Here, we shall discuss several operations on sets that allow to construct new sets from the given ones. We will present the notion of union, intersection, and complement. The family of sets with these operations is called "an algebra of sets". We shall start with the definition of union.

Definition 1.3.1

By the union of two given sets A and B we shall understand the set consisting of all elements of A and all elements of B. in symbolic notation A ∪ B. We shall write formally

x ∈ A ∪ B iff x ∈ A or x ∈ B.

Assuming A is the blue circle and B the dark blue circle, then A ∪ B, is a part of the plane that consists of all blue and dark blue points, see Figure 1.3.1

Let A = {2k : k ∈ N} and B = {3n : n ∈ N}. Then A ∪ B is a set of all natural numbers divisible by two or by three,

The number 15 ∈ A ∪ B, since it is divisible by 3, and 8 ∈ A ∪ B, since it is divisible by 2. The number 6 belongs to A ∪ B, since it is divisible by 2 as well as by 3. On the other hand, 5 does not belong to A ∪B since it is not divisible by 2 nor by 3.

Question 1.3.1 What is the necessary and sufficient condition for x ∉ A ∪ B?

Union is a two-argument operation that has several properties similar to the arithmetic sum. Below we shall mention some of them.

Theorem 1.3.1

The following equalities are valid for all sets A, B, C:

∅ ∪ A = A

A ∪ A = A (law of idempotence)

A ∪ B = B ∪ A (law of symmetry)

(A ∪ B) ∪ C = A ∪ (B ∪ C) (law of commutativity)

We leave the proofs of the above to the reader. It is recommended to verify the above equalities by showing that any element x belongs to the set on the left side of eqality if and only if it belongs to the set on the right.

Obviously, both A and B are subsets of A ∪ B. As a consequence, we have a very useful monotonicity property.

Lemma 1.3.1

For arbitrary sets A, B, C, D, if A ⊆ C and B ⊆ D, then A ∪ B ⊆ C ∪ D.

Assume that (1) A ⊆ C and (2) B ⊆ D and let us take x ∈ A ∪ B. By the definition of union x ∈ A or x ∈ B. If x ∈ A, then by assumption (1), x ∈ C, and therefore x ∈ C ∪ D. If x ∈ B, then by assumption (2), x ∈ D. Thus x belongs to C ∪ D. Therefore, every element of the set A ∪ B is in the set C ∪ D. ♦

Lemma 1.3.2

For arbitrary sets A and B, A ⊆ B iff A ∪ B = B.

Assume that A ⊆ B. Let us suppose that x ∈ A ∪ α and x ∉ B, then it ought to be that x ∈ A. In that case however (see assumption) x ∈ B, we have a contradiction. Conversely, if x ∈ B, then x must belong to any bigger set, in particular x ∈ A ∪ B. Hence, A ⊆ B implies A ∪ B = B.

Let us assume now that A ∪ B = B and let x be an element such that, x ∈ A. We have x ∈ Α ∪ α . In that case we have by assumption that sets A ∪ B and B are equal, we have x ∈ B. Since x was an arbitrary element, thus A ⊆ B. ♦

Question 1.3.2 How many elements are there in the set X ∪{ ∅}, where X is an n-element subset of the set of natural numbers?