The following table summarizes the correct symbol (implication, equality or inclusion) for each statement:

$\Rightarrow$	$\Leftrightarrow$	$\Leftarrow$	$\subset$	$\supset$	=	neither
a b j	c	d l	e m	f h q	g i n o p	k

Most of these can be easily shown using diagrams similar to those on pages 10-11. However, points including cartesian product might be trickier for the case when one of the sets is empty.

(j) The right implication holds regardless of whether any set is empty (in fact, if $A\times B$ is non-empty, then all four sets $A$ , $B$ , $C$ and $D$ are non-empty).

(k), (l) Here is where the empty set can mess things up. We are asked about whether the left implication holds in (j). Assume $(A\times B)\subset(C\times D)$ , and suppose we try to show that $A\subset C$ . If $a\in A$ then, if we knew that there is some $b\in B$ , we could argue that $(a,b)\in A\times B$ , hence, $(a,b)\in C\times D$ , and $a\in C$ . But the problem is that $B$ can be empty, in which case, $A\times B=\emptyset\subset C\times D$ for any $C$ and $D$ , regardless of whether $A\subset C$ or not.

Overall from (j)-(l) we get the following statement: for any non-empty sets $A$ , $B$ , $C$ and $D$ , $A\subset C$ and $B\subset D$ iff $(A\times B)\subset(C\times D)$ .