Cardinality of Image of Mapping not greater than Cardinality of Domain

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

Let $f: S \to T$ be a mapping.

Let $\left\vert{S}\right\vert$ denote the cardinal number of $S$.

Let $S \sim \left\vert{S}\right\vert$.

Then:


 * $\left\vert{\operatorname{Im} \left({f}\right)}\right\vert \le \left\vert{S}\right\vert$

Proof
By Restriction of Mapping to Image is Surjection, the mapping:
 * $f: S \to \operatorname{Im} \left({f}\right)$

is a surjection.

Let $h$ be a mapping such that:
 * $h: \left\vert{S}\right\vert \to S$

is a bijection.

By Composite of Surjections is Surjection:
 * $f \circ h : \left\vert{S}\right\vert \to \operatorname{Im}\left({f}\right)$

is a surjection.

Construct a set $R$ such that:


 * $R = \left\{ {x \in \left\vert{S}\right\vert : \forall y \in x: f \left({h \left({x}\right)}\right) \ne f \left({h \left({y}\right)}\right)}\right\}$

It follows that:
 * $R \subseteq \left\vert{S}\right\vert$

By Subset of Ordinal implies Cardinal Inequality:
 * $\left\vert{R}\right\vert \le \left\vert{S}\right\vert$

Suppose:
 * $x \in R$
 * $y \in R$
 * $f \left({h \left({x}\right)}\right) = f \left({h \left({y}\right)}\right)$

Then, $x$ and $y$ are ordinals and by Ordinal Membership is Trichotomy:
 * $x < y \lor x = y \lor y < x$

If $x < y$, then:
 * $f \left({h \left({x}\right)}\right) \ne f \left({h \left({y}\right)}\right)$ by the definition of $R$.

Similarly, $y < x$ implies that:
 * $f \left({h \left({x}\right)}\right) \ne f \left({h \left({y}\right)}\right)$

Therefore, $x = y$.

It follows that the restriction $f \circ h \restriction_R : R \to \operatorname{Im} \left({f}\right)$ is an injection.

Finally, by the definition of surjection, for all $x \in \operatorname{Im} \left({f}\right)$, there is some $y \in \left\vert{S}\right\vert$ such that:
 * $f \left({h \left({y}\right)}\right) = x$

But since this is true for some $y$, the set:
 * $\left\{ {y \in \left\vert{S}\right\vert : f \left({h \left({y}\right)}\right) = x}\right\}$

has a minimal element.

For this minimal element $y$, it follows that:


 * $\forall z \in y: f \left({h \left({z}\right)}\right) \ne f \left({h \left({y}\right)}\right)$

since if it were equal, this would contradict the fact that $y$ is a $\in$-minimal element.

It follows that the restriction $f \circ h \restriction_R : R \to \operatorname{Im} \left({f}\right)$ is a bijection.

By the definition of set equivalence, $R \sim \operatorname{Im} \left({f}\right)$.

So: