Results 1 to 8 of 8

Math Help - Some Cartesian product questions

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    67
    Awards
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    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}} ).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    67
    Awards
    1
    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!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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)?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2008
    Posts
    67
    Awards
    1
    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2008
    Posts
    67
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    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}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cartesian product of A*A*A*A
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 14th 2011, 07:29 AM
  2. Cartesian product
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: August 19th 2011, 12:38 PM
  3. Cartesian product.
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: November 4th 2010, 02:59 PM
  4. Cartesian product
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 9th 2010, 03:13 AM
  5. Cartesian product
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: August 27th 2009, 09:51 AM

Search Tags


/mathhelpforum @mathhelpforum