Cartesian Products Intersection When Sets Share nn Elements

Cartesian Products Intersection When Sets Share nn Elements

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Sep 13, 2026

Let AA and BB be finite sets and let ∣A∩B∣=n|A\cap B|=n. We will show that the Cartesian products A×BA\times B and B×AB\times A have exactly 2n2^n elements in common.

A pair (x,y)(x,y) lies in both A×BA\times B and B×AB\times A exactly when:

  • x∈Ax\in A and y∈By\in B, and also
  • x∈Bx\in B and y∈Ay\in A.

Equivalently, this means x∈A∩Bx\in A\cap B and y∈A∩By\in A\cap B. Therefore the intersection satisfies (A×B)∩(B×A)=(A∩B)×(A∩B).(A\times B)\cap (B\times A)=(A\cap B)\times (A\cap B).

Now if ∣A∩B∣=n|A\cap B|=n, then the set (A∩B)×(A∩B)(A\cap B)\times (A\cap B) has n2n^2 elements (since each coordinate can be any of the nn common elements).
So the cardinality of the intersection is ∣(A×B)∩(B×A)∣=n2.\big|(A\times B)\cap (B\times A)\big|=n^2.

Important note (consistency check): The standard result gives n2n^2 common elements, not 2n2^n. The statement “2n2^n elements in common” matches the size of a power set (i.e., ∣P(A∩B)∣=2n|\mathcal P(A\cap B)|=2^n), but not the intersection of Cartesian products.

Cartesian Product Basics (Sets and Elements)

Derive the intersection $(A×B)∩(B×A)$

  1. 1
    Step 1

    A pair (x,y)∈(A×B)∩(B×A)(x,y)\in (A\times B)\cap (B\times A) iff (x,y)∈A×B(x,y)\in A\times B and (x,y)∈B×A(x,y)\in B\times A.

  2. 2
    Step 2

    (x,y)∈A×B(x,y)\in A\times B means x∈Ax\in A and y∈By\in B; (x,y)∈B×A(x,y)\in B\times A means x∈Bx\in B and y∈Ay\in A.

  3. 3
    Step 3

    From both conditions, x∈A∩Bx\in A\cap B and y∈A∩By\in A\cap B.

  4. 4
    Step 4

    Thus (A×B)∩(B×A)=(A∩B)×(A∩B)(A\times B)\cap (B\times A)=(A\cap B)\times(A\cap B).

  5. 5
    Step 5

    If ∣A∩B∣=n|A\cap B|=n, then ∣(A∩B)×(A∩B)∣=n⋅n=n2|(A\cap B)\times(A\cap B)|=n\cdot n=n^2.

About the $2^n$ claim

For Cartesian products, the intersection size becomes n2n^2. The expression 2n2^n is typical for power sets: ∣P(X)∣=2∣X∣|\mathcal P(X)|=2^{|X|}, not for (X×X)(X\times X).

To connect with the common “2n2^n” intuition, observe:

  • (A∩B)×(A∩B)(A\cap B)\times(A\cap B) counts ordered pairs of common elements, giving n2n^2 outcomes.
  • P(A∩B)\mathcal P(A\cap B) counts subsets of common elements, giving 2n2^n outcomes.

These are different constructions: one is a Cartesian product, the other a power set.

We can state this cleanly with keywords:

  • Cartesian product
  • Intersection of sets
  • Power set
  • Cardinality

Common-element counts: Cartesian products vs power set

For ∣A∩B∣=n|A\cap B|=n, the intersection size of Cartesian products is n2n^2, while 2n2^n is the power set size.

FAQ / Edge cases

Intersection of Cartesian Products

1 / 4
Question · Term

$(A\times B)\cap(B\times A)$ — what does $(x,y)$ have to satisfy?

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Answer · Definition

It must satisfy x∈Ax\in A and y∈By\in B, and also x∈Bx\in B and y∈Ay\in A; hence x,y∈A∩Bx,y\in A\cap B.

Knowledge Check

Question 1 of 4
Q1Single choice

If ∣A∩B∣=n|A\cap B|=n, what is ∣(A×B)∩(B×A)∣|(A\times B)\cap(B\times A)|?