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Let $(L, \leq)$ be a lattice defined using a partial order $\leq$. In this document, it is claimed without proof that any complete lattice is a Boolean algebra (example 2.9 (2)). In the same document, a Boolean algebra is defined as a distributive lattice in which every element $a$ has a complement $\neg a$, such that $a \land \neg a = 0$ and $a \lor \neg a = 1$.

I found this related MSE post, but no satisfying answer was given. One of the points made in this thread was that a complete lattice is not necessarily distributive. For my purposes, we also can add this assumption if needed (if we use the definition in the linked document).

I think that we can define $0$ as the meet of $L$ itself and define $1$ as the join of $L$, for $L$ is a complete lattice. I am unsure how to define $\neg a$ for a given $a \in L$. Is this example true? If so, how can I construct a Boolean algebra on $L$?

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    $\begingroup$ $0$ is the meet of $\varnothing$, not "of $L$ with itself". $\endgroup$
    – Anne Bauval
    Commented Aug 15 at 14:54
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    $\begingroup$ @AnneBauval No, $0$ is the meet of $L$. The meet of $\varnothing$ is $1$. $\endgroup$
    – David Gao
    Commented Aug 15 at 15:47

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That statement is simply wrong. For example, the three-element linear order is a complete (even completely distributive) lattice but not a Boolean algebra.

What is true is that any complete lattice must be bounded (= have a top and bottom element). Moreover, we can "approximate" complements: for any element $a$ of a complete lattice $L$, we can form the sup of the set of all elements meeting $a$ to $0$, and similarly form the inf of the set of all elements joining $a$ to $1$. In a Boolean algebra, these elements - call them "$a^*$" and "$a_*$" respectively - are equal, and the map $a\mapsto a^*$ satisfies some nice properties. But this doesn't hold in general complete lattices: consider for example the five element lattice $M_3$ (which is trivially complete, as are all finite nonempty lattices).

It's separately worth noting that if we have joins of arbitrary sets then we also have meets of arbitrary sets. This is a fun exercise, and when teaching it I call it "Arby's theorem:" if we're an upper-complete semilattice, then

we have the meets.

(Sorry. :P)

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    $\begingroup$ Namely, the join (or supremum) is the meet (or infimum) of the upper bounds, while the meet (or infimum) is the join (or supremum) of the lower bounds. $\endgroup$
    – Geoffrey Trang
    Commented Aug 15 at 14:43
  • $\begingroup$ I see, thank you! I will accept your answer as soon as MSE allows me to. $\endgroup$
    – hff1
    Commented Aug 15 at 14:45
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    $\begingroup$ +1 for the joke. A proof of it in French: exercise 2-9, question 4. $\endgroup$
    – Anne Bauval
    Commented Aug 15 at 14:53
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    $\begingroup$ One thing that is true is that a complete distributive lattice, where distributivity "is also complete" (i.e. $x \wedge \bigvee_{i\in I} y_i = \bigvee_{i\in I} (x \wedge y_i)$ even for infinite $I$), is a Heyting algebra. That's a structure similar to a Boolean algebra, but with a "relative pseudocomplement" operation $\rightarrow$ satisfying certain conditions, and it forms a model of intuitionistic logic similar to the way a Boolean algebra forms a model of classical logic. $\endgroup$
    – Daniel Schepler
    Commented Aug 15 at 15:57
  • $\begingroup$ @DanielSchepler Good idea, done! $\endgroup$
    – Noah Schweber
    Commented Aug 15 at 16:11

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