Equivalence of Definitions of Extreme Point of Convex Set

Theorem
Let $X$ be a vector space over $\R$.

Let $K$ be a convex subset of $X$.

Definition 1 implies Definition 2
Suppose that:


 * whenever $a = t x + \paren {1 - t} y$ for $t \in \openint 0 1$, we have $x = y = a$.

Then, if $x, y \in K \setminus \set a$ and $t \in \openint 0 1$, we have:


 * $t x + \paren {1 - t} y \ne a$

If $t = 0$, then we have:


 * $t x + \paren {1 - t} y = y \ne a$

and if $t = 1$ we have:


 * $t x + \paren {1 - t} y = x \ne a$

So if $x, y \in K \setminus \set a$ and $t \in \closedint 0 1$ we have:


 * $t x + \paren {1 - t} y \ne a$

Since $K$ is convex, we have:


 * $t x + \paren {1 - t} y \in K$

so we have:


 * $t x + \paren {1 - t} y \in K \setminus \set a$

so:


 * $K \setminus \set a$ is convex.

Definition 2 implies Definition 1
Suppose:


 * $K \setminus \set a$ is convex.

Then for all $x, y \in K \setminus \set a$ and $t \in \closedint 0 1$ we have:


 * $t x + \paren {1 - t} y \in K \setminus \set a$

In particular, if $t \in \openint 0 1$ and $x, y \in K \setminus \set a$, we have:


 * $t x + \paren {1 - t} y \ne a$

We also have that:


 * $t a + \paren {1 - t} a = a$

Hence, if $x, y \in K$ and $t \in \openint 0 1$ have:


 * $t x + \paren {1 - t} y = a$

then:


 * $x = y = a$