# 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 $S$.

Let $S \sim \card S$.

Then:

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

## Proof

$f: S \to \Img f$

is a surjection.

Let $h$ be a mapping such that:

$h: \card S \to S$

is a bijection.

$f \circ h: \card S \to \Img f$

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:

$R \subseteq \card S$
$\card R \le \card S$

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.

$\Box$

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:

 $\displaystyle \Img f$ $=$ $\displaystyle \card R$ $\quad$ Equivalent Sets have Equal Cardinal Numbers $\quad$ $\displaystyle$ $\le$ $\displaystyle \card S$ $\quad$ Subset implies Cardinal Inequality $\quad$

$\blacksquare$