Cardinal Inequality implies Ordinal Inequality

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Theorem

Let $T$ be a set.

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

Let $x$ be an ordinal.


Then:

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


Proof

Sufficient Condition

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

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

$\Box$


Necessary Condition

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

By Subset of Ordinal implies Cardinal Inequality:

$\left|{\left({\left|{T}\right|}\right)}\right| \le \left|{x}\right|$.

Therefore, by Cardinal of Cardinal Equal to Cardinal:

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


By the Rule of Transposition:

$\neg \left|{T}\right| \le \left|{x}\right| \implies \neg \left|{T}\right| \le x$

By Ordinal Membership is Trichotomy:

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

$\blacksquare$


Sources