Finite iff Cardinality Less than Aleph Zero

Theorem
Let $X$ be a set.

Then $X$ is finite $\card X < \aleph_0$

where:
 * $\card X$ denotes the cardinality of $X$
 * $\aleph_0 = \card \N$ by Aleph Zero equals Cardinality of Naturals.

Sufficient Condition
Let $X$ be finite.

By definition of finite set:
 * $\exists n \in \N: X \sim \N_n$

where:
 * $\sim$ denotes the set equivalence
 * $\N_n$ denotes the initial segment of natural numbers less than $n$.

By the von Neumann construction of natural numbers:
 * $\N_n = n$

By definition of cardinality:
 * $\card X = n$

By the von Neumann construction of natural numbers:
 * $\forall i \in \N: i \subseteq \N$

Then:
 * $n + 1 \subseteq \N$

By Subset implies Cardinal Inequality:
 * $n + 1 = \card {n + 1} \le \card \N = \aleph_0$

Also:
 * $n < n + 1$

Thus:
 * $\card X < \aleph_0$

Necessary Condition
Let $\card X < \aleph_0$.

By definition of aleph mapping:
 * $\aleph_0 = \omega$

By the von Neumann construction of natural numbers:
 * $\N = \omega$

By definition of ordinal:
 * $\card X \in \N$

By definition of cardinal:
 * $\exists n \in \N: X \sim n$

By the von Neumann construction of natural numbers:
 * $\exists n \in \N: X \sim \N_n$

Thus by definition:
 * $X$ is finite.