Take the $n-1$-simplex, $\Delta$, whose vertices are the $n$ standard unit vectors and let $\mu \in$ int $\Delta$. Define a probability distribution with finite support and barycenter $\mu, P_k$, to be a collection of points $x_1, \ldots, x_k$ (this is the support of $P_k$, i.e., $\operatorname{supp} P_k=\left\{x_1, \ldots, x_k\right\}$ ) with $x_i \in \Delta$ for all $i=1, \ldots, k$, and probabilities $p_1, \ldots, p_k$ with $p_i \in[0,1]$ for all $i=1, \ldots, k, \sum_{i=1}^k p_i=1$, and $\sum_{i=1}^k p_i x_i=\mu$. That is,
$P_k\equiv\{(x_1, \ldots, x_k, p_1, \ldots, p_k): x_i \in \Delta, p_i \in[0,1] \forall i, \sum_{i=1}^k p_i=1, \sum_{i=1}^k p_i x_i=\mu\}$
Define $\mathcal{P}:=\{P_k: k \in \mathbb{N}\}$
The question asks to show that $\mathcal{P}$ convex.
What I understand is the following:
1- Each $P_k$ is a probability simplex.
2- $\mathcal{P}$ has (possibly) many elements for each $k \in \mathbb{N}$, that is for $k=2$ there are different $P_k$s and same goes for $k > 3$.
3- Since the coordinates do not match for different $k$s, the convex combinations are only possible for fixed $k$.
4- Taking $x^{\lambda}$ and $p^{\lambda}$ to be convex combinations of partial elements of any $P_k', P_k''$ with the associated $\lambda \in (0,1)$, is the convex combination element $P_k^{\lambda} \overset{?}{=} (x^{\lambda}, p^{\lambda})$? If not, if you can let me know which elements to combine and the reason, that would be sufficient.
Thank you!