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For Cardinal numbers A, B, C prove that

( AB ) C = A ( BC )

I've read that

A bijection between A × (B×C) and (A×B) × C can be given by (x,(y,z))↦((x,y),z)

from

Overview of basic results on cardinal arithmetic

but I cannot understand why there is a bijection between the two. Can you please explain this? I appreciate if you make it as simple as possible.

Thanks.

Community
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raffy cee
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  • Exactly what don’t you understand? The quoted line tells you exactly what the bijection is. – Brian M. Scott Oct 16 '15 at 19:29
  • I was able to make an illustration on how the bijection looks like..

    ((x1, y1), z1), ((x1, y1), z2), ((x1, y1), z3), ((x1, y1), z4), .... ((x2, y1), z1), ((x2, y1), z2), ((x2, y1), z3), ((x2, y1), z4), .... ((x3, y1), z1), ((x3, y1), z2), ((x3, y1), z3), ((x3, y1), z4), .... ((x4, y1), z1), ((x4, y1), z2), ((x4, y1), z3), ((x4, y1), z4), .... ... ... ... ...

    I just created a pattern that will pass through all the elements of the set which shows a bijection. Is this somehow, correct?

     – raffy cee Oct 17 '15 at 20:15
  • I’m sorry, but what you have there makes no sense at all. You want a function $f$ from $A\times(B\times C)$ to $(A\times B)\times C$, so for each $\langle a,\langle b,c\rangle\rangle\in A\times(B\times C)$ you want to know what element of $(A\times B)\times C$ is $f(\langle a,\langle b,c\rangle\rangle)$. The quoted line near the top of your question tells you exactly what it should be.\ – Brian M. Scott Oct 17 '15 at 20:18

1 Answers1

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The bijection is: $$ (x,y),z) \mapsto (x,(y,z)) \colon (A \times B) \times C \to A \times (B \times C) $$

BrianO
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