Is this property true:

For any set $A$ and $B$:

$$\left(A\times B\right)^n=A^n\times B^n?$$

Where $A\times B$ is the Cartesian product and $A^n=\underbrace{A\times A\times \cdots\times A}_{n\, \text{times}}$.

  • 2
    $\begingroup$ I don't think so. The order is important when we work with cartesian products. A typical element in the set on the LHS of the equality is of the form $(a_1, b_1, a_2, b_2, \ldots, a_n, b_n)$ while a typical element in the set on the RHS is of the form $(a_1, a_2, \ldots a_n, b_1, b_2, \ldots, b_n)$... $\endgroup$
    – Amateur
    May 5 '14 at 13:05

Not quite, but almost.

Note that $(A\times B)^n$ is a set of $n$-tuples of ordered pairs, whereas $A^n\times B^n$ is a set of ordered pairs made of $n$-tuples.

However, there is a very natural bijection between $(A\times B)^n$ and $A^n\times B^n$.



$(A\times B)^n = \{ ((a_1,b_1),(a_2,b_2), \dots ,(a_n,b_n)) , \; a_i\in A, b_i \in B\}$\


$(A^n \times B^n) = \{\left((a_1,a_2,\dots, a_n),(b_1,b_2,\dots,b_n)\right), \; a_i\in A, b_i \in B \}$

So there is clearly a bijection ($a_i \to a_i , b_i \to b_i$), note that i gave appropriate names to the elements to automatically define the bijection


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