Example of an infinite complete lattice which is distributive but not complemented

Question

Which is an example of an infinite complete lattice which is distributive but not complemented? Is the set of natural numbers under the relation divides an example? Also is the set of natural numbers under the usual $<=$ (less than or equal to) an example?

Answer 1

Every complete lattice is necessarily bounded, since the set of all elements must have a join, and the empty set must have a meet. Your second example has no maximum element, so it’s not complete. If your $\Bbb N$ includes $0$, your first example is a bounded lattice, with $0$ as its maximum element; otherwise it’s not. (Note also that quite apart from the question of completeness, if you’re going to talk about lattices being complemented or not, you really ought to restrict yourself to bounded lattices anyway, so that complements are at least possible.)

You can turn $\Bbb N$ under the usual ordering into a bounded lattice by adding a top element $\top$: let $L=\Bbb N\cup\{\top\}$, where $n<\top$ for each $n\in\Bbb N$. Every linearly ordered lattice is distributive, so $L$ is distributive, and clearly no element of $L$ besides $0$ and $\top$ has a complement: if $n\in\Bbb Z^+$, then $n\land x=0$ if and only if $x=0$, while $n\lor x=\top$ if and only if $x=\top$. $L$ is also complete; the possible difficulty is with the join of an infinite set, and that’s always $\top$.

If your $\Bbb N$ does not $0$, you can simply add a top element, as I did above. The resulting lattice is complete, but it’s not complemented; e.g., there is no $n$ such that $2\land n=1$ and $2\lor n=0$. It is distributive, however, so it’s also an example.