Let $X$ be a finite set. $\mathcal{P}(X)$ denotes the set of all subsets of $X$.
Let $\Gamma$ be a sub-lattice of $\mathcal{P}(X)$, i.e. $\Gamma$ is a collection of subsets of $X$ closed under union and intersection. Suppose $\Gamma$ non-empty and $\Gamma$ not made only of the empty set.
Set $$n:=n(\Gamma):=\min\{m\geq1\,|\, \Gamma\text{ contains at least one set of cardinality }m\}$$ and suppose $n\geq2$. I would like to find a lattice $\Gamma'$ isomorphic to $\Gamma$, but with $n(\Gamma')=1$.
With this purpose I thought to define a map $\Phi\!:\Gamma\to\Phi(\Gamma)=:\Gamma'$ doing the following operations:
- from each set $A\in\Gamma$ with cardinality $|A|=n$, choose $n-1$ elements and delete them; (Notice that all these sets $A$'s are pairwise disjoint, by the minimality of $n$ and since $\Gamma$ is closed under intersection)
- call $Y$ the set of all deleted elements;
- from each set $B\in\Gamma$ with cardinality $|B|>n$, delete the elements of $Y$.
Clearly in this way one obtains $\Gamma'\subseteq\mathcal{P}(X')$, where $X':=X\smallsetminus Y$, and $n(\Gamma')=1$. My questions are:
- is $\Gamma'$ a sub-lattice of $\mathcal P(X')$ ?
- is the map $\Phi$ an isomorphism of lattices (i.e. $\Phi$ bijection, $\Phi(A\cup B)=\Phi(A)\cup\Phi(B)\,$, $\,\Phi(A\cap B)=\Phi(A)\cap\Phi(B)\,$) ?
Edit. It is trivial that $\Phi$ is surjective (by definition of $\Gamma'$). It seems also clear that $\Phi$ is an homomorphism of lattices: it suffices to write $\Phi(A)=A\smallsetminus Y$ and apply elementary facts of set theory. Therefore the only thing that remains to prove it that $\Phi$ is injective.