Consider $k+1$ affinely independent vectors $\left\{p_0,p_1, \dots, p_k \right \}$ in $n$-dimensional euclidean vector space $n>k$ and consider their convex hull. It is known that each point $x$ of it can be written in form $x = \sum_{i=0}^k \lambda_ip_i, \sum_{i=0}^k \lambda_i = 1, \lambda_i \geq 0, 0 \leq i \leq k$. Interior of it is defined requiring for $\lambda_i, 0 \leq i \leq k$ to be positive. How could I prove that this interior is open in relative topology?
My try:
Pick $x = \sum_{i=0}^k \lambda_ip_i, \sum_{i=0}^k \lambda_i = 1, \lambda_i > 0, 0 \leq i \leq k$.
I want to find $\epsilon$ ball around it s.t. $\forall t\in B_\epsilon (x), t=\sum_{i=0}^k \mu_ip_i, \sum_{i=0}^k \mu_i = 1, \mu_i > 0, 0 \leq i \leq k $
We first perturbe $\lambda_0$: Take $\epsilon_0 = \frac{1}{2}\min\left\{\lambda_0, 1-\lambda_0, 1-\sum_{i=1}^k \lambda_i\right\}>0$ (motivation for taking such is the fact $0<\lambda_0<1$)
Pick arbitrary $\mu_0$ in $[\lambda_0 - \epsilon_0,\lambda_0 + \epsilon_0]$. It is easy to show that in this case $0<\mu_i<1$ and $\mu_0+\sum_{i=1}^k \lambda_i <1$.
Idea is to continue this procedure taking $\epsilon_1 = \frac{1}{2}\min\left\{\lambda_1, 1-\lambda_1, 1-\sum_{i=2}^k \lambda_i - \mu_0 \right\}>0$. In this step I easily show that for $\mu_1$ in $[\lambda_1 - \epsilon_1,\lambda_1 + \epsilon_1]$ we have $0<\mu_1<1$, but struggle to prove $\mu_0 + \mu_1 + \sum_{i=2}^k \lambda_i < 1$.
If I eventually prove this and continue to go on like this (changing lambdas), I would then take my $\epsilon = \min_{i=0,...,k} \epsilon_i$ which in max-norm should fit to prove the desired fact.
Am I completely wrong? Does anyone have advice/more elegant idea? Since I don't have much theory developed, I have to go to find $\epsilon$-ball around $x$ with such properties, but didn't manage to find more sophisticated thing.
OK let $C$ be your convex hull, $H=\{\lambda\in\mathbb R^{k+1}\vert \sum_i\lambda_i=1\}$ (affine subspace of $\mathbb R^{k+1}$). Let $f:H\cap\mathbb R_+^{k+1}\longrightarrow C$ such that $f(\lambda)=\sum_i\lambda_ip_i$. Notice that $f$ is continuous.
Now since the $p_i$ are affinely independent, you can calculate $f^{-1}$ by solving a linear system. Since solving a linear system only involves continuous operations, $f^{-1}$ is continuous.
Hence $f$ is a homeomorphism.
This means that $f(R_+^{*^{k+1}})=\mathring C$ is open in $C$ since $R_+^{*^{k+1}}$ is open in $R_+^{k+1}$.
Qed