Let $L$ be a complete lattice. An element of lattice $x$ is called $\wedge$-irreducible if $x=y\wedge z$ implies $x=y$ or $x=z$. Similarly, it is completely $\wedge$-irreducible if $x=\bigwedge x_i$ implies $x=x_i$ for some $i$. $\beta\subseteq L$ is a basis for $L$ if any element of $L$ can be expressed as an infimum of some subset of $\beta$. Any algebraic lattice has basis of both $\wedge$ irreducibles and completely $\wedge$ irreducibles. I have two questions about these notions:
(1) Is there some weaker (than algebraicity) sufficient condition for completele lattice to have either of these basis?
(2) Is there some additional condition $(\alpha)$ such that it would hold: whenever $L$ has both ($\alpha$) and basis of $\wedge$ irreducibiles, then it has a basis of completely $\wedge$- irreducibles?
I'm trying to answer the first of your questions.
Let $M(L)$ be the set of completely meet-irreducible elements from $L$.
In Formal Concept Analysis, from Ganter and Wille, page 35 (English version), there is a concept lattice of classes of complete lattices.
There, they call $M(L)$ infimum-dense to those lattices which may not be algebraic but in which every element is a (possibly infinite) meet of elements from the set $M(L)$.
This class is certainly different from the one of algebraic lattices.
From what you've pointed out, it is clear to you that this class contains the one of algebraic lattices.
I'm not sure if this was one of your questions, but an example of a complete lattice $L$ which is $M(L)$ infimum-dense but not algebraic can be as follows.
$L = \{\varnothing\} \cup \mathfrak{P}_{cofin}(\mathbb{N})$, that is, the co-finite subsets of the natural numbers (the subsets whose complements are finite), together with the empty set, ordered by inclusion.
In this case, $M(L) = \{ \mathbb{N}\setminus\{n\} : n \in \mathbb{N} \}$ and the only compact element is $\varnothing$, so that $L$ is not algebraic.