Shadow Posted March 5, 2012 Share Posted March 5, 2012 ...no three of them lie on one line. Say in 2D euclidean space. What would it look like? Link to comment Share on other sites More sharing options...

DrRocket Posted March 5, 2012 Share Posted March 5, 2012 ...no three of them lie on one line. Say in 2D euclidean space. What would it look like? The extreme points of any closed convex set would satisfy that criteria, so in particular the points on a circle would be one such set. Link to comment Share on other sites More sharing options...

uncool Posted March 6, 2012 Share Posted March 6, 2012 The extreme points of any closed convex set would satisfy that criteria, so in particular the points on a circle would be one such set. You need more than that; a square is closed and convex, but doesn't follow the criterion. You need strictly convex, but that barely gives any more information than was originally given. =Uncool- Link to comment Share on other sites More sharing options...

DrRocket Posted March 6, 2012 Share Posted March 6, 2012 You need more than that; a square is closed and convex, but doesn't follow the criterion. You need strictly convex, but that barely gives any more information than was originally given. =Uncool- The extreme points of a square are the corner points. No three of them are colinear. Link to comment Share on other sites More sharing options...

uncool Posted March 6, 2012 Share Posted March 6, 2012 The extreme points of a square are the corner points. No three of them are colinear. Ah. I apologize; recently I've been working a lot with convexity vs strict convexity vs uniform convexity and misread your post. =Uncool- Link to comment Share on other sites More sharing options...

imatfaal Posted March 6, 2012 Share Posted March 6, 2012 There are countless shapes which are not convex that will also have vertices no three of which are collinear (you can imagine forming by taking a convex shape joining all "diagonals" and then placing another internal vertex anywhere not on a diagonal. Are we limited to giving individual sets (or instructions to make sets) that fulfil the requirement - or is there a single set that encompasses all. It almost seems that the only definition of the whole set of sets of answers is the definition itself. If you are given two points - the set of points not collinear is everything else that doesn't lie on the line that connects the two points. But if you are given one or zero points then is there a way to define the set of answers other than by re-iterating the question. I am really grasping now (and it's clear I was above - just less so) - but to show all the points "at once" you would need far more that two axes on your graph. For every pair of points (4 inputs) there exists a group (on a plane) that are not collinear - does that mean you need 6 axes to show the answer in one fell swoop. Does the concept of a shape in 6d make any sense as an answer? Link to comment Share on other sites More sharing options...

DrRocket Posted March 6, 2012 Share Posted March 6, 2012 There are countless shapes which are not convex that will also have vertices no three of which are collinear (you can imagine forming by taking a convex shape joining all "diagonals" and then placing another internal vertex anywhere not on a diagonal. So what ? I made no claim that all sets that have the property that no three points are colinear are the extreme points of a convex set, only that the the extreme points of a convex set have that property. In fact, take a triangle. The vertices are the extreme points and no three are colinear. But you can also add to the set of extreme points the barycenter (or any interior point) and still have a set of points with no three being colinear. And that set is not the set of extreme points of a convex set. You can also take a shape with vertices which have that property, say a five-pointed star, and take the convex hull of those vertices, a regulat pentagon, and you are right back to the extreme points of a convex set. Are we limited to giving individual sets (or instructions to make sets) that fulfil the requirement - or is there a single set that encompasses all. It almost seems that the only definition of the whole set of sets of answers is the definition itself. If you are given two points - the set of points not collinear is everything else that doesn't lie on the line that connects the two points. But if you are given one or zero points then is there a way to define the set of answers other than by re-iterating the question. huh ? I am really grasping now (and it's clear I was above - just less so) - but to show all the points "at once" you would need far more that two axes on your graph. For every pair of points (4 inputs) there exists a group (on a plane) that are not collinear - does that mean you need 6 axes to show the answer in one fell swoop. Does the concept of a shape in 6d make any sense as an answer? The question was posed as a question in dimension 2, the plane. The notion of convexity extends to any vector space, even one of infinite dimension. However, higher dimensional examples do not appear to address the original question. I don't understand your statement regarding every set of 4 points. It is quite easy to find 4 colinear points -- start with a line and pick 4 points on it. If you intended to say that given any cardinal number between 3 and the cardinality of the real numbers, there is a subset of the plane having that cardinality having the property that no 3 points are colinear, then my example of a circle provides the necessary example as the circle or any subset of the circle with 3 or more elements has that property. Ah. I apologize; recently I've been working a lot with convexity vs strict convexity vs uniform convexity and misread your post. =Uncool- No problem. Link to comment Share on other sites More sharing options...

Shadow Posted March 7, 2012 Author Share Posted March 7, 2012 The extreme points of any closed convex set would satisfy that criteria, so in particular the points on a circle would be one such set. I thought as much, but I was thinking of a different type of set, although I'm not even sure if it'll make sense. Imagine this process. Two points are given. You chose a third (at random) that's not colinear with the two and repeat the process. In general, given n points, you chose the (n+1)st so that no three of them are colinear. If repeated indefinitely, what set will I get? Can the process even be repeated indefinitely? Link to comment Share on other sites More sharing options...

mathematic Posted March 7, 2012 Share Posted March 7, 2012 I thought as much, but I was thinking of a different type of set, although I'm not even sure if it'll make sense. Imagine this process. Two points are given. You chose a third (at random) that's not colinear with the two and repeat the process. In general, given n points, you chose the (n+1)st so that no three of them are colinear. If repeated indefinitely, what set will I get? Can the process even be repeated indefinitely? The process can continue indefinitely without any particular pattern and there is nothing to make it stop. Given n points which are not collinear, there are n(n-1)/2 lines determined by these points. There is plenty of empty space to put in an additional point. Link to comment Share on other sites More sharing options...

DrRocket Posted March 7, 2012 Share Posted March 7, 2012 (edited) I thought as much, but I was thinking of a different type of set, although I'm not even sure if it'll make sense. Imagine this process. Two points are given. You chose a third (at random) that's not colinear with the two and repeat the process. In general, given n points, you chose the (n+1)st so that no three of them are colinear. If repeated indefinitely, what set will I get? Can the process even be repeated indefinitely? I doubt that you will get any single well-defined set using that method. You could, for instance simply start selecting points from among the extreme points of a given convex set, say a circle. That could continue indefinitely and you could wind up, in the limit, with any contable subset of the circle. You could do exactly the same thing with a parabola. I have not thought about this in depth, but I think the following might also work. Take an equilateral triangle and consider the barycenters. Then take the barycentric subdivision and the consider the vertices. Continue. I think you will wind up with a dense subset of the interior with no 3 points being colinear. You would have to check this one rigorously as it may be wrong. Edited to correct the last (tentative) idea -- replaced vertices with barycenters. Edited March 7, 2012 by DrRocket Link to comment Share on other sites More sharing options...

Shadow Posted March 7, 2012 Author Share Posted March 7, 2012 (edited) The process can continue indefinitely without any particular pattern and there is nothing to make it stop. Given n points which are not collinear, there are n(n-1)/2 lines determined by these points. There is plenty of empty space to put in an additional point. Exactly. But the set can't ever contain all points in the plane. Thus, my question; what does it look like? EDIT: Ah, DrRocket replied while I was writing a reply. Thanks for the idea. EDIT2: That won't work; even after the first subdivision, every vertex will be colinear with two other vertices, for example the points {0}, {0, 1, 2}, {1, 2} in the following image: Edited March 7, 2012 by Shadow Link to comment Share on other sites More sharing options...

DrRocket Posted March 7, 2012 Share Posted March 7, 2012 Exactly. But the set can't ever contain all points in the plane. Thus, my question; what does it look like? EDIT: Ah, DrRocket replied while I was writing a reply. Thanks for the idea. EDIT2: That won't work; even after the first subdivision, every vertex will be colinear with two other vertices, for example the points {0}, {0, 1, 2}, {1, 2} in the following image: No, just add in the barycenters at each step, not the points that are clearly on line connecting two vertices. (Still may not work though). So in this case at the first step you only gain the point {0,1,2}. Link to comment Share on other sites More sharing options...

mathematic Posted March 8, 2012 Share Posted March 8, 2012 Exactly. But the set can't ever contain all points in the plane. Thus, my question; what does it look like? EDIT: Ah, DrRocket replied while I was writing a reply. Thanks for the idea. EDIT2: That won't work; even after the first subdivision, every vertex will be colinear with two other vertices, for example the points {0}, {0, 1, 2}, {1, 2} in the following image: The set won't look like anything in particular - just a random set of points in the plane. Link to comment Share on other sites More sharing options...

Shadow Posted March 9, 2012 Author Share Posted March 9, 2012 (edited) The set won't look like anything in particular - just a random set of points in the plane. Well, that can't be entirely true. As you said, there will always be plenty of space to put a new point, yet all the points in [math]\mathbb{R}^2[/math] can't possibly lie in [math]S[/math], to use the notation I used earlier. This in itself is enough for me to wonder what it would look like. It might just look like a black square, ie. it may appear that all the points in [math]\mathbb{R}^2[/math] are elements of [math]S[/math]. It might not. And DrRockets idea, if correct, would certainly yield an interesting fractal. I haven't gotten around to trying it out yet, I'll post here when I do. But thank you, and everyone else, for your inputs, they're much appreciated. Edited March 9, 2012 by Shadow Link to comment Share on other sites More sharing options...

mathematic Posted March 10, 2012 Share Posted March 10, 2012 Well, that can't be entirely true. As you said, there will always be plenty of space to put a new point, yet all the points in [math]\mathbb{R}^2[/math] can't possibly lie in [math]S[/math], to use the notation I used earlier. This in itself is enough for me to wonder what it would look like. It might just look like a black square, ie. it may appear that all the points in [math]\mathbb{R}^2[/math] are elements of [math]S[/math]. It might not. And DrRockets idea, if correct, would certainly yield an interesting fractal. I haven't gotten around to trying it out yet, I'll post here when I do. But thank you, and everyone else, for your inputs, they're much appreciated. What I said is entirely true. I never claimed that every point would be in the set - far from it. As long as the set is being created sequentially, there will be only a finite number of points. Looking like a black square seems impossible since you would then have many points lying on a given line. Even the idea of fractal doesn't make too much sense, since the points (unless you describe a different approach to creating the set) form a discrete set in the plane. Link to comment Share on other sites More sharing options...

Shadow Posted March 11, 2012 Author Share Posted March 11, 2012 I know you didn't. I'm just saying that's the part that has me interested in this topic; I don't know what the set will look like, if you keep on iterating. Or, to make use of a limit metaphor, what the set will "approach" when the number of iterations goes to infinity. The fractal part was aimed solely at what DrRocket suggested. Link to comment Share on other sites More sharing options...

John Posted March 25, 2012 Share Posted March 25, 2012 I would imagine the graph would look completely full. Given the line between two points, no matter how close a third point is to that line, a fourth point can be placed even closer without being on the line. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now