# Thread: Dense subset of R^n

1. ## Dense subset of R^n

Show that $\mathbb{R}^n$ has a dense subset with no three collinear points, for $n>1$.

2. Originally Posted by Bruno J.
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?

3. What else ?

4. Originally Posted by Moo
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.

5. 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 ?

6. Originally Posted by Bruno J.
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 $\mathbb{R}^n$.

7. Originally Posted by Opalg
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?

8. 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.

9. Originally Posted by Bruno J.
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).

10. That's a tad too technical for me just yet! But I appreciate that you could solve it in another fashion.