Prove that the equalities $A(B\cap C)=AB\cap AC$ and $(A\cap B)C=AC\cap BC$ do not generally hold.

Question

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?

Answer 1

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, $\;x\in\displaystyle\bigcap_{A\in\mathcal A}\big(Y\star A\big)\,.$

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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 ,

$\big(Y\star A_1\big)\bigcap\big(Y\star A_2\big)\not\subseteq Y\star\big(A_1\bigcap A_2\big)\,.$

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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

$\displaystyle\bigcap_{A\in\mathcal A}\big(A\star Y\big)\not\subseteq\bigg(\bigcap_{A\in\mathcal A}A\bigg)\star Y\,.$

Answer 2

Let $X$ be the real numbers under multiplication, $A=\{0\}$, $B=\{x\}$, and $C=\{y\}$ with $x \ne y$.

Then, $A(B \cap C)=A(\emptyset)=\emptyset$, while $AB \cap AC=\{0\} \cap \{0\}=\{0\}$.

A similar counterexample applies to the other equation, where $A=\{x\}$, $B=\{y\}$, and $C=\{0\}$ with $x \ne y$.

The problem here is that multiplication by $0$ is not injective.

For a counterexample where the binary operation is cancellative (i.e., multiplication by any fixed element on either side is injective) and $A$ has more than one element, let $X$ be the positive integers under multiplication, $A=\{1,2\}$, $B=\{1\}$, and $C=\{2\}$.

Then, $A(B \cap C)=A(\emptyset)=\emptyset$, while $AB \cap AC=\{1,2\} \cap \{2,4\}=\{2\}$.

For the other equation, just rename $A$, $B$, and $C$ above to $C$, $A$, and $B$ respectively.