# Set Theory

## Set Operations

The intersection of sets A and B, written $$A \cap B$$ is the set of elements common to both $$A$$ and $$B$$, represented as

$A \cap B = \{x|x\in A \,and\,x\in B\}$

Example: Find the intersection of the following sets, $$A = \{1, 2, 3, 5, 8 \}, \,B=\{1, 2, 3, 4 \}$$

Solution: $$A \cap B =\{1, 2, 3 \}$$ Elements that are in both sets A and B are in the intersection.

The union of sets and B, written $$A \cup B$$, is the set of elements belonging to either of the sets, represented as:

$A \cup B = \{x|x\in A \,or\,x\in B\}$

Example: Find the union of the following sets, $$A = \{1, 2, 3, 5, 8 \}, \,B=\{1, 2, 3, 4 \}$$

Solution: $$A \cup B =\{1, 2, 3, 4, 5, 8 \}$$ All elements that in sets A and B are in the union.

The difference of sets A and B, represented $$A - B$$, is the set of elements belonging to set A and not to set B,

$A - B = \{x|x\in A \,and\,x\notin B\}$

Example: Find the difference of the sets, $$A = \{1, 2, 3, 5, 8 \}, \,B=\{1, 2, 3, 4 \}$$, $$A - B$$, and $$B - A$$

Solution: $$A - B =\{5, 8 \}$$ All elements that are in set A but not in set B.

$$B - A =\{4\}$$ All elements that are in set B but not in set A.

## Ordered Pairs

In the ordered pair $$\left(a,b\right)$$, a is called the first component and b is called the second component. Switching components does not give you the same statement. $$\left(a,b\right)\ \neq \left(b,a\right)$$.

The Cartesian product of sets A and B, written $$A \times B$$, is: $A \times B = \{(a, b)|a\in A \,and\, b\in B\}$

Example: Let $$A = \{1, 2 \}, \, and \,B=\{a, b, c \}$$

Then $$A \times B = \{(1,a), (1,b), (1,c), (2,a), (2,b), (2,c) \}$$

The cardinal number of a Cartesian product is the product of the cardinal numbers of each set.

If $$n\left(A\right)=a$$ and $$n\left(B\right)=b$$, then

\begin{align} n\left(A \times B\right) &= n\left(B \times A\right) \\ &= n\left(A \right) \cdot n\left(B \right) = n\left(B \right) \cdot n\left(A \right) \\ &= ab = ba \end{align}

Example: Let $$A = \{1, 2 \}, \, and \,B=\{a, b, c \}$$, then $$n(A) = 2$$ and $$n(B) = 3$$. $$n(A \times B) = 6$$