Closure of Pointwise Operation on Algebraic Structure
Theorem
Let $S$ be a set such that $S \ne \O$.
Let $\struct {T, \circ}$ be an algebraic structure.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let $f, g \in T^S$, that is, let $f: S \to T$ and $g: S \to T$ be mappings.
Let $\oplus: T^S \to T^S$ be the pointwise operation on $T^S$ induced by $\circ$.
Then $\oplus$ is closed on $T^S$ if and only if $\struct {T, \circ}$ is closed.
Proof
Necessary Condition
Let $\struct {T, \circ}$ be closed.
Let $x \in S$ be arbitrary.
Then:
\(\ds \forall f, g \in T^S: \, \) | \(\ds \map {f \oplus g} x\) | \(=\) | \(\ds \map f x \circ \map g x\) | Definition of Pointwise Operation | ||||||||||
\(\ds \) | \(\in\) | \(\ds T\) | as $\struct {T, \circ}$ is closed |
So $\oplus$ is closed on $T^S$.
$\Box$
Sufficient Condition
Let $\oplus$ is closed on $T^S$.
Aiming for a contradiction, suppose $\struct {T, \circ}$ is not closed.
Then:
- $(1): \quad \exists s, t \in T: s \circ t \notin T$
By definition, $T^S$ is the set of all mappings from $S$ \to $T$.
As $S \ne \O$ it follows that $\exists x \in S$.
Thus, let $x \in S$ be arbitrary.
Let $f, g \in T^S$ such that:
- $(2): \quad \map f x = s, \map g x = t$
Then:
\(\ds \map {f \oplus g} x\) | \(=\) | \(\ds \map f x \circ \map g x\) | Definition of Pointwise Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds s \circ t\) | from $(2)$ | |||||||||||
\(\ds \) | \(\notin\) | \(\ds T\) | from $(1)$ |
That is, $\oplus$ is not closed on $T^S$.
From that contradiction it follows that $\struct {T, \circ}$ is closed.
$\blacksquare$