If $*$ is a binary operation on a set $X$ then it is custom to define
$$
\tag{1}\label{1}A\star B:=\{x\in X:x=a*b\text{ with } (a,b)\in A\times B\}
$$
for any $A,B\in\mathcal P(X)$.
So my algebra text ask as exercise to prove that the equality
$$
A\star(B\cup C)=(A\star B)\cup(A\star C)
$$
and equality
$$
(A\cup B)\star C=(A\star C)\cup(B\star C)
$$
holds
but it seemed to me that actually a more general equalities holds: indeed, it seem to me that if $\mathcal A$ is collection in $X$ then for any $Y$ in $\mathcal P(X)$ the equality
$$
\tag{2}\label{2}Y\star\left(\bigcup_{A\in\mathcal A}A\right)=\bigcup_{A\in\mathcal A}(Y\star A)
$$
and the equality
$$
\tag{3}\label{3}\left(\bigcup_{A\in\mathcal A}A\right)\star Y=\bigcup_{A\in\mathcal A}(A\star Y)
$$
hold and I even tried to prove it as to follow.
So if $x$ is in $Y\star\left(\bigcup_{A\in\mathcal A} A\right)$ then there exist $y$ in $Y$ and $a$ in $\bigcup_{A\in\mathcal A}A$ such that
$$
x=y*a
$$
but by axiom of union there exists $A$ in $\mathcal A$ containing $a$ so that $x$ is in $Y\star A$ and thus the inclusion
$$
\tag{4}\label{4}Y\star\left(\bigcup_{A\in\mathcal A}A\right)\subseteq\bigcup_{A\in\mathcal A}(Y\star A)
$$
holds. Analogously if $x$ is in $\bigcup_{A\in\mathcal A}(Y\star A)$ then there exists $A$ in $\mathcal A$ such that $x$ is in $Y\star A$ and so there exist $a$ in $A$ and $y$ in $Y$ such that the equality
$$
\tag{5}\label{5}x=y*a
$$
holds: however, $a$ is obviously in $\bigcup_{A\in\mathcal A}A$ so that by eq. \eqref{5} $x$ is in $Y\star\left(\bigcup_{A\in\mathcal A}A\right)$ and thus even the inclusion
$$
\tag{6}\label{6}\bigcup_{A\in\mathcal A}(Y\star A)
$$
holds. Finally, by incl. \eqref{4} and incl. \eqref{6} we conclude that eq. \eqref{2} holds and moreover by analogous arguments it is possible to prove that even \eqref{5} holds. So we conclude that $\star$ is surely distributive over union but I am wondering to know if $\star$ is distributive over intersection since I tried to prove it but I failed so that afer some time I started to suspect it is not: so first of all I ask if \eqref{2} and \eqref{3} hold since I found them by myself and so I want be sure I did not make a blunder; second I ask to prove or disprove (right and left) distributivity of $\star$ over intersection. So could someone help me, please?
Best Answer
It is possible to prove that
$\displaystyle Y\star\bigg(\bigcap_{A\in\mathcal A}A\bigg)\subseteq\bigcap_{A\in\mathcal A}\big(Y\star A\big)\,.\\[10pt]$
Proof :
For any $\;x\in\displaystyle Y\star\bigg(\bigcap_{A\in\mathcal A}A\bigg)\;,\;$ there exist $\,y\in Y\,$ and $\;a\in\displaystyle\bigcap_{A\in\mathcal A}A\;$ such that $\;x=y*a\;.\quad\color{blue}{(1)}$
Since $\;a\in A\;$ for all $\,A\in\mathcal A\,,\,$ from the equality $\,(1)\,,\,$ it follows that
$x\in Y\star A\;$ for all $\;A\in\mathcal A\;.$
Consequently, $\;x\in\displaystyle\bigcap_{A\in\mathcal A}\big(Y\star A\big)\,.$
Whereas the inclusion
$\displaystyle\bigcap_{A\in\mathcal A}\big(Y\star A\big)\subseteq Y\star\bigg(\bigcap_{A\in\mathcal A}A\bigg)$
is not correct in general, not only if $\,\displaystyle\bigcap_{A\in\mathcal A}A=\varnothing\,,\,$ but also in the cases that $\,\displaystyle\bigcap_{A\in\mathcal A}A\neq\varnothing\,.$
For example,
if $\,\star\,$ is the usual multiplication in $\,X=\Bbb R\,,\,$ $\,Y=[0,1]\,,\,$ $A_1=\{1,2\}\,,\,$ $\,A_2=\{1,3\}\,,\,$ we get that
$\big(Y\star A_1\big)\bigcap\big(Y\star A_2\big)=\big[0,2\big]\bigcap\big[0,3\big]=\big[0,2\big]\;,$
$Y\star\big(A_1\bigcap A_2\big)=\big[0,1\big]\star\big\{1\big\}=\big[0,1\big]\;,$
hence ,
$\big(Y\star A_1\big)\bigcap\big(Y\star A_2\big)\not\subseteq Y\star\big(A_1\bigcap A_2\big)\,.$
Analogously, you can prove that
$\displaystyle\bigg(\bigcap_{A\in\mathcal A}A\bigg)\star Y\subseteq\bigcap_{A\in\mathcal A}\big(A\star Y\big)$
but in general
$\displaystyle\bigcap_{A\in\mathcal A}\big(A\star Y\big)\not\subseteq\bigg(\bigcap_{A\in\mathcal A}A\bigg)\star Y\,.$