Empty set does not belong to any cartesian product? I am reading from Halmos naive set theory, for Cartesian product, defined as:
$$
A\times B=\left\{x: x \in P(P(A\cup B))\,\wedge\,\exists a \in A,\exists b \in B,\, x=(a,b)\right\}
$$
Empty set belongs to $P(P(AUB))$, but since in $(a,b)$ for some $a$ in $A$ and some $b$ in $B$ , because of existence quantifier empty set doesn't belong to $A\times B$ right? But still empty set could be an ordered pair.
in Halmos, $(a,b)=\{\{a\},\{a,b\}\}$  , so empty set also an ordered pair but no cartesian product has it as an element. am i correct?

With the definition,
$$A \times B = \{ X \in P(P (A \cap B)) : \exists a \in A, \exists b \in B[X = (a,b)]\}$$
One should note that we are defining with $\in$, which does not consider the empty set, but $\{\varnothing\} \in A$ is certainly possible, i.e. the set containing the empty set can be an element (also note that there are set-theoretic definition of $0$ such as $\{\varnothing\})$. However, the empty set is still a subset of $A \times B$, which differs from being an element.
