# Cardinality of Cartesian Product of Finite Sets/General Result

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## Theorem

Let $\ds \prod_{k \mathop = 1}^n S_k$ be the cartesian product of a (finite) sequence of sets $\sequence {S_n}$.

Then:

- $\ds \card {\prod_{k \mathop = 1}^n S_k} = \prod_{k \mathop = 1}^n \card {S_k}$

This can also be written:

- $\card {S_1 \times S_2 \times \ldots \times S_n} = \card {S_1} \times \card {S_2} \times \ldots \times \card {S_n}$

### Corollary

Let $S$ be a finite set.

Let $S^n$ be a cartesian space on $S$.

Then:

- $\card {S^n} = \card S^n$

## Proof

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Proof by mathematical induction.

Let:

- $\ds \map P n = \paren {\card {\prod_{k \mathop = 1}^n S_k} = \prod_{k \mathop = 1}^n \card {S_k} }$

For $n = 1$:

- $\ds \card {\prod_{k \mathop = 1}^1 S_k} = \card{S_1} = \prod_{k \mathop = 1}^1 \card {S_k}$

These equalities follow directly from the definition of a finite cartesian product.

Thus $\map P n$ is true for $n = 1$.

For $n = m + 1$ we have:

- $\ds \card {\prod_{k \mathop = 1}^{m + 1} S_k} = \card {\paren {\prod_{k \mathop = 1}^m S_k} \times S_{m + 1} }$

Applying Cardinality of Cartesian Product of Finite Sets:

- $\ds \card {\paren {\prod_{k \mathop = 1}^m S_k} \times S_{m + 1} } = \card {\prod_{k \mathop = 1}^m S_k} \times \card {S_{m + 1} }$

Applying the induction step for $n = m$:

- $\ds \card {\prod_{k \mathop = 1}^m S_k} \times \card {S_{m + 1} } = \prod_{k \mathop = 1}^m \card {S_k} \times \card {S_{m + 1} } = \prod_{k \mathop = 1}^{m + 1} \card {S_k}$

Hence for $n = m + 1$, assuming $\map P m$:

- $\ds \card {\prod_{k \mathop = 1}^{m + 1} S_k} = \prod_{k \mathop = 1}^{m + 1} \card {S_k}$

which completes the proof.

$\blacksquare$