Can someone help me prove that $$X \times (Y \times Z) \sim (X \times Y) \times Z$$ I know that there is supposed to be a bijection between these two. The first one will contain elements like $(x, (y, z))$ and the second one $((x, y) z)$. I just need some instructions, I believe I can manage on my own from that.

  • 3
    $\begingroup$ The way you've written the elements of each set should be really suggestive of what the bijection could be. $\endgroup$ – user296602 Jan 9 '16 at 17:10
  • $\begingroup$ If you only need a bijection between sets (you didn't write what $X,Y, Z$ are), why not $(x,(y,z))\mapsto ((x,y),z)$? $\endgroup$ – user302982 Jan 9 '16 at 17:12
  • $\begingroup$ yeah, but how am i going to proove that that's in fact a bijection? $\endgroup$ – Stefan Jan 9 '16 at 17:13
  • $\begingroup$ Well, you need to prove it's injective and surjective. Here's an element of (X x Y) x Z: ((a, b), c). Can you write down something that is mapped to this element? $\endgroup$ – user296602 Jan 9 '16 at 17:19
  • $\begingroup$ @DietrichBurde A stray * from **bold formatting** $\endgroup$ – user147263 Jan 9 '16 at 17:54

$$X \times (Y \times Z) = \{(x,(y,z)) \mid x \in X, y \in Y, z \in Z\}$$ $$ (X \times Y) \times Z = \{ ((x,y),z) \mid x \in X, y \in Y, z \in Z \}$$

So consider the maps $(x,(y,z)) \mapsto ((x,y),z)$ and $((x,y),z) \mapsto (x,(y,z))$. One immediately verifies that these are indeed functions and they are clearly inverses, thus bijections.

  • $\begingroup$ Okay guys, thanks a lot $\endgroup$ – Stefan Jan 10 '16 at 11:57

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