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I had the following on an exam today.

Let the universe = $Z^+$, $Z = (1,2,3,....100)$

$|A| = 10$

$|B| = 10$

$|A\cap B| = 0$

(a) What is $|A^c \cup B^c|$

I thought about it like this.

Let set A be dogs

Let set B be cats.

Or let A be even numbers and B be odd numbers. Sticking with dogs and cats though.

A-compliment is cats

B-compliment is dogs

their union is the whole Z so $|A^c \cup B^c|= 100$ is this correct?

(b) What is $|A^c \cap B^c|$ I said this was 0 using the same logic above

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    @Ross Millikan yes it is as you assumed.2011-03-10

2 Answers 2

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Let $A^c$ and $B^c$ be the complements of $A$ and $B$ respectively. We know that $|A \cap B| = 0$, that is, $A$ and $B$ have no elements in common, which implies that $A \subset B^c$ and $B \subset A^c$.

So $A^c$ contains all elements, except $A$. $B^c$ contains all elements except $B$. But $A^c$ contains $B$ and $B^c$ contains $A$, so $A^c \cup B^c = Z$, thus $|A^c \cup B^c| = 100$.

Your reasoning is not correct, however. If you mean that $A$ is all dogs in $Z$ and $B$ is all cats in $Z$, then there are 80 other things in $Z$ which are neither cats or dogs, say elephants. Thus $A^c$ is the set of all cats and elephants and $B^c$ is the set of all dogs and elephants.

If you mean that $A$ consist of only dogs, but not necessarily all dogs in $Z$, $B$ consists only of cats, but not necessarily all cats in $Z$, and $Z$ contains only dogs and cats, the situation is a bit different. Then three things are possible: $A^c$ contains dogs, $B^c$ contains cats, or both.

As for the b-part. We know that $|A^c| = 90$ and $|B^c| = 90$. As per above, we know that $A \subset B^c$ and $B \subset A^c$, but $A$ is not part of $A^c$ and $B$ is not part of $B^c$. However, "the rest" of $A^c$ and $B^c$ is the same, so $|A^c \cap B^c| = 80$.

Using the cats, dogs and elephants: $A^c$ is all cats and elephants, $B^c$ is all dogs and elephants. The elements in common are the elephants.

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    Your answer is the correct way to think about it. $T$hanks2011-03-10
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The reasoning is more or less correct.

Your analogies to other $A$ and $B$ (cats/dogs/even/odd) are misleading, because they have $A$ and $B$ as complementary sets themselves. Because $|A|=10$ and $|B|=10$ and there are 100 elements in the universe, there are at least 80 other elements in the universe that are not in $A$ and $B$.

As to part (a), I tend to think about it this way: Since $A$ and $B$ have no common elements, all of $B$ is contained in the complement of $A$, so the union of a set that contains all of $B$ with the complement of $B$ has to be the whole universe.

Beyond that, Calle's answer does a good job of answering the question.

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    @lampShade: Yes, (b) is wrong. Since you know the universe is the integers 1 to 100, it might help to consider specific sets $A$ and $B$ consisting of integers in the range 1 to 100 that satisfy the given information.2011-03-10