# Images of Elements under Repeated Composition of Injection form Equivalence Classes

## Theorem

Let $S$ be a set.

Let $f: S \to S$ be an injection.

- $f^0, f^1, f^2, \ldots, f^n, \ldots$

be defined as:

- $\forall n \in \N: \map {f^n} x = \begin {cases} x & : n = 0 \\ \map f x & : n = 1 \\ \map f {\map {f^{n - 1} } x} & : n > 1 \end{cases}$

Let $\mathcal R \subseteq S \times S$ be the relation on $S$ defined as:

- $\mathcal R = \set {\tuple {a, b} \in S \times S: \exists k \in \Z: b = \map {f^k} a \lor \exists j \in \Z: a = \map {f^j} b}$

Then $\mathcal R$ is an equivalence relation.

## Proof

Checking in turn each of the criteria for equivalence:

### Reflexivity

By definition, $f^0$ is the identity mapping.

So by definition of $f^0$:

- $\forall a \in S: a = \map {f^0} a$

That is:

- $a \mathop {\mathcal R} a$

So $\mathcal R$ has been shown to be reflexive.

$\Box$

### Symmetry

Let $a \mathop {\mathcal R} b$.

That is:

- $b = \map {f^k} a$

for some $k \in \Z$.

Let $g_k$ be the restriction of $f^k$ to its image.

From Injection to Image is Bijection, $g^k$ is a bijection.

From Inverse of Bijection is Bijection:

- $\map {\paren {g^k}^{-1} } b = a$

Consider the extension $\paren {f^k}^{-1}$ of $\paren {g^k}^{-1}$ to its codomain which is $S$.

From Inverse of Injection is Many-to-One Relation, $\map {\paren {f^k}^{-1} } b = a$ is well-defined.

By interpreting $\map {\paren {f^k}^{-1} } b = a$ as $\map {f^{-k} } b = a$, it follows that we can set $j = -k$ and so:

- $\exists j \in \Z: \map {f^j} b = a$

That is:

- $b \mathop {\mathcal R} a$

So $\mathcal R$ has been shown to be symmetric.

$\Box$

### Transitivity

Let $a \mathop {\mathcal R} b$ and $b \mathop {\mathcal R} c$.

That is:

- $b = \map {f^{k_1} } a$
- $c = \map {f^{k_2} } b$

where $k_1, k_2 \in \Z$.

That is:

- $c = \map {f^{k_2} } {\map {f^{k_1} } a}$

By Composition of Repeated Compositions of Injections:

- $c = \map {f^{k_2} } {\map {f^{k_1} } a} = \map {f^{k_1 + k_2} } a$

and so:

- $a \mathop {\mathcal R} c$

So $\mathcal R$ has been shown to be transitive.

$\Box$

$\mathcal R$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $3$