Why we can conclude immediately that $x \in B$, if $(x, y) \in A \times B = B \times A$

Question

The following statements are part of a proof involving cartesian products, specifically involving this theorem:

$A \times B = B \times A \iff$ either $A = \emptyset$, $B= \emptyset$, or $A = B$

Let $x$ be an arbitrary element of $A$. Since $B \neq \emptyset$ we can choose some $y$ in $B$ and $(x, y) \in A \times B = B \times A$, and so $x \in B$. Thus, $A \subseteq B$

What I am not exactly understanding is why it follows immediately that $x \in B$ in the above statement?

Answer 1

You have assumed that $A \times B = B \times A$, so in particular $A \times B \subset B \times A$. This means all members of $A \times B$ are also members of $B \times A$.

Therefore, since $(x,y) \in A \times B$, $(x,y) \in B \times A$ as well.

e: To be more specific, the snippet you have posted appears to me to be part of a proof where you are proving the "$\implies$" part of the theorem, beginning by assuming that the products are equal, and then proving that the right side is true. At this stage of the overall proof, I am also inferring that the cases where either $A$ or $B$ is empty have been handled, and we now are showing that if they both are nonempty we can conclude they are equal.

Answer 2

Because by deffinition $$(x,y)\in B\times A\iff x\in B\land y\in A$$

Answer 3

That's the definition of $B\times A$.

$(x, y)$ is in $B\times A$, means that $x\in B$ and $y\in A$.

Answer 4

Maybe this will help:

Since the assumption is that $A \times B=B \times A$ and because you know that B is not empty, it follows that $(x,y) \in A \times B$, $(x,y) \in B \times A$ because you are using the definition of cartesian cross products.

That is, you can think of these this way: $$(x,y) \in A \times B$$ $$and$$ $$(x,y) \in B \times A$$

I think this distinction will make it easy to see why $X \in B$ since the definition of cartesian cross product is:

$$A \times B ={(a,b)| a \in A, b \in B}$$