In set theory, one of the subjects we can address is the Cartesian Product of two Sets. In it we talk about ordered pairs, the pair axiom and the Cartesian product.

Euclid's work.

Ordered pair

Before we address the subject of the Cartesian product of two sets, we need to talk about ordered pairs.

Euclid's work.

Let $A$ and $B$ be two arbitrary sets. For $a \in A$ and $b \in B,$ and using the specification axiom we can construct the set

$$\\{a, b\\} = \\{x;\quad x = a \text{ ou } x = b\\}.$$
Note that as sets $\\{a, b\\} = \\{b, a\\}.$

For a notion of ordered pair, it is necessary in a set with two given elements, to know which is the first and which is the second element.

For this, we need to be sure that this pair is also an element. This is by the following axiom.

Axiom of the pair: For any two sets there is a set to which they both belong.

Euclid's work.

This axiom guarantees the existence of the set defined below:

Definition (Ordered pair): the ordered pair of $a$ and $b,$ denoted by $(a, b)$, with first coordinate $a$ and second coordinate $b$ is the set

$$(a, b) = \\{a, \\{a, b\\}\\}.$$

Cartesian product of two sets

Definition (Cartesian product of two sets): Given two sets $A$ and $B,$ the Cartesian product of $A$ and $B,$ denoted by $A\times B$, is the set

$$A \times B = \\{x;\quad x = (a, b) \text{ para algum } a \in A \text{ e algum } b \in B \\}.$$
Euclid's work.
Note that in general $(a, b) \neq (b, a)$ and $A \times B \neq B \times A.$

Given these definitions, let’s see how this new operation between sets behaves.

Result for Cartesian product

Euclid's work.

Theorem Let $A, B$ and $C$ be any sets. So we have:

  • (a) $A \times \emptyset = \emptyset × A = \emptyset.$
  • (b) $A \times (B \cap C) = (A \times B) \cap (A \times C).$
  • (c) $A \times (B \cup C) = (A \times B) \cup (A \times C).$
  • (d) $A \times (B − C) = (A \times B) − (A \times C).$

Demo:

  • (a) By the definition of the Cartesian product, we have
$$ A \times \emptyset = \\{(a, b); a \in A \text{ e } b \in \emptyset \\}.$$

As there is no $b \in \emptyset,$ we have that there is no ordered pair whose second coordinate is $b,$ like $A \times \emptyset = \emptyset.$

The other equality is analogous.

  • (b) Note that here we can assume that the 3 sets are different from the empty one, otherwise the proof follows easily from item (a). For $a \in A$ and $x \in (B \cap C),$ we have from the definition of Cartesian product that:
$$(a, x) \in A \times (B \cap C) \quad \Longleftrightarrow \quad (a \in A) \wedge (x \in B \cap C)$$

From the definition of intersection we have:

$$(a \in A) \wedge (x \in B \wedge x \in C)$$

From the associativity property:

$$(a \in A) \wedge (x \in B) \wedge (x \in C),$$

From the cancellation and commutativity property of $\wedge$:

$$(a ∈ A) ∧ (x ∈ B) ∧ (a ∈ A) ∧ (x ∈ C)$$

From the associativity property:

$$[(a \in A) \wedge (x \in B)] \wedge [(a \in A) \wedge (x \in C)]$$

From the definition of product. Cartesian:

$$[(a, x) \in A \times B] \wedge [(a, x) \in A \times C]$$

From the definition of intersection:

$$(a, x) \in (A × B) ∩ (A × C)$$

which shows the equality of item (b).

The demonstrations of the equality of items (c) and (d) are left as an exercise. 😂😂😂

$\square$

Euclid's work.

This is the summary of Cartesian Product of Sets. Wait for the next posts.