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Thread: Dense subset of R^n

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

    Show that $\displaystyle \mathbb{R}^n$ has a dense subset with no three collinear points, for $\displaystyle n>1$.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Show that $\displaystyle \mathbb{R}^n$ has a dense subset with no three collinear points, for $\displaystyle n>1$.
    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 $\displaystyle \mathbb{R}^n$, 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 $\displaystyle \mathbb{R}^n$ has a dense subset with no three collinear points, for $\displaystyle n>1$.
    Easy solution by inductive construction!

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

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

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

    Spoiler:

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

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

    Non-colinearity: for any distinct $\displaystyle i,j,k$, $\displaystyle P(X_k\in (X_i X_j))=0$ because the line $\displaystyle (X_i X_j)$ has zero Lebesgue measure (since $\displaystyle n\geq 2$), and $\displaystyle X_k$ has a density, and is independent of $\displaystyle X_i,X_j$. There are countably many triplets $\displaystyle i,j,k$, 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; Jan 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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