# Definition:Closed under Mapping

## Definition

Let $f: S \to T$ be a mapping.

Let $S' \subseteq S$.

Then $S'$ is closed under $f$ if and only if:

$f \sqbrk {S'} \subseteq S'$

where $f \sqbrk {S'}$ is the image of $S'$ under $f$.

That is:

$x \in S' \implies \map f x \in S'$

### Arbitrary Product

Let $\phi: X^I \to T$ be a mapping or a partial mapping, taking $I$-indexed families as arguments.

Denote with $\Dom \phi$ the domain of $\phi$ (if $\phi$ is a mapping, this is simply $X^I$).

A set $S$ is closed under $\phi$ if and only if:

$\forall \family {s_i}_{i \mathop \in I} \in S^I \cap \Dom \phi: \map \phi {\family {s_i}_{i \mathop \in I} } \in S$

Phrased in terms of image of a mapping, this translates to:

$\map \phi {S^I \cap \Dom \phi} \subseteq S$

Thus, in words, $S$ is closed under $\phi$, if and only if:

Whenever $\phi$ is defined for an $I$-indexed family from $S$, it maps that indexed family into $S$ again.

### Class Theoretical Definition

Let $A$ and $B$ be classes such that $A$ is a subclass of $B$.

Let $g: B \to B$ be a mapping on $B$.

Then $A$ is closed under $g$ if and only if:

$\forall x \in A: \map g x \in A$

## Also known as

A mapping $f$ such that $S'$ is closed under $f$ can itself be referred to as being closed in $S'$, but care needs to be taken to distinguish between this and the concept of a closed mapping in the context of topology.

## Also see

• Results about closedness under mappings can be found here.