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Math Help - Some Cartesian product questions

  1. #1
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    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.
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  2. #2
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    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}} ).
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  3. #3
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    Quote Originally Posted by clic-clac View Post
    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!
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  4. #4
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    Quote Originally Posted by clic-clac View Post
    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)?
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    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?
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  6. #6
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    Quote Originally Posted by HTale View Post
    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.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    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.
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  8. #8
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    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}
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