I was wondering if anything is known about this problem. Fix $0\leq m\leq k$. We are given a graded poset and we fix an element $x$ of rank $k$. Is it possible to estimate the number of elements $y$ of rank $k$ such that $x\wedge y$ has rank $m$? (I suppose the poset must be a meet-semilattice for this to be well-defined.)

I have kept the formulation general, but I'd be happy to hear any results with extra hypotheses on the poset. In fact, my use case is a lattice of certain subsets of a finite ground set. I think another formulation would be in terms of shadows in a set system. Is there a bound on the cardinality of the $(k-m)$th upper shadow of the $(k-m)$th lower shadow of a singleton in $X^{(k)}$? I apologise in advance if this question turns out to be trivial.

__Edit.__ Based on the answer below, it seems there isn't something one can really say in general, so I'll just post my motivating example. Consider the poset of all subsets of $\{1,2,\ldots,n\}$ that are a (nonempty) arithmetic progression and fix a progression $P$ of length $k$. How many progressions of length $k$ have $m$ elements in common with $P$, for $0\leq m\leq k$?

I'd also be interested in any references that deal with any interesting properties of this poset. For example, the size of this poset seems to be $$1+n+\sum_{m=1}^{n-1}\sum_{k=1}^{n-1} \bigg\lfloor{m\over k}\bigg\rfloor,$$
according to the [OEIS][1].


  [1]: https://oeis.org/A051336