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.
Tiny answer: Yes.
Short answer: use Birkhoff's representation theorem.
Long answer: Since $\Gamma$ is a sublattice of $\mathcal P(X)$, it is a distributive lattice. The proof of the Birkhoff's representation theorem basically says that the categories of finite distributive lattices and finite posets are dually equivalent. When you follow Birkhoff's construction from distributive lattices to posets and then back, you obtain a corrolary:
Corollary Every finite distributive lattice is isomorphic to a lattice of sets, with lattice operations $\cup,\cap$.
This corollary is sometimes called Birkhoff's representation theorem. Moreover, it turns out that the Birkhoff's construction gives you exactly what you need.
Let me show how this thing works. Assume that you are given a finite poset $P$. Call a subset $A$ of $P$ a downset if it has the following property: if $x\in A$ and $y\leq x$, then $y\in A$. Write $J(P)$ for the set of all downsets of $P$. Clearly, a $J(P)$ is closed with respect to both $\cup$ and $\cap$, so $J(P)$ is a distributive lattice. The bounds of this lattice are $\emptyset$ and $P$.
For the opposite direction, suppose you are given a distributive lattice $\Gamma$. Call an element $a$ join-irreducible if $b\vee c=a$ implies $a\in\{b,c\}$. Write $A(\Gamma)$ for the set of all nonzero join irreducible elements. ($0$ is always join-irreducible, but we have to throw it out because it would spoil the fun). It is easy to spot the elements of $A(\Gamma)$ if you draw the Hasse diagram of $\Gamma$: they are the elements of $\Gamma$ that cover exactly one element. If $\Gamma$ is a Boolean algebra, $A(\Gamma)$ are the atoms of $\Gamma$. If $\Gamma$ is a chain, then $A(\Gamma)=\Gamma\setminus\{0\}$. When you equip the set $A(\Gamma)$ with the partial order inherited from $\Gamma$, you obtain a finite poset.
And now the coolest part: for every finite poset $P$, $A(J(P))\simeq P$ and, for every finite distributive lattice $\Gamma$, $J(A(\Gamma))\simeq\Gamma$. The second isomorphism gives you the Birkhoff's representation theorem. I think this is really nice.
Back to your question: the set $A(\Gamma)$ is your $X'$ and $J(A(\Gamma))$. To see that the set system $J(A(\Gamma))$ contains a singleton, observe that the finite poset $A(\Gamma)$ must contain a minimal element $m$ and then $\{m\}\in J(A(\Gamma))$.
Example of $\Gamma$, $A(\Gamma)$ and $J(A(\Gamma))$.
Let me tell you a couple more facts: