I would like to prove the following properties for the standard $n$-simplex whose vertices are $e_i$ in $\mathbb{R^{n+1}}$ and then use the properties of an affine transformation $T$ to generalize to any simplex. Can you help me simplify the more general proof (in the answer section) and prove the existence of $T$?
Let the set $S=\left \{ a_{0}, a_{1}, a_{2}, …, a_{n} \right \} \in \mathbb{R}^{n+1}$ be a geometrically independent set and let $\sigma$ be the simplex spanned by $S$.
Prove
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The n-simplex $\sigma$ is the union of all line segments joining $a_0$ to points of the simplex $s$ spanned by $\left \{ a_{1}, a_{2}, …, a_{n} \right \}$. Two such line segments intersect only at $a_0$.
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The $n$-simplex $\sigma$ is a compact, convex set in $\mathbb{R}^{n+1}$ which equals the convex hull of $\left \{ a_{0}, a_{1}, a_{2}, …, a_{n} \right \}$.
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Given $\sigma$, there exists a unique set of geometrically independent points (vertices) spanning $\sigma$.
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The interior of $\sigma$ is convex and is open in the plane $P$; its closure is $\sigma$. $Interior(\sigma)$ is the union of all open line segments joining $a_0$ to the points of $Interior(s)$ where $s$ is the face of $\sigma$ opposite $a_0$.
Best Answer
Now we show that $\sigma$ is in the union of the line segments. Given a point $\sum\limits_{i=0}^n t_ia_i$ in $\sigma$ with $t_0 \neq 1$, set $s_i=\frac{t_i}{1-t_0}$ for $i=1,...,n$. This shows that every point in the simplex spanned by $a_0,...,a_n$, except $a_0$, is in the union of the line segments. Clearly $a_0$ is also in the union, since it is in each line segment. Therefore $\sigma$ is the union of all line segments from $a_0$ to the simplex spanned by $a_1,...,a_n$.
Furthermore, we show that two such line segments intersect only at $a_0$. Clearly two line segments from $a_0$ to the simplex spanned by $a_1,...,a_n$ do intersect at $a_0$. Now suppose they intersect at some other point $y$. Then the two segments must lie on the same line. The other endpoints of the line segments lie on the simplex spanned by $a_1,...,a_n$. But then the line must be contained in the simplex spanned by $a_1,...,a_n$. This contradicts the fact that $a_0$ does not lie on the simplex spanned by $a_1,...,a_n$, since $\left \{a_0,a_1,...,a_n \right \}$ are geometrically independent.
For convexity, suppose that $\left(t_0,..,t_n\right)$ and $\left(s_0,...,s_n\right)$ are two distinct $n$-tuples satisfying $t_i \geq 0$, $s_i \geq 0$, $\sum\limits_{i}{t_i}=1$ and $\sum\limits_{i}{s_i}=1$. Then $\lambda \left(t_0,...,t_n\right) + (1-\lambda)\left(s_0,...,s_n\right)$ parametrizes a line segment between these two points for $\lambda \in \left[ 0,1 \right]$. The corresponding curve \begin{equation} \lambda\sum\limits_{i=0}^n{t_ia_i}+(1-\lambda)\sum\limits_{i=0}^n{s_ia_i}=\sum\limits_{i=0}^n{(\lambda t_i + (1-\lambda) s_i) a_i} \end{equation} is a line segment connecting $\sum\limits_{i=0}^n{t_ia_i}$ to $\sum\limits_{i=0}^n{s_ia_i}$. Since $t_i$ and $s_i$ are non-negative and sum to $1$, and since $\lambda \in \left[0,1\right]$, we have $\lambda t_i + (1-\lambda) s_i$ is also non-negative and sums to $1$. So $\sigma$ is convex.
To show that $\sigma$ is the convex hull of $\left\{a_0,...,a_n\right\}$, we apply Theorem \ref{thm:convexsets} and show that $\sigma$ is the intersection of all convex sets containing $\left\{a_0,...,a_n \right\}$. Suppose $S$ is a convex set containing $a_0,...,a_n$. Then for every point $x=\sum\limits_{i=0}^n{t_ia_i} \in \sigma$, we can show that $x \in S$ as follows. By convexity, since $a_0$ and $a_1$ are in $S$, we have $b_1=\frac{t_0}{t_0+t_1}a_0+\frac{t_1}{t_0+t_1}a_1$ must also be in $S$. Therefore, \begin{equation} b_2=\frac{t_0+t_1}{t_0+t_1+t_2}b_1+\frac{t_2}{t_0+t_1+t_2}a_2 \end{equation} \begin{equation} b_2=\frac{t_0}{t_0+t_1+t_2}a_0 + \frac{t_1}{t_0+t_1+t_2}a_1 + \frac{t_2}{t_0+t_1+t_2}a_2 \end{equation} is also in $S$. Continuing in this manner, we find \begin{equation} b_n=\frac{t_0}{t_0+...+t_n}a_0 + ... + \frac{t_n}{t_0+...+t_n}a_n \in S, \end{equation} but $\sum\limits_{i=0}^n{t_i}=1$, so $x \in S$. Then $\sigma$ is contained in every convex set containing $\left\{a_0,...,a_n \right\}$
Next we'll show that the vertices are not on any open line segments contained in $\sigma$.
If any vertex, say $a_0$, were on an open segment contained in $\sigma$, then there would exist points $x,y\in \sigma$ with $a_0 = tx+(1-t)y$. If $x=\sum\limits_{i=0} ^n t_i a_i$ and $y=\sum\limits_{i=0}^n s_i a_i$, then this means $a_0 = \sum\limits_{i=0}^n \left( t t_i + (1-t) s_i\right) a_i$. This violates the geometric independence of $a_0, ..., a_n$, unless $t_0=s_0=1$ and the others are zero, in which case $x=y$. Thus $a_0$ is not on an open line segment.
The interior of $\sigma$ is convex since the convex combination $(1-t)x+ty$ of any $x,y \in Int(\sigma)$ is in $Int(\sigma)$. \begin{equation} (1-t)x+ty\, =\, (1-t)\sum\limits_{i=0}^nt_ia_i + t \sum\limits_{i=0}^n s_ia_i \, =\, \sum\limits_{i=0}^n ((1-t)t_i+ t s_i)a_i \end{equation} and if $t_i$ and $s_i$ are positive, so is $(1-t)t_i+ t s_i$.
Since $\sigma$ is closed, the closure of $Int(\sigma)$ is a subset of $\sigma$. On the other hand if a point $x \in Bd(\sigma)$ is not in the closure of $Int(\sigma)$, one can construct a sequence of points in $Int(\sigma)$ which converges to $x$.
Without loss of generality, assume that $x= \sum\limits_{i=0}^n t_i a_i,$ and $ t_i(x) > 0$ for $0 \le i \le k$ and $ t_{k+1}(x)= ... = t_n(x)=0.$
Let $r=\min\{ t_i(x)\, \vert \, 0\le i \le k\, \}/2$ and let $x_m\, =\, \sum\limits_{i=0}^k (t_i(x) - \frac{r}{m(k+1)})a_i + \sum\limits_{i=k+1}^n \frac{r}{m(n-k)}$. \newline Then $x_m \in Int(\sigma)$ and the sequence converges to $x$.
Finally, we show that $Int(\sigma)$ equals the union of open line segments which join a vertex $a_0$ to points in the interior of the opposite face. Let $x$ be an interior point of $\sigma$ so that $x = \sum\limits_{i=0}^{n} t_ia_i$ with $t_i > 0$ and $\sum{t_i} =1.$
Rewrite this as $x= t_0a_o + (1-t_0) \sum\limits_{i=1}^{n} \frac{t_i}{1- t_0} a_i.$ Therefore $x$ is on an open line segment joining $a_0$ to a point on the face opposite $a_0$.