Countable version of Erdös-Lovasz-Faber conjecture

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Let $X$ be an infinite set, and let $(A_n)_ninomega$ be a collection of subsets of $X$ with the following properties:



  1. $|A_mcap A_n| leq 1$ for $mneq nin omega$, and

  2. $|A_n|=aleph_0$ for all $nin omega$.

We consider the following statement:




(EFL$_omega$:) There is $f:Xto omega$ such that for all $ninomega$ the restriction $f|_A_n:A_ntoomega$ is a bijection.




Questions. Is (EFL$_omega$) true? Or does (EFL$_omega$) imply the original Erdös-Faber-Lovasz conjecture?






co.combinatorics graph-theory graph-colorings infinite-combinatorics erdos







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    Let $X$ be an infinite set, and let $(A_n)_ninomega$ be a collection of subsets of $X$ with the following properties:



    1. $|A_mcap A_n| leq 1$ for $mneq nin omega$, and

    2. $|A_n|=aleph_0$ for all $nin omega$.

    We consider the following statement:




    (EFL$_omega$:) There is $f:Xto omega$ such that for all $ninomega$ the restriction $f|_A_n:A_ntoomega$ is a bijection.




    Questions. Is (EFL$_omega$) true? Or does (EFL$_omega$) imply the original Erdös-Faber-Lovasz conjecture?






    co.combinatorics graph-theory graph-colorings infinite-combinatorics erdos







    share|cite|improve this question























      up vote
      5
      down vote

      favorite









      up vote
      5
      down vote

      favorite











      Let $X$ be an infinite set, and let $(A_n)_ninomega$ be a collection of subsets of $X$ with the following properties:



      1. $|A_mcap A_n| leq 1$ for $mneq nin omega$, and

      2. $|A_n|=aleph_0$ for all $nin omega$.

      We consider the following statement:




      (EFL$_omega$:) There is $f:Xto omega$ such that for all $ninomega$ the restriction $f|_A_n:A_ntoomega$ is a bijection.




      Questions. Is (EFL$_omega$) true? Or does (EFL$_omega$) imply the original Erdös-Faber-Lovasz conjecture?






      co.combinatorics graph-theory graph-colorings infinite-combinatorics erdos







      share|cite|improve this question













      Let $X$ be an infinite set, and let $(A_n)_ninomega$ be a collection of subsets of $X$ with the following properties:



      1. $|A_mcap A_n| leq 1$ for $mneq nin omega$, and

      2. $|A_n|=aleph_0$ for all $nin omega$.

      We consider the following statement:




      (EFL$_omega$:) There is $f:Xto omega$ such that for all $ninomega$ the restriction $f|_A_n:A_ntoomega$ is a bijection.




      Questions. Is (EFL$_omega$) true? Or does (EFL$_omega$) imply the original Erdös-Faber-Lovasz conjecture?






      co.combinatorics graph-theory graph-colorings infinite-combinatorics erdos




      co.combinatorics graph-theory graph-colorings infinite-combinatorics erdos






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      asked Sep 5 at 6:00









      Dominic van der Zypen

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          If I understand it correctly, it's false. Let $x notin A_0 = 1,2,dots $. Then let $A_i$ all meet at $x$, and also each meet $A$ at $i$ (add extra elements as necessary; they should be irrelevant). Then $f(x) neq f(i)$ for any $i$, so $f(x) notin f(A)$.



          This didn't work in the finite case because the sets meeting $A$ cannot exhaust $A$, as the number of such sets is strictly less than the number of elements of $A$. When working over infinite sets, this is no longer true.






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            Nice construction, thanks! (There is one typo, I guess by $A$ you mean $A_0$?)
            – Dominic van der Zypen
            Sep 5 at 8:04










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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes








          up vote
          7
          down vote



          accepted










          If I understand it correctly, it's false. Let $x notin A_0 = 1,2,dots $. Then let $A_i$ all meet at $x$, and also each meet $A$ at $i$ (add extra elements as necessary; they should be irrelevant). Then $f(x) neq f(i)$ for any $i$, so $f(x) notin f(A)$.



          This didn't work in the finite case because the sets meeting $A$ cannot exhaust $A$, as the number of such sets is strictly less than the number of elements of $A$. When working over infinite sets, this is no longer true.






          share|cite|improve this answer


















          • 1




            Nice construction, thanks! (There is one typo, I guess by $A$ you mean $A_0$?)
            – Dominic van der Zypen
            Sep 5 at 8:04














          up vote
          7
          down vote



          accepted










          If I understand it correctly, it's false. Let $x notin A_0 = 1,2,dots $. Then let $A_i$ all meet at $x$, and also each meet $A$ at $i$ (add extra elements as necessary; they should be irrelevant). Then $f(x) neq f(i)$ for any $i$, so $f(x) notin f(A)$.



          This didn't work in the finite case because the sets meeting $A$ cannot exhaust $A$, as the number of such sets is strictly less than the number of elements of $A$. When working over infinite sets, this is no longer true.






          share|cite|improve this answer


















          • 1




            Nice construction, thanks! (There is one typo, I guess by $A$ you mean $A_0$?)
            – Dominic van der Zypen
            Sep 5 at 8:04












          up vote
          7
          down vote



          accepted







          up vote
          7
          down vote



          accepted






          If I understand it correctly, it's false. Let $x notin A_0 = 1,2,dots $. Then let $A_i$ all meet at $x$, and also each meet $A$ at $i$ (add extra elements as necessary; they should be irrelevant). Then $f(x) neq f(i)$ for any $i$, so $f(x) notin f(A)$.



          This didn't work in the finite case because the sets meeting $A$ cannot exhaust $A$, as the number of such sets is strictly less than the number of elements of $A$. When working over infinite sets, this is no longer true.






          share|cite|improve this answer














          If I understand it correctly, it's false. Let $x notin A_0 = 1,2,dots $. Then let $A_i$ all meet at $x$, and also each meet $A$ at $i$ (add extra elements as necessary; they should be irrelevant). Then $f(x) neq f(i)$ for any $i$, so $f(x) notin f(A)$.



          This didn't work in the finite case because the sets meeting $A$ cannot exhaust $A$, as the number of such sets is strictly less than the number of elements of $A$. When working over infinite sets, this is no longer true.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 5 at 6:35

























          answered Sep 5 at 6:28









          user44191

          2,059720




          2,059720







          • 1




            Nice construction, thanks! (There is one typo, I guess by $A$ you mean $A_0$?)
            – Dominic van der Zypen
            Sep 5 at 8:04












          • 1




            Nice construction, thanks! (There is one typo, I guess by $A$ you mean $A_0$?)
            – Dominic van der Zypen
            Sep 5 at 8:04







          1




          1




          Nice construction, thanks! (There is one typo, I guess by $A$ you mean $A_0$?)
          – Dominic van der Zypen
          Sep 5 at 8:04




          Nice construction, thanks! (There is one typo, I guess by $A$ you mean $A_0$?)
          – Dominic van der Zypen
          Sep 5 at 8:04

















           

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