N-Union and Intersection Set Operations

The union and intersection set operations were introduced in a previous post using two sets, \(a\) and \(b\). These set operations can be generalized to accept any number of sets.

Arbitrary Set Unions Operation

Consider a set of infinitely many sets:

$$ A = \large{\{b_0, b_1, b_2, \cdots \} \large} $$

It would be very tedious and unnecessary to repeat the union statement repeatedly for any non-trivial amount of sets, for example, the first few unions would be written as:

$$ \large{b_0 \cup b_1 \cup b_2 \cup b_3 \cup b_4 \cup b_5\large} $$

Thus a more general operation for performing unions is needed. This operation is denoted by the ⋃ symbol. For example, the set \(A\) above and the desired unions of the member sets can be generalized to the following using the new notation:

$$ \large{\bigcup A = \bigcup_i b_i} $$

We can then state the following definition: For a set \(A\), the union ⋃\(A\) of \(A\) is defined by:

$$ \large{\bigcup A = \{x \space | \space (\exists b \in A) \space x \in b \} \large} $$

For example, consider the three sets:

$$ \large{a = \{2, 4, 6 \} \qquad b = \{3, 5, 7\} \qquad c = \{2, 3, 8 \} \large}$$

The union of the three sets is written as:

$$ \large{\bigcup \Big\{\{2,4,6\}, \{3,5,7\}, \{2,3,8\} \Big\} = \{2,3,4,5,6,7,8\}} $$

Recalling our union axiom from a previous post, the union axiom states for two sets \(A\) and \(B\), there is a set whose members consist entirely of those belonging to sets \(A\) or \(B\), or both. More formally, the union axiom is stated as:

$$ \large{\forall a \space \forall b \space \exists B \space \forall x (x \in B \Leftrightarrow x \in a \space \vee \space x \in b)} $$

As we are now dealing with an arbitrary amount of sets, we need an updated version of the union axiom to account for the change.

Restating the union axiom:

For any set \(A\), there exists a set \(B\) whose members are the same elements of the elements of \(A\). Stated more formally:

$$ \large{\forall x \bigg[ x \in B \space \Leftrightarrow \space (\exists b \in A) \space x \in b \bigg] }$$

The definition of ⋃\(A\) can be stated as:

$$ \large{x \in \bigcup A \Leftrightarrow (\exists b \in A) \space x \in b} $$

For example, we can demonstrate the updated axiom with the union of four sets \({a, b, c, d}\):

$$ \large{\bigcup \{a, b, c, d \} = \big\{(\exists B \in A) \space x \in \{a, b, c, d\}\big\} \large}$$
$$ \large{ \bigcup \{a, b, c, d \} = a \cup b \cup c \cup d \large}$$

We can implement the set operation for an arbitrary amount of sets by expanding upon the function we wrote previously.

set.unions <- function(a, b, ...) {
  u <- a
  for (i in 1:length(b)) {
    if (!(b[i] %in% u)) {
      u <- append(u, b[i])

  s <- list(...)

  for (i in s) {
    for (j in i) {
      if (!(j %in% u)) {
        u <- append(u, j)

Perform the set union operation of four sets:

$$ \large{a = \{1,2,3\} \qquad b = \{3,4,5\} \qquad c = \{1,4,6\} \qquad d =\{2,5,7\} \large}$$
a <- c(1,2,3)
b <- c(3,4,5)
c <- c(1,4,6)
d <- c(2,5,7)

set.unions(a, b, c, d)

## [1] 1 2 3 4 5 6 7

Intersections of an Arbitrary Number of Sets

The intersection set operation can also be generalized to any number of sets. Consider the previous set containing an infinite number of sets.

$$ \large{A = \{b_0, b_1, b_2, \cdots \}} $$

As before, writing out all the intersections would be tedious and not elegant. The intersection can instead be written as:

$$ \large{\bigcap A = \bigcap_i b_i} $$

As before in our previous example of set intersections, there is no need for a separate axiom for intersections, unlike unions. Instead, we can state the following theorem, for a nonempty set \(A\), a set \(B\) exists that such for any element \(x\):

$$ \large{x \in B \Leftrightarrow x \in \forall A} $$

Consider the following four sets:

$$ \large{a = \{1,2,3\} \qquad b = \{1,3,5\} \qquad c = \{1,4,5,3\} \qquad d = \{2,6,1,3\}} $$

The intersection of the sets is written as:

$$ \large{\bigcap \big\{\{1,2,3,5\}, \{1,3,5\}, \{1,4,5,3\}, \{2,5,1,3\}\big\}} $$
$$ = \large{\{1,2,3,5\} \cap \{1,3,5\} \cap \{1,4,5,3\} \cap \{2,5,1,3\} = \{1,3,5\}} $$

We can write another function to implement the set intersection operation given any number of sets.

set.intersections <- function(a, b, ...) {

  intersect <- vector()
  for (i in a) {
    if (i %in% b) {
      intersect <- append(intersect, i)

  s <- list(...)

  for (i in s) {
    for (j in i) {
      if (j %in% intersect) {
        intersect <- append(intersect, j)
  intersect <- unique(intersect)


Perform set intersections of the four sets specified earlier.

a <- c(1,2,3,5)
b <- c(1,3,5)
c <- c(1,4,5,3)
d <- c(2,5,1,3)

set.intersections(a, b, c, d)

## [1] 1 3 5


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