Product of Subset with Union/Proof 2

Theorem

Let $\struct {G, \circ}$ be an algebraic structure.

Let $X, Y, Z \subseteq G$.

Then:

$X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$

$\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$

where $X \circ Y$ denotes the subset product of $X$ and $Y$.

Proof

Consider the relation $\RR \subseteq G \times G$ defined as:

$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$

Then:

$\forall S \subseteq G: X \circ S = \map \RR S$

Then:

\(\ds X \circ \paren {Y \cup Z}\)

\(=\)

\(\ds \map \RR {Y \cup Z}\)

\(\ds \)

\(=\)

\(\ds \map \RR y \cup \map \RR Z\)

Image of Union under Relation

\(\ds \)

\(=\)

\(\ds \paren {X \circ Y} \cup \paren {X \circ Z}\)

Next, consider the relation $\RR \subseteq G \times G$ defined as:

$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists h \in X$

Then:

$\forall S \subseteq G: S \circ X = \map \RR S$

Then:

\(\ds \paren {Y \cup Z} \circ X\)

\(=\)

\(\ds \map \RR {Y \cup Z}\)

\(\ds \)

\(=\)

\(\ds \map \RR Y \cup \map \RR Z\)

Image of Union under Relation

\(\ds \)

\(=\)

\(\ds \paren {Y \circ X} \cup \paren {Z \circ X}\)

$\blacksquare$