# Let $A = \{2,3,4\}$ and $B = \{a,b\}$. List elements of $A\times B$. [closed]

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Is the correct answer to this:

$A \times B = \{(2a),(3a),(4a),(2b),(3b),(4b)\}$

?

Thanks

• Given the sets $A$ and $B$, how is $A\ast B$ defined? Nov 22 '15 at 12:48
• @JoãoVictorBateliRomão the title of this question is all I am provided with in the test paper I am working on Nov 22 '15 at 12:50
• Isn't the notatin $A\times B$? Nov 22 '15 at 12:52
• Yes, apologies. My head stays in programming language arithmetic operators. Nov 22 '15 at 12:53
• For one thing, the elements of $A\times B$ are tuples, not products; e.g. $(2,a)$, not $(2a)$. Nov 22 '15 at 20:44

The terms of the ordered pairs must be separated by a comma. The right answer is $\{(2,a),(3,a),(4,a),(2,b),(3,b),(4,b)\}$.

From comments I see you meant $\times$, which is the standard notation for Cartesian product defined for two (or more) sets.

So for sets $A,B$ $$A \times B = \{(a\color{red}{,}b) \mid \forall a \in A, \forall b \in B\}.$$

To be more precise, what does $(a,b)$ mean? It means that the pair $a,b$ is ordered, that is, if $a\neq b$, then $(a,b) \neq (b,a)$. But in set theory, what means ordered, e.g. $\{1, 2, 3\} = \{3, 1, 2\}$. There are probably many (infinitely many, I suppose) ways to define $(a, b)$ so that is satisfies the given condition, the standard way I met (called Kuratowski, thanks wiki) is the following $$(a,b) := \{\{a\},\{a,b\}\}.$$ And it is true if $a\neq b$, that $$\{\{a\},\{a,b\}\} = (a,b) \neq (b,a) = \{\{b\},\{a,b\}\}.$$
The second note is about associativity of $\times$. The question is, whether it is true that $A \times (B \times C) = (A \times B) \times C$, for $A,B,C$ sets. Suppose now $a \in A, b \in B, c \in C$, and look at both sides. To be precise, on the left we get $$(a,(b,c)),$$ because first we do $B \times C$, and we get $(b,c)$ and then $(a,(b,c))$. However, on the right side we get $$((a,b),c).$$ We see, that $(a,(b,c)) \neq ((a,b),c)$, not even for $a=b=c$. But we see, that there is natural correspondence between those two, meaning, that there is natural bijection between $A\times(B\times C)$ and $(A\times B)\times C$, which is given by $$(x,(y,z)) \mapsto ((x,y),z).$$ And therefore we can see $A\times(B\times C)$ and $(A\times B)\times C$ as the same thing, and instead writing $(a,(b,c)), ((a,b),c)$ we write $(a,b,c)$ . (We used three sets $A,B,C$, but this can be done for infinitely many.)