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Convex sets - How do we get (1−t)x+ty

  1. Jan 9, 2014 #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?
  2. jcsd
  3. Jan 9, 2014 #2
    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.
  4. Jan 9, 2014 #3
    Draw two vectors [itex]x,y[/itex] on a copy of [itex]\mathbb R^2[/itex]. Or on a piece of paper, if that's all you have available.

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

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

    This is the sense in which [itex]\{x+tz: t\in[0,1]\}[/itex] parametrizes the line segment between [itex]x[/itex] and [itex]y[/itex]. Finally, notice that [itex]x+tz = (1-t)x + ty[/itex].
  5. Jan 10, 2014 #4
    Thank you very much! I actually drew the lines on the paper and am trying to figure it out. Excellent explanation!
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