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?
 A: The empty set is not an ordered pair; neither intuitively  (what things would form the ordered if you just have an empty set?) nor in the formalization you recall. 
Note that $(a,b)= \{\{a\}, \{a,b\}\}$ implies $\{a\} \in \{\{a\}, \{a,b\}\}$. 
Whence the set $(a,b)$ cannot be empty, as it has $\{a\}$ as element. 
Let me add though that a Cartesian product can be equal to the empty set. 
Namely $ A \times B = \emptyset $ when $A= \emptyset $ or $B = \emptyset$. 
Yet even in this case the empty set is not an ordered pair, as in this case there are just no ordered pairs at all.  
So, $ \emptyset= A \times B $ is possible,
 $\emptyset \in A \times B$ is impossible.
A: 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.
