# 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 an ordered set.

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 an injection.

Thus:

- $\forall S \in \mathbb S: S \preccurlyeq S$

So $\preccurlyeq$ is reflexive.

$\Box$

### 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.

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

So $\preccurlyeq$ is transitive.

$\Box$

### Antisymmetry

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

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

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

$\Box$

Hence the result.

$\blacksquare$

## Sources

- 1964: Steven A. Gaal:
*Point Set Topology*... (previous) ... (next): Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability - 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*: $\S 8$: Theorem $8.2$ (Corollary)