Construction of Sets

Intersection

Let $A$ and $B$ be two given sets. The intersection is the set:$A\cap B=\{x:x\in A\wedge x\in B\}$

Union

The union of sets $A$ and $B$ is the set:$A\cup B=\{x:x\in A\vee x\in B\}$

Disjoint Union

The disjoint union of $A$ and $B$ is the set:$A\bigsqcup B=\{(x,a):x\in A\}\cup \{(y,b):y\in B\}$

Difference

The difference of $A$ from $B$ is the set $A\setminus B=\{x:x\in A\wedge x\notin B\}$In the circumstance that sets are considered within a fixed set $X$, for any subset $A$ of $X$ we write $X\setminus A$ as $A^c$ and call it the complement of $A$ in $X$.

Proposition

Let $A$, $B$, $C$ and $B_i(i\in I)$ be sets. The following properties hold:

• Commutativity: $A\cap B=B\cap A$, $A\cup B=B\cup A$.
• Associativity: $(A\cap B)\cap C=A\cap (B\cap C)$, $(A\cap B)\cap C=A\cap (B\cap C)$.
• Distributivity: $A\cap \left({\bigcup }_{i\in I}{B}_i\right)={\bigcup }_{i\in I}(A\cap B_i)$, $A\cup \left({\bigcap }_{i\in I}{B}_i\right)={\bigcap }_{i\in I}(A\cup B_i)$.
• de Morgan’s laws: ${\left({\bigcap }_{i\in I}{B}_i\right)}^c={\bigcup }_{i\in I}{B}_i^c$, ${\left({\bigcup }_{i\in I}{B}_i\right)}^c={\bigcap }_{i\in I}{B}_i^c$.

Proof These are direct consequences of the arithmetic of logical connectives. $\square$

Cartesian Product

Cartesian Product

Let $X_1,\dots ,X_n$ be sets. The set$\{(x_1,x_2,\dots ,x_n):x_1\in X_1,\dots ,x_n\in X_n\}$is called the Cartesian product of $X_1,\dots ,X_n$ and is written as$X_1\times \cdots \times X_n\text{or}{\prod }_{i=1}^n{X}_i$where each $(x_1,\dots ,x_n)$ is an ordered $n$-tuple and the $x_i$ is called the $i$-th coordinate.