# Kleene Closure is Monoid

## Theorem

Let $S$ be a set, and let $S^*$ be its Kleene closure.

Let $*$ denote concatenation of ordered tuples.

Then $\struct {S^*, *}$ is a monoid.

## Proof

First, to prove that $\struct {S^*, *}$ is a semigroup.

That is, to prove $*$ is associative.

Let $s, s', s'' \in S^*$ be sequences of lengths $n, n', n''$, respectively.

Then:

 $\ds \map {s * \paren {s' * s''} } i$ $=$ $\ds \begin {cases} \map s i & \text {if 1 \le i \le n} \\ \map {s' * s''} {i - n} & \text {if n < i \le n + n' + n''} \end {cases}$ $\ds$ $=$ $\ds \begin {cases} \map s i & \text {if 1 \le i \le n} \\ \map {s'} {i - n} & \text {if n < i \le n + n'} \\ \map {s''} {i - n - n'} & \text {if n + n' < i \le n + n' + n''} \end {cases}$ $\ds \map {\paren {s * s'} * s''} i$ $=$ $\ds \begin {cases} \map {s * s'} i & \text {if 1 \le i \le n + n'} \\ \map {s''} {i - n - n'} & \text {if n + n' < i \le n + n' + n''} \end {cases}$ $\ds$ $=$ $\ds \begin {cases} \map s i & \text {if 1 \le i \le n} \\ \map {s'} {i - n} & \text {if n < i \le n + n'} \\ \map {s''} {i - n - n'} & \text {if n + n' < i \le n + n' + n''} \end {cases}$

Hence, by Equality of Mappings:

$s * \paren {s' * s''} = \paren {s * s'} * s''$

that is, $*$ is associative.

Now, to prove $\struct {S^*, *}$ has an identity $e$.

It follows immediately from the length of a concatenation that $e$ must have length $0$.

That is, the only choice for $e$ is the empty sequence.

Now, for any $s \in S^*$:

$\map {e * s} i = \begin {cases} \map e i & \text {if$1 \le i \le 0$} \\ \map s i & \text {if$0 < i < 0 + n$} \end {cases}$

from which we see that $e * s = s$.

Also:

$\map {s * e} i = \begin {cases} \map s i & \text {if$1 \le i \le n$} \\ \map e i & \text {if$n < i < n + 0$} \end {cases}$

which shows $s * e = s$.

So indeed the empty sequence is an identity element of $\struct {S^*, *}$.

Hence $\struct {S^*, *}$ is a monoid.

$\blacksquare$