Convex sets - How do we get (1−t)x+ty

  • Thread starter Thread starter bhagwad
  • Start date Start date
  • Tags Tags
    Convex Sets
Click For Summary
SUMMARY

The discussion centers on the concept of convex sets as defined in Rudin's text, specifically definition 2.17, which states that a set E is convex if for any two points x and y in E, the point (1−t)x + ty belongs to E for 0 < t < 1. Participants explore the intuitive understanding of this definition by visualizing the line segment between points x and y in \(\mathbb{R}^2\). The parametrization of the line segment is expressed as {x + tz: t ∈ [0, 1]}, where z = y - x, providing a clear geometric interpretation of convexity.

PREREQUISITES
  • Understanding of convex sets and their properties
  • Familiarity with vector notation and operations
  • Basic knowledge of parametrization in geometry
  • Ability to visualize concepts in \(\mathbb{R}^2\)
NEXT STEPS
  • Study the properties of convex sets in higher dimensions
  • Learn about vector operations and their geometric interpretations
  • Explore the implications of convexity in optimization problems
  • Investigate the relationship between convex sets and linear combinations
USEFUL FOR

Mathematicians, students of linear algebra, and anyone interested in geometric interpretations of mathematical concepts, particularly in the context of convex analysis.

bhagwad
Messages
28
Reaction score
1
In definition 2.17 of Rudin's text, he says that a set E is convex if for any two points x and y belonging to E, (1−t)x+ty belongs to E when 0<t<1.

I learned that this means the point is between x and y. But I'm not able to see this intuitively. Can anyone help me "see" this?
 
Physics news on Phys.org
How would you parametrize the line segment between the points ##x## and ##y##? If you know how to do that, I think you'll see the answer.
 
Draw two vectors x,y on a copy of \mathbb R^2. Or on a piece of paper, if that's all you have available.

Draw an arrow from x to y on your paper. Its tail should be at x and its head at y. We can think of the arrow as the vector z:= y-x. Indeed, if we moved the arrow so that its tail is at the origin, the head would lie at z.

-Imagine an ant is sitting at x and you want it to travel to y in a straight line. What should it do? It should travel all the way along z. This would bring it to x + z, also known as y.
-Now, imagine the ant is sitting at x again, and you want it to travel \frac13 of the way to y in a straight line. What should it do now? It should travel \frac13 of the way along z. This would bring it to x + \frac13z.
-And what if you want it to travel (again, from x) \frac57 of the way to y in a straight line? Now, it should go to x + \frac57z

This is the sense in which \{x+tz: t\in[0,1]\} parametrizes the line segment between x and y. Finally, notice that x+tz = (1-t)x + ty.
 
  • Like
Likes 1 person
economicsnerd said:
Draw two vectors x,y on a copy of \mathbb R^2. Or on a piece of paper, if that's all you have available.

Draw an arrow from x to y on your paper. Its tail should be at x and its head at y. We can think of the arrow as the vector z:= y-x. Indeed, if we moved the arrow so that its tail is at the origin, the head would lie at z.

-Imagine an ant is sitting at x and you want it to travel to y in a straight line. What should it do? It should travel all the way along z. This would bring it to x + z, also known as y.
-Now, imagine the ant is sitting at x again, and you want it to travel \frac13 of the way to y in a straight line. What should it do now? It should travel \frac13 of the way along z. This would bring it to x + \frac13z.
-And what if you want it to travel (again, from x) \frac57 of the way to y in a straight line? Now, it should go to x + \frac57z

This is the sense in which \{x+tz: t\in[0,1]\} parametrizes the line segment between x and y. Finally, notice that x+tz = (1-t)x + ty.

Thank you very much! I actually drew the lines on the paper and am trying to figure it out. Excellent explanation!
 
As shown by this animation, the fibers of the Hopf fibration of the 3-sphere are circles (click on a point on the sphere to visualize the associated fiber). As far as I understand, they never intersect and their union is the 3-sphere itself. I'd be sure whether the circles in the animation are given by stereographic projection of the 3-sphere from a point, say the "equivalent" of the ##S^2## north-pole. Assuming the viewpoint of 3-sphere defined by its embedding in ##\mathbb C^2## as...

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K