# Posets of finite sequences are highly connected

I need the following result for an example in a paper I'm writing. It's easy enough to prove, but I'd prefer to just give a reference. Does anyone know one?

Fix $1 \leq k \leq n$. Define $X_{n,k}$ to be the following poset. The elements of $X_{n,k}$ are ordered sequences $\omega = (x_1,\ldots,x_m)$, where the $x_i$ are distinct elements of the $n$-element set $\{1,\ldots,n\}$ and $m \geq k$. The order relation is that $\omega_1 \leq \omega_2$ if $\omega_1$ is a subsequence of $\omega_2$. The theorem then is that the geometric realization of $X_{n,k}$ is $(n-1-k)$-connected.

-
what's the geometric realization of a poset ? –  Suresh Venkat Oct 13 '10 at 20:47
The set of chains is an abstract simplicial complex, so the geometric realization is just the realization of this complex. –  Richard Kent Oct 13 '10 at 20:54
ah ok. i was wondering if that is what it was. –  Suresh Venkat Oct 13 '10 at 21:44