Productively countably tight spaces

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Productively countably tight spaces

Leandro F. Aurichi¹ Joint work with Angelo Bella

ICMC-USP

1Sponsored by FAPESP

Leandro F. Aurichi Joint work with Angelo Bella (ICMC-USP) Productively countably tight spaces 1 / 19

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Productively countably tight spaces

Recall that a topological spaceX has countable tightness at a pointx∈X if, for everyA⊂X such thatx∈A, there is a countable subsetB ⊂Asuch thatx ∈B.

We say that a spaceX is productively countably tight atx ∈X if, for every space Y and everyy ∈Y such thatY has countable tightness aty,X×Y has

countable tightness at (x,y).

IfX is productively countably tight at every pointx, we simply say thatX is productively countably tight.

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Productively countably tight spaces

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Productively countably tight spaces

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Here comes the game

The game that we will use is the following:

At then-th inning, player I choosesDn⊂Xr{x}, such thatx ∈Dn.

Then player II choosesdn∈Dn. We say that player II wins ifx ∈ {dn:n∈ω}. We use the notation G1(Ωx,Ωx) for this game.

(Ωx ={D⊂Xr{x}:x∈D})

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Here comes the game

At then-th inning, player I choosesDn⊂Xr{x}, such thatx ∈Dn. Then player II choosesdn∈Dn.

We say that player II wins ifx ∈ {dn:n∈ω}. We use the notation G1(Ωx,Ωx) for this game.

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Here comes the game

At then-th inning, player I choosesDn⊂Xr{x}, such thatx ∈Dn. Then player II choosesdn∈Dn. We say that player II wins ifx ∈ {dn:n∈ω}.

We use the notation G1(Ωx,Ωx) for this game. (Ωx ={D⊂Xr{x}:x∈D})

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Here comes the game

We use the notation G1(Ωx,Ωx) for this game.

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Here comes the game

We use the notation G1(Ωx,Ωx) for this game.

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A picture is worth a thousand words

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A picture is worth a thousand words

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A picture is worth a thousand words

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A picture is worth a thousand words

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A picture is worth a thousand words

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A picture is worth a thousand words

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A picture is worth a thousand words

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The first theorem

Theorem

If player II has a winning strategy for the gameG1(Ωx,Ωx)played over X , then X is productively countably tight at x .

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The idea for the proof

According to Arhangel’skii [1], we need to show the following:

For every collectionP of centered families of countable subsets ofX such that, for any neighborhoodOx ofx, there existsB ∈ P andB ∈ Bsuch that B⊂Ox, there is a subfamily{Bn:n< ω} ⊂ P such that for every choiceBn∈ Bn, x∈S

{Bn:n< ω}.

Actually, we will do a little better. We will find a subfamily (Bs)_s∈ω^<ω ofP such that, for everyf :ω−→ω, (Bs)s⊂f is a collection of sets containing the answers from playerII using the winning strategy for the game G1(Ωx,Ωx). Thus, x∈S

s⊂f Bs.

The key idea is: “how can we make this in such way that all the branches in this tree follow the winning strategy?”.

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The idea for the proof

{Bn:n< ω}.

s⊂f Bs.

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The idea for the proof

{Bn:n< ω}.

s⊂f Bs.

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The idea for the proof

{Bn:n< ω}.

s⊂f Bs.

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What would Galvin do?

The idea is to get families that we will know for sure that, no matter how we go over the tree, we can find an answer from player II that fits our needs. We can do this using the following:

Lemma

Let X be a space and let F be a strategy for player II in the gameG₁(Ω_x,Ω_x)for some x∈X . Then, for every sequence D0, ...,Dn∈Ωx, there is an open set A such that x ∈A and, for every a∈A\ {x}, there is a Da ∈Ωx such that F(D0, ...,Dn,Da) =a.

Proof.

LetB={y∈X \ {x}: there is noD∈Ωx such thatF(D0, ...,Dn,D) =y}. Note thatB∈/Ωx since, otherwise,F(D0, ...,Dn,B)∈B which is a contradiction. Thus, there is an open setAsuch thatx ∈AandA∩D=∅.

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What would Galvin do?

Lemma

Let X be a space and let F be a strategy for player II in the gameG₁(Ω_x,Ω_x)for some x∈X . Then, for every sequence D₀, ...,D_n∈Ω_x, there is an open set A such that x ∈A and, for every a∈A\ {x}, there is a Da ∈Ωx such that F(D0, ...,Dn,Da) =a.

Proof.

LetB={y∈X \ {x}: there is noD∈Ωx such thatF(D0, ...,Dn,D) =y}. Note thatB∈/Ωx since, otherwise,F(D0, ...,Dn,B)∈B which is a contradiction. Thus, there is an open setAsuch thatx ∈AandA∩D=∅.

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What would Galvin do?

Lemma

Let X be a space and let F be a strategy for player II in the gameG₁(Ω_x,Ω_x)for some x∈X . Then, for every sequence D₀, ...,D_n∈Ω_x, there is an open set A such that x ∈A and, for every a∈A\ {x}, there is a Da ∈Ωx such that F(D0, ...,Dn,Da) =a.

Proof.

LetB={y∈X \ {x}: there is noD∈Ωx such thatF(D0, ...,Dn,D) =y}.

Note thatB∈/Ωx since, otherwise,F(D0, ...,Dn,B)∈B which is a contradiction.

Thus, there is an open setAsuch thatx ∈AandA∩D=∅.

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Playing it

Then we only have to pick eachBs ⊂As where As is the appropriate open set given by the Lemma.

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The second Theorem

Theorem

Let X be a space. If X is productively countably tight at x , thenS1(Ωx,Ωx)holds.

This is done by proving that, ifX×Sc has countable tightness at (x,0), then S₁(Ω_x,Ω_x) holds (S_c is the sequential fan space of cardinalityc).

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The second Theorem

Theorem

Let X be a space. If X is productively countably tight at x , thenS1(Ωx,Ωx)holds.

This is done by proving that, ifX×Sc has countable tightness at (x,0), then S1(Ωx,Ωx) holds (Scis the sequential fan space of cardinalityc).

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An example of application

Recall the following theorems:

Theorem (Uspenskii [5])

For any Tychonoff space X , Cp(X)is productively countably tight if, and only if, Xδ is Lindel¨of.

Theorem (Scheepers [4])

Let X be a Tychonoff space. Player II has a winning strategy inG₁(Ω,Ω)played on X if, and only if, player II has a winning strategy inG₁(Ω₀,Ω₀)played on Cp(X).

Theorem (Sakai [3])

Let X be a Tychonoff space. The following are equivalent:

1 Cp(X)satisfiesS1(Ω0,Ω0);

2 X satisfiesS₁(Ω,Ω);

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An example of application

Theorem (Uspenskii [5])

Theorem (Scheepers [4])

Theorem (Sakai [3])

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An example of application

Theorem (Uspenskii [5])

Theorem (Scheepers [4])

Theorem (Sakai [3])

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An example of application

Theorem (Uspenskii [5])

Theorem (Scheepers [4])

Theorem (Sakai [3])

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A diagram

C_p(X) X

II↑G₁(Ω₀,Ω₀) ←→ II↑G₁(Ω,Ω)

↓

productively countably tight ←→ X_δ is Lindel¨of

↓

S1(Ω0,Ω0) ←→ S1(Ω,Ω)

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Bibliography

A. V. Arhangel’skii.

The frequency spectrum of a topological space and the product operation.

Trans. Moscow Math. Soc., 40(2):163–200, 1981.

L. F. Aurichi and A. Bella.

Topological games and productively countably tight spaces.

arXiv:1307.7928, 2013.

M. Sakai.

Property $C”$ and function spaces.

Proceedings of the American Mathematical Society, 104(3):917–919, 1988.

M. Scheepers.

Remarks on countable tightness.

arXiv:1201.4909, pages 1–23, 2013.

V. V. Uspenskii.

Frequency spectrum of functional spaces.

Vestnik. Mosk. Universita, Ser. Matematica, 37(1):31–35, 1982.

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Trivia

Some years ago, some undergrad students of the University of S˜ao Paulo created a table for the level of difficulty of an example of the form “a space that has the propertiesA1, ...,An but does not have the propertiesB1, ...,Bm”.

The students defined that the example is

Ofelia-easy if you ask Ofelia about it and she answers immediately Ofelia-medium if you ask Ofelia about it

and she thinks for a second and then answer it Ofelia-hard if you ask Ofelia,

she thinks for a few seconds

and then she says “I will answer it tomorrow”.

The legend says that once a student made an Ofelia-hard question. This fact was very celebrated afterwards.

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Trivia

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Trivia

Ofelia-easy if you ask Ofelia about it and she answers immediately

Ofelia-medium if you ask Ofelia about it

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Trivia

and she thinks for a second and then answer it

Ofelia-hard if you ask Ofelia,

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Trivia

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Trivia

The legend says that once a student made an Ofelia-hard question.

This fact was very celebrated afterwards.

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