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2 total messages Started by Claudio Gutierrez Wed, 11 Nov 1992 16:40
combinatorial set theory
#125
Author: Claudio Gutierrez
Date: Wed, 11 Nov 1992 16:40
16 lines
1186 bytes
Keywords: 

I need the following result (that can be of comon sense)

Exists a family of sets *A* = { Ai}_{i \in \omega}
such that  there is a family
*F* of infinite subsets of *A* with |*F*|=2^{\omega}
with the properties:
  i) Each B \in *F* has F.I.P. (The intersection of a 
	finite numbers of elements of each B \in *F*  is non empty.
 ii) For all B \in *F* , the intersection of all the
	elements of B is empty.
iii) For all B={B1,B2,...} \in *F*, B'={B'12,B'2,...} \in *F*
	there exists n \in |N  such that
	Intersection{B1,...,Bn} and  Intersection{B'1,..,B'n}
	are dosjoint.
Re: combinatorial set theory
#129
Author: W.Jose Castrellon G.
Date: Thu, 12 Nov 1992 22:07
33 lines
2459 bytes
In article <1992Nov11.164014.14768@tolten.puc.cl> cgutierr@mat.puc.cl (Claudio Gutierrez) writes:

>
>I need the following result (that can be of comon sense)
>
>Exists a family of sets *A* = { Ai}_{i \in \omega}
>such that  there is a family
>*F* of infinite subsets of *A* with |*F*|=2^{\omega}
>with the properties:
>  i) Each B \in *F* has F.I.P. (The intersection of a 
>	finite numbers of elements of each B \in *F*  is non empty.
> ii) For all B \in *F* , the intersection of all the
>	elements of B is empty.
>iii) For all B={B1,B2,...} \in *F*, B'={B'12,B'2,...} \in *F*
>	there exists n \in |N  such that
>	Intersection{B1,...,Bn} and  Intersection{B'1,..,B'n}
>	are dosjoint.
>

Yes, there is such a family: Let  A_(i,n) be the set of numbers  = i (mod 2^n),
and call  A  the (countable) set of all such  A_(i,n) 's .  Now to get the
uncountable family  B , take all the subsets of  A  that are totally ordered
by inclusion (one can take just the ones that are maximal chains, to make it
easier to visualize).  To see that there are  2^Aleph_0  of them, and that
i, ii and iii hold, note that  N  is the disjoint union of two elements of A
(even and odd numbers), and each of these is again the disjoint union of two
elements of  A ... so one can go along  2^Aleph_0 many paths and choose one
of the two possible sets to get the elements of  B : ordered by inclusion, 
finite intersection non-empty, total intersection empty  and two different 
elements of  B  will start to have pairwise disjoint elements from the moment 
their paths differ, forcing  iii to hold.

I wonder what application you have in mind for this, is it a forcing argument?
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