Equivalence of Definitions of Simplex

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Theorem

The following definitions of the concept of Simplex are equivalent:

Definition 1

A simplex $S$ in $\R^n$ with vertices $\family {\alpha_i}_{i \mathop = 0}^n$ is a set such that:

$S = \set {\ds \sum_{i \mathop = 0}^n \theta_i \alpha_i}$

where:

$\sequence {\alpha_i}_{i \mathop = 0}^n$ is a sequence of $n + 1$ affinely independent points in $\R^n$
$\sequence {\theta_i}_{i \mathop = 0}^n$ is a sequence of arbitrary real numbers such that:
$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \hointr 0 \to$
$\ds \sum_{i \mathop = 0}^n \theta_i = 1$

Definition 2

A simplex $S$ in $\R^n$ with vertices $\family {\alpha_i}_{i \mathop = 0}^n$ is a set such that:

$S = \set {\ds \sum_{i \mathop = 0}^n \theta_i \alpha_i}$

where:

$\sequence {\alpha_i}_{i \mathop = 0}^n$ is a sequence of $n + 1$ affinely independent points in $\R^n$
$\sequence {\theta_i}_{i \mathop = 0}^n$ is a sequence of arbitrary real numbers such that:
$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \closedint 0 1$
$\ds \sum_{i \mathop = 0}^n \theta_i = 1$


Proof

Let $S_1$ denote the set obtained by definition 1.

Let $S_2$ denote the set obtained by definition 2.


Let $x \in S_2$ for some $x \in \R^n$.

Then, by definition 2, a sequence of real numbers $\sequence {\theta_i}_{i \mathop = 0}^n$ exists such that:

$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \closedint 0 1$
$\ds \sum_{i \mathop = 0}^n \theta_i = 1$
$x = \ds \sum_{i \mathop = 0}^n \theta_i \alpha_i$

We have that $\sequence {\theta_i}_{i \mathop = 0}^n$ also satisfies the three conditions:

$\ds \sum_{i \mathop = 0}^n \theta_i = 1$
$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \hointr 0 \to$
$x = \ds \sum_{i \mathop = 0}^n \theta_i \alpha_i$

Hence $x \in S_1$ by definition 1.

Then by definition of subset:

$S_2 \subseteq S_1$

$\Box$


Let $x \in S_1$ for some $x \in \R^n$

Then, by definition 1, a sequence of real numbers $\sequence {\theta_i}_{i \mathop = 0}^n$ exists such that:

$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \hointr 0 \to$
$\ds \sum_{i \mathop = 0}^n \theta_i = 1$
$x = \ds \sum_{i \mathop = 0}^n \theta_i \alpha_i$


Aiming for a contradiction, suppose:

$\exists j \in \set {0, 1, 2, \ldots, n}: \theta_j > 1$

Then:

$\ds \sum_{i \mathop = 0}^n \theta_i \ge \theta_j > 1$

which is a contradiction.


So we see that:

$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \le 1$

Because:

$\ds \sum_{i \mathop = 0}^n \theta_i = 1$

and:

$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \hointr 0 \to$

it follows by definition 2 that:

$x \in S_2$

Then by definition of subset:

$S_1 \subseteq S_2$

$\Box$


Thus we have:

$S_1 \subseteq S_2$

and:

$S_2 \subseteq S_1$

and by definition of set equality:

$S_1 = S_2$

$\blacksquare$