Operations on Sets

1.9 Operations on Sets

In earlier classes, you learned how to add, subtract, multiply, and divide numbers. Each operation took a pair of numbers and produced a single number. For instance, adding 5 and 13 gives 18; multiplying them gives 65.

Sets have their own operations. When we perform an operation on two sets, we get another set. From now on, we assume every set we discuss is a subset of some fixed universal set, which we denote by $U$.

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Union of Sets

The union of two sets $A$ and $B$ is the set of all elements that belong to $A$ or to $B$ (or to both). We write it as $A \cup B$.

In set-builder notation:

The word "or" here is inclusive — it means "at least one of the two sets contains the element."

Example. Let $A = \{2, 4, 6, 8\}$ and $B = \{6, 8, 10, 12\}$. Then $A \cup B = \{2, 4, 6, 8, 10, 12\}$. The elements 6 and 8 appear in both sets, but we list them only once in the union.

Example. If $A = \{1, 3, 5, 7\}$ and $B = \{2, 4, 6, 8\}$, then $A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8\}$.

Example. Let $X = \{a, b, c, d\}$ and $Y = \{c, d, e, f\}$. Then $X \cup Y = \{a, b, c, d, e, f\}$.

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Intersection of Sets

The intersection of two sets $A$ and $B$ is the set of all elements that belong to both $A$ and $B$. We write it as $A \cap B$.

In set-builder notation:

Example. Let $A = \{2, 4, 6, 8\}$ and $B = \{6, 8, 10, 12\}$. Then $A \cap B = \{6, 8\}$.

Example. If $A = \{1, 3, 5, 7\}$ and $B = \{2, 4, 6, 8\}$, then $A \cap B = \varnothing$ (the empty set). Sets with no common elements are called disjoint sets.

Example. Let $X = \{a, b, c, d\}$ and $Y = \{c, d, e, f\}$. Then $X \cap Y = \{c, d\}$.

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Difference of Sets

The difference of two sets $A$ and $B$, written as $A - B$ or $A \setminus B$, is the set of all elements that belong to $A$ but not to $B$.

In set-builder notation:

Similarly, $B - A$ is the set of elements in $B$ but not in $A$.

Example. Let $A = \{2, 4, 6, 8\}$ and $B = \{6, 8, 10, 12\}$. Then:

• $A - B = \{2, 4\}$ (remove 6 and 8 from $A$)
• $B - A = \{10, 12\}$ (remove 6 and 8 from $B$)

Example. If $A = \{1, 2, 3, 4, 5, 6\}$ and $B = \{2, 4, 6, 8\}$, then $A - B = \{1, 3, 5\}$ and $B - A = \{8\}$.

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Complement of a Set

Let $U$ be the universal set. The complement of a set $A$ (relative to $U$) is the set of all elements of $U$ that are not in $A$. We denote it by $A'$ or $A^c$.

In set-builder notation:

Equivalently, $A' = U - A$.

Example. Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and $A = \{1, 3, 5, 7, 9\}$. Then $A' = \{2, 4, 6, 8, 10\}$.

Example. If $U = \mathbb{N}$ (the set of natural numbers) and $A = \{1, 2, 3\}$, then $A' = \{4, 5, 6, 7, \dots\}$, the set of all natural numbers greater than 3.

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Properties of Complement

The textbook lists several important properties of the complement operation. Each one is proved below.

›Proof

Property 1: Complement Laws

1. $A \cup A' = U$
2. $A \cap A' = \varnothing$ Proof of (i): Let $x \in A \cup A'$. Then $x \in A$ or $x \in A'$. If $x \in A$, then $x \in U$ (since $A \subseteq U$). If $x \in A'$, then $x \in U$ by definition of complement. So every element of $A \cup A'$ is in $U$, meaning $A \cup A' \subseteq U$. Conversely, let $x \in U$. Then either $x \in A$ or $x \notin A$. If $x \in A$, then $x \in A \cup A'$. If $x \notin A$, then $x \in A'$, so again $x \in A \cup A'$. Thus $U \subseteq A \cup A'$. Since both inclusions hold, $A \cup A' = U$. Proof of (ii): Suppose $x \in A \cap A'$. Then $x \in A$ and $x \in A'$. But $x \in A'$ means $x \notin A$. An element cannot simultaneously belong to $A$ and not belong to $A$. This contradiction shows that no such $x$ exists. Hence $A \cap A' = \varnothing$.

›Proof

Property 2: Law of Double Complementation

$(A')' = A$

Proof:

Let $x \in (A')'$. Then $x \notin A'$. But $A'$ contains all elements of $U$ that are not in $A$. So $x \notin A'$ means $x$ is not one of those elements — that is, $x \in A$. Hence $(A')' \subseteq A$.

Conversely, let $x \in A$. Then $x \notin A'$ (since $A'$ contains only elements not in $A$). Therefore $x \in (A')'$. So $A \subseteq (A')'$.

Both inclusions give $(A')' = A$.

›Proof

Property 3: Laws of Empty Set and Universal Set

1. $\varnothing' = U$
2. $U' = \varnothing$ Proof of (i): $\varnothing' = \{ x \in U : x \notin \varnothing \}$. Since $\varnothing$ contains no elements, every $x \in U$ satisfies $x \notin \varnothing$. Thus $\varnothing' = U$. Proof of (ii): $U' = \{ x \in U : x \notin U \}$. No element of $U$ can satisfy $x \notin U$, so $U'$ has no elements. Hence $U' = \varnothing$.

›Proof

Property 4: De Morgan's Laws

1. $(A \cup B)' = A' \cap B'$
2. $(A \cap B)' = A' \cup B'$ Proof of (i): Let $x \in (A \cup B)'$. Then $x \notin A \cup B$. This means $x$ is not in $A$ and $x$ is not in $B$ (if $x$ were in either, it would be in the union). So $x \in A'$ and $x \in B'$, hence $x \in A' \cap B'$. Thus $(A \cup B)' \subseteq A' \cap B'$. …