# How do I take a power set product and find 5 elements such that the intersection between them is an empty set?

Here is the problem:

List 5 elements of this set:

$${\{(X, Y)\}\in\\} \mathcal {P}\{1,2,3\} \times \mathcal{P}\{1,2,3\}:X\cap Y= \emptyset\ \}$$

I know that: $${\mathcal{P }(\{1,2,3\}= {\{\emptyset,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}}}$$

Does this involve taking massive Cartesian products? The intersection between two sets is disjoint when they have no other like elements, but I don't get how to find the intersection between the product of power sets as an empty set or disjoint?

• $|A\times B| = |A|\times |B|$ regardless of how complicated or simple of sets $A,B$ happen to be, so you have $|\mathcal{P}(\{1,2,3\})\times \mathcal{P}(\{1,2,3\})| = 2^3\times 2^3 = 64$. That is not entirely relevant here though... – JMoravitz Oct 19 '18 at 18:17
• $(\varnothing,\{1,2,3\})$ and $(\{1,2\},\{3\})$ are both elements of the set. Do see why? Can you add $3$ pairs yourself? Then you are ready. – drhab Oct 19 '18 at 18:17
• You just want to list out five ordered pairs where each entry in the ordered pair is itself a subset of $\{1,2,3\}$ and their intersection is empty... Here are two to get you started: $(\{1\},\{2\}), (\{1,2\},\{3\})$ – JMoravitz Oct 19 '18 at 18:18
• Now... as for the question of how large your set in question actually is, for each of $1,2,3$ choose whether it appears in the set on the left exclusively, appears in the set on the right exclusively, or does not appear at all. Applying multiplication principle you'll find that there are $3^3 = 27$ valid ordered pairs in your set. You were only tasked with listing five. – JMoravitz Oct 19 '18 at 18:20
• Here is another similar example with a slightly different condition. – JMoravitz Oct 19 '18 at 18:23