What are the extreme points of this polytope? Let $n, k$ be positive integers with $n\geq k$.
Let $P_{n,k}$ be the set of vectors $x$ in $[0,1]^n$ for which
$$ \sum_{i=1}^n x_i = k $$
$P_{n,k}$ is defined by linear equations, so it is a polytope. What are its corners (extreme points)?
My guess is that the corners are the "$k$-binary" vectors - the vectors with exactly $k$ ones and $n-k$ zeros. To prove this, it is sufficient to prove (I think) that every vector in $P_{n,k}$ is a convex combination of $k$-binary vectors. Is this correct?
Note: when $k=1$, $P_{n,1}$ is just the standard simplex in $\mathbb{R}^n$. Does it have a name when $k > 1$?

To find its corners, let $\mathbf{x} = (x_i)_{i=1}^n \in P_{n,k}$.  Let the elementary vector $\mathbf{e}_i \in \mathbb{R}^n$ be a vector whose $i$-th component is $1$, and $0$ elsewhere.
It's easy to see that if there exists $i$ such that $x_i \in (0,1)$, then $\mathbf{x}$ isn't a corner, as we also have another index $i' \ne i$ such that $x_{i'} \in (0,1)$.  (Otherwise, if $i$ is the only non-integer index, the sum of components $\sum_i x_i \notin \mathbb{Z}$, contradicting $\sum_i x_i = k \in \mathbb{Z}$.)  It remains to construct the interval containing $\mathbf{x}$.  Choose a sufficiently small $\epsilon > 0$ such that $\mathbf{x}$ is the midpoint of the interval $[\mathbf{x} - \epsilon \mathbf{e}_i + \epsilon \mathbf{e}_{i'}, \mathbf{x} + \epsilon \mathbf{e}_i - \epsilon \mathbf{e}_{i'}] \subseteq P_{n,k}$.  That's possible as $\mathbf{x}$ has finitely many components.
As a result, we can restrict the possible candidates to $\{0,1\}^n$.  The equality constraint $\sum_i x_i = k$ tells us everything, so that the proof is finished.
N.B. If $x_i = 1$, then the $i$-th component of $\mathbf{x} + \epsilon \mathbf{e}_i - \epsilon \mathbf{e}_{i'}$ would be $x_i + \epsilon > 1$, contradicting our choice of $\mathbf{x} \in [0,1]^n$.
