Is the cardinality of AxBxC different to that of Ax(BxC), since AxBxC gives a 3 tuple, but Ax(BxC) gives a two tuple? but in such case, what would be the formula for calculating the cardinality of Ax(BxC) ? since in the sets A = {1,2} B = {3,4} C={5,6} |AxBxC| = 2*2*2 = 8, but |Ax(BxC)| would be 2*(2*2) = 2*(4) = also 8...
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$\begingroup$There's an obvious bijection between the sets $A \times B \times C$ and $A \times (B \times C)$: $$f(a,b,c) = (a, (b,c))$$ For $a \in A$, $b \in B$, and $c \in C$.
Once you've checked that this is a bijection, then you'll see the sets have the same cardinality.
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