Why do the different probability independence rules imply each other?

Question

"The events A and B are independent if any one of the following three equivalent conditions hold:

1) P(A ∩ B) = P(A)P(B)

2) P(A|B) = P(A)

3) P(B|A) = P(B) "

Why does 2) imply 3) and vice versa?

Also, a related question: I know that P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B)

But in practice (real-life situations), how can we assume that P(A)P(B|A) = P(B)P(A|B) ?

Answer 1

Imagine condition $2$ holds. Then we have $P(A)=P(A|B)=\frac{P(A\cap B)}{P(B)}$. From here you get that $P(A)P(B)=P(A\cap B)$ and hence $\frac{P(A\cap B)}{P(A)}=P(B|A)=P(B)$ so you get condition $3$. The other direction is the same.

Answer 2

Let $A=\Omega$ (the whole space) and $B=\varnothing$

Then $P(B\mid A)=0=P(B)$ but there is no formal definition of $P(A\mid B)$ so that we cannot observe that $P(A\mid B)=P(A)$.

In that sense it is incorrect to state the 2) implies 3).