# Is a chain-complete lattice a complete lattice without the axiom of choice?

This question is inspired by this question. Consider the following result:

Let $(L,\leq)$ be a chain-complete lattice. Then $(L,\leq)$ is a complete lattice.

Can this result be proven without using the axiom of choice?

For completeness, here is a proof that the result holds under the axiom of choice: Let $S\subseteq L$. We want to show that $S$ has an upper bound. Since $\emptyset$ is a chain, we can assume $S\neq\emptyset$. Let $\preceq$ be a well ordering of $S$. Fix $s\in S$. Define a transfinite sequence in the following way: Let $s_0=s$. For a successor ordinal $\alpha+1$ we let $s_{\alpha+1}=s_\alpha\vee m$ with $m$ being the $\preceq$-smallest element in $S$ that is not smaller than $s_\alpha$. For a limit ordinal $\beta$, let $s_\alpha=\sup_{\beta<\alpha}s_\alpha$ which exists as the supremum of a chain. The range of this sequence is clearly a chain and its supremum is the supremum of $S$. That $S$ has an infimum follows in the usual way by taking the infimum to be the supremum of all lower bounds.

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I am going to be in Vienna from the 5th to the 19th, if you wanna meet for a cup of beer, let me know. I'll be visiting the Kurt Godel center. – Asaf Karagila Aug 28 '12 at 17:14
@Asaf: I'll be probably much of the time in Innsbruck, but I'm sure to be in Vienna some time during the period. I'll drop you an email a few days before. – Michael Greinecker Aug 28 '12 at 17:26
This is still one of my favorite answers. – Asaf Karagila Sep 27 '12 at 23:39

Let $A$ be an amorphous set, that is an infinite set that every subset is either finite or co-finite. Let $L$ be the collection of finite subsets of $A$, ordered by inclusion.

From Jech's The Axiom of Choice, Chapter 4 exercise 10, we have that $A$ is T-finite, namely every chain has a maximal element. This maximal element is indeed the supremum of the chain.

However note that this is obviously not a complete lattice, since the collection of singletons clearly does not have an upper bound in this lattice.

For sake of completeness, let us prove the above fact on the chains. Suppose that $D$ is an infinite chain, in particular there is at most one element of every finite cardinality. The union, then, is a countable subset of $A$ in contradiction to its Dedekind finiteness.

Thinking about this for a little while, we can easily generalize this proof to show that $DC_\kappa$ is not enough to prove the equivalence you are looking for:

We say that $A$ is $\kappa$-amorphous if every subset of $A$ is of size $<\kappa$, or its complement is. In this terminology amorphous is $\omega$-amorphous.

It is consistent to have $A$ which is $\kappa$-amorphous along with $DC_{<\kappa}$. In this setting the collection of sets of cardinality $<\kappa$ ordered by inclusion has the same properties as above (replacing maximal element by supremum). But there are not $\kappa$-chains. Again the singletons prove the lack of upper-bound.

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Thank you! I found the exercise in Jech's "Axiom of Choice" as Exercise 10 in the chapter on permutation models. – Michael Greinecker Apr 6 '12 at 14:24
You're welcome. It was a fun exercise! :-) – Asaf Karagila Apr 6 '12 at 14:26