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.


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