0

Prove that the cardinality of the intersection of the power sets of two sets is of the form $2^n$ for some positive $n$.

My Thoughts:

Let the 2 sets be A and B

$$ A = \{1,2,3,4,5\} $$

$$ B = \{3,4,5,6,7\} $$

so,

$A\cap B = \{3, 4, 5\}$

$P(A\cap B ) = \{\{\varnothing\} ,\{3\},\{4\},\{5\},\{3, 4\},\{3 ,5\},\{4, 5\}, \{3, 4, 5\}\}$

$|P (A \cap B )| = 2^n$, where $n$ is the number of elements in the set

$|P (A\cap B )| = 2^3 = 8$

I proved 1 case. But how do I prove for all cases?

becozlah
  • 359
  • 1
    You did not really prove 1 case. "Intersection of the power sets" has a different meaning than "the power set of an intersection"; the problem was about the former, while you dealt with the latter. – Jonas Meyer Nov 21 '12 at 17:33
  • @JonasMeyer They are in fact equal, and the proof follows from that pretty quickly. That's the main thing which should be shown though. – Cocopuffs Nov 21 '12 at 17:35
  • @Cocopuffs: Yes, I edited my wording before (or while) you commented. Thank you. – Jonas Meyer Nov 21 '12 at 17:36

1 Answers1

6

Well, if we know that $P(A)\cap P(B)=P(A\cap B)$, then by knowing that $P(X)$ has $2^n$ elements we are done.

To see this note the following:

$X\in P(A)\cap P(B)$ then $X\subseteq A$ and $X\subseteq B$, therefore $X\subseteq A\cap B$ and so $X\in P(A\cap B)$. Similarly in the opposite direction, we have $P(A)\cap P(B)=P(A\cap B)$.

Now we use the fact that if $X$ has $n$ members then $P(X)$ has $2^n$ members to deduce that $P(A)\cap P(B)$ has $2^{|A\cap B|}$ elements.

Note however that the question is false. If $A\cap B=\varnothing$ then $P(A)\cap P(B)=\{\varnothing\}$ whose size is $2^0$, and $0$ is never a positive number.

Asaf Karagila
  • 393,674
  • Oh, in that case my answer above should answer your question. I'm still not 100% sure why you'd ask it in the comments. But the question is generally not stupid. – Asaf Karagila Nov 08 '14 at 21:04
  • One inclusion is easy to prove in a similar fashion. The other, which was trivial in the above case, is impossible to prove. – Asaf Karagila Nov 08 '14 at 21:09
  • I don't know you at all to say otherwise. It's generally not making people take your things less lightly just because you're stupid. You think about what I had to say, and also why $P(A\cup B)$ is not a subset of $P(A)\cup P(B)$, unless of course $A\subseteq B$ or $B\subseteq A$. – Asaf Karagila Nov 08 '14 at 21:19