Disjoint sets

Two sets $A$ and $B$ are disjoint sets if they have no elements in common — their intersection is the empty set.

where: $A$ and $B$ are the two sets being compared, $\cap$ is the intersection operator (elements in both sets), and $\varnothing$ is the empty set (no elements at all). If $A \cap B = \varnothing$, then no element of $A$ is also an element of $B$, and no element of $B$ is also an element of $A$. On a Venn diagram this has an exact visual meaning: disjoint sets are two circles drawn completely apart, with no shared region, because there is nothing to place in an overlap. This is different from sets that merely look unrelated by name — disjointness is decided purely by comparing actual elements, never by how the sets are described in words. Disjointness is also symmetric: if $A$ is disjoint from $B$, then $B$ is automatically disjoint from $A$, since $A \cap B$ and $B \cap A$ are the same set. And testing it never requires listing every element of an infinite set — for finite sets described in roster form, a direct element-by-element comparison is enough, which is exactly the method used throughout this topic. It is worth lining disjointness up against two situations students often confuse it with:

• Disjoint sets: $A \cap B = \varnothing$ — circles do not touch.
• Overlapping (non-disjoint) sets: $A \cap B \neq \varnothing$, but neither set contains the other — circles cross, sharing a middle region.
• Subset relationship: $A \subseteq B$ means $A \cap B = A$, which is non-empty whenever $A$ is non-empty — one circle sits entirely inside the other, so subsets are generally not disjoint.

A special and very common example: a set $A$ and its complement $A'$ within a universal set $U$ are always disjoint, since $A \cap A' = \varnothing$ by definition of a complement — an element cannot both belong to $A$ and be outside $A$ at the same time.

Let $A = \{1, 3, 5, 7, 9\}$, the odd numbers up to $9$, and $B = \{2, 4, 6, 8, 10\}$, the even numbers up to $10$. Check whether $A$ and $B$ are disjoint, and if so, find $n(A \cup B)$ two ways.

Step 1 — Test for common elements. Compare each element of $A$ against every element of $B$: $1, 3, 5, 7, 9$ are all odd, while $2, 4, 6, 8, 10$ are all even, so no number appears in both lists.

Step 2 — Apply the definition. Since no element is shared,

so $A$ and $B$ are disjoint sets. On a Venn diagram, circles $A$ and $B$ would be drawn completely separate, with $1, 3, 5, 7, 9$ inside circle $A$ and $2, 4, 6, 8, 10$ inside circle $B$.

Step 3 — Find $A \cup B$ by listing. Combine both lists without repetition: $A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, which has $10$ elements.

Step 4 — Verify using the counting rule. Here $n(A) = 5$ and $n(B) = 5$. Since $A$ and $B$ are disjoint, the derivation above gives

which matches the direct count from Step 3 exactly. The two sets are disjoint, and $n(A \cup B) = \mathbf{10}$.

Competitive stretch

At a school event, $X = \{\text{Ravi, Sita, Tara}\}$ are in the Debate Club and $Z = \{\text{Sita, Wasim}\}$ are in the Art Club. Here $X \cap Z = \{\text{Sita}\}$, which is not empty, so $X$ and $Z$ are not disjoint — the shortcut $n(X \cup Z) = n(X) + n(Z) = 3 + 2 = 5$ would be wrong here, since Sita is counted once in each set but should appear only once in the union: the true value is $n(X \cup Z) = 4$.

• Assuming different-looking sets are automatically disjoint. Disjointness depends only on shared elements, not on how differently the sets are named or described — always compare elements before deciding.
• Confusing an empty intersection with an empty set. $A \cap B = \varnothing$ means the intersection is empty, not that $A$ or $B$ itself has no elements; disjoint sets can each be large, as long as they share nothing.
• Drawing overlapping circles for disjoint sets, or vice versa. The Venn diagram must match the actual intersection: separate circles only when $A \cap B = \varnothing$, overlapping circles whenever even one element is shared.
• Using $n(A \cup B) = n(A) + n(B)$ when the sets are not disjoint. This addition shortcut only works for disjoint sets; for overlapping sets it double-counts the shared elements, as the competitive stretch above shows.

Ask the tutor: Want to test a few of your own set pairs for disjointness, or practise the difference between disjoint, overlapping, and subset cases with the AI tutor?