3
$\begingroup$

It's been known for 35 years that every finite lattice can be embedded in a finite partition lattice (Pudlak and Tuma, Algebra Universalis 1980, Volume 10, Issue 1, pp. 74--95).

I don't follow the construction in the proof enough to understand what size partition lattice it produces in the case of the power set lattice, and if it does, what is its size. Is the embedding known earlier for the power set lattice?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

By Exercise V.4.7 of Lattice theory: foundation by George Grätzer, we can take $M=N+1$.

As for possible sharpness of this result, note that $P(N)$ has sizes 1,2,4,8 for $N=0,1,2,3$, whereas PART$(N+1)$ has sizes 1,2,5,15 (the Bell numbers).

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ The result is sharp for a simple reason: you cannot embed P(N) to PART(N) because the former has length N and the latter has length N-1. As pointed out by Richard Stanley in a later question mathoverflow.net/questions/286725/… $\endgroup$
    – Jukka Kohonen
    Commented Feb 16, 2021 at 7:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .