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

Consider the following statement:

If $L$ contains an ideal $J$ such that $\bigvee_L J \notin J$, then there is a complete lattice $K$ and an incomplete lattice homomorphism $f: L\to K$.

Is this true? (The converse is true.)

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  • $\begingroup$ Since incomplete lattice homomorphisms are only required to not preserve all least upper bounds, I suggest calling these homomorphisms join-incomplete instead of incomplete. $\endgroup$
    – Joseph Van Name
    Commented Jul 9, 2015 at 18:08
  • $\begingroup$ Here is question 1. mathoverflow.net/q/211063/22277 $\endgroup$
    – Joseph Van Name
    Commented Jul 9, 2015 at 18:11
  • $\begingroup$ +1 since this question gives a more categoric theoretic formulation of when a complete lattice satisfies the ascending chain condition. $\endgroup$
    – Joseph Van Name
    Commented Jul 9, 2015 at 21:10

1 Answer 1

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Yes. Suppose that $J$ is an ideal in $L$ such that $\bigvee^{L}J\not\in J$. Let $\mathcal{Id}(L)$ be the set of all ideals of $L$. Then $\mathcal{Id}(L)$ is a complete lattice. Define a mapping $V:L\rightarrow\mathcal{Id}(L)$ by letting $V(x)=\downarrow x$ for each $x\in L$. Then $V$ is a lattice homomorphism. However, since $\bigvee^{X}J\not\in J$, we have $\bigvee V[J]=J$, but $V[\bigvee J]\neq J$, so $V$ is an incomplete lattice homomorphism.

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  • $\begingroup$ Great idea to look at the (complete) lattice of ideals! - I agree that join-incomplete is the better terminology. $\endgroup$
    – Dominic van der Zypen
    Commented Jul 10, 2015 at 6:57

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