## 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)}

## Complement

_

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