# 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. – Autumn 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

## 1 Answer

If I understand your question correctly, an answer should appear in the paper "On lexicographically shellable posets" of Anders Bj\"orner and Michelle Wachs, in Transactions of the AMS 277, pp. 323-341.

• Yes, this paper gives me exactly what I want. Thanks! – Andy Putman Oct 14 '10 at 1:39