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Is it possible to construct a family of sets $\{A_{ij}\}_{i,j=1}^\infty$ and $\{B_{ij}\}_{i,j=1}^\infty$ such that:

$(1)$For any positive integer $i\geq1$,$A_{ij}\searrow\emptyset$ and $B_{ij}\searrow\emptyset$ as $j\to\infty$;

$(2)\bigcup_{i=1}^\infty A_{i1}=\bigcup_{i=1}^\infty B_{i1}$;

$(3)A_{11},A_{21},A_{31}...$are pairwise disjoint and $B_{11},B_{21},B_{31}...$are pairwise disjoint;

$(4)$For any positive integers $i$,there exist positive integers $j_1,j_2$ and sequences of positive integers $\{i_k\}_{k=1}^\infty,\{j_k\}_{k=1}^\infty$ such that $A_{ij_1}\bigcap (\bigcup_{k=1}^\infty B_{ki_k})=\emptyset$ and $B_{ij_2}\bigcap (\bigcup _{k=1}^\infty A_{kj_k})=\emptyset$;

$(5)$For any sequences of positive integers $\{i_k\}_{k=1}^\infty$ and $\{j_k\}_{k=1}^\infty$, $(\bigcup_{k=1}^\infty A_{ki_k})\bigcap (\bigcup_{k=1}^\infty B_{kj_k})\neq\emptyset$.

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  • $\begingroup$ Sorry,there were some mistakes in my question and I have revised it. $\endgroup$
    – user40096
    Feb 26, 2014 at 10:25
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    $\begingroup$ The new version is indeed more interesting. Could you please give a bit of motivation for the question? where does it come from? $\endgroup$ Feb 26, 2014 at 21:33

2 Answers 2

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This answer corresponds to an earlier version of the question:

No, it is not possible. Given $\{A_{ij}\}_{i,j=1}^\infty$ and $\{B_{ij}\}_{i,j=1}^\infty$ satisfying (1) and (4), construct sequences $\{i_k\}_{k=1}^\infty$ and $\{j_k\}_{k=1}^\infty$ violating (5) as follows: for each positive integer $n$, use (4) $n$ times to find $l_1,m_1,\dots,l_n,m_n$ such that $A_{k1} \cap B_{nl_k}=B_{k1}\cap A_{nm_k}=\emptyset$ for $k=1,\dots,n$, then let $i_n=max\{m_1,\dots,m_n\}$ and $j_n=max\{l_1,\dots,l_n\}$.

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  • $\begingroup$ Hello,Ramiro de la Vega,I think the sets you contructed did not satisfy $(4)$.Because for any sequence of positive integers $\{i_k\}_{k=1}^\infty$,$\bigcup_{k=1}^\infty B_{ki_k}$ is dense in $C$. $\endgroup$
    – user40096
    Feb 28, 2014 at 7:23
  • $\begingroup$ @user40096: You are right I posted that too fast. I have deleted it now. $\endgroup$ Feb 28, 2014 at 13:41
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I found the contruction satisfying the properties above do not exist.Suppose for any positive integer $k$,the positive integers $i_k$,$j_k$ satisfy there exist sequences of positive integers $\{i_{k,s}\}$ and $\{j_{k,s}\}$ such that $A_{k,i_k}\bigcap (\bigcup_{s=1}^\infty B_{s,i_{k,s}})=\emptyset$ and $B_{kj_k}\bigcap (\bigcup _{s=1}^\infty A_{sj_{k,s}})=\emptyset$. Then consider $\{C_{k,i}=A_{k,i_k+i-1}\}_{k,i=1}^\infty$ and $\{D_{k,i}=B_{k,j_k+i-1}\}_{k,i=1}^\infty$,they satisfy:

$(1)$For any positive integer $k\geq1$,$C_{k,i}\searrow\emptyset$ and $D_{k,i}\searrow\emptyset$ as $j\to\infty$;

$(2)$For any positive integers $k,j$,there exist positive integers $i_1,i_2$ such that $C_{k,1}\bigcap D_{j,i_1}=\emptyset$ and $D_{k,1}\bigcap C_{j,i_2}=\emptyset$.

Then use Ramiro de la Vega's answer for my earlier version of the question and we can get it is impossible to construct the families of sets.

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