Thread: Dense subset of R^n

Show that has a dense subset with no three collinear points, for .

Originally Posted by Bruno J.

Show that has a dense subset with no three collinear points, for .

Dense in the sense that the set's closure is the full space?

What else ?

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.

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 , which is not here.

Okay Bruno J., I'll bite to Drexel's wondering : what do you call a dense set ?

Originally Posted by Bruno J.

Show that has a dense subset with no three collinear points, for .

Easy solution by inductive construction!

Spoiler:

Start with a dense countable sequence (such as an ordering of all the points whose coordinates are all rational). Now construct by induction a sequence as follows. Let . Suppose that have been chosen so that no three of them are collinear, and for each j. There are only finitely many lines containing a pair of these points (there are lines, to be precise). Let be any point in the open ball of radius 1/k centred at that is not on any of these lines. Then obviously satisfies the conditions for the inductive construction, and it is almost equally obvious that the set is dense in .

Originally Posted by Opalg

Easy solution by inductive construction!

Spoiler:

Start with a dense countable sequence (such as an ordering of all the points whose coordinates are all rational). Now construct by induction a sequence as follows. Let . Suppose that have been chosen so that no three of them are collinear, and for each j. There are only finitely many lines containing a pair of these points (there are lines, to be precise). Let be any point in the open ball of radius 1/k centred at that is not on any of these lines. Then obviously satisfies the conditions for the inductive construction, and it is almost equally obvious that the set is dense in .

Would your idea work if wasn't separable?

Good job Oplag!

And Moo : of course has a natural topology. For example it can be defined recursively as the product of the topologies of and .

A set is dense in if every nonempty open set of contains a point of . Many alternative definitions are possible and they are all equivalent.

Originally Posted by Bruno J.

Show that has a dense subset with no three collinear points, for .

Here is another less constructive way...

Spoiler:

Let be a family of independent -dimensional standard Gaussian random variables (or any other continuous distribution supported by ). Almost surely, this family answers your question.

Density: comes as a direct consequence of Borel-Cantelli lemma (for any ball , hence , and these events are independent, so that infinitely many terms of the sequence fall in almost-surely)

Non-colinearity: for any distinct , because the line has zero Lebesgue measure (since ), and has a density, and is independent of . There are countably many triplets , 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).

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