# Thread: Some Cartesian product questions

1. ## Some Cartesian product questions

Hello guys,

A question goes as follows: "Let $S$ be the set of all sequences $\{a_n\}_n$ satisfying $|a_1|<1, |a_3|<2, |a_4|<1$. Express $S$ as a cartesian product of subsets of the real numbers" I know in this instance that $a_1 \in [-1,1], a_3 \in [-2,2], a_4 \in [-1,1]$. But I'm confused as to how to put this into a cartesian product. I mean, is this correct: $[-1,1] \times \mathbb{R} \times [-2,2] \times [-1,1] \times \mathbb{R} \times \cdots \times \mathbb{R}$? That's probably very wrong.

Thanks a lot in advance.

HTale.

2. Your idea is correct ( just a little thing, if inequalities are strict, then it will be $\{(a_n)_{n\geq 1}\in\mathbb{R}^{\mathbb{N}};\ |a_1|<1,|a_3|<2,|a_4|<1\}=$ $]-1,1[\times\mathbb{R}\times ]-2,2[\times ]-1,1[\times\mathbb{R}^{\mathbb{N}}$ ).

3. Originally Posted by clic-clac
Your idea is correct ( just a little thing, if inequalities are strict, then it will be $\{(a_n)_{n\geq 1}\in\mathbb{R}^{\mathbb{N}};\ |a_1|<1,|a_3|<2,|a_4|<1\}=$ $]-1,1[\times\mathbb{R}\times ]-2,2[\times ]-1,1[\times\mathbb{R}^{\mathbb{N}}$ ).
Thanks for pointing out the silly mistake!

4. Originally Posted by clic-clac
Your idea is correct ( just a little thing, if inequalities are strict, then it will be $\{(a_n)_{n\geq 1}\in\mathbb{R}^{\mathbb{N}};\ |a_1|<1,|a_3|<2,|a_4|<1\}=$ $]-1,1[\times\mathbb{R}\times ]-2,2[\times ]-1,1[\times\mathbb{R}^{\mathbb{N}}$ ).
What bothers me is that if $f\in S$ then $f = \{ (n,a_n) : n\geq 1,a_n\in \mathbb{R}, |a_1|<1,|a_3|<2,|a_4|<1\}$. Thus, $f$ is a set of ordered pairs. But if $S = ]-1,1[\times\mathbb{R}\times ]-2,2[\times ]-1,1[\times\mathbb{R}^{\mathbb{N}}$ then it means $f = (a_1,a_2,a_3,a_4,g)$ where $a_j \in \mathbb{R}, |a_1|<1,|a_3|<2,|a_4|<1$ and $g$ is a function $\mathbb{N}\to \mathbb{R}$. How can you have $f = (a_1,a_2,a_3,a_4,g)$?

5. Originally Posted by ThePerfectHacker
What bothers me is that if $f\in S$ then $f = \{ (n,a_n) : n\geq 1,a_n\in \mathbb{R}, |a_1|<1,|a_3|<2,|a_4|<1\}$. Thus, $f$ is a set of ordered pairs. But if $S = ]-1,1[\times\mathbb{R}\times ]-2,2[\times ]-1,1[\times\mathbb{R}^{\mathbb{N}}$ then it means $f = (a_1,a_2,a_3,a_4,g)$ where $a_j \in \mathbb{R}, |a_1|<1,|a_3|<2,|a_4|<1$ and $g$ is a function $\mathbb{N}\to \mathbb{R}$. How can you have $f = (a_1,a_2,a_3,a_4,g)$?
My understanding is that each sequence can be treated as an n-tuple. I'm not entirely sure of your argument, however. Why do you see it as a problem?

6. Originally Posted by HTale
My understanding is that each sequence can be treated as an n-tuple. I'm not entirely sure of your argument, however. Why do you see it as a problem?
Ignore what I said. It will only confuse you.
If you understand what was done above then good.
I understood that you were thinking of sequences as n-tuples.
But I was thinking of sequences as a set of ordered pairs.
And I was wondering if the two approaches are consistent with eachother.
But again ignore what I was wondering.

7. Originally Posted by ThePerfectHacker
Ignore what I said. It will only confuse you.
If you understand what was done above then good.
I understood that you were thinking of sequences as n-tuples.
But I was thinking of sequences as a set of ordered pairs.
And I was wondering if the two approaches are consistent with eachother.
But again ignore what I was wondering.
Having re-read it, I do see what you're saying now. I think you could define a further (bijective) function on $f$, from $A_1 \times A_2 \times A_3 \times \cdots \to A_1 \times (A_2 \times A_3 \times \cdots)$. Maybe, I don't know.

8. We can consider $S$ as a subset of $\mathcal{P}(\mathbb{N}\times\mathbb{R}),$ no problem with that.

Of course, $card(\mathcal{P}(\mathbb{N}\times\mathbb{R}))>card (\mathbb{R}^{\mathbb{N}})$, but we can imagine a bijection between $S\subset\mathcal{P}(\mathbb{N}\times\mathbb{R})$ and $S^{'}=]-1,1[\times\mathbb{R}\times ]-2,2[\times ]-1,1[\times\mathbb{R}^{\mathbb{N}}\subset\mathbb{R}^{\m athbb{N}}:$

$S\rightarrow S^{'}:\ \{(n,a_n),n\geq 1\}\mapsto(a_k)_{k\geq 1}$