Finite iff Cardinality Less than Aleph Zero

Theorem
Let $X$ be a set.

Then $X$ is finite $\left\vert{X}\right\vert < \aleph_0$

where:
 * $\left\vert{X}\right\vert$ denotes the cardinality of $X$
 * $\aleph_0 = \left\vert{\N}\right\vert$ 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 construction of natural numbers
 * $\N_n = n$

By definition of cardinality:
 * $\left\vert{X}\right\vert = n$

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

Then
 * $n+1 \subseteq \N$

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

Also:
 * $n < n+1$

Thus
 * $\left\vert{X}\right\vert < \aleph_0$

Necessary Condition
Let $\left\vert{X}\right\vert < \aleph_0$

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

By construction of natural numbers
 * $\N = \omega$

By definition of ordinal:
 * $\left\vert{X}\right\vert \in \N$

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

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

Thus by definition:
 * $X$ is finite.