A non-algebraic complete lattice

Question

Do you have an example of a complete lattice which is not algebraic‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌?

Answer 1

Consider $L = \mathbb{R} \cup \{-\infty, +\infty\}$ with usual ordering on $\mathbb{R}$ and $-\infty, +\infty$ are the least and the greatest elements of $L$ respectively. This lattice is complete.

Let $x \in \mathbb{R}$ and $C = (x-1,x)$ so $x \leqslant \bigvee C = x$, but for any finite $B \subset C$ we have $x > \bigvee B$, hence $x$ is not compact. Same argument shows that $+\infty$ is not compact too. So the only compact element of $L$ is $-\infty$ and since $\bigvee\{-\infty\} = -\infty$, there is no element in $L \setminus \{-\infty\}$ which can be represented as a join of compact elements. In particular, this lattice is not algebraic.

Answer 2

Let $X$ be a topological space and let $\Omega$ be the set of open subsets of $X$, partially ordered by inclusion. Then $\Omega$ is a complete lattice, and it is not hard to see that a compact element of $\Omega$ is precisely a compact open subset of $X$. If $X$ is a connected Hausdorff space, the only compact open subsets of $X$ are $\emptyset$ and $X$ itself, so $\Omega$ is an algebraic lattice if and only if $X$ is the one-point space.

In particular, the lattice of open subsets of, say, $\mathbb{R}$ is complete but not algebraic.