Ordered Set of All Mappings is Ordered Set

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Theorem

Let $L = \struct {S, \preceq}$ be an ordered set.

Let $X$ be a set.


Then $L^X$ is also an ordered set.


Proof

By definition of ordered set of all mappings:

$L^X = \struct {S^X, \precsim}$

where

$\forall f, g \in S^X: f \precsim g \iff f \preceq g$
$\preceq$ denotes the ordering on mappings,
$S^X$ denotes the set of all mappings from $X$ into $S$.


Reflexivity

Let $f \in S^X$.

By definition of reflexivity:

$\forall x \in X: \map f x \preceq \map f x$

By definition of ordering on mappings:

$f \preceq f$

Thus by definition of $\precsim$:

$f \precsim f$

$\Box$


Transitivity

Let $f, g, h \in S^X$ such that

$f \precsim g$ and $g \precsim h$

By definition of $\precsim$:

$f \preceq g$ and $g \preceq h$

By definition of ordering on mappings:

$\forall x \in X: \map f x \preceq \map g x$

and

$\forall x \in X: \map g x \preceq \map h x$

By definition of transitivity:

$\forall x \in X: \map f x \preceq \map h x$

By definition of ordering on mappings:

$f \preceq h$

Thus by definition of $\precsim$:

$f \precsim h$

$\Box$


Antisymmetry

Let $f, g \in S^X$ such that

$f \precsim g$ and $g \precsim f$

By definition of $\precsim$:

$f \preceq g$ and $g \preceq f$

By definition of ordering on mappings:

$\forall x \in X: \map f x \preceq \map g x$

and

$\forall x \in X: \map g x \preceq \map f x$

By definition of antisymmetry:

$\forall x \in X: \map f x = \map g x$

Thus by Equality of Mappings:

$f = g$

$\blacksquare$


Sources

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