# Convex sets - How do we get (1−t)x+ty

1. Jan 9, 2014

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. Jan 9, 2014

### spamiam

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.

3. Jan 9, 2014

### economicsnerd

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 $x+tz = (1-t)x + ty$.

4. Jan 10, 2014