5

This is probably going to turn out to be an embarrassment for me (took me an hour to get the courage to post this), but I can't figure this out.

Using the definition of cartesian product from wikipedia, we have $$\mathbb R^3=\big\{(x_1,x_2,x_3); x_1,x_2,x_3\in \mathbb R \big\}$$ $$\mathbb R^2=\big\{(x_1,x_2); x_1,x_2\in \mathbb R \big\}$$ $$\mathbb R^2 \times \mathbb R^3 = \big\{\big((x_1,x_2),(x_3,x_4,x_5)\big); (x_1,x_2) \in \mathbb R^2, (x_3,x_4,x_5) \in R^3 \big\} $$

Which does not seem to be equal to $\mathbb R^5$, which has $(x_1,x_2,x_3,x_4,x_5)$ as elements instead of ordered pairs of 2-tuples and 3-tuples.

Please make the universe functional again, thanks!

Zev Chonoles
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Dahn
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    This has definitely been discussed at least twice before. – Asaf Karagila Dec 25 '15 at 20:14
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    Which is another reason why this will be an embarrassement. The problem is, how do you search for things like this? I tried a couple of searches but couldn't find anything. Perhaps improving my searching skills rather than set theoretic knowledge is what I am after :) – Dahn Dec 25 '15 at 20:15
  • http://math.stackexchange.com/questions/338319/associativity-of-cartesian-product should include an answer. – Asaf Karagila Dec 25 '15 at 20:16
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    @DahnJahn: Part of what helps the effectiveness of searches in general is using titles that are not comprised solely of MathJax. – Zev Chonoles Dec 25 '15 at 20:17
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    Also http://math.stackexchange.com/questions/1313958/some-questions-about-the-cartesian-product (and I'm sure others too, but too lazy to find them after finding at least two occasions! :-)) – Asaf Karagila Dec 25 '15 at 20:18
  • @AsafKaragila Ah, I didn't think of seeing it as an problem with the associative property! Thanks – Dahn Dec 25 '15 at 20:18
  • @ZevChonoles True, I will think about that from now on. Any suggestions for a better title? – Dahn Dec 25 '15 at 20:19
  • There is a natural bijection \begin{align}\mathbf R^2\times\mathbf R^3&\longrightarrow\mathbf R^5\ \bigl((x_1,x_2),(x_3,x_4,x_5)\bigr)&\longmapsto(x_1,x_2,x_3, x_4,x_5) \end{align} which is an isomorphism. – Bernard Dec 25 '15 at 20:21

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