# Can this set-theoretic vector space be endowed with this operation so that everything is fine?

Mike asked in this question to find a model of vector space with additional operation $$\wedge$$ so that we have $$a \wedge (a+b) = (a+b) \wedge b = a \wedge b$$ and $$\wedge$$ is non-distributive.

I built set-theoretic vector space like this:

Let $$(X,P(X))$$ be some nonempty set $$X$$ and $$P(X)$$ the set of all subsets of $$X$$.

Define $$+(A,B)$$ to be the symmetric difference of the sets $$A$$ and $$B$$.

Let the field be a field with two elements: $$0$$ and $$1$$.

$$+$$ is commutative and associative.

Zero vector is the empty set.

$$-A=A$$

Define $$0 \cdot A$$ to be the empty set and $$1 \cdot A$$ to be $$A$$.

And now the problem is how to define $$\wedge$$ so that $$a \wedge (a+b) = (a+b) \wedge b = a \wedge b$$ is true and is non-distributive.

But, I know of only few operations on arbitrary sets: of union, intersection, difference, symmetric difference, complementation, and that´s all.

My question would be:

Can we define on this set-theoretic vector space an operation $$\wedge$$ so that we have $$a \wedge (a+b) = (a+b) \wedge b = a \wedge b$$ and $$\wedge$$ is non-distributive?

• Doesn't the union operator work? $A\cup (A\Delta B)=A\cup B=(A\Delta B)\cup B$, but $A\cup B\neq (A\cup A)\Delta (A\cup B)=B\setminus A$ – Vsotvep Jun 18 at 9:11
• @Vsotvep I am not sure, I have to put it on the paper. – Grešnik Jun 18 at 9:15