Definition:Simplex

Definition
A simplex is an $n$-dimensional generalization of a triangle and tetrahedron, for $n \in \Z_{>0}$.

A $k$-simplex is a $k$-dimensional polytope which is the convex hull of its $k + 1$ vertices.

1. definition for Euclidean space $\R^n$
A simplex $S$ in $\R^n$ with vertices $\family {\alpha_i}_{i \mathop = 0}^n$ ($n + 1$ points, which must be affinely independent) is a set $S = \set {\ds \sum_{i \mathop = 0}^n o_i \alpha_i; \ds \sum_{i \mathop = 0}^n o_i = 1 \land \forall i \in \set {0, 1, 2, \ldots, n}: o_i \in [0,\infty) }$.

In the above set $\{o_i\}_{i \mathop = 0}^n$ are arbitrary real numbers that satisfy the condition in the description (each of them is nonnegative and their sum is $1$).

2. definition for Euclidean space $\R^n$
A simplex $S$ in $\R^n$ with vertices $\family {\alpha_i}_{i \mathop = 0}^n$ ($n + 1$ points, which must be affinely independent) is a set $S = \set {\ds \sum_{i \mathop = 0}^n o_i \alpha_i; \ds \sum_{i \mathop = 0}^n o_i = 1 \land \forall i \in \set {0, 1, 2, \ldots, n}: o_i \in [0,1] }$.

In the above set $\{o_i \}_{i \mathop = 0}^n$ are arbitrary real numbers, such that each of them is between $0$ and $1$ and their sum is $1$.

Proof
We want to prove that a simplex $S \subset \R^n$ with vertices $\family {\alpha_i}_{i \mathop = 0}^n$ obtained by the 1. definition is the same as the one obtained by the 2.

Let $S_1$ denote the set obtained by the 1. definition and $S_2$ denote the set obtained by the second.

Let for some $x \in \R^n$ be true that $x \in S_2$. Then by the 2. definition a set of real numbers $\{o_i\}_{i \mathop = 0}^n$ exists, such that $\forall i \in \set {0, 1, 2, \ldots, n}: o_i \in [0,1] $ and $\ds \sum_{i \mathop = 0}^n o_i = 1$. That same set $\{o_i\}_{i \mathop = 0}^n$ satisfies next two conditions: $\ds \sum_{i \mathop = 0}^n o_i = 1$ and $\forall i \in \set {0, 1, 2, \ldots, n}: o_i \in [0,\infty)$. And so by 1. definition $x \in S_1$.

$(x \in S_2 \implies x \in S_1) \iff S_2 \subseteq S_1$

Let for some $x \in \R^n$ be true that $x \in S_1$. Then by the 1. definition a set of real numbers $\{o_i\}_{i \mathop = 0}^n$ exists, such that $\forall i \in \set {0, 1, 2, \ldots, n}: o_i \in [0,\infty) $ and $\ds \sum_{i \mathop = 0}^n o_i = 1$. Suppose $\exists j \in \set {0, 1, 2, \ldots, n}: o_j > 1$. Then $\ds \sum_{i \mathop = 0}^n o_i \ge o_j > 1$, which is a contradiction. So we see that $\forall i \in \set {0, 1, 2, \ldots, n}: o_i \le 1$. Because $\ds \sum_{i \mathop = 0}^n o_i = 1$ and $\forall i \in \set {0, 1, 2, \ldots, n}: o_i \in [0,1] $, by the 2. definition $x \in S_2$.

$(x \in S_1 \implies x \in S_2) \iff S_1 \subseteq S_2$

$S_1 \subseteq S_2 \land S_2 \subseteq S_1 \implies S_1 = S_2$ $\square$

Also see

 * Definition:Pentatope