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The problem (taken from Kaplansky's Set theory and metric spaces, p. 40) is as follows

Let $L$ be a lattice in which every chain has a least upper bound and a greatest lower bound. Prove that $L$ is complete.

Using the fact that $L$ is a lattice in which every chain has an upper bound I was able to prove that $L$ has a unique maximal element (with Zorn's lemma). With this I can prove that an empty set and any subset of $L$ containing its maximal element have a least upper bound.

But I get stuck trying to prove that a non-empty subset of $L$ not containing the maximal element has a least upper bound. How can I do this? Note that I can not use the axiom of choice here, only Zorn's Lemma.

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  • $\begingroup$ What's the difference between using the axiom of choice and Zorn's lemma? $\endgroup$ – Asaf Karagila Oct 13 '16 at 20:51
  • $\begingroup$ The only difference is that at this stage in the book Kaplansky hasn't shown what the axiom of choice is yet, so I'm supposed to be able to solve this exercise without it. $\endgroup$ – tadejsv Oct 13 '16 at 21:26
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I'll show that Kaplansky's assumptions imply that every subset $X$ of your lattice $L$ has a greatest lower bound.

Step 1: Let $Y$ be the set of all lower bounds of $X$. I claim that, if $C$ is a chain in $Y$, then $C$ has an upper bound (in fact a least upper bound) in $Y$. By assumption, $C$ has a least upper bound $b\in L$. Because $b$ is least, every element of $X$, being an upper bound for $C$, is $\geq b$. This means $b\in Y$, as required.

Step 2: In view of Step 1, we can apply Zorn's Lemma to conclude that $Y$ has a maximal element $m$.

Step 3: I claim that $m$ is $\geq$ all elements of $Y$. To see this, consider any $y\in Y$. Each element of $X$ is $\geq$ both $m$ and $y$ (since they're both in $Y$) and therefore also $\geq m\lor y$. So $m\lor y$ is an element of $Y$ that is $\geq m$. By maximality of $m$, we conclude that $m\lor y=m$, which is equivalent to $y\leq m$.

So $m$ is the largest element of the set $Y$ of lower bounds of $X$, i.e., $m$ is the greatest lower bound of $X$.

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  • $\begingroup$ But how do you know that every element of $X$ is $\geq b$? Obviously it can not be $\leq b$, but could it be the case that there is simply no relation between some element of $X$ and $b$? $\endgroup$ – tadejsv Oct 13 '16 at 21:29
  • $\begingroup$ Nevermind, I see it now :) $\endgroup$ – tadejsv Oct 13 '16 at 21:38
  • $\begingroup$ How is this not assuming the conclusion? i.e. how do we know that the set $X$ has a lower bound at all? $\endgroup$ – tmpys Mar 10 '18 at 3:56
  • $\begingroup$ @tmpys (1) The OP says "I can prove that an empty set ... [has] a least upper bound". That gives you a smallest element in $L$ (because everything is an upper bound for the empty set), which is then a lower bound for any $X$. (2) The hypothesis of the problem includes the fact that every chain has a least upper bound; apply that to the empty chain. (3) I didn't use or assume that $Y$ is nonempty until after Step 1, in which this nonemptiness is (along with more information) proved. $\endgroup$ – Andreas Blass Mar 10 '18 at 4:09
  • $\begingroup$ The empty set having a lub comes from the assumption about chains having lub? $\endgroup$ – tmpys Mar 10 '18 at 4:21
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HINT: Let $A$ be a non-empty subset of $L$. Let $L_A=\{x\in L:\exists X\subseteq A\,(x=\sup X)\}$. Show that if $C$ is a chain in $L_A$, then $\sup_LC\in L_A$. Your argument that $L$ has a maximum element can now be applied to $L_A$.

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