# Equivalence of Definitions of Extreme Point of Convex Set

## Theorem

The following definitions of the concept of Extreme Point of Convex Set are equivalent:

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

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

### Definition 1

We say that $a$ is an extreme point of $K$ if and only if:

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

### Definition 2

We say that $a$ is an extreme point of $K$ if and only if:

$K \setminus \set a$ is convex.

## Proof

### 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.

$\Box$

### 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:

$x = y = a$

$\blacksquare$