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Does the cartesian product have an associative property such that $M_1\times(M_2\times M_3) = (M_1\times M_2)\times M_3$, or does the different order result in different ordered pairs?

dietervdf
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Daniel Valland
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1 Answers1

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The cartesian product is non-associative because the elements of $A \times (B \times C)$ are in the form $(a,(b,c))$ while the elements of $(A \times B) \times C$ are in the form $((a,b),c)$. These are not the same.

On the other hand, there's a bijection between them. And this bijection can preserve a lot of structure. For example, if $A$, $B$ and $C$ are groups and $A \times B$ has its group product defined pointwise, then $\times$ would be associative up to isomorphism.

wlad
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  • Thanks, so in general, the statement M1x(M2xM3) = (M1xM2)xM3 would be false? – Daniel Valland May 27 '15 at 23:07
  • @DanielValland Yes – wlad May 27 '15 at 23:07
  • What about M1x(M2xM3) = M1 x M2 x M3? – Daniel Valland May 27 '15 at 23:09
  • what do you mean by M1 x M2 x M3? – wlad May 27 '15 at 23:10
  • Would the statement Ax(BxC) = AxBxC also be false... – Daniel Valland May 27 '15 at 23:11
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    @DanielValland, "$A\times B\times C$" is ambiguous, similarly to how "$3-2-5$" is ambiguous: in the absence of an order of operations assumption, $3-2-5=(3-2)-5=-4$ and $3-2-5=3-(2-5)=6$ are both possible interpretations. When writing colloquially we do allow expressions such as "$A\times B\times C$" under one of two circumstances: if there's an underlying assumption like "$\times$ associates to the right", or if the results are understood to not depend on how $\times$ associates. Usually, by the way, it doesn't matter which we use, but sometimes it really does. – Noah Schweber May 27 '15 at 23:20