True or false? $(A\setminus B)\times(A\setminus B)=(A\times A)\setminus(B\times B)$ and $(A\cup B)\setminus(B\cap A)=(A\cap B)\setminus (A \cup B)$

Question

$\forall A,B \in \mathcal U:$

$$\begin{align} (A\setminus B) \times (A\setminus B) &= (A \times A)\setminus (B \times B) &\text{True or False?} \\ (A \cup B)\setminus (B \cap A) &= (A \cap B)\setminus (A \cup B) &\text{True or False?} \end{align}$$

After learning the set operands, I am supposed to research and understand the problems above. Could anyone help?

Answer 1

For $A=\{1,2\}$ and $B=\{2\}$ we have $A\setminus B=\{1\}$, so $$(A\setminus B)^2 = \{(1,1)\}$$ while $$A^2\setminus B^2 = \{(1,1), (1,2), (2,1), (2,2)\} \setminus \{(2,2)\} = \{(1,1), (1,2), (2,1)\}$$ so the first statement is false.

For the second statement check $A=\{0\}$ and $B=\{1\}$.

Answer 2

How to proceed: one method is to use set builder notation to verify if an equality exists (or not).

$\begin{split}(A\smallsetminus B)\times(A\smallsetminus B) &= \{(x,y)\in\mathcal U^2: (x\in A\smallsetminus B)\wedge (y\in A\smallsetminus B)\} & \text{def'n cartesian product}\\ &= \{(x,y)\in\mathcal U^2: (x\in A\wedge x\notin B)\wedge (y\in A\wedge y\notin B)\}&\text{def'n set minus}\\ &=\{(x,y)\in\mathcal U^2: (x\in A\wedge y\in A)\wedge (x\notin B\wedge y\notin B)\} ~& \text{association&commutivity}\end{split}$

Now, does this equal $(A\times A)\smallsetminus(B\times B)$ ?

Likewise try the other.

Answer 3

The second should be obvious.

$A\cap B \subset A\cup B$ so $(A\cap B)\setminus (A\cup B) = \emptyset$ whereas whenever $A \cap B \subsetneq A \cup B$ then $(A\cup B) \setminus (A\cap B)\ne \emptyset$.

As for the first. If $(x,y) \in (A\setminus B)\times (A\setminus B)$ then $x,y\in A$ and $x,y \not \in B$ so $(x,y) \in A\times A$ and $(x,y) \not \in B\times B$ so $(x,y) \in (A\times A)\setminus (B\times B)$. SO $ (A\setminus B)\times (A\setminus B) \subset (A\setminus B)\times (A\setminus B)$.

But if $(x,y) \in (A\times A)\setminus (B\times B)$ then $x\in A, y\in A$ and it is not the case that $x$ and $y$ are both in $B$. But one could be and the other not be.

Example. If $(1, \frac 12) \in (\mathbb Q \times \mathbb Q) \setminus (\mathbb N \times \mathbb N)$ because $(1, \frac 12) \in (\mathbb Q \times \mathbb Q)$ but $(1, \frac 12)\not \in (\mathbb N \times \mathbb N)$. But $1 \not \in \mathbb Q\setminus \mathbb N$ so $(1,\frac 12) \not \in (\mathbb Q\setminus \mathbb N)\times (\mathbb Q\setminus \mathbb N)$.

So $(A\times A)\setminus (B\times B)$ need not be a subset of $(A\setminus B)\times (A\setminus B)$ and need not be equal.