# 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 $$\mathbb{R}$$.

• They're not equal; there's just a bijection between them for finite products. – Malice Vidrine Nov 3 '18 at 16:45
• Thank your for your answer. In my example, if $I = \{0, 1, 2\}$, can you give me a bijection between $\mathcal{F}(I, \mathbb{R})$ and $\mathbb{R}^{3}$ ? – deeppinkwater Nov 3 '18 at 16:59

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:

1. 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.

2. 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}

3. 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 $$0$$".

• Thank you for your help ! – deeppinkwater Nov 3 '18 at 18:38
• @deeppinkwater It's a pleasure :) – Scientifica Nov 3 '18 at 18:38

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 $$\{(1,a),(2,b),(3,c)\}$$.

• Thank you for your answer ! – deeppinkwater Nov 3 '18 at 18:39