## Ordered and Unordered Pairs

A pair set is a set with two members, for example, $${2, 3}$$, which can also be thought of as an unordered pair, in that $${2, 3}={3, 2}$$. However, we seek a more a strict and rich object that tells us more about two sets and how their elements are ordered. Call this object ⟨2, 3⟩, which specifies that 2 is the first component and 3 is the second component. We also make the requirement that ⟨2, 3⟩≠⟨3, 2⟩. We can also represent this object, generalized as ⟨$$x$$, $$y$$⟩, as:

$$\large{\langle x, y\rangle = \langle u, v \rangle}$$

Therefore $$x = u$$ and $$y = v$$. This property is useful in the formal definition of an ordered pair, which is stated here but not explored in-depth. The currently accepted definition of an ordered pair was given by Kuratowski in 1921 (Enderton, 1977, pp. 36), though there exist several other definitions.

$$\large{\langle x, y \rangle = \big\{\{x\}, \{x, y\} \big\}}$$

The pair ⟨$$x, y$$⟩ can be represented as a point on a Cartesian coordinate plane.

## Cartesian Product

The Cartesian product $$A × B$$ of two sets $$A$$ and $$B$$ is the collection of all ordered pairs ⟨$$x, y$$⟩ with $$x ∈ A$$ and $$y ∈ B$$. Therefore, the Cartesian product of two sets is a set itself consisting of ordered pair members. A set of ordered pairs is defined as a 'relation.'

For example, consider the sets $$A = {1, 2, 3}$$ and $$B = {2, 4, 6}$$. The Cartesian product $$A × B$$ is then:

$$A × B = {{1, 2},{1, 4},{1, 6},{2, 2},{2, 4},{2, 6},{3, 2},{3, 4},{3, 6}}$$

Whereas the Cartesian product $$B × A$$ is:

$$B × A = {{2, 1},{2, 2},{2, 3},{4, 1},{4, 2},{4, 3},{6, 1},{6, 2},{6, 3}}$$

The following function implements computing the Cartesian product of two sets $$A$$ and $$B$$.

cartesian <- function(a, b) {
axb <- list()
k <- 1
for (i in a) {
for (j in b) {
axb[[k]] <- c(i,j)
k <- k + 1
}
}
return(axb)
}


Let's use the function to calculate the Cartesian product $$A × B$$ and $$B × A$$ to see if it aligns with our results above.

a <- c(1,2,3)
b <- c(2,4,6)

as.data.frame(cartesian(a, b))

##   c.1..2. c.1..4. c.1..6. c.2..2. c.2..4. c.2..6. c.3..2. c.3..4. c.3..6.
## 1       1       1       1       2       2       2       3       3       3
## 2       2       4       6       2       4       6       2       4       6

as.data.frame(cartesian(b, a))

##   c.2..1. c.2..2. c.2..3. c.4..1. c.4..2. c.4..3. c.6..1. c.6..2. c.6..3.
## 1       2       2       2       4       4       4       6       6       6
## 2       1       2       3       1       2       3       1       2       3


Both outputs agree to our previous results. One could also simply use the expand.grid() function like so to get the same result for the Cartesian product.

t(expand.grid(a, b))

##      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
## Var1    1    2    3    1    2    3    1    2    3
## Var2    2    2    2    4    4    4    6    6    6


## Some Cartesian Product Theorems

We can state some theorems related to the Cartesian product of two sets. The first theorem states:

If $$A$$ is a set, then $$A × ⌀ = ⌀$$ and $$⌀ × A = ⌀$$.

We can demonstrate this theorem with our cartesian() function.

cartesian(a, c()) # c() represents the empty set.

## list()

cartesian(c(), a)

## list()


The outputs are an empty list which is equivalent to the empty set ⌀ for our purposes of demonstration.

The next theorem involves three sets $$A$$, $$B$$, $$C$$.

• $$A × (B ∩ C)=(A × B)∩(A × C)$$
• $$A × (B ∪ C)=(A × B)∪(A × C)$$
• $$(A ∩ B) × C = (A × C) ∩ (B × C)$$
• $$(A ∪ B) × C = (A × C) ∪ (B × C)$$

We can demonstrate each in turn with a combination of our cartesian() from above, and the set.union() and set.intersection() functions from a previous post on set unions and intersections. The base R functions union() and intersect() can be used instead of the functions we defined previously.

a <- c(1,2,3)
b <- c(2,4,6)
c <- c(1,4,7)


The first identity \$A × (B ∩ C) = (A × B) ∩ (A × C)%.

ident1.rhs <- cartesian(a, set.intersection(b, c)) # Right-hand Side
ident1.lhs <- set.intersection(cartesian(a, b), cartesian(a, c)) # Left-hand Side

isequalset(ident1.rhs, ident1.lhs)

## [1] TRUE

as.data.frame(ident1.rhs)

##   c.1..4. c.2..4. c.3..4.
## 1       1       2       3
## 2       4       4       4

as.data.frame(ident1.lhs)

##   c.1..4. c.2..4. c.3..4.
## 1       1       2       3
## 2       4       4       4


The second identity $$A × (B ∪ C)=(A × B) ∪ (A × C)$$.

ident2.rhs <- cartesian(a, set.union(b, c))
ident2.lhs <- set.union(cartesian(a, b), cartesian(a, c))

isequalset(ident2.rhs, ident2.lhs)

## [1] TRUE

as.data.frame(ident2.rhs)

##   c.1..2. c.1..4. c.1..6. c.1..1. c.1..7. c.2..2. c.2..4. c.2..6. c.2..1.
## 1       1       1       1       1       1       2       2       2       2
## 2       2       4       6       1       7       2       4       6       1
##   c.2..7. c.3..2. c.3..4. c.3..6. c.3..1. c.3..7.
## 1       2       3       3       3       3       3
## 2       7       2       4       6       1       7

as.data.frame(ident2.lhs)

##   c.1..2. c.1..4. c.1..6. c.2..2. c.2..4. c.2..6. c.3..2. c.3..4. c.3..6.
## 1       1       1       1       2       2       2       3       3       3
## 2       2       4       6       2       4       6       2       4       6
##   c.1..1. c.1..7. c.2..1. c.2..7. c.3..1. c.3..7.
## 1       1       1       2       2       3       3
## 2       1       7       1       7       1       7


The third identity $$(A ∩ B) × C = (A × C) ∩ (B × C)$$.

ident3.rhs <- cartesian(set.intersection(a, b), c)
ident3.lhs <- set.intersection(cartesian(a, c), cartesian(b, c))

isequalset(ident3.rhs, ident3.lhs)

## [1] TRUE

as.data.frame(ident3.rhs)

##   c.2..1. c.2..4. c.2..7.
## 1       2       2       2
## 2       1       4       7

as.data.frame(ident3.lhs)

##   c.2..1. c.2..4. c.2..7.
## 1       2       2       2
## 2       1       4       7


We finish the post with the fourth identity $$(A ∪ B) × C = (A × C) ∪ (B × C)$$.

ident4.rhs <- cartesian(set.union(a,b), c)
ident4.lhs <- set.union(cartesian(a,c), cartesian(b,c))

isequalset(ident4.rhs, ident4.lhs)

## [1] TRUE

as.data.frame(ident4.rhs)

##   c.1..1. c.1..4. c.1..7. c.2..1. c.2..4. c.2..7. c.3..1. c.3..4. c.3..7.
## 1       1       1       1       2       2       2       3       3       3
## 2       1       4       7       1       4       7       1       4       7
##   c.4..1. c.4..4. c.4..7. c.6..1. c.6..4. c.6..7.
## 1       4       4       4       6       6       6
## 2       1       4       7       1       4       7

as.data.frame(ident4.lhs)

##   c.1..1. c.1..4. c.1..7. c.2..1. c.2..4. c.2..7. c.3..1. c.3..4. c.3..7.
## 1       1       1       1       2       2       2       3       3       3
## 2       1       4       7       1       4       7       1       4       7
##   c.4..1. c.4..4. c.4..7. c.6..1. c.6..4. c.6..7.
## 1       4       4       4       6       6       6
## 2       1       4       7       1       4       7


## References

Enderton, H. (1977). Elements of set theory (1st ed.). New York: Academic Press.