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Now I'm taking a look at cardinality exercises. My understanding is very basic: it is the number of elements in a set, and the cardinality of a power set like $P(A)$ is $2^{|A|}$.

So, I found this exercise:

Given |A| = 10, |B| = 5, |A-B| = 7, determine

  1. $|A \triangle B|$

  2. $|P(A \times B)|$

  3. $|P(A \cup (A \times B))|$

Now, I solved them (I think), but my basic understanding of this doesn't make me very confident - can you help me see if my answers are right?


First of all, drawing a Venn diagram, it seems to me that $A$ has $7$ exclusive elements, $B$ has $2$ and their intersection has $3$.


$|A \triangle B|$

Well, $A$ has $7$ exclusive elements and $B$ has $2$, so the answer is $9$.


$|P(A \times B)|$

Hmph. I'm guessing that $A \times B$ means that I got to multiply the amount of elements in $A$ by $B$'s cardinality. So $|A \times B| = 50$. The power set of a set with $50$ elements should be $2^{50}$.

I'm not sure here: when doing $A \times B$, should I have multiplied $10$ by $5$, or should I have multiplied only their exclusive elements? ($7$ by $2$).


$|P(A \cup (A \times B))|$

Again, I'm guessing that $|A \times B|$ is $50$. Now I have to know $A \cup (A \times B)$.

I'm a bit puzzled. Thinking about it, technically, the result of a set shouldn't have elements in the cartesian product with another set, so I guess that $A \cap A \times B = \emptyset$ right? Therefore, I can simply add $|A| + |A \times B|$, which should be $60$.

Finally, the power set of a set with cardinality $60$ should be $2^{60}$.


Is my cardinality reasoning valid?

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$|A\times B|$ is indeed $10\cdot5=50$; the fact that $A$ and $B$ have some elements in common doesn’t matter. Look at a smaller example: say $A=\{1,2,3\}$, and $B=\{1,2\}$. Then

$$A\times B=\{\langle 1,1\rangle,\langle 1,2\rangle,\langle 2,1\rangle,\langle 2,2\rangle,\langle 3,1\rangle,\langle 3.2\rangle\}$$

has $3\cdot2=6$ elements, as you can see by inspection.

$A\cap(A\times B)$ need not be empty: we might, for instance, have $A=\{0,\langle 0,0\rangle\}$ and $B=\{0\}$, so that $A\times B=\left\{\langle 0,0\rangle,\big\langle\langle 0,0\rangle,0\big\rangle\right\}$, and $A\cup(A\times B)=\left\{0,\langle 0,0\rangle,\big\langle\langle 0,0\rangle,0\big\rangle\right\}$ has cardinality $3$ instead of $2+2\cdot1=4$. In fact you haven’t enough information to answer (3). I suspect, however, that you were supposed to assume that $A\cap(A\times B)=\varnothing$, in which case your calculations are all correct.

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  • $\begingroup$ Awesome! Never would've thought of that one. Thank you. $\endgroup$ – Zol Tun Kul Sep 28 '13 at 5:41
  • $\begingroup$ @Omega: You’re welcome! $\endgroup$ – Brian M. Scott Sep 28 '13 at 5:41

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