Venn diagrams and operations on sets

To compute with sets, we need operations , ways of combining two sets to make a new one. The three core operations are union, intersection and difference. Each has a clean Venn-diagram picture and obeys algebraic laws that mirror the laws of arithmetic.

Definitions

Let $A$ and $B$ be subsets of a universal set $U$.

• Union: $A \cup B = \{x : x \in A \text{ or } x \in B\}$.
• Intersection: $A \cap B = \{x : x \in A \text{ and } x \in B\}$.
• Difference: $A - B = A \setminus B = \{x : x \in A \text{ and } x \notin B\}$.
• Symmetric difference: $A \triangle B = (A - B) \cup (B - A) = (A \cup B) - (A \cap B)$.

Two sets are disjoint if $A \cap B = \varnothing$.

Venn diagrams

A Venn diagram draws the universal set $U$ as a rectangle and each set inside it as a region (usually a circle). The picture shows which elements lie in which set.

• $A \cup B$: the entire shaded region of either circle.
• $A \cap B$: the overlap.
• $A - B$: the part of $A$ outside $B$.
• $B - A$: the part of $B$ outside $A$.

A Venn diagram never proves a set identity (the picture might miss a case), but it is a powerful guide , if a claimed identity fails on the picture, it is wrong; if it holds on the picture, you should try to prove it algebraically.

The algebra of sets

For all sets $A, B, C \subseteq U$:

Commutative: $A \cup B = B \cup A$, $A \cap B = B \cap A$.

Associative: $(A \cup B) \cup C = A \cup (B \cup C)$, $(A \cap B) \cap C = A \cap (B \cap C)$.

Identity: $A \cup \varnothing = A$, $A \cap U = A$.

Domination: $A \cup U = U$, $A \cap \varnothing = \varnothing$.

Idempotent: $A \cup A = A$, $A \cap A = A$.

Distributive: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C),$ $A \cap (B \cup C) = (A \cap B) \cup (A \cap C).$

Absorption: $A \cup (A \cap B) = A$, $A \cap (A \cup B) = A$.

Each law can be proved by the double-inclusion method. Let us prove one in detail.

Proof of distributivity: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

($\subseteq$) Let $x \in A \cap (B \cup C)$. Then $x \in A$ and $x \in B \cup C$. So $x \in B$ or $x \in C$.

• If $x \in B$: combined with $x \in A$, we get $x \in A \cap B$, hence $x \in (A \cap B) \cup (A \cap C)$.
• If $x \in C$: similarly $x \in A \cap C \subseteq (A \cap B) \cup (A \cap C)$.

Either way, $x \in (A \cap B) \cup (A \cap C)$.

($\supseteq$) Let $x \in (A \cap B) \cup (A \cap C)$. Then $x \in A \cap B$ or $x \in A \cap C$.

• If $x \in A \cap B$: $x \in A$ and $x \in B \subseteq B \cup C$, so $x \in A \cap (B \cup C)$.
• If $x \in A \cap C$: $x \in A$ and $x \in C \subseteq B \cup C$, so $x \in A \cap (B \cup C)$.

Either way $x \in A \cap (B \cup C)$. By antisymmetry, equality holds. $\qed$

Disjoint sets

If $A \cap B = \varnothing$, then $|A \cup B| = |A| + |B|$. Disjoint unions are easy to count. When the intersection is non-empty, we must subtract the overlap , leading to the inclusion–exclusion formula in a later subtopic.

Worked examples

Example 1. Let $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$. Find $A \cup B$, $A \cap B$, $A - B$, $B - A$, $A \triangle B$.

$A \cup B = \{1, 2, 3, 4, 5, 6\}$, $A \cap B = \{3, 4\}$, $A - B = \{1, 2\}$, $B - A = \{5, 6\}$, $A \triangle B = \{1, 2, 5, 6\}$.

Example 2. Verify the distributive law for $A = \{1, 2\}$, $B = \{2, 3\}$, $C = \{1, 3\}$.

LHS: $B \cup C = \{1, 2, 3\}$, so $A \cap (B \cup C) = \{1, 2\}$. RHS: $A \cap B = \{2\}$, $A \cap C = \{1\}$, so $(A \cap B) \cup (A \cap C) = \{1, 2\}$. They match.

Example 3. Prove $A \cup (A \cap B) = A$.

Clearly $A \subseteq A \cup (A \cap B)$ (every set is a subset of itself unioned with anything). For the reverse, take $x \in A \cup (A \cap B)$. Then $x \in A$ or $x \in A \cap B$. In the first case $x \in A$; in the second, $x \in A \cap B \subseteq A$. Either way $x \in A$. $\qed$

Example 4. If $A \subseteq B$, show $A \cup B = B$ and $A \cap B = A$.

For the first, $A \subseteq B$ gives $A \cup B \subseteq B$ (since both $A$ and $B$ are subsets of $B$). Also $B \subseteq A \cup B$ trivially. So $A \cup B = B$. For the second, $A \cap B \subseteq A$ always, and if $x \in A$ then $x \in B$ too (since $A \subseteq B$), so $x \in A \cap B$. Hence $A \subseteq A \cap B$ and equality holds.

Example 5 (harder). Prove $A \triangle B = (A \cup B) - (A \cap B)$.

Take $x \in A \triangle B = (A - B) \cup (B - A)$. Then $x$ is in exactly one of $A, B$, so $x \in A \cup B$ but $x \notin A \cap B$. Hence $x \in (A \cup B) - (A \cap B)$.

Conversely, if $x \in (A \cup B) - (A \cap B)$, then $x \in A$ or $x \in B$, but not both. So $x$ is in exactly one of $A, B$, i.e. $x \in (A - B) \cup (B - A) = A \triangle B$. $\qed$

Try it yourself

1. Let $A = \{a, b, c, d\}$, $B = \{c, d, e, f\}$. Find $A \cup B$, $A \cap B$, $A - B$, $B - A$.
2. Show $A - B = A \cap B'$ where $B'$ is the complement in some $U$ containing both.
3. Verify $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ for $A = \{1, 2\}$, $B = \{2, 3\}$, $C = \{3, 4\}$.
4. Prove $(A - B) \cap B = \varnothing$.
5. Draw Venn diagrams for $A \cap B \cap C$ and $A \cup B \cup C$.
6. Prove $(A \cup B) - C = (A - C) \cup (B - C)$.
7. If $A \subseteq B$, prove $A - C \subseteq B - C$.
8. Find $A \triangle A$ and $A \triangle \varnothing$.
9. Prove $A - (A - B) = A \cap B$.
10. Show $A \cap (B - A) = \varnothing$.
11. Find sets $A, B$ such that $A - B = B - A$. What does this force?
12. Prove the absorption law $A \cap (A \cup B) = A$.

Pitfalls / Tricks

• $A - B$ keeps only what is in $A$ but not in $B$. Order matters: $A - B \ne B - A$ in general.
• "or" in $A \cup B$ is inclusive: $x$ can be in $A$, in $B$, or in both. Set-theoretic "or" is never exclusive.
• Venn diagrams help intuition but are not proofs. Always confirm an identity by the double-inclusion argument.
• Insight. Set operations mirror logic: $\cup$ is "or", $\cap$ is "and", $-$ is "and not", $'$ is "not". Every law of sets is a law of logic in disguise.