Product of Subset with Union

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Theorem

Let $\left({G, \circ}\right)$ be an algebraic structure.

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

Then:

$X \circ \left({Y \cup Z}\right) = \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$
$\left({Y \cup Z}\right) \circ X = \left({Y \circ X}\right) \cup \left({Z \circ X}\right)$

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


Proof 1

Let $x \circ t \in X \circ \left({Y \cup Z}\right)$.

We have $x \in X, t \in Y \cup Z$ by definition of subset product.

By definition of set union, it follows that $t \in Y$ or $t \in Z$.

So we also have $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$.

That is:

$x \circ t \in \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$

and so:

$X \circ \left({Y \cup Z}\right) \subseteq \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$


Now let $x \circ t \in \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$.

By definition of set union, it follows that $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$.

So $x \in X$, and $y \in Y$ or $y \in Z$.

That is, $x \in X$, and $y \in Y \cup Z$ by definition of set union.

Hence:

$x \circ t \in X \circ \left({Y \cup Z}\right)$

and so:

$\left({X \circ Y}\right) \cup \left({X \circ Z}\right) \subseteq X \circ \left({Y \cup Z}\right)$


That is:

$X \circ \left({Y \cup Z}\right) = \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$


The result:

$\left({Y \cup Z}\right) \circ X = \left({Y \circ X}\right) \cup \left({Z \circ X}\right)$

follows similarly.

$\blacksquare$


Proof 2

Consider the relation $\mathcal R \subseteq G \times G$ defined as:

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

Then:

$\forall S \subseteq G: X \circ S = \map {\mathcal R} S$

Then:

\(\displaystyle X \circ \paren {Y \cup Z}\) \(=\) \(\displaystyle \map {\mathcal R} {Y \cup Z}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\mathcal R} y \cup \map {\mathcal R} Z\) Image of Union under Relation
\(\displaystyle \) \(=\) \(\displaystyle \paren {X \circ Y} \cup \paren {X \circ Z}\)


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

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

Then:

$\forall S \subseteq G: S \circ X = \map {\mathcal R} S$

Then:

\(\displaystyle \paren {Y \cup Z} \circ X\) \(=\) \(\displaystyle \map {\mathcal R} {Y \cup Z}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\mathcal R} Y \cup \map {\mathcal R} Z\) Image of Union under Relation
\(\displaystyle \) \(=\) \(\displaystyle \paren {Y \circ X} \cup \paren {Z \circ X}\)

$\blacksquare$


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