Is the meet of complemented elements complemented? We are not supposing anything about the lattice. This question raised because a paper that I'm reading and says:
Define $({A})_{c}=\uparrow({A}\cap{C}_L)$
where $\mathcal{C}_{L}$ is the set of all complemented elements of the lattice $L$.
If $\mathcal{A}$ is a filter, then $(\mathcal{A})_{c}$ is a filter again.
I don't see how that could be true without ensuring that $\mathcal{A}\cap\mathcal{C}_{L}$ is meet closed. The paper doesn't said explicitly what it means with $\uparrow$.
Not in general.
Take the following lattice, for example
Here, $a$ and $b$ are complemented, with complements $a'$ and $b'$, respectively.
However, there is no complement for $a \wedge b$, as it can be easily checked.
Generally, it is common to define, for a poset $P$, $${\uparrow}A = \{x \in P : a \leq x, \text{ for some } a \in A\}.$$ With the lattice above, take $A = \{a,b,a\wedge b, 1\}$;
then $A \cap C_L = \{a,b,1\}$, and ${\uparrow}(A\cap C_L)= \{a,b,1\}$ is not a filter.
It may be that that author gives another meaning to $\uparrow$, a different definition of filter, or that (s)he thinking of a particular class of lattices.
For example, that property holds for distributive lattices; actually, semi-distributivity is enough.
It will also hold for other lattices, but I don't know of any defining formula more general that semi-distributivity which entails that property.