Largest proper subfamily of $P(S)$ closed under unions and intersections 
Take a set $x$ with $10$ distinct elements.
Every time you have two subsets $A$ and $B,$ you also have $A \cup B$ and $A \cap B.$ What is the maximum number of subsets you can have such that you don't have all subsets in $P(x)$

I thought it would be $2^{n-1}=2^9,$ but learned that this was wrong because it was distinct and you can not have all subsets in $P(x).$
I later concluded that this might be $2^8$ to make it more stricter. I am not sure about this. Does anyone have a proof or explanation of how to solve this problem.
 A: In other words, you want the maximum size of a proper "sublattice" of $P(x)$, i.e., a proper subset of $P(x)$ which is closed under $\cup$ and $\cap$. Here's how you can get a proper sublattice with $3\cdot2^8$ elements: let $a,b\,$ be two distinct elements of $x$, and take all sets $A$ in $P(x)$ such that $A\cap\{a,b\}\ne\{a\}$. This is in fact the maximum. More generally:
Theorem. Let $X$ be an $n$-element set, $n\ge2$. The maximum size of a proper sublattice of $P(X)$ is $3\cdot2^{n-2}$; in other words, a proper sublattice can contain $3/4$ of the elements of $P(X)$, but no more.
Proof. I've already shown how to construct a sublattice containing exactly $3/4$ of the elements of $P(X)$. I will prove by induction on $n$ that, for an $n$-element set $X$, any proper sublattice of $P(X)$ contains at most $3/4$ of the elements of $P(X)$.
Suppose $L$ is a proper sublattice of $P(X)$. Choose $a\in X$ so that $\{a\}\notin L$.
Case 1. There is no set $Y\in L$ such that $a\in Y\ne X$.
That is, $L\subseteq\{Y\in P(X):a\notin Y\}\cup\{X\}$. In that case, 
$|L|\le2^{n-1}+1\le3\cdot2^{n-2}$ and we're done.
Case 2. There is a set $Y\in L$ such that $a\in Y\ne X$.
Now $L'=L\cap P(Y)$ is a proper sublattice of $P(Y)$; by the induction hypothesis, it contains at most $3/4$ of the elements of $P(Y)$. Since $Y\in L$, we have $L\subseteq\{A\in P(X):A\cap Y\in L'\}$, whence $L$ contains at most $3/4$ of the elements of $P(X)$.
