I derived this definition by searching for a representation of a family of sets. I am quite sure that someone should have thought to this before, because it seems to be quite straightforward given a family of sets. However, I did not find anything suitable on google or on wikipedia.

Let a family of sets, say $A_1, \ldots, A_n$, be given. To avoid misunderstanding I will call them modules. This family induces a unique partition on the union set $A = \bigcup_{i=1}^{n} A_i$ in the following way: I call building block a maximal subset $B$ of $A$ such that do not exist 2 different modules $A_i$ and $A_j$ with:

$B \cap A_i \not= \emptyset$,

$B \cap A_j \not= \emptyset$ and

$B\not\subseteq A_i \cap A_j$.

For example, if my family consists of 2 different overlapping modules $A_1$ and $A_2$, I can partition the set $A = A_1 \cup A_2$ as:

(elements in $A_1 \cap A_2$);

(elements in $A_1$ but not in $A_2$);

(elements in $A_2$ but not in $A_1$).

I know that in logic there is something similar, but I am searching for something in set theory. Moreover, I want to underline the dependency of this uniquely derived partition from the family of sets I am given.

See also http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics#Given_a_family_of_sets.2C_I_can_generate_a_partition_of_their_union_set.2C_by_looking_at_their_overlapping...

Thanks to all!

A newcomer


1 Answer 1


Your building blocks are known as the atoms in the Boolean algebra or field of sets generated by the $A_i$. Each building block will consist of points that have the same pattern of answers for membership in the various $A_i$. To see this, observe first that by maximality any building block will respect this equivalence and therefore be a union of such atoms. Conversely, if a set has points with two different patterns of answers, then it will contain points from two $A_i$ without being contained in their intersection. (Specifically, if $x,y\in B$ and $x\in A_i$ but $y\notin A_i$, then pick $j$ such that $y\in A_j$, and observe that $B$ meets both $A_i$ and $A_j$, buit is not contained in $A_i\cap A_j$.) So the building blocks are the atoms.

If your family is finite as you indicated, then the Boolean algebra generated by the $A_i$ consists precisely of the unions of blocks. This is the representation theorem showing that every finite Boolean algebra is isomorphic to a finite power set---the power set of the atoms.

In the infinite case, however, the Boolean algebra generated by the $A_i$ may be atomless---it may have no atoms at all, and this is a fascinating case. Nevertheless, your blocks still form a partition, and are precisely the atoms in the infinitary-generated field of sets, still determined by my argument above by the patttern-of-answers to membership in the $A_i$.

  • $\begingroup$ Thank you! I have to look closer at it, but it seems to be potentially inspiring! Anyway, I do not have a Boolean algebra, because the family is not closed under union and intersection - for example, with my notation, the union of a module is not in general a module, but possibly I have to add some other building blocks. Does something like this already exists? I would be really interested in properties that such a "structure" have! (Otherwise I have to figure out by myself if there are any...) $\endgroup$
    – klaraspina
    Commented Jun 18, 2010 at 10:06
  • $\begingroup$ My proposal was that you should close your modules under intersection, union and complement---this would be the generated Boolean algebra---but that after doing so, you will have the same building blocks, and these are the atoms of that Boolean algebra. $\endgroup$ Commented Jun 18, 2010 at 11:03
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    $\begingroup$ You could also think in terms of the Venn diagram (en.wikipedia.org/wiki/Venn_diagram). Your buiilding blocks are the smallest spaces appearing in the appropriate Venn diagram of your modules. $\endgroup$ Commented Jun 18, 2010 at 11:06
  • $\begingroup$ Closing the algebra is not a good thing for me. First of all because I want to keep in memory which are the given modules. Second, because if I close I have an exponential blowup of sets, and since I want to apply this representation in computer science, it would not be a good thing to do... $\endgroup$
    – klaraspina
    Commented Jun 18, 2010 at 11:07
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    $\begingroup$ It isn't necessary to actually calculate and represent the whole Boolean algebra. You can describe and index each building block by the in-out membership pattern of points in the modules. $\endgroup$ Commented Jun 18, 2010 at 11:16

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