Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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 Answer 1

up vote 6 down vote accepted

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.

share|improve this answer
    
How do we prove that $P(A) \cup P(B)$ = $P(A \cup B)$? –  nbro Nov 8 at 20:59
    
@Broly: We don't? Because it's false? Why are you asking here, instead of finding the several posts that were made on this topic on the main site? –  Asaf Karagila Nov 8 at 21:00
    
Ok. I mistook writing, I meant $\cap$ instead of $\cup$, but I have just seen that your answer answers also my stupid question, right? –  nbro Nov 8 at 21:03
    
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 8 at 21:04
    
By the way, lol, why my first equality is not valid? Following your same reasoning (steps) in your answer, it seems to be correct, but it is not, because I am wrong, right? –  nbro Nov 8 at 21:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.