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Sets.
1. Prevelance of sets.
a is a member of a set A a ∈A either a is in A or not in A a ∉A. A 0/1 situation.
e.g. A = {a, e, i, o, u} A = { x:xeveninteger, 4 |x}
A set B is a subset of A if every element of B is an element of A. B ⊆ A
These are called normal sets (Boolean sets) as opposed to Fuzzy sets in which a membership value need not be restricted to 0/1.
Let A = {set of young men} Is bruce? Depends on how we see A. Is 30 years a young age? Is it not so
∈ A
young?
Perhaps we need a type of sets in which elements belong to subsets in some degree. The age “30 years” may be 0.25 in “rather young” set, but 0.67 for “not old” set?
2. Complement of a set. Every set A is a subset of a
universal set U. The complement set of A is Ac whose members are in U but not in A.
A: {set of all even integers divisible by 4} U: {set of all even integers}
A c : {x, x=4n + 2 with n ≥ 0}
Relative complement of B with respect to A, difference A\B or A-B or A ~ B, is the set of those who are members of A but not in B.
A \B={ x:x∈A,x∉B }
examples:
resources = {hardware, software} hardware = {CPU, memory, bus, peripherals} software = resources/hardware memory = {RAM, ROM, hard_disk} RAM = {SRAM, DRAM} DRAM = {FPM DRAM, EDO DRM, BEDO DRAM, SDRAM, RDRAM}
Some typical sets:
U universal set (includes all relevant things)
Φ null set (empty set)
R set of real numbers Q set of rational numbers Z set of integers N set of positive integers
4. Fundamental set operations:
We’ll use the notations for an “and” and ∨ for an “or”. would be used to indicate “there exists” and
∀ to indicate to represent “for all”.
Union: U as in. The members of C will be members of both A and B.
A UB= C
A UB={x∈A∨ x∈B }
Intersection: I as in. The members of C will be only those who are both members of A and B.
A IB= C
A IB={x∈A∧ x∈ B}
Symmetric difference: as in. The members of C will be those of A who are not in B and those of B who are not in A.
⊕ A ⊕B=C
A ⊕B=(AUB)(AIB )
5. Algebra of sets and dualities.
A typical set equation (an equation over sets) is an expression with U s and I s interposing between sets and their complements. Example:
Distributive law:
A U (BIC)=(AUB)I(AUC )
DeMorgan’s Law:
( AU B)c^ = AcIB^ c
If we discover any such set identity then its dual could be found by interchanging U s with I s, and vice versa.
So, if A U (BIC)=(AUB)I(AUC) is valid then its
Dual
A I (BUC)=(AIB)U(AIC) is also valid.
Dual symmetry is an essential property of set algebras.
6. Finite sets. Counting principle.
For A finite set A, its cardinality n(A) is the number of elements it contains.
n( A)=|A|=card(A )
some sets are infinite. Like set of positive integers {2,4,6, …}
ΓP^ = [ {2,4,6}, {4,6,8}, {2,4,8}, {2,6,8}]
Power sets. From a given set S, we form all possible subsets of S, including the null set. The cardinality of a power set is
n( power_setof S)= 2 n(s)
Note that power set would also include itself.
A partition of S is a collection of its elements in subsets such that no two subsets assume any overlap (disjoint subsets).
e.g.
S ={ 1 , 2 , 3 ..., 5 , 6 }
Its one of possible partitions is the arrangement
[ {1,2], {3,6,5}, {4}]
If is a subset within a partition spawned from the
members of a set S, then
A i
Ui Ai = S
Ai IAj =Φ,∀Ai,Aj ⊂ S
The number of possible partitions of a set is called bell- number.
There are 5 ways the set S = {1,2,3} could be partitioned. They are
[{1}, {2}, {3}]
[{1}, {2,3}]
[{1,2}, {3}]
[{1,3}, {2}]
[{1,2,3}]
B n is the Bell-number of partitioning an n-integer set.
The first ten Bell-numbers are
1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975 …
A typical partition problem. How to cut a rectangular object into different sized circles with a minimum wastage?
8. Proofs on set-theoretic properties using method of induction.
Proposition: Prove that on some property p, a function f (n ) becomes g(n).
- Prove the prposition holds for the smallest number we care about given the property p.
- Assume it holds for an arbitrary number N
Observation. If we assume it is true up to p, then it must be true for (p+1).
Conclusion: the proposition P(n) is true for all positive integers n.
Another one.
P(n): Prove that ∀ n >= 4 , 2 n^ >n^2
a. Is it true for n = 4? b. Assume it is true for any integer greater than or equal to 4, less than or equal to p. Therefore, we have
2 p^ > p^2 c. What about 2 p+^1?
2 p^ +^1 = 2 × 2 p and 2 p^ > p^2
Therefore, 2 p^ +^1 > 2 p^2
But ( p + 1 )^2 = p^2 + 2 p+ 1 < 2 p^2
(prove that p (^2) > 2 p+ (^1) ).
Observation: 2 p^ +^1 >(p+ 1 )^2
Conclusion: The theorem is correct.
9. Proof by contrapositive argument.
Attacking a proof backward. Assume the result is false, and then show that it violates one of the assumptions. This is based on logical law of contrapositive:
If A → B is equivalent to if ( ~ B) → ( ~A). Also called as proof by contradiction.
To prove the If and only if type proposition, as in A iff B:
a. we prove that A→B by showing that ( ~ B) → ( ~A). b. we prove next that B→A by showing ( ~ A) → ( ~B).
