Dominance Relation is Ordering

Theorem
Let $$S$$ and $$T$$ be cardinals.

Let $$S \preccurlyeq T$$ denote that $$S$$ is dominated by $$T$$.

Let $$\mathbb S$$ be any set of cardinals.

Then the relational structure $$\left({\mathbb S, \preccurlyeq}\right)$$ is a poset.

That is, $$\preccurlyeq$$ is an ordering (at least partial) on $$\mathbb S$$.

Proof
From the definition, a cardinal is a set, so standard set theoretic results apply.

So, checking in turn each of the criteria for an ordering:

Reflexivity
For any cardinal $$S$$, the identity mapping $$I_S: S \to S$$ is a is an injection, so $$\forall S \in \mathbb S: S \preccurlyeq S$$.

So $$\preccurlyeq$$ is reflexive.

Transitivity
Let $$S_1, S_2, S_3 \in \mathbb S$$ such that $$f: S_1 \to S_2$$ and $$g: S_2 \to S_3$$ be injections.

Then from Composite of Injections is an Injection, $$g \circ f$$ is an injection and so $$S_1 \preccurlyeq S_3$$.

So $$\preccurlyeq$$ is transitive.

Antisymmetry
Suppose $$S \preccurlyeq T$$ and $$T \preccurlyeq S$$.

Then from the Cantor-Bernstein-Schroeder Theorem, $$S \sim T$$ and, as $$S$$ and $$T$$ are cardinals, $$S = T$$ by definition.

So $$\preccurlyeq$$ is antisymmetric on a set of cardinals.

Hence the result.