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