A two-set Venn diagram has four regions: in A only, in B only, in both, in neither. Easy to eyeball. A three-set Venn diagram is a different animal — three overlapping circles partition the universe into eight regions, one for every combination of "in" and "out" across A, B, C. Most students draw the picture, stare at it, and never quite internalise which region is which.
The fix is a bit-string reading of the diagram. Each region corresponds to a three-bit code — one bit per set, reading "1" if the element is in the set and "0" if it is not. Eight codes, eight regions. Once you see the mapping, every set-builder formula you meet (unions, intersections, complements) lights up a specific collection of zones and nothing else.
The picture
The bit-string decoder
Label the three sets A, B, C in order. For any element x in the universe, write a three-bit code (\text{bit}_A, \text{bit}_B, \text{bit}_C) where each bit is 1 if x is in the corresponding set and 0 if not.
| Code | Meaning | Set-builder formula | Zone |
|---|---|---|---|
| 000 | in none of the three | A' \cap B' \cap C' | outside all circles |
| 100 | in A only | A \cap B' \cap C' | left crescent |
| 010 | in B only | A' \cap B \cap C' | right crescent |
| 001 | in C only | A' \cap B' \cap C | bottom crescent |
| 110 | in A and B, not C | A \cap B \cap C' | top lens |
| 101 | in A and C, not B | A \cap B' \cap C | bottom-left lens |
| 011 | in B and C, not A | A' \cap B \cap C | bottom-right lens |
| 111 | in all three | A \cap B \cap C | central triple overlap |
Eight codes, eight zones, bijection. Every element of the universe lives in exactly one zone, identified by its three-bit code.
A worked example
Let U = \{1, 2, 3, \dots, 12\}, A = \{\text{multiples of } 2\}, B = \{\text{multiples of } 3\}, C = \{\text{multiples of } 4\}.
- A = \{2, 4, 6, 8, 10, 12\}
- B = \{3, 6, 9, 12\}
- C = \{4, 8, 12\}
Now place each element of U by its code.
| x | \in A? | \in B? | \in C? | Code | Zone |
|---|---|---|---|---|---|
| 1 | no | no | no | 000 | outside |
| 2 | yes | no | no | 100 | A only |
| 3 | no | yes | no | 010 | B only |
| 4 | yes | no | yes | 101 | A \cap C \setminus B |
| 5 | no | no | no | 000 | outside |
| 6 | yes | yes | no | 110 | A \cap B \setminus C |
| 7 | no | no | no | 000 | outside |
| 8 | yes | no | yes | 101 | A \cap C \setminus B |
| 9 | no | yes | no | 010 | B only |
| 10 | yes | no | no | 100 | A only |
| 11 | no | no | no | 000 | outside |
| 12 | yes | yes | yes | 111 | A \cap B \cap C |
The 12 elements distribute across zones as: outside = \{1, 5, 7, 11\}, A only = \{2, 10\}, B only = \{3, 9\}, C only = \{\,\}, A \cap B \setminus C = \{6\}, A \cap C \setminus B = \{4, 8\}, B \cap C \setminus A = \{\,\}, A \cap B \cap C = \{12\}. Total: 4 + 2 + 2 + 0 + 1 + 2 + 0 + 1 = 12. Check. Every element landed in exactly one zone.
Why the counts have to add to |U|: the eight zones form a partition of U — disjoint (no element has two different codes) and exhaustive (every element has some code). Any partition of U has cardinalities summing to |U|.
Reading unions, intersections, and complements off the diagram
Every set-builder expression built from A, B, C using \cup, \cap, and ' corresponds to a collection of zones — the zones where the expression holds. Shade the zones, and the picture is the set.
- A \cup B \cup C: every zone except the outside. Seven zones shaded.
- A \cap B \cap C: only the central triple overlap. One zone shaded.
- A': every zone where the first bit is 0 — i.e., B only, C only, B \cap C \setminus A, and the outside. Four zones shaded.
- (A \cup B) \cap C: zones where (bit_A = 1 or bit_B = 1) and bit_C = 1. Codes 101, 011, and 111. Three zones shaded.
- A \triangle B (symmetric difference): zones where exactly one of bit_A, bit_B is 1. Codes 10\_ and 01\_ — that is, 100, 101, 010, 011. Four zones shaded.
Every formula you meet in this chapter translates, via bit strings, to a specific selection of zones. The diagram is not just a picture; it is a complete calculator for three-set Boolean expressions.
Inclusion-exclusion from the diagram
The three-set inclusion-exclusion formula
is exactly the statement that if you sum |A| + |B| + |C|, you double-count the pairwise lenses (each appears in two of the three sums) and triple-count the central lens (it appears in all three). Subtracting the three pairwise intersections overcorrects for the central lens — which you then add back. The formula is the zone-count arithmetic with every double-count undone.
Run the example above:
- |A| + |B| + |C| = 6 + 4 + 3 = 13
- |A \cap B| = |\{6, 12\}| = 2; |A \cap C| = |\{4, 8, 12\}| = 3; |B \cap C| = |\{12\}| = 1
- |A \cap B \cap C| = |\{12\}| = 1
- |A \cup B \cup C| = 13 - 2 - 3 - 1 + 1 = 8
Verify: |A \cup B \cup C| = |\{2, 3, 4, 6, 8, 9, 10, 12\}| = 8. Match.
The misconception to dodge
A common error is to forget that the outside zone exists. Students write "A \cup B \cup C = U" as if the three circles always cover the whole universe. They do not — in the example above, \{1, 5, 7, 11\} are in U but in none of the three sets. The eighth zone is as real as the other seven; ignoring it breaks complement and inclusion-exclusion calculations.
Another mistake: labelling the lenses correctly but forgetting that the entire circle for A is the union of four zones — the A-only crescent, plus the three lenses where A overlaps with others. When a question asks for |A|, you have to add those four zones, not just the crescent.
Related
Set Operations · Venn Diagrams from Scratch · Sets — Introduction · Cardinality Meter