Subset implies Cardinal Inequality

Theorem
Let $S$ and $T$ be sets such that $S \subseteq T$.

Furthermore, let $T \sim |T|$.

Then:


 * $|S| \le |T|$

Proof
By Union with Superset is Superset, it follows that $S \cup T \sim |S \cup T|$.

Therefore, by Condition for Set Union Equivalent to Associated Cardinal Number, $S \sim |S|$

Let $f : T \to |T|$ be a bijection.

It follows that $f \restriction_S : S \to |T|$ is an injection.

The image of $S$ under $f$ is a subset of $|T|$ and thus is a subset of an ordinal.

By Unique Isomorphism between Ordinal Subset and Unique Ordinal, there is a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi : x \to f\left({S}\right)$ is an order isomorphism.

It follows that $S \sim x$ by the definition of order isomorphism.

Furthermore, $\phi$ is a strictly increasing mapping from ordinals to ordinals.

By Strictly Increasing Ordinal Mapping Inequality, $y \in x \implies y \le \phi\left({y}\right)$.

But $\phi\left({y}\right) < |T|$, so since $|T|$ is an ordinal, it follows that $y < |T|$.

Therefore, $y \in x \implies y \in |T|$ and $x \le |T|$ by the definition of subset.

But $|S| \le x$ by Cardinal Number Less than Ordinal.

So $|S| \le |T|$ by the fact that Subset Relation is Transitive.