# Cardinality of Finite Set is Well-Defined

## Theorem

Let $S$ be a finite set.

Then there is a unique natural number $n$ such that $S \sim \N_n$, where:

- $\sim$ represents set equivalence

and:

- $\N_n = \left\{{0, 1, \dots, n - 1}\right\}$ is the initial segment of $\N$ determined by $n$.

## Proof

By the definition of finite set, there is an $n \in \N$ such that $S \sim \N_n$.

Suppose $m \in \N$ and $S \sim \N_m$.

It follows from Set Equivalence is Equivalence Relation that $\N_n \sim \N_m$.

Thus by Equality of Natural Numbers, $n = m$.

Therefore the cardinality of a finite set is well-defined.

$\blacksquare$

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): $\S 2.4$: Corollary $2.19.1$ - 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 3.7$. Similar sets - 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.5$