So it starts like this:
Given a discrete set $N$, the probability simplex over $N$, denoted $∆(N)$ is defined to be:
$$Δ(N) = \left\{ x \in \mathbb{R}^{|N|} \;:\; x_{i} \geq 0 \text{ for all } i, \text{ and } \sum_{i=1}^{|N|} x_i = 1\right\}$$
I guess I kind of understand what it tries to convey, but in practice something doesn't add up.
For example, suppose, that we have $x= \{x_1,x_2\}$ with dimentionality $|N| = 3$ according to the specification above.
Say:
that $x_1 = \{\mathbb{R}_1, \mathbb{R}_2, \mathbb{R}_3\}$ and $x_2 = \{\mathbb{R}_4, \mathbb{R}_5, \mathbb{R}_6\}$.
How does this summation $\sum_{i=1}^{|N|} x_i = 1$ work?
I think, this is a kind of question that should be answered via examples.
For any set $N_2=\{a,b\}$ with 2 elements the following vectors $x_1, x_2, x_3$ belong to $\Delta(N_2)$:
$$ x_1 = \begin{bmatrix} 0.2 \\ 0.8 \end{bmatrix}, \quad x_2 = \begin{bmatrix} 0.4 \\ 0.6 \end{bmatrix}, \quad x_3 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. $$ For each of them we have $x_i \in \Delta(N_2)$. Of course there are many more such vectors.
For any set $N_3=\{a,b,c\}$ with 3 elements the following vectors $x_1, x_2, x_3$ belong to $\Delta(N_3)$:
$$ x_1 = \begin{bmatrix} 0.2 \\ 0.3 \\ 0.5 \end{bmatrix}, \quad x_2 = \begin{bmatrix} 0.4 \\ 0.4 \\ 0.2 \end{bmatrix}, \quad x_3 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}. $$ For each of them we have $x_i \in \Delta(N_3)$. Again, there are many more elements of $\Delta(N_3)$.
Note that the set $N$ is actually of no matter for constructing $\Delta(N)$, it is only its size $|N|$ that matters.