The only reaction to [this question](http://math.stackexchange.com/q/1867404/214353) on math.SE was 21 views in 4 days, so I decided to repost it here. I am not changing anything.

There was an interesting question on MO which OP removed by some reason. Here is a (more or less) equivalent form.

Take a finite cartesian product of finite linear orders, and remove top and bottom. What is the homotopy type of the obtained poset?

When all linear orders have two elements, this is a sphere. In that question, it was conjectured that if at least one of them has more than two elements, then it is contractible. I think in fact from $3\times2\times2\times2$ one gets a (thickened) 2-sphere but I am not sure.

In general, is it known which homotopy types may occur when one removes top and bottom from a finite lattice?

**Later**:

In the course of answers, the following additional question became relevant (I think).

Call an element of a bounded lattice dense if it has nonbottom meet with every nonbottom element. Top is obviously such; call a dense element nontrivial if it is not top.

It is known that a finite distributive lattice (more generally a not necessarily finite [pseudocomplemented](https://en.wikipedia.org/wiki/Pseudocomplement) lattice) does not have nontrivial dense elements if and only if it is a Boolean algebra.

Is a generalization of this known for arbitrary lattices? That is, which general lattices do not have any nontrivial dense elements?

**Still later:**

I've posted a question about the last one, http://mathoverflow.net/q/245366/41291; it was answered by [Tri](http://mathoverflow.net/users/51389/tri). So seems like correct type of lattices to consider in this context are the [geometric lattices](https://en.wikipedia.org/wiki/Geometric_lattice).