Does the equation $A⋆B=\{a*b:(a,b)\in A×B\}$ define an operation in $\mathscr P(X)$ and is $\{x_0\}$ absorbing for $⋆$ if $x_0$ is absorbing for $*$?

Question

So if $(X,*)$ is a gruppoid then we say that $x_0$ in $X$ is absorbing with respect $\bot$ if for any $x$ in $X$ the equality $$ x_0*x=x_0=x*x_0 $$ holds. So with respect the last definition it is immediately to see that the absorbing element is unique: indeed, if $x'_0$ was another absorbing element for $*$ then the equality $$ x'_0=x'_0*x_0=x_0 $$ would hold. However, now let's we put $$ \tag{1}\label{1}A\star B:=\big\{x\in X:\text{$x=a*b$ with $(a,b)\in A\times B$}\big\} $$ for any $A$ and $B$ in $\mathscr P(X)$.

Well, I would like if actually \eqref{1} defines an operation on $\mathscr P(X)$ since what I observed to follow confused me a lot.

So if $x_0$ in $X$ was absorbing for $*$ then the equality holds $$ \{x_0\}\star U= \big\{x\in X:\text{$x=x_0*u$ with $u\in U$}\big\}= \{x\in X:x=x_0\}= \{x_0\}= \{x\in X:x=x_0\}= \big\{x\in X:\text{$x=u*x_0$ with $u\in U$}\big\}= U\star\{x_0\} $$ so that $\{x_0\}$ would be absorbing for $\star$.

Now if $A\star\emptyset$ was not empty then there would be exist $x$ in $A\star\emptyset$ so that by \eqref{1} therewould be exist $(a,y)\in A\times\emptyset$ such that $$ x=a*y $$ but this is impossibile and thus $A\star\emptyset$ is empty; analogously even $\emptyset\star A$ is empty: thus we conclude that for any $A$ in $\mathscr P(X)$ the equality $$ A\star\emptyset=\emptyset=\emptyset\star A $$ holds so that $\emptyset$ is always absorbing for $\star$.

So by uniqueness of absorbing element we conclude that the equality holds $$ \emptyset=\{x_0\} $$ and this is clearly absurd!

So I would have clarification about: I would like to understand if \eqref{1} defines an operation in $\mathscr P(X)$ or simply in $\mathscr P(X)\setminus\{\emptyset\}$ and so I would like to understand if $\{x_0\}$ is absorbing for $\star$ if $x_0$ is it for $*$.

Could someone help me, please?

Answer 1

Of course $(1)$ defines an operation on $\mathscr P(X)$. For any $A, B \in \mathscr P(X)$ we have $A\star B =\{a*b\mid (a,b)\in A\times B\}$ is also a subset of $X$.

Your mistake happens when you calculated $\{x_0\}\star U$:

$\{x\in X: x=x_0*u \text{ with }u\in U\}\color{red}{=} \{x\in X:x=x_0\}$

This equality happens only if $U\ne\emptyset$ because you need at least one $u\in U$ for the LHS to be non-empty.

So you have $\{x_0\}\star U = \begin{cases}\{x_0\}&\text{ if }U\ne\emptyset\\ \emptyset &\text{ if }U=\emptyset. \end{cases}$

Answer 2

$(1)$ defines indeed an operation on $\mathscr P(X)$ but $\{x_0\}$ is not absorbing for it because $\{x_0\}\star\varnothing=\varnothing\ne\{x_0\}.$

However, this operation on $\mathscr P(X)$ restricts to an operation on $\mathscr P(X)\setminus\{\varnothing\}$ and for that restriction, $\{x_0\}$ is absorbing.