Why isn't the empty set an element of $A \times B$, while it is a relation from $A$ to $B$? Let $A$ be $\{1,2\}$, let $B$ be $\{x,y\}$.
According to the information I get from most textbooks, 
$$A \times B = \{(a,b): a\in A\text{ and } b\in B\}$$
$$A \times B = \{(1,x),(1,y),(2,x),(2,y)\}$$
But the smallest relation from $A$ to $B$ is the empty set. Presumably this is because 'the empty set is a subset of any set' is vacuously true.
But I don't understand how is the empty set not a member of $A \times B$, while it is a subset of $A \times B$. Two reasons why I am confused.


*

*The empty set can be an ordered pair, because 'All elements of the empty set is an ordered pair' is vacuously true. 


So surely 'All elements of the empty set is a member of $A$ and $B$' is vacuously true as well, thus making it an element of $A \times B$?


*If the empty set is a subset of $A \times B$, then, surely, it must be true that 'all element of the empty set is a member of $A$ and $B$' is true, be it vacuously so or not. If so, shouldn't the empty set be a member of $A \times B$?


Or am I confused by the idea between membership and subset?
 A: You're confused. Your statement 1 is incorrect; a correct version would say 

The empty set can be a set of ordered pairs, because 'Every element of the empty set is an ordered pair' is vacuously true.

I mean, if I say "$S$ is a set all of whose elements are purple", then it follows that $S$ is a set of purple elements (perhaps vacuously, if $S$ is empty), but it does not follow that $S$ is purple.
The empty set is not itself an ordered pair.
As for statement 2, here is a general fact: if $C\subseteq A$, and $D\subseteq B$, then $C\times D\subseteq A\times B$. Now, you've correctly stated that $\varnothing\subseteq A$, and $\varnothing\subseteq B$, for all sets $A$ and $B$, which you phrased as 'every element of the empty set is a member of $A$ and $B$', but this implies that $\varnothing\times\varnothing\subseteq A\times B$. This doesn't tell us anything new, since $\varnothing\times\varnothing=\varnothing$, and we know that $\varnothing\subseteq S$ for every set $S$.
Even if it were the case that $\varnothing\in A$ and $\varnothing\in B$, that would tell you that $(\varnothing,\varnothing)\in A\times B$, but the ordered pair whose entries are both $\varnothing$ is not at all the same as $\varnothing$ itself.

Let's tackle this another way. You'd like to know whether $\varnothing\in A\times B$, where $A=\{1,2\}$ and $B=\{x,y\}$. Then the question you need to ask is:
$$\text{Is there some $\alpha\in A$ and $\beta\in B$ such that $\varnothing=(\alpha,\beta)$?}$$
The answer is clearly no, there is no such $\alpha$ and $\beta$.
A: Yes, you are confused between membership and subset.  You have listed $A \times B$ in your third line.  As $|A|=|B|=2$, it has $4$ elements.  Are any of them $\emptyset$?  No, so $\emptyset \not \in A \times B$.  It is true that $\emptyset \subset A \times B$.  Under $1$, it is not true that the empty set is an ordered pair.  The line starting "So surely" demonstrates that the empty set is a subset of $A \times B$, not an element of $A \times B$. As a relation between is a subset of $A \times B$, and the empty set is a subset of $A \times B$ it is a relation.
A: The statement you made between 1 and 2. Shows that the empty set is a subset of $A\times B$ not an element of it.
