Let U be a "universal" set for a given discussion
Then define A^c = A-complement = U - A = {x | x in U ^ x not in A}.
So x in A^c iff x not in A (where U is the domain for x) 

DeMorgan's Laws:  (A int B)^c = A^c u B^c, etc

       example of proof by element nethod:
       x in (A int B)^c iff x not in (A int B)
                        iff ~(x in A int B) iff ~(x in A & x in B)
                        iff x not in A  v  x not in B
                        iff x in A^c  v  x in B^c

Equality of sets: two sets A and B are said to be equal if they have
the very same elements:  
          A = B iff ForAll x (x in A iff x in B). 
This can also be expressed as
          A = B iff (A subst of B, and B subst of A)

Sets give is a new look as "if...then...":  ForAll x (Px --> Qx) is
regarded as true if 
          {x | Px} subst {x | Qx}.  
Any x *not* in P is irrelevant to this.  This means that in taking
ForAll x (Px --> Qx) to be true, we are saying that no matter what x
is, Px --> Qx is true, even when x is not in {x | Px}, ie., even when
Px is false.  That is, when Pb fails, we need to take Pb --> Qb to be
true, since we don't want such a b (one not in {x | Px}) to be able to
have any negative affect on the truth of ForAll x (Px --> Qx}.
EXAMPLE: Every prime integer > 2 is odd.  We can write this as 
ForAll x (PrimeInt(x) --> Odd(x)).  
And we want it to come out true, literally, for ALL x. So what about
when x=4, say? We want PrimeInt(4 --> Odd(4) to be true. And the same
for PrimeInt(9) --> Odd(9).  So this FORCES us to take the truth table
for if-then (ie, for the conditional connective -->) to be the way we
did in chapter 2.

More on cardinality:
         | A u B | = |A| + |B| if A and B are disjoint (A int B = empty)
         In general: |A u B| = |A| + |B| - |A int B|, as a picture quickly shows.

         Similarly, |A u B u C| = |A| + |B| + |C| - |AnB| - |BnC| - |AnC| + |AnBnC|
  
         |P(A)| = 2^|A|
         Proof (easy way; will do it again later with induction)
         Let A = {a1,a2,...,an}
         To form a  subset we need to decide, for each ai, whether to
         include it or not. This is a sequence of decisions:  

                       start
      a1        yes        no                2 paths so far
      a2       y   n      y   n              2^2 paths to this level
     ...
      an     yn yn yn yn ... yn yn yn    2^n paths; 

          and each path gives a unique seq of choices, and every subst
          can be formed that way. So there are 2^n subsets of a set
          with n elements: |P(A)| = 2^|A|. 


Partitions: Let S be nonempty. A partition of S is a collection (set)
of subsets A1, A2, .., An of S such that

     (i) S = A1 u A2 u ... u An
    (ii) Si int Sj = empty if i=/=j.

Think of S in terms of maps. Let S be a geographic population
(pictured as spread out across a large territory). Then if we divide S
up into countries, so the people in each of those countries is a
subset of S, each person in S is in exactly one country (one
subset). Their union is all of S, and no two countries overlap.

Examples:
      Z = integers = 2Z u (1+2Z) = Evens u Odds
      Z = 3Z u (1+3Z) u (2+3Z)
      Z = 4Z u (1+4Z) u (2+4Z) u (3+4Z)
      etc
      N = non-neg integers = Primes u Composites U {1}


Infinite sets; Russell's paradox; Turing and the Halting Prob