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 $\card S$ denote the cardinal number of $S$.

Then:


 * $\card {\Img f} \le \card S$

Proof
By Restriction of Mapping to Image is Surjection, the mapping:
 * $f: S \to \Img f$

is a surjection.

Let $h$ be a mapping such that:
 * $h: \card S \to S$

is a bijection.

By Composite of Surjections is Surjection:
 * $f \circ h: \card S \to \Img f$

is a surjection.

Construct a set $R$ such that:


 * $R = \set {x \in \card S: \forall y \in x: \map f {\map h x} \ne \map f {\map h y} }$

It follows that:
 * $R \subseteq \card S$

By Subset of Ordinal implies Cardinal Inequality:
 * $\card R \le \card S$

Suppose:
 * $x \in R$
 * $y \in R$
 * $\map f {\map h x} = \map f {\map h y}$

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

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

Similarly, $y < x$ implies that:
 * $\map f {\map h x} \ne \map f {\map h y}$

Therefore, $x = y$.

It follows that the restriction $f \circ h \restriction_R : R \to \Img f$ is an injection.

Finally, by the definition of surjection, for all $x \in \Img f$, there is some $y \in \card S$ such that:
 * $\map f {\map h y} = x$

But since this is true for some $y$, the set:
 * $\set {\map y \in \card S: \map f {\map h y} = x}$

has a minimal element.

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


 * $\forall z \in y: \map f {\map h z} \ne \map f {\map h y}$

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

It follows that the restriction $f \circ h \restriction_R : R \to \Img f$ is a bijection.

By the definition of set equivalence, $R \sim \Img f$.

So: