ )) respectively, pro-vide counterexamples to the conclusion of Proposition 18.They also provide counterexamples to the conclusion of Proposition 19. Thepoint is that compact spaces with countable network weight are metrizable.Any compact strong S -space Y refutes the conclusion of Proposition 17.( C p ( Y )) n will be hereditarily Lindel¨of, but C p ( Y ) does not have a countablenetwork, else Y would, but then Y would be hereditarily Lindel¨of.The hypothesis for Example 1 seems too strong; ♦ ought to suffice. I con-jecture that ♦ implies there is a “strong Ostaszewski space”, i.e. a strong S -space X which is countably compact, perfectly normal, but not compact. C p ( X ) would then have finite products hereditarily Lindel¨of, but would not OUNTABLE TIGHTNESS & GROTHENDIECK 9 be Grothendieck since X would be embedded in C p ( C p ( X )), violating The-orem 29. Notice we are not using perfect normality, so even CH mightsuffice. Remark. [OR07] appears to have some misprints. I believe that their ref-erence 2 should actually be our [Arh92]. Their Question 0.5 is asserted tobe essentially the same as Problem IV.1.8 in [Arh92] but the latter problemapparently has nothing to do with the former, so this may be a misprint.They call “centered” what is normally called “linked” [KT79]. On the otherhand, their use of “surlindel¨of” for the concept [Arh92] calls “suplindel¨of”is correct. Professor Arhangel’ski˘ı has informed me that “suplindel¨of” wasa mistranslation by the translator of the Russian original.
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