Is a chain complete lattice complete?

Question

If every chain in a lattice is complete (we take the empty set to be a chain), does that mean that the lattice is complete? If yes, why?

My intuition says yes, and the reasoning is that we should somehow be able to define a supremum of any subset of the lattice to be the same as the supremum of some chain related to that lattice, but I've not abled to make more progress on the same. ANy suggestions?

Answer 1

If $L$ is not complete, it has a subset with no join; among such subsets let $A$ be one of minimal cardinality, say $A=\{a_\xi:\xi<\kappa\}$ for some cardinal $\kappa$. For each $\eta<\kappa$ let $A_\eta=\{a_\xi:\xi\le\eta\}$; $|A_\eta|<\kappa$, so $A_\eta$ has a join $b_\eta$. Clearly $b_\xi\le b_\eta$ whenever $\xi\le\eta<\kappa$, so $\{b_\xi:\xi<\kappa\}$ is a chain; indeed, with a little more argumentation we can assume that the chain is a strictly increasing $\kappa$-sequence. Now let $b=\bigvee_{\xi<\kappa}b_\xi$ and show that $b=\bigvee A$ to get a contradiction.

Answer 2

By completeness, I mean every nonempty subset has a sup and inf. I am not sure if a chain-complete poset is complete and my personal guess is negative. Here is a proof showing a nonempty chain-complete lattice $L$ is complete, with the lattice property emphasised.

By Zorn's lemma, $L$ has a maximal element $M$. It is the maximum: For every $x\in L$, $x\vee M$ exists because $L$ is a lattice. Now $x\vee M\ge M$, so $M=x\vee M$ by maximality of $M$. Then $M\ge x$. So, $M=\max(L)$. Idem, $\min(L)$ exists.

For every nonempty subset $A\subset L$, let $B:=\{x\in L:x\le a,\forall a\in A\}$. Then $\min(L)\in B$, so $B$ is nonempty. For every chain $C\subset B$, $\sup_L(C)$ exists by chain-completeness. For every $a\in A$, $c\in C$, we have $c\le a$, so $\sup_L(C)\le a$ and then $\sup_L(C)\in B$. By Zorn's lemma again, $B$ has a maximal element $b$. Again, it is $\max(B)$: For every $b'\in B$, $b\vee b'$ exists because $L$ is a lattice. For every $a\in A$, $b\le a$ and $b'\le a$ give $b\vee b'\le a$. So, $b\vee b'\in B$. As $b$ is maximal in $B$ and $b\vee b'\ge b$, we get $b=b\vee b'\ge b'$. So, $b=\max(B)=\inf_L(A)$. Idem, $\sup_L(A)$ exists. The proof is completed.