I'm attempting to prove that the set $\Delta X = \{π ∈ \mathbb{R}^n |\sum \pi_i=1, \pi_i \ge 0, \forall i\}$ is a convex subset of the vector space $\mathbb{R}^n$. It is specified that $X$ is a finite set of size $n$ and $\Delta X$ is the space of probabilities on $X$.
That $\Delta X$ needs to be convex is not immediately clear to me, as I do not see what contradiction is made by letting $\pi_0 = \alpha (\pi_1) + (1-\alpha)\pi_2 \not \in \Delta X$. If the probability space is of size $n=2$, it could be that $ X = \{x_1, x_2\}$ where $P(x_1)= \frac{1}{3}$ and $P(x_2)= \frac{2}{3}$. I know that $\{\frac{1}{3}, \frac{2}{3}\}$ is not convex.
I do know this is trivially true for $n=1$, as $ X = \{x'\}$ where $\Delta X=\{1\}$, but I am unsure of how to expand this to the nontrivial case.
This may be as simple as me misunderstanding the question, any tips are appreciated.