If $A, B, C$ are nonempty sets prove that there exists a bijection between $(A \times B) \times C$ and $A \times (B \times C)$. If there is a bijection between $A$ and $B$ and between $B$ and $C$ prove that there is a bijection between $A$ and $C$.

How can I prove this?

The Cartesian product of two sets $A$ and $B$ is defined to be the set of all ordered pairs $(a,b)$ where $a \in A$ and $b \in B.$

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    $\begingroup$ These are two separate questions! Please only ask one question per post. $\endgroup$ – walcher Sep 12 '13 at 0:08
  • $\begingroup$ @walcher Should I separate the posts then to two questions? $\endgroup$ – Tom Sep 12 '13 at 0:12
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    $\begingroup$ Yes, that would be very considerate. $\endgroup$ – walcher Sep 12 '13 at 0:14

For the first part: The set $(A\times B)\times C$ has elements of the from $((a,b),c)$, whereas $A\times (B\times C)$ has elements of the form $(a,(b,c))$. Can you think of a natural map between these two sets?

For the second part: If $f:A\to B, g:B\to C$ are bijective, then what about $g\circ f:A\to C$?

  • $\begingroup$ I think I got the second part now from the hint you gave, thanks for that. But what natural map can I make between those two sets? $\endgroup$ – Tom Sep 12 '13 at 0:43
  • $\begingroup$ What about the map $((a,b), c) \mapsto (a, (b, c))$ would that give you a bijection? $\endgroup$ – Ryan Sullivant Sep 12 '13 at 0:45
  • $\begingroup$ @RyanSullivant I was thinking about that. OKay since $((a,b), c) \mapsto (a, (b, c))$ is a map then I must show that it is one-one and onto? And I will be complete? $\endgroup$ – Tom Sep 12 '13 at 0:49
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    $\begingroup$ Yep, that's correct! And remember a key property of ordered pairs (or ordered 3 tuples in this case) is that $(x,y,z) = (x',y',z') \Leftrightarrow x = x', y =y', z = z'$ $\endgroup$ – Ryan Sullivant Sep 12 '13 at 0:52
  • $\begingroup$ @RyanSullivant thank you very much! $\endgroup$ – Tom Sep 12 '13 at 1:59

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