Can ZF prove that every set admits a lattice order?

Question

Can ZF set theory without the axiom of choice prove that every set $S$ can be endowed with a partial order $L$ which is a lattice? And if so, can ZF also prove that every set can be endowed with a partial order which is a distributive lattice? I know that ZF can't prove that every set can be linearly ordered. I would be very interested in knowing the relative strengths of ZF + "every set can be lattice ordered", ZF + "every set can be distributive lattice ordered", and ZF + "every set can be linearly ordered".

Answer 1

Working in ZF.

It has been pointed out in comments that any set can be ordered as a complete modular lattice, namely, with all elements incomparable except the top and bottom elements.

It is consistent that there are sets which can't be distributive-lattice ordered, namely, any amorphous set. An amorphous set is an infinite set which does not contain two disjoint infinite subsets. It's consistent with ZF that such sets exist.

Proposition. No amorphous lattice is distributive.

Proof. Let $L$ be an amorphous lattice. We call $x$ a successor of $y$ if $x\gt y$, a predecessor of $y$ if $x\lt y$; "almost all" means "all but finitely many."

Suppose infinitely many (therefore almost all) elements have infinitely many successors. Let $S$ be the (cofinite) set of all elements with infinitely many successors (and therefore only a finite number of predecessors). Then each $x\in S$ has a successor $y\in S$, and $y$ has more predecessors than $x$. Thus the number of predecessors of an element takes infinitely many different (finite) values, which is impossible since $L$ is amorphous. We conclude that almost all elements have finitely many successors.

Let $A$ be the (cofinite) set of all elements with finitely many successors. Let $A_n$ be the set of all elements with exactly $n$ successors. Since $A$ is amorphous, one of the sets $A_n$ is cofinite; fix that $n$. Let $F=L\setminus A_n$.

Plainly $A_n$ is an antichain; thus, if $x,y\in A_n$, $x\ne y$, then $x\land y\in F$ and $x\lor y\in F$. Since $F$ is a finite set, it follows by the finite Ramsey theorem that there are three distinct elements $x,y,z\in A_n$ with $x\land y=x\land z=y\land z$ and $x\lor y=x\lor z=y\lor z$. That is, $L$ embeds the nondistributive diamond lattice $M_3$, and $x\land(y\lor z)\gt(x\land y)\lor(x\land z)$.

Comment. This is only a partial answer, as I have no idea whether "every set can be distributive-lattice ordered" is weaker than "every set can be linearly ordered."