Union

S ∪ T = {x | (x ∈ S ) ∨ (x ∈ T )}

Intersection

S ∩ T = {x | (x ∈ S) ∧ (x ∈ T)}

Set Difference

Set of all elements in S but not T.

S - T = {x | (x ∈ S) and ∉ T} = {x | (x ∈ S) ∧ (x ∉ T)}

_
S = {x | x ∉ S}

CartesianProduct

S × T ={ (x,y) |x ∈ S,y ∈ T}

What common sets do the following sets refer to?
A {m ∈ $$\mathbb{Z}$$:2|m} ∪ {2k + 1:k ∈ Z}
B {m ∈ $$\mathbb{Z}$$:2|m} ∩ {2k + 1:k ∈ Z}

A This is the set of all integers

B This is an empty set

A set S is called a subset of a set T, written S ⊆ T, when the implication “if x ∈ S then x ∈ T” is true for every x in the universe of discourse.
To prove S ⊆ T , we prove the implication (x ∈ S) => (x ∈ T).

Prove:

{n ∈ $$\mathbb{N}$$ : 5 | n + 2} ⊆ {2k + 1 : k ∈ $$\mathbb{N}$$}

when n = 8

8 ∈ {n ∈ $$\mathbb{N}$$ : 5 | n + 2}

but 8 ∉ {2k + 1 : k ∈ $$\mathbb{N}$$}

Converse of DIC (Divisibility of Integer Combinations)

If for any integers x and y, a | (bx + cy)
Then a | b and a | c

Proof: Assume for any integer x,y a | (bx + cy)
then when x = 1, y = 0
a | (b(1) + c(0)), i.e. a | b
Use specific scenarios when it is true
then when x = 0, y = 1
a | (b(0) + c(1)), i.e. a | c

Converse in Sets

To prove S ⊆ T we prove the implication
x ∈ S and x ∈ T
Two sets S and T are equal i.e. S = T,
S and T have the same elements

[(x ∈ S) ∧ (x ∈ T) AND (x ∈ T) ∧ (x ∈ S)]

i.e. we prove
(1) S ⊆ T
(2) T ⊆ S