Proof by contradiction attempt involving the cartesian product of two non-empty sets

Question

I was wondering if the way I have tried to prove by contradiction the following claim is correct:

Given two nonempty sets $A$ and $B$, show that if $(A \times B) \cup (B \times A) = A \times A$, then $A \subseteq B$.

My proof by contradiction attempt:

Assume that $(A \times B) \cup (B \times A) = A \times A$, then $A \nsubseteq B$.
Now suppose that $(x, y) \in(A \times B) \cup (B \times A)$.
$\Rightarrow$ ($x \in A$ and $y \in B$) or ($x \in B$ and $y\in A$)

In both cases, we get a contradiction since both $x \in A$ and $x \in B$, or $y \in B$ and $y \in A$, which implies that sets $A$ and $B$ share at least one common element. Well, that is a contradiction since we assumed that $A \nsubseteq B$ (which means that set $A$ and $B$ share no common elemens). $\square$

Is this sufficient to prove this claim by contradiction?

Answer 1

To do a proof by contradiction rather than a direct proof you must use the assumption $A\not \subset B$ which you didn't. $A$ and $B$ having a common element doesn't contradict $A$ not being a subset. Consider $A = \{$ all even numbers $\}$ and $B = \{$ all perfect squares$\}$. They share all even squares but neither is a subset of the other.

The negation of $A\subset B$ is that there is a $a\in A$ but $a \not \in B$.

If so note that $(a,a) \in A\times A$ so if $(A\times B)\cup (B\times A)=A\times A$ then $(a,a)\in (A\times B)\cup (B\times A)$ so either $(a,a) \in A\times B$ and $a \in B$, a contradiction. Of $(a,a)\in B\times A$ and $a \in B$, a contradiction.

Answer 2

No this proof doesn't make sense.

Why do you say that $x \in A $and $x \in B$ or $y\in B$ and $y \in A$?

And then you conclude that $A \subseteq B$ because A and B share an element which is a false assumption. Were you confusing this with $A \cap B = \emptyset$ ?

Answer 3

$A\subseteq B\overset{\mathrm{definition}}{\iff}\forall x(x\in A\Rightarrow x\in B)$

$A\nsubseteq B\iff\exists x(x\in A\ \mathrm{and}\ x\notin B)$

So by assuming $A\nsubseteq B$, we get an element $x\in A$ that is not in $B$.

We could directly prove $A\subseteq B$. Let $x\in A$. Then $(x,x)\in A\times A$. And since $A\times A=(A\times B)\cup(B\times A)$, $(x,x)\in(A\times B)\cup(B\times A)$.