MHB Cartesian Product of Non-Real Sets

[Dethrone]

Suppose we have the sets $A=\left\{2,3\right\}$ and $B=\left\{5\right\}$, then $A$ X $B$ is defined as $\left\{(x,y)|x \in A, y\in B\right\}=\left\{(2,5), (3,5)\right\}$. But what happens when $A$ contains elements that are not in $\Bbb{R}$?

Example:
$A=\left\{(2,3),(3,4)\right\}\subset \Bbb{R}^2$ and $B=\left\{(3,2,5)\right\}\subset \Bbb{R}^3$, then following the same definition as above, we have $A$ X $B=\left\{((2,3),(3,2,5)), ((3,4),(3,2,5))\right\}$, but my book tells me that $A$ X $B$ should have elements in $\Bbb{R}^5$. Did I make a mistake?

 

[Euge]

Hi Rido12,

You haven't made a mistake. Your elements belong to $\Bbb R^5$: $((2,3), (3,2,5)) = (2,3,3,2,5)$ and $((3,4),(3,2,5)) = (3,4,3,2,5)$.

 

[I like Serena]

Hey Rido! (Happy)

Strictly speaking they are not the same, but they are isomorphic (literally meaning same shape with symbol ≅).

Note that:
$$\mathbb R^2 \times \mathbb R^3 = (\mathbb R \times \mathbb R) \times (\mathbb R \times \mathbb R \times \mathbb R)
≅ \mathbb R \times \mathbb R \times \mathbb R \times \mathbb R \times \mathbb R
= \mathbb R^5
$$
Anyway, it boils down to the same thing and the distinction is often not made.

 

[Euge]

Hey Rido! (Happy)

Strictly speaking they are not the same, but they are isomorphic (literally meaning same shape with symbol ≅).

Note that:
$$\mathbb R^2 \times \mathbb R^3 = (\mathbb R \times \mathbb R) \times (\mathbb R \times \mathbb R \times \mathbb R)
≅ \mathbb R \times \mathbb R \times \mathbb R \times \mathbb R \times \mathbb R
= \mathbb R^5
$$
Anyway, it boils down to the same thing and the distinction is often not made.

Thanks for making this note. I assumed that in the book, a tuple made of up an $m$-tuple and an $n$-tuple is an $(m + n)$-tuple. However, that needs to be made explicitly clear in the text, for otherwise we consider them as different objects.

 

[Dethrone]

Hey ILS and Euge,

Thanks for clarifying my confusion! That is interesting to know. :D