Let $A \subset \mathbb R^n$ be a set, and we define $B$ to be the set of all convex combinations of elements of $A$.
By convex combination I mean $\sum_{i=1}^{p}t_ix_i$ where $x_i \in A$, $t_i \geq 0, \sum_{i=1}^p t_i = 1$ for some $p$.
I want to understand this proof that $B$ is convex:
Take $x,y \in B, \lambda \in [0, 1]$:
$\displaystyle \lambda x + (1-\lambda)y = \lambda \sum_{i=1}^pt_ix_i + (1-\lambda)\sum_{i=1}^rs_iy_i = \sum_{i=1}^{p+r}\xi_iz_i$
Where $\xi_i = \lambda t_i + (1-\lambda)s_i, \xi_i \geq0, \sum_{i=1}^{p+r}\xi_i = 1$
I can't understand what $z_i$ is and why $z_i \in A$.
It is sloppy.
Suppose $x= \sum_{i=1}^p t_i x_i$ and $y=\sum_{j=1}^r s_j y_j$ with $t,s$ satisfying the relevant constraints.
Let $z=(x_1,...,x_p,y_1,...,y_r)$, $\bar{t} = (t_1,...,t_p,0,...,0)$ and $\bar{s} = (0,...,0,s_1,...,s_r)$, then we can write $x=\sum_{k=1}^{p+r} \bar{t_k}z_k $, $y=\sum_{k=1}^{p+r} \bar{s_k}z_k $.
Then $\lambda x+(1-\lambda)y = \sum_{k=1}^{p+r} (\lambda\bar{t_k} +(1-\lambda) \bar{s_k}) z_k $.
Letting $\xi=(t_1,...,t_p,s_1,...,s_r)$, we have $\lambda x+(1-\lambda)y = \sum_{k=1}^{p+r} \xi_k z_k $.
It is straightforward to check that $\xi$ satisfies the summation and non negativity constraints.