$x$ is a vector in the unit simplex in $\mathbb{R}^n$, i.e: $$x = (x_1,\dots,x_n)\,\,\,\,\,\,\,\,;\,\,\,\,\forall i: x_i\geq 0\,\,\,\,;\,\,\,\,\,\,\,\,\sum_{i=1}^n x_i = 1$$ Initially, $x=(0,0,\dots,0,1)$. You are allowed to transform $x$ using the following actions:
- Permute the elements of $x$.
- Halve the first $n-1$ elements of $x$ while modifying the $n$-th element accordingly, i.e: \begin{align*} x_1'&\leftarrow x_1/2 \\ x_2'&\leftarrow x_2/2 \\ \dots \\ x_{n-1}'&\leftarrow x_{n-1}/2 \\ x_n'&\leftarrow 1-(x_1'+x_2'+\cdots+x_{n-1}') \end{align*}
You are allowed to make infinitely many such actions, in any order. Here is an example for $n=3$:
$(0,0,1)\to (1,0,0)\to (0.5,0,0.5)\to (0.25,0,0.75)\to (0.25,0.75,0)\to(0.125,0.375,0.5)\to\dots$
What is the set of vectors that can be attained as the limit of an infinite sequence of such actions?
Particularly, can you create all vectors in the unit simplex?
Note that after a doubling move
$$x_n'=x_n+\sum_{k=1}^{n-1}x_k'\;,$$
which is necessarily at least $\frac{1}2$. Thus, every point that can be obtained has a coordinate that is at least $\frac{1}2$. For $n\ge 3$ this implies that the closure of the set of obtainable points is not the entire unit simplex.
Matters are different for $n=2$: we can obtain precisely the points $\langle x,1-x\rangle$ such that $x$ is a dyadic rational in $[0,1]$, and the closure of this set is the unit simplex. To see this, let $x$ be a dyadic rational in $[0,1]$; we want to obtain $\langle x,1-x\rangle$. Since $x$ is dyadic, it has a finite binary expansion $0.b_1\ldots b_m$ such that $b_1=0$ and $b_m=1$, and we may as well assume that $0<x<\frac{1}2$. Let $\ell$ be minimal such that $b_\ell=1$; using reverse moves we can double $x$ $\ell-1$ times to obtain $\langle 2^{\ell-1}x,1-2^{\ell-1}x\rangle$, whose first coordinate has the binary expansion $0.b_\ell\ldots b_m$.
If $\ell=m$, we're done: $\left\langle\frac{1}2,\frac{1}2\right\rangle$ is certainly obtainable, so $\langle x,1-x\rangle$ is as well. If not, we reverse the coordinates to get $\langle 1-2^{\ell-1}x,2^{\ell-1}x\rangle$, whose first coordinate has a binary expansion with a leading $0$ and only $m-\ell+1<m$ digits to the right of the binary point. Again we use reverse moves to double the first coordinate until it's at least $\frac{1}2$, and if it exceeds $\frac{1}2$, we again transpose the coordinates. Clearly the procedure eventually terminates at the obtainable point $\left\langle\frac{1}2,\frac{1}2\right\rangle$, since the binary expansions of the coordinates are decreasing in length, so $\langle x,1-x\rangle$ is obtainable.
A similar argument shows that in the general case we can obtain precisely the points of the unit simplex whose coordinates are all dyadic rationals, at least one of which is at least $\frac{1}2$; for the closure of this set, just delete the restriction to dyadic rationals.