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Math Help - Dense subset of R^n

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Dense subset of R^n

    Show that \mathbb{R}^n has a dense subset with no three collinear points, for n>1.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Show that \mathbb{R}^n has a dense subset with no three collinear points, for n>1.
    Dense in the sense that the set's closure is the full space?
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    Moo
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    What else ?

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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Moo View Post
    What else ?

    Did you just aim a comment at me with a picture of George Clooney, haha?

    Regardless, I've never seen Bruno post anything to do with topology before, so I wasn't sure if he was using the term for something else.
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    Moo
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    Because of that : http://www.swica.ca/images/nespresso...%20Clooney.jpg

    There's a beginning to everything, like you doing algebraic topology
    Also, there has to be a topology defined on \mathbb{R}^n, which is not here.

    Okay Bruno J., I'll bite to Drexel's wondering : what do you call a dense set ?
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    Quote Originally Posted by Bruno J. View Post
    Show that \mathbb{R}^n has a dense subset with no three collinear points, for n>1.
    Easy solution by inductive construction!

    Spoiler:
    Start with a dense countable sequence \bigl(a_k\bigr)_{k=1}^\infty\subset\mathbb{R}^n (such as an ordering of all the points whose coordinates are all rational). Now construct by induction a sequence (x_k) as follows. Let x_1=a_1,\ x_2=a_2. Suppose that x_j\ (1\leqslant j\leqslant k-1) have been chosen so that no three of them are collinear, and \|x_j-a_j\|<1/j for each j. There are only finitely many lines containing a pair of these points x_j (there are \textstyle {k-1\choose2} lines, to be precise). Let x_k be any point in the open ball of radius 1/k centred at a_k that is not on any of these lines. Then x_k obviously satisfies the conditions for the inductive construction, and it is almost equally obvious that the set \{x_k:1\leqslant k<\infty\} is dense in \mathbb{R}^n.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Opalg View Post
    Easy solution by inductive construction!

    Spoiler:
    Start with a dense countable sequence \bigl(a_k\bigr)_{k=1}^\infty\subset\mathbb{R}^n (such as an ordering of all the points whose coordinates are all rational). Now construct by induction a sequence (x_k) as follows. Let x_1=a_1,\ x_2=a_2. Suppose that x_j\ (1\leqslant j\leqslant k-1) have been chosen so that no three of them are collinear, and \|x_j-a_j\|<1/j for each j. There are only finitely many lines containing a pair of these points x_j (there are \textstyle {k-1\choose2} lines, to be precise). Let x_k be any point in the open ball of radius 1/k centred at a_k that is not on any of these lines. Then x_k obviously satisfies the conditions for the inductive construction, and it is almost equally obvious that the set \{x_k:1\leqslant k<\infty\} is dense in \mathbb{R}^n.
    Would your idea work if \mathbb{R}^n wasn't separable?
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    MHF Contributor Bruno J.'s Avatar
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    Good job Oplag!

    And Moo : of course \mathbb{R}^n has a natural topology. For example it can be defined recursively as the product of the topologies of \mathbb{R}^{n-1} and \mathbb{R}.

    A set S is dense in \mathbb{R}^n if every nonempty open set of \mathbb{R}^n contains a point of S. Many alternative definitions are possible and they are all equivalent.
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    Quote Originally Posted by Bruno J. View Post
    Show that \mathbb{R}^n has a dense subset with no three collinear points, for n>1.
    Here is another less constructive way...

    Spoiler:

    Let (X_i)_{i\geq 0} be a family of independent n-dimensional standard Gaussian random variables (or any other continuous distribution supported by \mathbb{R}^n). Almost surely, this family answers your question.

    Density: comes as a direct consequence of Borel-Cantelli lemma (for any ball B, \forall i,\ P(X_i\in B)=p>0 hence \sum_i P(X_i\in B)=\infty, and these events are independent, so that infinitely many terms of the sequence fall in B almost-surely)

    Non-colinearity: for any distinct i,j,k, P(X_k\in (X_i X_j))=0 because the line (X_i X_j) has zero Lebesgue measure (since n\geq 2), and X_k has a density, and is independent of X_i,X_j. There are countably many triplets i,j,k, hence the probability that there exists one satisfying the alignment property is 0. (A union of countably many sets of zero measure has zero measure).
    Last edited by Laurent; January 27th 2010 at 01:56 PM.
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  10. #10
    MHF Contributor Bruno J.'s Avatar
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    That's a tad too technical for me just yet! But I appreciate that you could solve it in another fashion.
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