Set Identities

Commutativity

We know that the union and intersection operations are commutative. We also know that the set difference operation is not commutative. There’s a difference between keeping the elements of $A$ that are not in $B$ and keeping the elements of $B$ that are not in $A$. Both sets can have elements that are not common. Therefore, we can say that ${A\setminus B} \ne {B\setminus A}$. But if we look at the symmetric difference, we can see that it’s commutative. This is because the union operation is commutative, as exhibited below:

Further, the Cartesian product is not commutative. That is, for arbitrary sets $A$ and $B$, $A\times B \ne B\times A$. In other words:

Associativity

The union and intersection operations are associative. Now, let’s look at the other set operations and see if they are also associative.

Set difference

For the set difference operation, we use $(A\setminus B)\setminus C \ne A \setminus (B \setminus C)$. To see the reason, let’s construct a counterexample. Take the sets $X = \{1,2,3,4,5\}$, $Y=\{2,3,4,5\}$, and $Z=\{3,4,5\}$.

We can note that $(X\setminus Y)\setminus Z =\{1\}$. Whereas on the other side, we have $X \setminus (Y \setminus Z) = \{1,3,4,5\}$. This concludes the argument.

The relation above can be proved through Venn diagrams for both the LHS and the RHS, as shown below:

Symmetric difference

Let’s revisit the following definition of the symmetric difference between the two sets $A$ and $B$:

Let’s see the associativity for their symmetric difference. Are the sets $(A\oplus B)\oplus C$ and $A\oplus (B\oplus C)$ equal?

To solve this problem, let $x \in (A\oplus B)\oplus C$. This implies the following:

$$\begin{gathered} (x\in (A\oplus B)\wedge x\in ̸C)\vee (x\in ̸(A\oplus B)\wedge x\in C) \\ \Rightarrow (((x\in A\wedge x\in ̸B)\vee (x\in ̸A\wedge x\in B))\wedge x\in ̸C)\vee (((x\in A \end{gathered}$$ ...