Why a closed bounded convex set in $\mathbb{R}^{n}$ always has an extreme point?

Let $X\subseteq \mathbb{R}^{n}$ is closed, bounded convex set. How to prove that $X$ contains such point $x$ that we can't represent as $x=\frac{1}{2}x_{1}+\frac{1}{2}x_{2}$ where $x_1\in X$ and $x_2\in X$ and $x_1 \neq x_2$?

For example closed square in $\mathbb{R}^{2}$ has $4$ such points.

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I don't quite understand your question. It seems to me that $X$ has to be either a one-point set or an empty set. – Tunococ Jan 19 '13 at 13:36
@Tunococ what about square? – Ashot Jan 19 '13 at 13:44
@Ashot The square is not finite. A finite convex set in $\mathbb{R}^n$ can contain one point at most. – Ayman Hourieh Jan 19 '13 at 13:46
@AymanHourieh I edited question – Ashot Jan 19 '13 at 13:47
Pages 37 and 38 (Proposition 3.1.1 and relevant definitions) here may help. – David Mitra Jan 19 '13 at 13:53

Let $R$ be the diameter of $X$. Choose any $x, y \in X$ such that $\lVert x - y\rVert = R$; prove by contradiction that $x$ is the desired point.

Basically, suppose that $x = \frac{1}{2}x_1 + \frac{1}{2}x_2$ for some distinct $x_1, x_2 \in X$. Note that either $\lVert x_1 - y\rVert$ or $\lVert x_2 - y\rVert$ must exceed $R$.

Detailed Proof:

Suppose that $x = \frac{1}{2}x_1 + \frac{1}{2}x_2$ for some distinct $x_1, x_2 \in X$. Let $x_1 = x + v$ and $x_2 = x - v$. Since $x_1 \neq x_2$, $v \neq 0$. \begin{align} \lVert x_2 - y \rVert &\leq R\\ \implies \lVert x_2 - y \rVert^2 &= ((x - y) - v) \cdot ((x - y) - v)\\ &= R^2 + \lVert v \rVert^2 - 2v \cdot (x - y)\\ &\leq R^2 \end{align}

Hence, $2v \cdot (x - y) \geq \lVert v \rVert^2 \geq 0$ and thus $R^2 + \lVert v \rVert^2 + 2v \cdot (x - y) \geq R^2$ ($\ast$).

Similarly, \begin{align} \lVert x_1 - y \rVert &\leq R\\ \implies \lVert x_1 - y \rVert^2 &= ((x - y) + v) \cdot ((x - y) + v)\\ &= R^2 + \lVert v \rVert^2 + 2v \cdot (x - y)\\ &\leq R^2 \end{align}

Together with ($\ast$), this gives us $R^2 + \lVert v \rVert^2 + 2v \cdot (x - y) = R^2$, which implies that $\lVert x_2 - y \rVert = \lVert x_2 - y \rVert = R$.

Here comes the part where the Euclidean metric is crucial. Since $x_1$ and $x_2$ lie on the sphere of radius $R$ and centered at $y$, $x$ is the midpoint of the chord joining $x_1$ and $x_2$, so $x - y$ is perpendicular to $v$. The Pythagorean Theorem gives us $R^2 = \lVert x - y\rVert^2 = R^2 - \lVert v\rVert^2 < R^2$. Contradiction!

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