# standard n-simplex

We know that a n-simplex is a convex hull of n+1 affinely independents points in $$\mathbb{R}^n$$, i,e, let $$x_0,x_1, \ldots ,x_n$$ affinely independents points in $$\mathbb{R}^n$$ then the n-simplex determinated is :

$$C=\lbrace \lambda_0 x_0 + \lambda_1 x_1+ \ldots + \lambda_n x_n / \lambda_i \geq 0, \sum_{i=0}^{n}\lambda_i =1 \rbrace$$

Also a standard n-simplex, $$\Delta^n$$, is the set :

$$\Delta^n=\lbrace (t_0,t_1,\ldots,t_n) \in \mathbb{R}^{n+1}/t_i \geq 0, \sum_{i=0}^{n}\lambda_i =1 \rbrace .$$

It is noted that the standard n-simplex is the convex hull of the canonical vectors $$e_0,e_1, \ldots, e_n \in \mathbb{R}^{n+1}$$

We also see that the set $$X=\lbrace 0,e_0,e_1,\ldots,e_n \rbrace$$ is affinely independent in $$\mathbb{R}^{n+1}$$ so following the definition of n+1-simplex we should have that $$\Delta^n$$ is a $$n+1$$-simplex because the convex hull of $$X$$ is $$\Delta^n$$.

Why do the n-simplex standard define it that way if it is really a n + 1 simplex?

The convex hull of $$X$$ is an ($$n + 1$$)-simplex, that's true, but it's not the same as $$\Delta^n$$. For example, take $$n = 1$$. $$\Delta^n$$ is the line segment from $$(0, 1)$$ to $$(1, 0)$$, but the convex hull of $$X = \{(0, 0), (1, 0), (0, 1)\}$$ is a solid triangle.