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Let $\text{Part}(X)$ denote the collection of all partitions of $X$. For $A, B\in \text{Part}(X)$ we set $A\leq B$ if $A$ refines $B$, that is for all $a\in A$ there is $b\in B$ such that $a\subseteq b$. This relation defines a lattice structure on $\text{Part}(X)$.

Is there a distributive lattice $L$ such that for no set $X$ there is a surjective lattice homomorphism $s:\text{Part}(X)\to L$?

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    $\begingroup$ When X has more than 3 elements, the lattice of partitions should be non modular. I would expect all but very small distributive lattices L to avoid being homomorphic images of any partition lattice. Gerhard "Should Be A Known Result" Paseman, 2017.01.16. $\endgroup$
    – Gerhard Paseman
    Commented Jan 16, 2017 at 8:13

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In the paper

Ore, Oystein, Theory of equivalence relations, Duke Math. J. 9, (1942), 573–627

it is proved that $\textrm{Part}(X)$ is simple. (This is very easy to prove.)

So the only nontrivial quotient of $\textrm{Part}(X)$ is $\textrm{Part}(X)$ itself, which is not distributive when $|X|>2$. Therefore the answer to the question is: any distributive lattice of more than 2 elements is a suitable $L$, and these are the only suitable $L$'s.

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Yes. Since $\text{Part}(X)$ has a least element and a greatest element, just let $L$ not have that, e.g., let $L$ be the ordering of the integers.

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