Set-Theoretic Properties and Combinatorics Let $G$ be a family of subsets of $[n] = \{1,2,...,n\}$ such that for every
$C \in G$, its complement $[n]\setminus C$ is also in $G$ and for every
$A,B \in G$, $A \cap B \in G \wedge A \cup B \in G.$ I want to figure out the
number of possible sizes of $G$ assuming that $G$ is non-empty.
To me, this feels like a problem that could be handled using some form of
induction on $n$. That being said, I am still in the explorative phase of
figuring out what is the statement to induct upon. If $n = 1$ it is apparent
that $\{\{1\},\emptyset\}$ is a
legal candidate,  and thus I thought
that $1$ possible size for this.
When $n = 2$, I see that the only possible set that works in this case is
the full family: $\{\{1\},\{2\},\{1,2\}, \emptyset\}$, and thus I thought that it
would be $1$ possible size as well for this.
However, as I scale up $n$, I begin to find that this search is somewhat
unweildy. Any recommendations on how to go about finding the statement to
induct upon? In fact, is it possible that I may be better off proving this
directly instead of in an inductive manner? Any recommendations would be
appreciated.
 A: The collection $G$ is partially ordered by inclusion. Let $M$ be the set of minimal elements.
Prove $M$ is a partition of $G$ and every $c\in G$ is a union of $m\in M$.
Determine $|G|$ in terms of $|M|$.
A: In fact, such sets $G\subset 2^X$ ($X$ is a finite set) are characterized by the set of their atoms which are the equivalence classes of the relation 
$$
(x\equiv_G y) \Longleftrightarrow_{def} (\forall B\in G)(x\in B \Longleftrightarrow y\in B)
$$ 
or one can see the class of $x$ as the set $\cap_{B\in G\atop x\in B}B$.Then your sets $G$ are unions of their atoms, and the atoms are exactly set-partitions of $X$ (i.e. partitions of $X$ in non-empty blocks). 
To take your example with $n=2$, there exists two set-partitions :  $\{\{1\},\{2\}\}$ (of type 1^2) and $\{\{1,2\}\}$ (of type 2) and $G$ is the one you have given and $\{\emptyset,\{1,2\}\}$.  
With $n=3$, you have $\{\{1\},\{2\},\{3\}\}$ (of type 1^3), $\{\{x\},\{y,z\}\}$ hence 3 set-partitions (of type 1.2) and $\{\{1,2,3\}\}$ (of type 3). The types of set partitions are denoted 
$$
1^{\alpha_1}.2^{\alpha_2}.3^{\alpha_3}.4^{\alpha_4} ... 
$$
which means $\alpha_1$ singletons, $\alpha_2$ doubletons (or 2-blocks), $\alpha_3$ 3-blocks and so on ... Of course, one has 
$$
\sum_i (i\times \alpha_i)=n
$$ 
The number of set-partitions of $[n]$ is the n-th Bell number this number is filtered by the coefficients of the Bell polynomials. In the sequence of Bell numbers given by the OEIS you can check that $B(3)=5$ (our example above).
