Equivalence of Definitions of Simplex

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:
 * $S_2 \subseteq S_1$

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


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

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:
 * $S_1 \subseteq S_2$

Thus we have:
 * $S_1 \subseteq S_2$

and:
 * $S_2 \subseteq S_1$

and by definition of set equality:
 * $S_1 = S_2$