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:

$\N_n = \set {0, 1, \dotsc, n - 1}$ 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 behaves like 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 $6.5$
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