Cardinal Inequality implies Ordinal Inequality

Theorem
Let $T$ be a set.

Let $\left|{ T }\right|$ denote the cardinal number of $T$.

Let $x$ be an ordinal.

Then:


 * $x < \left|{ T }\right| \iff \left|{ x }\right| < \left|{ T }\right|$

Sufficient Condition
By Cardinal Number Less than Ordinal, it follows that $\left|{ x }\right| \le x$.

So if $x < \left|{ T }\right|$, then $\left|{ x }\right| < \left|{ T }\right|$.

Necessary Condition
Suppose $\left|{ T }\right| \le x$.

By Subset of Ordinal implies Cardinal Inequality, it follows that $\left|{ \left({ \left|{ T }\right| }\right) }\right| \le \left|{ x }\right|$.

Therefore, by Cardinal of Cardinal Equal to Cardinal, $\left|{ T }\right| \le \left|{ x }\right| $

By contraposition, $\neg \left|{ T }\right| \le \left|{ x }\right| \implies \neg \left|{ T }\right| \le x$.

By Ordinal Membership Trichotomy, $x < \left|{ T }\right| \iff \left|{ x }\right| < \left|{ T }\right|$.