Dominates is Equivalent to Subset

Theorem

Let $S$ and $T$ be sets.

Then $S$ is dominated by $T$ iff $S$ is equivalent to a subset of $T$:

$S \le T \iff S \sim T_1 \subseteq T$

Thus, the cardinality of $S$ is at most as large as that of $T$.

The use of the term large is justified by the fact that Dominates is an Ordering.

Proof

Let $S \le T$.

Then by definition, $\exists f: S \to T$ injective.

From Image is Subset of Codomain, $f \left({S}\right) \subseteq T$.

Now consider the mapping $f_1: S \to f \left({S}\right)$, defined as:

$\forall s \in S: f_1 \left({s}\right) = f \left({s}\right)$

Clearly $f_1: S \to f \left({S}\right)$ is a bijection.

By definition of equivalent sets:

$S \sim f \left({S}\right) \subseteq T$

$\Box$

Let $S \sim S_1 \subseteq T$.

Then by definition, $\exists f: S \to S_1$ is bijective.

Now consider the mapping $f_1: S \to f \left({S}\right)$, defined as:

$\forall s \in S: f_1 \left({s}\right) = f \left({s}\right)$

Clearly $f_1: S \to T$ is injective.

Thus $S \le T$, by definition.

$\blacksquare$

Alternative treatments

Some sources use this equivalence to define dominance.

Also see

• Set equivalence
• Cardinality

Sources

• Steven A. Gaal: Point Set Topology (1964)... (previous)... (next): Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability