# Math Help - Sets

1. ## Sets

Let $P(S)$ be denoted as the power set of $S$ and,

Let $S \oplus T$ be denoted as the symmetric diference of the sets $S$ and $T$ and,

Let $|S|$ be denoted as the # of elements in set S.

Then, given $|A| = 10$ and $|B| = 3$,

a.) Determine the largest value possible of $|P(A \oplus B)|$

b.) Determine the smallest value possible of $|P(A \oplus B)|$

c.) Determine the largest value possible of $|P(A) \oplus P(B)|$

d.) Determine the smallest value possible of $|P(A) \oplus P(B)|$

2. In the first two you should consider:
$\begin{array}{l}
a)\;A \cap B = \emptyset \\
b)\;B \subseteq A \\
\end{array}
$
.

3. Originally Posted by Plato
In the first two you should consider:
$\begin{array}{l}
a)\;A \cap B = \emptyset \\
b)\;B \subseteq A \\
\end{array}
$
.
Hmm, well initially the problem was $n$ and $m$ ( $n > m > 0$) but then it was changed to $10$ and $3$, respectively for $n$ and $m$ to make it easier. Not sure how |B| would be a subset |A|. This whole prob is confusing w/ power sets/symmetric stuff.

4. (a) The symmetric difference between two sets is largest when both sets are disjoint. Then $|\mathcal{P}(A \oplus B)|=|\mathcal{P}(A\dot{\cup}B)|=2^{|A\dot{\cup}B|} =2^{|A|+|B|}$. ( $\dot{\cup}$ indicates disjoint union.)

(b) The symmetric difference between two sets is when one set is a subset of the other. Here A can’t be a subset of B as A has more elements than B; so B must be a subset of A. Then $|\mathcal{P}(A \oplus B)|=|\mathcal{P}(A\dot{\backslash}B)|=2^{|A\dot{\b ackslash}B|}=2^{|A|-|B|}$. ( $\dot{\backslash}$ indicates subset exclusion.)

(c) If $A\cap B=\O$, then their power sets have nothing in common other than the empty set. In this case, $|\mathcal{P}(A) \oplus \mathcal{P}(B)|$ $=|\mathcal{P}(A) \cup \mathcal{P}(B)|-|\mathcal{P}(A) \cap \mathcal{P}(B)|=|\mathcal{P}(A)|+|\mathcal{P}(B)|-2|\mathcal{P}(A)\cap\mathcal{P}(B)|=2^{|A|}+2^{|B| }-2$.

(d) If $B\subseteq A$, then every subset of B is also a subset of A. In this case, $|\mathcal{P}(A) \oplus \mathcal{P}(B)|=|\mathcal{P}(A)\dot{\backslash}\ma thcal{P}(B)|=|\mathcal{P}(A)|-|\mathcal{P}(B)|=2^{|A|}-2^{|B|}$.

5. Thanks a lot! I would have never been able to come up with that...

Thanks again.

6. You’re welcome. Do you understand what I’ve written though? That’s much more important.

7. Originally Posted by JaneBennet
You’re welcome. Do you understand what I’ve written though? That’s much more important.
Yes, well everything but the "disjoint union"