Union and intersection

Union of two sets. The union of $A$ and $B$, written $A \cup B$, is the set of all elements belonging to $A$, to $B$, or to both:

where: $x$ is any element under consideration, $\in$ means "is a member of," and "or" is taken in its inclusive sense — an element counts even if it belongs to both sets. No element is ever written twice, because a set lists each of its members only once.

Intersection of two sets. The intersection of $A$ and $B$, written $A \cap B$, is the set of elements that belong to both $A$ and $B$ simultaneously:

where: the symbols mean the same as above, but "and" now demands membership in both sets at once. If no element satisfies this, $A \cap B = \varnothing$, the empty set — this happens whenever $A$ and $B$ share nothing, a situation with its own name: disjoint sets.

Reading a Venn diagram. $A$ and $B$ are drawn as two overlapping circles inside a rectangle that represents the universal set $\mathcal{U}$. Every point inside a circle stands for an element of that set. The union $A \cup B$ is the entire region covered by either circle, overlap included. The intersection $A \cap B$ is only the lens-shaped overlap — the part shared by both circles.

Basic properties (worth remembering, not proving): $A \cup B = B \cup A$ and $A \cap B = B \cap A$ — order does not matter. $A \cup A = A$ and $A \cap A = A$ — combining a set with itself changes nothing. $A \cup \varnothing = A$ but $A \cap \varnothing = \varnothing$ — the empty set adds nothing to a union and wipes out an intersection. If $A \subseteq B$ (every element of $A$ is also in $B$), then $A \cup B = B$ and $A \cap B = A$.

Union and intersection are not arbitrary rules — they follow directly from what "or" and "and" mean in everyday logic, applied to set membership. Ask "does $x$ belong to $A \cup B$?" The honest answer is yes precisely when $x$ passes at least one of two tests: is it in $A$? Is it in $B$? That is the everyday meaning of "or," and it is why the union can only be the same size as or larger than either original set — you are never throwing an element away, only adding elements from the other set that were missing. Ask instead "does $x$ belong to $A \cap B$?" Now $x$ must pass both tests together — it must clear the bar for $A$ and the bar for $B$. Demanding two conditions at once is always at least as strict as demanding one, so the intersection can only be the same size as or smaller than either original set. This is exactly why, in Worked example 1 below, $A \cap B$ ends up with far fewer elements than $A \cup B$. The Venn diagram is simply a picture of this logic: a point lies inside circle $A$ exactly when it satisfies "is in $A$," and inside the overlap exactly when it satisfies both conditions. Shading follows the logic automatically — you never need to memorise which region is which if you remember what "or" and "and" mean. This also explains the subset property from above. If every element of $A$ is already inside $B$ (that is, $A \subseteq B$), then the "or" test in the union adds nothing new beyond $B$ itself, so $A \cup B = B$; and the "and" test in the intersection is satisfied by every element of $A$ automatically (since it is also in $B$), so $A \cap B = A$.

Let $A = \{2, 4, 6, 8, 10\}$ and $B = \{3, 6, 9, 12\}$. Find $A \cup B$, $A \cap B$, and state whether the sets are disjoint.

Step 1 — Find $A \cup B$. List every element that is in $A$, in $B$, or in both, writing each element once. Going through $A$ first (2, 4, 6, 8, 10) then adding any new elements from $B$ (3, 9, 12 — since 6 is already listed):

This has $n(A \cup B) = 8$ elements. Notice 6 appears in both original sets but is written only once here.

Step 2 — Find $A \cap B$. Check each element of $A$ against $B$ in turn: $2 \notin B$, $4 \notin B$, $6 \in B$, $8 \notin B$, $10 \notin B$. Only 6 clears both tests.

This has $n(A \cap B) = 1$ element.

Step 3 — Interpret and check disjointness. Since $A \cap B \neq \varnothing$, the sets are not disjoint — they share exactly one element, 6. On a Venn diagram, 6 sits alone in the overlapping lens of two circles labelled $A$ and $B$; the remaining four elements of $A$ (2, 4, 8, 10) sit in the non-overlapping part of circle $A$, and the remaining three elements of $B$ (3, 9, 12) sit in the non-overlapping part of circle $B$.

Competitive stretch

Now add a third set, $C = \{6, 10, 14\}$, and find $(A \cup B) \cap C$.

Step 1. Reuse the union from above, $A \cup B = \{2, 3, 4, 6, 8, 9, 10, 12\}$.

Step 2. Intersect this with $C = \{6, 10, 14\}$: check each element of $C$ against $A \cup B$. Is $6 \in A \cup B$? Yes. Is $10 \in A \cup B$? Yes. Is $14 \in A \cup B$? No.

Working with three sets is done the same way — one operation at a time, from the innermost brackets outward, checking membership carefully at each step.

• Writing repeated elements twice in a union. If an element belongs to both sets, it still appears only once in $A \cup B$ — a set never lists a member more than once.
• Confusing "or" and "and." Union uses "or" and gives the larger combined set; intersection uses "and" and gives the smaller, shared set. Mixing these up is the single most common error in this topic.
• Forgetting that an intersection can be empty. If two sets share nothing, $A \cap B = \varnothing$ — not "nothing," not left blank — it is still a valid set, and this case is called disjoint sets.
• Shading the wrong Venn region. For $A \cap B$ shade only the overlap; for $A \cup B$ shade the whole of both circles, including the overlap, but only once.

Ask the tutor: Want to try union and intersection on three overlapping sets of your own, or check a Venn diagram you have drawn — ask the AI tutor for extra practice sets on this topic.