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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.

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  • $\begingroup$ They're not equal; there's just a bijection between them for finite products. $\endgroup$ – Malice Vidrine Nov 3 '18 at 16:45
  • $\begingroup$ 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 $\mathbb{R}^{3}$ ? $\endgroup$ – deeppinkwater Nov 3 '18 at 16:59
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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 $0$".

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  • $\begingroup$ Thank you for your help ! $\endgroup$ – deeppinkwater Nov 3 '18 at 18:38
  • $\begingroup$ @deeppinkwater It's a pleasure :) $\endgroup$ – Scientifica Nov 3 '18 at 18:38
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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 $\{(1,a),(2,b),(3,c)\}$.

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

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