The document discusses how many different sets can be constructed from a collection of sets A1, A2, ..., An using the operations of intersection, union, and complementation. It proves by induction that the number of possible sets is at most 2^2n. It provides an example where X is a unit cube in Rn divided into 2n subcubes by hyperplanes, showing that exactly 2^2n different sets can be constructed from the sets A1, ..., An defining the subcubes.

1. 1, A2, . . . , An ⊆ X. Using the operations ∩, ∪ and ·c
(complementation relative to X), one can construct at most 22n
different
sets from A1, A2, . . . , An.
(general distributive laws).
7. Let X be a set and A
7. Let H(X; A1, . . . , An) be the collection of those sets that can be obtained
from A1, . . . , An using the operations ∩, ∪, and ·c
(complementation with
respect to X). We have to show that
|H(X; A1, . . . , An)| ≤ 22n
. (1.1)
This is clearly true for n = 1, and we can proceed by induction. Thus, suppose
that (1.1) is true for an n. Note that H(X; A1, . . . , An) is nothing else than the
smallest set containing X; A1, . . . , An that is closed under union, intersection,
and complementation. Therefore, it immediately follows that
H(X; A1, . . . , An, An+1) = {S ∪ T},
where on the right we take all possible unions with S ∈ H(An+1; A1 ∩
An+1, . . . , An ∩ An+1) and T ∈ H(Ac
n+1; A1 ∩ Ac
n+1, . . . , An ∩ Ac
n+1). By the
induction hypothesis these latter sets have at most 22n
elements, so there are
that many choices for S and T. Thus, for S∪T we have at most 22n
·22n
= 22n+1
choices, and this proves (1.1) with n replaced by (n + 1).
8. Let
X = {(x1, . . . , xn) : 0 ≤ xi < 1, 1 ≤ i ≤ n}
be the unit cube of Rn
, and set
Ak = {(x1, . . . , xn) ∈ X : 1/2 ≤ xk < 1}.
Using the operations ∩, ∪, and ·c
(complementation with respect to X),
one can construct 22n
different sets from A1, A2, . . . , An.
9. Using the operations , ∩ and ∪ one can construct at most 2
8. The hyperplanes xi = 1/2 divide the unit cube into 2n
pairwise disjoint
subcubes C1, . . . , C2n of side length 1/2. Clearly, each of C1, . . . , C2n can be
obtained from the sets Ak using the operations ∩ and ·c
, and so taking the
union of any possible subcollection of C1, . . . , C2n (there are 22n
different such
subcollections), one can construct 22n
different sets from A1, A2, . . . , An.