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I'm having issues with a question regarding the cardinality of a cartesian product.

Question: Let $A=\{0, 1, 2, 3, 4, 5, 7\}$ and $B=\{0, 2, 4, −1, 12\}.$

How many elements are in $\{(a, b) ∈ A × B \; | \; a < 7 \text{ and } b < 4\}$?

I want this to be $6 \cdot 3 = 18$. but that seems to be wrong.

Levi
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Metal Mike
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3 Answers3

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You may begin by omitting all $a \in A$ such that $a\ge 7$ and all $b \in B$ such that $b \ge 4$ because we know ordered pairs with said elements will not be included in the Cartesian product we're looking for. So, after omission, we have $A = \{0,1,2,3,4,5\}$ and $B=\{0,2,-1\}$. Now, the cardinality of the Cartesian product between sets $A$ and $B$, denoted as $|A$ x $B|$, is given by $|A$ x $B|$ = $|A| \cdot |B| = 6 \cdot 3 = 18$. So you are correct! If your book says differently, then I highly suspect your book contains a mistake.

RyRy the Fly Guy
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Why does it seem wrong?

$\{(a,b)\in A\times B| a<7$ and $b< 4\}=$

$\{(a,b)| a\in \{k\in A|k < 7\}$ and $b\in \{j\in B|j < 4\}=$

$\{(a,b)| a \in \{0,1,2,3,4,5\}$ and $b \in \{-1,0,2\}\}=$

$\{0,1,2,3,4,5\} \times \{-1,0,2\}$

And if $C, D$ are finite then $|C\times D| = |C|\cdot |D|$.

And as $|\{0,1,2,3,4\}|= 6$ and $|\{-1,0,2\}| = 3$ we have

$|\{0,1,2,3,4,5\} \times \{-1,0,2\}= 6\cdot 3 = 18$

If you want we can even list them all:

$(0,-1),(0,0), (0,2),(1,-1),(1,0),(1,2), (2,-1),(2,0), (2,2),(3,-1),(3,0),(3,2),(4,-1),(4,0), (4,2),(5,-1),(5,0),(5,2)$.

.... I .... really have no idea why you think this seems wrong. There are very few things in my life right now that feel as right as this feels.

fleablood
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  • Oh.... by "it seems to be wrong" did you mean "apparently the text gave a different solution"? Or "this feels wrong"? – fleablood Oct 31 '19 at 21:57
  • Yes, the OP thinks the book's answer is wrong. He says "seems to be", not just "seems". – Ethan Bolker Oct 31 '19 at 22:01
  • If the books question is what the OP gave and the answer isn't what the OP gave, then the book answer IS wrong. (Or we've all managed to make the same absurd counting mistake and none of us have caught it, but if there are $6$ elements of $A$ less than $7$ and $3$ elements of $B$ less than $4$ and if $6 \times 3$ does actually equal $18$, I'm not sure how any of us can be wrong...) – fleablood Oct 31 '19 at 22:23
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If I've understood your question correctly, you'll begin by taking the appropriate subsets of A and B.

Since A = {0,1,2,3,4,5,7} and B = {0,2,4,−1,12}, and you're restricting your subsets to a < 7, and b < 4, this leaves the sets C = {0,1,2,3,4,5} and D = {0,2,-1}.

If you allow for repetition, which your question seems to assume, then there are 6 ways to select the first element, and 3 ways to select the second, implying a total of 18 ways to construct ordered pairs on the subsets.

Note that your textbook might be looking for un-ordered subsets of size 2, which is a different question.

Feynmanfan85
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  • This is the OPs answer, which the book says is wrong. The question asks explicitly about the cartesian product (so ordered pairs). – Ethan Bolker Oct 31 '19 at 21:50
  • And I'm suggesting that perhaps OP misread the question, which would change the answer to the question, making the book right, and implying that OP just needs to re-read the question. – Feynmanfan85 Nov 01 '19 at 01:37