# Subset Relation is Compatible with Subset Product

## Theorem

Let $\struct {S, \circ}$ be a magma.

Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.

Then the subset relation on $S$ is compatible with $\circ_\PP$.

That is:

 $\ds \forall X, Y, Z \in \powerset S: \,$ $\ds X \subseteq Y$ $\implies$ $\ds \paren {X \circ_\PP Z} \subseteq \paren {Y \circ_\PP Z}$ $\ds X \subseteq Y$ $\implies$ $\ds \paren {Z \circ_\PP X} \subseteq \paren {Z \circ_\PP Y}$

### Corollary 1

Let $A, B, C, D \in \powerset S$.

Let $A \subseteq B$ and $C \subseteq D$.

Then:

$A \circ_\PP C \subseteq B \circ_\PP D$

### Corollary 2

Let $A, B \subseteq S$.

Let $A \subseteq B$.

Then:

 $\ds \forall x \in S: \,$ $\ds x \circ A$ $\subseteq$ $\ds x \circ B$ $\ds A \circ x$ $\subseteq$ $\ds B \circ x$

## Proof

Let $x \in X, z \in Z$.

Then:

$x \circ z \in X \circ Z$ and $z \circ x \in Z \circ X$

Now:

$Y \circ Z = \set {y \circ z: y \in Y, z \in Z}$
$Z \circ Y = \set {z \circ y: y \in Y, z \in Z}$

But by the definition of a subset:

$x \in X \implies x \in Y$

Thus:

$x \circ z \in Y \circ Z$ and $z \circ x \in Z \circ Y$

and the result follows.

$\blacksquare$