Cardinal Inequality implies Ordinal Inequality

Theorem

Let $T$ be a set.

Let $\card T$ denote the cardinal number of $T$.

Let $x$ be an ordinal.

Then:

$x < \card T \iff \card x < \card T$

Proof

Sufficient Condition

By Cardinal Number Less than Ordinal, it follows that $\card x \le x$.

So if $x < \card T$, then $\card x < \card T$.

$\Box$

Necessary Condition

Suppose $\card T \le x$.

Then:

\(\ds \card T\)

\(\le\)

\(\ds x\)

\(\ds \leadsto \ \ \)

\(\ds \card {\paren {\card T} }\)

\(\le\)

\(\ds \card x\)

Subset of Ordinal implies Cardinal Inequality

\(\ds \leadsto \ \ \)

\(\ds \card T\)

\(\le\)

\(\ds \card x\)

Cardinal of Cardinal Equal to Cardinal

By the Rule of Transposition:

$\neg \card T \le \card x \implies \neg \card T \le x$

By Ordinal Membership is Trichotomy:

$x < \card T \iff \card x < \card T$

$\blacksquare$

Sources

• 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 10.31$