My problem is to show that any lower-bounded lattice satisfying the maximal condition is a complete lattice.
Let's call the lattice $L$. I'm having some trouble with this. I have tried to look at it from a maximal condition perspective. I saw two directions I could potentially go. One was to use the fact that the maximal condition is equivalent to the ACC in partial orders and see if I could do something with that, but I got nothing. The other direction I tried was to gather the maximal elements from each subset. I noticed that this was necessarily an anti-chain (maximal, no less). But I couldn't really see how this could help either.
I'm sure this is easier than I think. Any help is appreciated!
The maximal condition says that every non-empty subset has a maximal element; ie there's an totally-defined set-valued operation $max$ such that for any subset $S$ we have that $\max S \neq \emptyset$ (!) and $$\forall m \in \max S; s \in S :: m \not< s \text{ and } m \in S$$
To show completeness, it suffices to show that every subset has an infimum (or equivalently supremum).
Recall for a set $S$ and an element $i$, that:
$ \;\;\;\; i \text{ greatest lower bound of } S \\\equiv i \text{ lower bound of } S \text{ and greatest such bound} \\\equiv i \in lbd \, S \text{ and } \forall l \in lbd \, S :: l \leq i, \text{ where lower bounds: } lbd \, T := \{l \,|\, (\forall t \in T :: l \leq t)\} \\\equiv i \in lbd \, S \text{ and } \forall l \in lbd \, S :: i \not< l \text{ and } \forall l \in lbd \, S :: l \leq i \\\equiv i \in \max(lbd \, S) \text{ and } i \in ubd(lbd \, S), \text{ where uppers: } ubd \, T := \{u \,|\, (\forall t \in T :: t \leq u)\} \\\equiv i \in \max(lbd \, S) \,\cap\, ubd(lbd \, S) $
Now with the given, it remains to show that this set $E := \max(lbd \, S) \,\cap\, ubd(lbd \, S)$ is non-empty, if $S$ is nonempty, and is in fact a singleton set $\{ e \}$. Then we define, $\inf S := e$.
I've personally not made these checks; if you make them, please post them.
From the comments: Let L be a lower-bounded lattice satisfying the maximal condition and let X be an arbitrary subset of L. Since L is lower bounded, lbd(X) is non-empty, and hence has at least one maximal element. To see that max(lbd(X)) is a singleton, let a,b∈max(lbd(X)). Then, since L is a lattice, a∨b exists, call it c. We claim that c∈lbd(X). <-- Here is where I need help. If I can show that, then a=b=c and there would be no elements in lbd(X) that are greater than c, hence c must be ⋀X.
You've shown $max \ lbd \ X$ has at-least one element; ($\bot \in L \implies \bot \in lbd \ X \implies max \ lbd \ X \neq \emptyset$).
It remains to show that $max \ lbd \ X$ has at-most one element. Let $a,b$ be in that set, then
$\;\;\;a,b \in max \ lbd \ X \\\Rightarrow a,b \in lbd \ X \;\;\text{ by defn of max} \\\equiv \forall x \in X :: a,b \leq x \;\;\text{ defn of lbd} \\\Rightarrow \forall x \in X :: a \vee b \leq x \text{ since $x$ is upper bound of $a,b$ means it's at-least the least upper bound of $a,b$} \\\equiv a \vee b \in lbd \ X \;\;\text{ by defn of lbd } $
Now we have $a,b \in max \ lbd \ X$ and $a \vee b \in lbd \ X$. But by maximality we then have $a,b \not< a \vee b$, ie $a = a \vee b = b$. Hence, $max \ lbd \ X$ has exactly one element :)
Let us call this unique element $m$, then according to the calculation way above in this post, we need to show that $m \in ubd \ lbd \ X$ in order to conclude that $m$ is in-fact the glb of $X$.
(See difference between maximal element and greatest element )