Prove properties of a standard n-simplex 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

*

*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 $a_0$.


*The $n$-simplex $\sigma$ is a compact, convex set in $\mathbb{R}^{n+1}$ which equals the convex hull of $\left \{ a_{0}, a_{1}, a_{2}, ..., a_{n} \right \}$.


*Given $\sigma$, there exists a unique set of geometrically independent points (vertices) spanning $\sigma$.


*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 $a_0$.
 A: *

*First we show that the union of the line segments is in $\sigma$. Let $x=\left \{ \sum\limits_{i=1}^n s_ia_i | s_i \geq 0, \sum\limits_{i=1}^n s_i=1 \right \}$ be a point on the simplex spanned by $\left \{a_1,...,a_n \right \}$.  Then the line segment $\left \{t_0a_0+(1-t_0)\sum\limits_{i=1}^n s_ia_i | 0 \leq t_0 \leq 1 \right \}$ is in $\sigma$ because $t_0s_i \geq 0$ and $t_0+(1-t_0)(s_1+...+s_n)=1$.

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.


*We demonstrate compactness by showing that $\sigma$ is closed and bounded.  The simplex $\sigma$ is bounded since
\begin{equation}
\left \Vert \sum\limits_{i=0}^n t_ia_i \right \Vert \leq \max \left \{ t_0,...,t_n \right \} \max \left \{\|a_o\|,...,\|a_n\| \right \} \leq \max \left \{\|a_0\|,...,\|a_n\| \right \}
\end{equation}
The standard $n$-simplex is closed since it is the inverse image of $\left \{1 \right \}$ under the continuous map $\left(x_i \right) \rightarrow \sum\limits_{i}{x_i}$.

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 $\left\{a_0,...,a_n \right\}$


*We begin by showing that if  $x \in \sigma$ and $x \neq a_0,...,a_n$ then $x$ lies in some open line segment contained in $\sigma$.  Suppose $x=\sum\limits_{i=0}^n{t_ia_i}$ is not a vertex.  Then $t_i \neq 1$ for any $i$.  It follows that at least two of the $t_i$ are nonzero, say $t_0$ and $t_1$, without loss of generality.  Choose $\epsilon > 0$ such that $(t_0-\epsilon, t_0+\epsilon) \subset (0,1)$ and $(t_1-\epsilon, t_1+\epsilon) \subset (0,1)$.  Then for $t \in (-\epsilon, \epsilon)$, $(t_0+t)a_0+(t_1-t)a_1+\sum\limits_{i=2}^n{t_ia_i}$ is in $\sigma$.  Then it follows that every point in $\sigma$ which is not a vertex is on an open line segment contained in $\sigma$.

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 barycentric coordinates $t_i$ are continuous functions from the plane $P$ spanned by vertices of $\sigma$ to $\mathbb{R}$. The preimage of the open interval $(0, \infty)$ under each $t_i$ is therefore open in $P$. The interior of $\sigma$ is the intersection of these finitely many open sets and therefore it is open.

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 $a_0$.
