Cartesian product of a family of sets I have a question about the Cartesian product of a family of sets.
Let $X$ be a set and let $(S_{i})_{i \in I}$ be a family of sets indexed by an arbitrary index set $I$ such that $\forall i \in I, S_{i} \in \mathcal{P}(X)$.
The Cartesian product of $(S_{i})_{i \in I}$ is the set :
$$\prod_{i \in I} S_{i} = \{f \in \mathcal{F}(I, X) \ | \ [f : I \rightarrow \bigcup_{i \in I} S_{i}] \ \wedge \ [\forall i \in I, \ f(i) \in S_{i}]\} \text{.}$$
First, are we agree that, if we take an arbitrary set $S \in \mathcal{P}(X)$, we have :
$$\prod_{i \in I} S = \{f \in \mathcal{F}(I, X) \ | \ [f : I \rightarrow \bigcup_{i \in I} S] \ \wedge \ [\forall i \in I, \ f(i) \in S]\}$$
$$= \{f \in \mathcal{F}(I, X) \ | \ [f : I \rightarrow S] \ \wedge \ [\forall i \in I, \ f(i) \in S]\} = \mathcal{F}(I, S) \ \text{?}$$
If it's okay, my problem is the following. We have :
$$\mathbb{R}^{3} = \mathbb{R} \times \mathbb{R} \times \mathbb{R} = \{(x, y, z) \ | \ [x \in \mathbb{R}] \ \wedge \ [y \in \mathbb{R}] \ \wedge \ [z \in \mathbb{R}]\}$$
and, if we take $I \subset \mathbb{N}$ such that $|I| = 3$ :
$$\mathbb{R}^{3} = \prod_{i \in I} \mathbb{R} = \mathcal{F}(I, \mathbb{R}) \ \text{,}$$
but I cannot see clearly why do we have the equality between these sets.
I don't see why the set of $3$-uples of elements of $\mathbb{R}$ is equal to the set of mappings from $I$ (with $|I| = 3$) to $\mathbb{R}$.
Thank you for your help.
 A: You're right, and by the way $\mathcal{F}(I,X)$ is denoted by $X^I$ (you see that $\mathcal F(I,X)=\prod_{i\in I}X$ so you can "see" it as "$X$ times itself $I$ times).
Let us do the case $|I|=2$ it's simpler. As you said, formally speaking, $\mathbb{R}\times\mathbb{R}\neq\mathbb{R}^2$. Indeed, in set theory, $2$ is typically taken as $2=\{0,1\}$, where $0=\emptyset$ and $1=\{0\}$. So $\mathbb{R}^2=\mathcal{F}(2,\mathbb R)$. An element $f$ of $\mathbb R^2$ is a function from $2$ to $\mathbb{R}$. Let $x,y\in\mathbb R$ such that $f(0)=x$ and $f(1)=y$. Then, formally speaking, $f=\{(0,x),(1,y)\}$. Hence $\mathbb{R}^2=\{\{(0,x),(1,y)\}\mid x,y\in\mathbb R\}$.
On the other hand, $\mathbb{R}\times \mathbb R=\{(x,y)\mid x,y\in\mathbb R\}$. You see that $\mathbb R^2$ and $\mathbb R\times\mathbb R$ are different.
So, why do we take $\mathbb R^2=\mathbb R\times\mathbb R$? For two reasons:


*

*You can see it as a mere notational convinience, not to be confused with the $X^I$ for sets. See it like when you take powers of numbers.

*There's a nice bijection between the two sets:
\begin{align}
\psi:\mathbb R\times\mathbb R&\to \mathbb R^2\\
(x,y)&\mapsto\{(0,x),(1,y)\}
\end{align}

*The bijection above tells you that you can "see" $(x,y)$ as a function that maps $0$ to the first entry of $(x,y)$, and $1$ to the second entry of $(x,y)$. You may find mapping $0$ to first and $1$ to second is weird, but that's common use in computer science, and typically $\mathbb N=\omega=\{0,1,2,\dots\}$, so "counting starts from $0$".
A: They're not equal, but there is a bijection between them in the finite case.
If $I=\{1,2,3\}$ and we're taking the product of $\mathbb{R}$, the easy bijection is the function $\mathcal{F}(I,\mathbb{R})\to \mathbb{R}^3$ that takes $f$ to $(f(1),f(2),f(3))$. For the inverse, take $(a,b,c)\in\mathbb{R}^3$ to the function $\{(1,a),(2,b),(3,c)\}$.
