Product of Subset with Intersection

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

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

Then:


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

Proof 1
Let $x \in X, t \in Y \cap Z$.

By the definition of intersection, $t \in Y$ and $t \in Z$.


 * Consider $X \circ \left({Y \cap Z}\right)$.

We have $x \circ t \in X \circ \left({Y \cap Z}\right)$ by definition of subset product.

As $t \in Y$ and $t \in Z$, we also have $x \circ t \in X \circ Y$ and $x \circ t \in X \circ Z$.

The result follows.


 * Similarly, consider $\left({Y \cap Z}\right) \circ X$.

Then we have $t \circ x \in \left({Y \cap Z}\right) \circ X$ by definition of subset product.

As $t \in Y$ and $t \in Z$, we also have $t \circ x \in Y \circ X$ and $t \circ x \in Z \circ X$.

Again, the result follows.

Proof 2
Consider the relation $\mathcal R \subseteq G \times G$ defined as:


 * $\forall g, h \in G: \left({g, h}\right) \in \mathcal R \iff \exists g \in X$

Then:
 * $\forall S \subseteq G: X \circ S = \mathcal R \left({S}\right)$

Then:

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


 * $\forall g, h \in G: \left({g, h}\right) \in \mathcal R \iff \exists h \in X$

Then:
 * $\forall S \subseteq G: S \circ X = \mathcal R \left({S}\right)$

Then:

Note
It is not always the case that:
 * $X \circ \left({Y \cap Z}\right) \supseteq \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$
 * $\left({Y \cap Z}\right) \circ X \supseteq \left({Y \circ X}\right) \cap \left({Z \circ X}\right)$

so this result can not be expressed as an equality.

Example
Let $a \in G$ such that $a \ne a^{-1}$.

Let $X = \left\{{a, a^{-1}}\right\}, Y = \left\{{a}\right\}, Z = \left\{{a^{-1}}\right\}$.

Then: so clearly $X \circ \left({Y \cap Z}\right) \ne \left({X \circ Y}\right) \cap \left({X \circ Z}\right)$.
 * $X \circ \left({Y \cap Z}\right) = X \circ \varnothing = \varnothing$
 * $\left({X \circ Y}\right) \cap \left({X \circ Z}\right) = \left\{{a^2, e}\right\} \cap \left\{{e, a^{-2}}\right\} \ne \varnothing$