# Union of Sets — Definition, Symbol, and Formula  
  
## What Is The Union Of Sets?  
The **union of sets** AAA and BBB is the set containing every element that belongs to AAA, to BBB, or to both. It is written A∪BA \cup BA∪B, and the symbol ∪\cup∪ reads as "or" in the inclusive sense.  
In set-builder notation:  
A∪B=,x:x∈A or x∈B,A \cup B = {, x : x \in A \text{ or } x \in B ,}A∪B=,x:x∈A or x∈B,  
The key rule: a set never repeats an element, so anything sitting in both sets appears once in the union. The union is the _largest_ of the standard combinations — it never has fewer elements than the bigger of the two starting sets.  
  
### How do you find the union of two sets?  
List every element of the first set, then add any element of the second set that is not already written down.  
A=1,2,3,4,B=3,4,5,6A = {1, 2, 3, 4}, \quad B = {3, 4, 5, 6}A=1,2,3,4,B=3,4,5,6A∪B=1,2,3,4,5,6A \cup B = {1, 2, 3, 4, 5, 6}A∪B=1,2,3,4,5,6  
The 333 and 444 live in both sets, so they appear once. That single rule, writing shared elements once, is the whole skill.  
  
## The Union Formula And Cardinality  
When you only need the _count_ of elements in a union, not the listing, use the cardinality formula:  
n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)n(A∪B)=n(A)+n(B)−n(A∩B)  
| Symbol | Meaning |  
| --- | --- |  
| n(A)n(A)n(A) | number of elements in set AAA |  
| n(B)n(B)n(B) | number of elements in set BBB |  
| n(A∩B)n(A \cap B)n(A∩B) | number of elements in both (the intersection) |  
| n(A∪B)n(A \cup B)n(A∪B) | number of elements in the union |  
**Why subtract the intersection?** Add n(A)n(A)n(A) and n(B)n(B)n(B) and the shared elements get counted twice — once in each set's total. Subtracting n(A∩B)n(A \cap B)n(A∩B) removes the double-count, leaving each element counted exactly once. This is the two-set form of the inclusion-exclusion principle.  
For three sets the same logic extends:  
n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)  
  
## Properties Of The Union Of Sets  
- Commutative: A∪B=B∪AA \cup B = B \cup AA∪B=B∪A. Order does not change the union.  
- Associative: (A∪B)∪C=A∪(B∪C)(A \cup B) \cup C = A \cup (B \cup C)(A∪B)∪C=A∪(B∪C). Grouping does not matter.  
- Idempotent: A∪A=AA \cup A = AA∪A=A. Uniting a set with itself changes nothing.  
- Identity (empty set): A∪∅=AA \cup \varnothing = AA∪∅=A. The empty set is the neutral element for union, the way 000 is for addition.  
- Universal set: A∪U=UA \cup U = UA∪U=U. Uniting with everything gives everything.  
- Subset rule: A⊆(A∪B)A \subseteq (A \cup B)A⊆(A∪B) and B⊆(A∪B)B \subseteq (A \cup B)B⊆(A∪B). Both starting sets sit inside their union.  
  
## Examples of Union of Sets  
### Example 1  
**Find A∪BA \cup BA∪B for A=1,3,5A = {1, 3, 5}A=1,3,5 and B=2,4,6B = {2, 4, 6}B=2,4,6.**  
The sets share no elements, so the union is every element listed once.  
A∪B=1,2,3,4,5,6A \cup B = {1, 2, 3, 4, 5, 6}A∪B=1,2,3,4,5,6  
Final answer: 1,2,3,4,5,6{1, 2, 3, 4, 5, 6}1,2,3,4,5,6.  
### Example 2 (where the first instinct goes wrong)  
**Find A∪BA \cup BA∪B for A=1,2,3A = {1, 2, 3}A=1,2,3 and B=2,3,4B = {2, 3, 4}B=2,3,4.**  
The first instinct is to glue the lists end to end: 1,2,3,2,3,4{1, 2, 3, 2, 3, 4}1,2,3∪2,3,4=1,2,3,2,3,4. Look at what that says — it claims 222 and 333 are each in the set twice. A set cannot hold the same element more than once; repetition is meaningless in set notation.  
Drop the duplicates:  
A∪B=1,2,3,4A \cup B = {1, 2, 3, 4}A∪B=1,2,3,4  
Final answer: 1,2,3,4{1, 2, 3, 4}1,2,3,4. The "list once" rule is the entire point of union.  
### Example 3  
**Use the formula to find n(A∪B)n(A \cup B)n(A∪B) when n(A)=12n(A) = 12n(A)=12, n(B)=9n(B) = 9n(B)=9, and n(A∩B)=4n(A \cap B) = 4n(A∩B)=4.**  
n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)n(A∪B)=n(A)+n(B)−n(A∩B)n(A∪B)=12+9−4=17n(A \cup B) = 12 + 9 - 4 = 17n(A∪B)=12+9−4=17  
Final answer: 171717 elements.  
### Example 4  
**Find the union of the set of even numbers and the set of odd numbers from 111 to 101010.**  
E=2,4,6,8,10,O=1,3,5,7,9E = {2, 4, 6, 8, 10}, \quad O = {1, 3, 5, 7, 9}E=2,4,6,8,10,O=1,3,5,7,9E∪O=1,2,3,4,5,6,7,8,9,10E \cup O = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}E∪O=1,2,3,4,5,6,7,8,9,10  
Every whole number from 111 to 101010 is either even or odd, so the union is all ten. Final answer: 1,2,…,10{1, 2, \dots, 10}1,2,…,10.  
### Example 5  
**A café surveys 505050 customers: 303030 order coffee, 252525 order tea, and 121212 order both. How many order coffee or tea?**  
This asks for n(C∪T)n(C \cup T)n(C∪T).  
n(C∪T)=n(C)+n(T)−n(C∩T)n(C \cup T) = n(C) + n(T) - n(C \cap T)n(C∪T)=n(C)+n(T)−n(C∩T)n(C∪T)=30+25−12=43n(C \cup T) = 30 + 25 - 12 = 43n(C∪T)=30+25−12=43  
Final answer: 434343 customers order coffee or tea. (The other 777 ordered neither.)  
### Example 6  
**Three clubs have memberships n(A)=20n(A) = 20n(A)=20, n(B)=15n(B) = 15n(B)=15, n(C)=10n(C) = 10n(C)=10; pairwise overlaps n(A∩B)=6n(A \cap B) = 6n(A∩B)=6, n(B∩C)=4n(B \cap C) = 4n(B∩C)=4, n(A∩C)=5n(A \cap C) = 5n(A∩C)=5; and n(A∩B∩C)=2n(A \cap B \cap C) = 2n(A∩B∩C)=2. How many students are in at least one club?**  
Apply the three-set formula, one term per line:  
n(A∪B∪C)=20+15+10n(A \cup B \cup C) = 20 + 15 + 10n(A∪B∪C)=20+15+10−6−4−5+2n(A∪B∪C)=45−15+2=32n(A \cup B \cup C) = 45 - 15 + 2 = 32n(A∪B∪C)=45−15+2=32  
Final answer: 323232 students belong to at least one club.  
## Why the union of sets matters  
The union answers "everything in either group" — and that question runs under more systems than it first appears.  
- **Merging without duplicates** is the core of database queries (the SQL `UNION` keyword does exactly this) and of de-duplicating mailing lists.  
- **Counting combined groups** uses the cardinality formula directly — surveys, audience overlap, and probability all lean on n(A∪B)n(A \cup B)n(A∪B).  
- **Probability of "A or B"** is built on union: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)P(A∪B)=P(A)+P(B)−P(A∩B) is the same inclusion-exclusion idea, now with probabilities instead of counts. That is the destination — the union you learn here becomes the addition rule of probability later.  
## Where Students Lose Marks On The Union Of Sets  
### Mistake 1: Writing duplicate elements  
**Where it slips in:** Whenever the two sets share elements.  
**Don't do this:** 1,2,3∪2,3,4{1, 2, 3} \cup {2, 3, 4}1,2,3∪2,3,4=1,2,3,2,3,4. Look at what that says — it claims 222 and 333 are each in the set twice. A set cannot hold the same element more than once; repetition is meaningless in set notation.  
**The correct way:** List each element once: 1,2,3,4{1, 2, 3, 4}1,2,3,4. A set has no repeated members. The first instinct on overlapping sets is to concatenate the lists; the fix is to scan the second set and add only what is new.  
### Mistake 2: Confusing union with intersection  
**Where it slips in:** Reading the symbol fast — ∪\cup∪ versus ∩\cap∩ — under time pressure.  
**Don't do this:** Answering A∪B=3,4A \cup B = {3, 4}A∪B=3,4 for A=1,2,3,4A = {1, 2, 3, 4}A=1,2,3,4, B=3,4,5,6B = {3, 4, 5, 6}B=3,4,5,6 (that is the intersection).  
**The correct way:** Union (∪\cup∪, "or") collects _everything_: 1,2,3,4,5,6{1, 2, 3, 4, 5, 6}1,2,3,4,5,6. The memorizer who pairs "∪\cup∪ = cup = holds more" with "∩\cap∩ = cap = the small lid on top" stops swapping them. Intersection is the small shared part; union is the whole.  
### Mistake 3: Forgetting to subtract the intersection in the count  
**Where it slips in:** Counting problems where the sets overlap.  
**Don't do this:** n(A∪B)=n(A)+n(B)n(A \cup B) = n(A) + n(B)n(A∪B)=n(A)+n(B) when the sets share elements.  
**The correct way:** Subtract the overlap: n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)n(A∪B)=n(A)+n(B)−n(A∩B). Adding alone double-counts everyone in both groups. The habit that fixes this is to ask "do these sets overlap?" before adding — if yes, the −,n(A∩B)-,n(A \cap B)−,n(A∩B) term is not optional.  
## Conclusion  
- The **union of sets** A∪BA \cup BA∪B collects every element in AAA, BBB, or both, with each element listed once.  
- The symbol is ∪\cup∪ ("or"), and union is commutative, associative, and idempotent.  
- The cardinality formula is n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)n(A∪B)=n(A)+n(B)−n(A∩B) — subtract the overlap to avoid double-counting.  
- The most common mistakes are writing duplicate elements, swapping union with intersection, and forgetting to subtract the intersection in counts.  
- Union becomes the addition rule of probability and the `UNION` operation in databases.  
## Practice Questions on the Union of Sets  
Try these, then check your answers below.  
1. Find a,b,c∪c,d{a, b, c} \cup {c, d}a,b,c∪c,d.  
2. Compute n(A∪B)n(A \cup B)n(A∪B) when n(A)=18n(A) = 18n(A)=18, n(B)=14n(B) = 14n(B)=14, and n(A∩B)=7n(A \cap B) = 7n(A∩B)=7.  
3. Verify the absorption law A∪(A∩B)=AA \cup (A \cap B) = AA∪(A∩B)=A for A=1,2A = {1, 2}A=1,2, B=2,3B = {2, 3}B=2,3.  
4. Find A∪BA \cup BA∪B for A=2,4,6,8A = {2, 4, 6, 8}A=2,4,6,8 and B=1,2,3,4B = {1, 2, 3, 4}B=1,2,3,4.  
5. A club has n(A)=12n(A) = 12n(A)=12 swimmers and n(B)=9n(B) = 9n(B)=9 runners, with 555 who do both. How many do at least one?  
**Answers**  
1. a,b,c,d{a, b, c, d}{1, 2, 3, 4}.  
2. n(A∪B)=18+14−7=25n(A \cup B) = 18 + 14 - 7 = 25n(A∪B)=18+14−7=25.  
3. A∩B=2A \cap B = {2}A∩B=2, so A∪(A∩B)=1,2∪2=1,2=AA \cup (A \cap B) = {1, 2} \cup {2} = {1, 2} = AA∪(A∩B)=1,2∪2=1,2=A. The law holds.  
4. 1,2,3,4,6,8{1, 2, 3, 4, 6, 8}, where the shared 222 and 444 appear once.  
5. n(A∪B)=12+9−5=16n(A \cup B) = 12 + 9 - 5 = 16n(A∪B)=12+9−5=16 members do at least one.