Cardinality of Finite Set is Well-Defined

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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