# Subset of Ordinal implies Cardinal Inequality

## Theorem

Let $S$ be a set.

Let $x$ be an ordinal such that $S \subseteq x$.

Then:

$\left|{S}\right| \le \left|{x}\right|$

where $\left|{S}\right|$ denotes the cardinality of $S$.

## Proof

Since $x$ is an ordinal, it follows that $x \sim \left|{x}\right|$ by Ordinal Number Equivalent to Cardinal Number.

This satisfies the hypothesis for Subset implies Cardinal Inequality.

Therefore:

$\left|{S}\right| \le \left|{x}\right|$

$\blacksquare$