Maximality with respect to the splitting property

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Let $X$ be a set and $cal P(X)$ its powerset. We say that $cal F subseteq cal P(X)$ has the splitting property (SP) if there is $Ain cal P(X)$ such that for all $Fin cal F$ we have $$F cap A neq emptyset neq Fcap (Xsetminus A).$$



Let $textSP(X)$ denote the collection of all subsets of $cal P(X)$ with (SP), and we order it with $subseteq$.



If $X$ is infinite, and $cal Fin textSP(X)$, is there $cal MintextSP(X)$ such that $cal M$ is maximal in $(textSP(X),subseteq)$ and $cal Fsubseteq cal M$?



(The obvious tool to try to use, Zorn's Lemma seems to be of no help in this, but I might be wrong.)




co.combinatorics set-theory infinite-combinatorics




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  • You could just say that $mathcal F$ is $2$-colorable, i.e., the hypergraph $(X,mathcal F)$ has chromatic number $le2.$
    – bof
    Aug 13 at 11:46















up vote
2
down vote

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Let $X$ be a set and $cal P(X)$ its powerset. We say that $cal F subseteq cal P(X)$ has the splitting property (SP) if there is $Ain cal P(X)$ such that for all $Fin cal F$ we have $$F cap A neq emptyset neq Fcap (Xsetminus A).$$



Let $textSP(X)$ denote the collection of all subsets of $cal P(X)$ with (SP), and we order it with $subseteq$.



If $X$ is infinite, and $cal Fin textSP(X)$, is there $cal MintextSP(X)$ such that $cal M$ is maximal in $(textSP(X),subseteq)$ and $cal Fsubseteq cal M$?



(The obvious tool to try to use, Zorn's Lemma seems to be of no help in this, but I might be wrong.)




co.combinatorics set-theory infinite-combinatorics




share|cite|improve this question




















  • You could just say that $mathcal F$ is $2$-colorable, i.e., the hypergraph $(X,mathcal F)$ has chromatic number $le2.$
    – bof
    Aug 13 at 11:46













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $X$ be a set and $cal P(X)$ its powerset. We say that $cal F subseteq cal P(X)$ has the splitting property (SP) if there is $Ain cal P(X)$ such that for all $Fin cal F$ we have $$F cap A neq emptyset neq Fcap (Xsetminus A).$$



Let $textSP(X)$ denote the collection of all subsets of $cal P(X)$ with (SP), and we order it with $subseteq$.



If $X$ is infinite, and $cal Fin textSP(X)$, is there $cal MintextSP(X)$ such that $cal M$ is maximal in $(textSP(X),subseteq)$ and $cal Fsubseteq cal M$?



(The obvious tool to try to use, Zorn's Lemma seems to be of no help in this, but I might be wrong.)




co.combinatorics set-theory infinite-combinatorics




share|cite|improve this question












Let $X$ be a set and $cal P(X)$ its powerset. We say that $cal F subseteq cal P(X)$ has the splitting property (SP) if there is $Ain cal P(X)$ such that for all $Fin cal F$ we have $$F cap A neq emptyset neq Fcap (Xsetminus A).$$



Let $textSP(X)$ denote the collection of all subsets of $cal P(X)$ with (SP), and we order it with $subseteq$.



If $X$ is infinite, and $cal Fin textSP(X)$, is there $cal MintextSP(X)$ such that $cal M$ is maximal in $(textSP(X),subseteq)$ and $cal Fsubseteq cal M$?



(The obvious tool to try to use, Zorn's Lemma seems to be of no help in this, but I might be wrong.)





co.combinatorics set-theory infinite-combinatorics





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asked Aug 13 at 6:36









Dominic van der Zypen

12.6k43167




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  • You could just say that $mathcal F$ is $2$-colorable, i.e., the hypergraph $(X,mathcal F)$ has chromatic number $le2.$
    – bof
    Aug 13 at 11:46

















  • You could just say that $mathcal F$ is $2$-colorable, i.e., the hypergraph $(X,mathcal F)$ has chromatic number $le2.$
    – bof
    Aug 13 at 11:46
















You could just say that $mathcal F$ is $2$-colorable, i.e., the hypergraph $(X,mathcal F)$ has chromatic number $le2.$
– bof
Aug 13 at 11:46





You could just say that $mathcal F$ is $2$-colorable, i.e., the hypergraph $(X,mathcal F)$ has chromatic number $le2.$
– bof
Aug 13 at 11:46











1 Answer
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Yes, any family of subsets of $X$ with the splitting property can be extended to a maximal such family. The axiom of choice is not needed for this.



Suppose $mathcal Fin operatornameSP(X),$ and let $Ainmathcal P(X)$ be such that for all $Finmathcal F$ we have
$$Fcap Aneemptysetne Fcap(Xsetminus A).$$



We may assume that $mathcal Fneemptyset,$ so that $emptysetne Ane X.$ Now it is easy to see that the set
$$mathcal M=Finmathcal P(X):Fcap Aneemptysetne Fcap(Xsetminus A)$$
has the desired properties.






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    Yes, any family of subsets of $X$ with the splitting property can be extended to a maximal such family. The axiom of choice is not needed for this.



    Suppose $mathcal Fin operatornameSP(X),$ and let $Ainmathcal P(X)$ be such that for all $Finmathcal F$ we have
    $$Fcap Aneemptysetne Fcap(Xsetminus A).$$



    We may assume that $mathcal Fneemptyset,$ so that $emptysetne Ane X.$ Now it is easy to see that the set
    $$mathcal M=Finmathcal P(X):Fcap Aneemptysetne Fcap(Xsetminus A)$$
    has the desired properties.






    share|cite|improve this answer


























      up vote
      5
      down vote



      accepted










      Yes, any family of subsets of $X$ with the splitting property can be extended to a maximal such family. The axiom of choice is not needed for this.



      Suppose $mathcal Fin operatornameSP(X),$ and let $Ainmathcal P(X)$ be such that for all $Finmathcal F$ we have
      $$Fcap Aneemptysetne Fcap(Xsetminus A).$$



      We may assume that $mathcal Fneemptyset,$ so that $emptysetne Ane X.$ Now it is easy to see that the set
      $$mathcal M=Finmathcal P(X):Fcap Aneemptysetne Fcap(Xsetminus A)$$
      has the desired properties.






      share|cite|improve this answer
























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        Yes, any family of subsets of $X$ with the splitting property can be extended to a maximal such family. The axiom of choice is not needed for this.



        Suppose $mathcal Fin operatornameSP(X),$ and let $Ainmathcal P(X)$ be such that for all $Finmathcal F$ we have
        $$Fcap Aneemptysetne Fcap(Xsetminus A).$$



        We may assume that $mathcal Fneemptyset,$ so that $emptysetne Ane X.$ Now it is easy to see that the set
        $$mathcal M=Finmathcal P(X):Fcap Aneemptysetne Fcap(Xsetminus A)$$
        has the desired properties.






        share|cite|improve this answer














        Yes, any family of subsets of $X$ with the splitting property can be extended to a maximal such family. The axiom of choice is not needed for this.



        Suppose $mathcal Fin operatornameSP(X),$ and let $Ainmathcal P(X)$ be such that for all $Finmathcal F$ we have
        $$Fcap Aneemptysetne Fcap(Xsetminus A).$$



        We may assume that $mathcal Fneemptyset,$ so that $emptysetne Ane X.$ Now it is easy to see that the set
        $$mathcal M=Finmathcal P(X):Fcap Aneemptysetne Fcap(Xsetminus A)$$
        has the desired properties.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 13 at 11:43

























        answered Aug 13 at 7:23









        bof

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