Join-incomplete lattice endomorphisms

Question

Let $L$ be an complete lattice. A lattice homomorphism $f: L\to L$ is said to be join-incomplete if there is an infinite set $S \subseteq L$ such that $f(\bigvee_L S) \neq \bigvee_L f(S).$

Suppose $L$ contains an ideal $J$ such that $\bigvee_L J \notin J$. Does this imply that there is a join-incomplete lattice homomorphism $f: L\to L$?

(Joseph Van Name gave a positive answer to a similar question. The difference to this question here is that we considered homomorphisms $f:L\to K$ where $K$ was any complete lattice, whereas here we focus on endomorphisms $f:L\to L$.)

Answer 1

Here is a way to build examples of complete lattices with no join-incomplete endomorphisms. Suppose $L$ is

1. complete,

2. simple, and

3. each proper principal ideal of $L$ has ACC, but $L$ does not have ACC.

Then $L$ has a nonprincipal ideal but no join-incomplete endomorphism.

Reason: If $L$ does not have ACC, then the ideal generated by any increasing $\omega$-chain is nonprincipal. Suppose that $f\colon L\to L$ is an endomorphism. By item 2, $f$ is constant or an embedding. If $f$ is constant it preserves complete joins, so we are done. If $f$ is an embedding, then it must preserve $1$, since $1$ is the only element that dominates an increasing $\omega$-chain. Now, assuming still that $f$ is an embedding, if $J\subseteq L$ is a nonprincipal ideal it contains an increasing $\omega$-chain, so by item 3 the join of $J$ must $1$, but by the same argument the set $f(J)$ has join $1$. Thus $f$ preserves infinite joins.

What remains to show is that there really is a lattice satisfying 1, 2 and 3.

Start with the lattice $L_0$ of finite dimensional subspaces of an infinite dimensional vector space. $L_0$ satisfies items 2 and 3. Add three new elements to $L_0$: $1$ (a top element), and $a$ and $b$ which are incomparable with all other elements except $0$ (bottom of $L_0$) and $1$. I want $a$ and $b$ to be incomparable with each other, as well. $L$ has the property that, for every element $\ell<1$, there is a finite bound on the lengths of chains below $\ell$. This implies that $L$ has DCC. Since $L$ has DCC and a top element, it is complete. $L$ inherits item 3 from $L_0$. I leave it to you to check item 2, that $L$ is a simple lattice.

Answer 2

Inspired by Keith's example, let me write down an easier to understand (but essentially the same) lattice:

Let $L = \omega^{<\omega}\cup \{\omega\}$ where $\omega^{<\omega}$ denotes the set of finite subsets of $\omega$, ordered by $\subseteq$. To $L$ we add elements $a,b$ as in the example above: they are incomparable with everything except $\emptyset$ and $\omega$. Then $L$ has uncountably many non-principal ideals: for every infinite $A\subseteq \omega$, $\{x\in\omega^{<\omega}: x\subseteq A\}$ is a non-principal ideal. But every lattice homomorphism $f:L\to L$ is join-complete.