Operators with certain $\cup$-like properties

Question

I'm wondering whether there is an operation distinct from set union that has the following properties. Consider an operation $O$ on a field of sets such that:

1. $O$ is commutative, associative, and idempotent
2. $O(\emptyset,X)=X$
3. $X\cup Y\subseteq O(X,Y)$
4. $O(X,\{x\})=X$ whenever $x\in X$
5. If $x\in X$, then there is a unique $Y$ such that $x\not\in Y$ and $O(Y, \{x\}) = X$

Does it follow that $O(X,Y)=X \cup Y$?

(EDIT: I originally asked about $O$ defined on the full powerset of some carrier set.)

Answer 1

No, it does not follow that $O(X,Y)=X\cup Y$.

Consider the field of sets $(\mathbb{N},F)$ where $X\in F$ iff $X$ is either:

1. finite,
2. co-finite,
3. contains finitely many odds and co-finitely many evens, or
4. contains finitely many evens and co-finitely many odds.

Let $C: F\to \{1,2,3,4\}$ categorize members of $F$ as above. For $X,Y\in F$, let $O(X,Y)=\mathbb{N}$ if $\{C(X),C(Y)\}\in\{\{2,3\},\{2,4\},\{3,4\}\}$ and $=X\cup Y$ otherwise.

$O$ is clearly commutative and idempotent. To see that it is associative, consider two cases. Case 1: $|\{C(X),C(Y),C(Z)\}\cap\{2,3,4\}|\leq 1$. Then $O(X,O(Y,Z)) = O(O(X,Y),Z) = X \cup Y \cup Z$. Case 2: $|\{C(X),C(Y),C(Z)\}\cap\{2,3,4\}|\geq 2$. Then $O(X,O(Y,Z)) = O(O(X,Y),Z) = \mathbb{N}$.

The remaining conditions on $O$ are clearly satisfied: the second, fourth, and fifth because $O(X,Y)=X\cup Y$ whenever one of $X$ and $Y$ is finite, and the third is obvious.