6
$\begingroup$

Let $X$ be an infinite set, and consider the lattice $\left(\mathscr P(X), \cap, \cup, \emptyset, X\right)$.

The following is well known:

Lemma Every Ultrafilter is either Principal or contains the Frechet Filter (consisting of all cofinite sets).

Proof.

The intersection of all elements in an ultrafilter can either be nonempty or empty. In the first case, it is easy to argue that by maximality, the intersection must consist of one element, hence we have a principal ultrafilter. In the second case, we have that for every element $x\in X$, there is a filter Element $S$ not containing $X$. By upward closedness, $X \setminus \{x\}\supset S$ is also contained. Hence, Our Ultrafilter must contain the Filter generated by $\left\{ X\setminus\{x\}\mid x\in X \right\}$, which is the Frechet Filter.

However, we're not always interested in $\mathscr P(X)$ – sometimes we take sublattices like the lattice of neighborhoods around a point in a topological space. There, all of our Ultrafilters restrict to ultrafilters again, but principal ones need not be principal anymore! So in some sense, the sublattice does not see the principal-ness of (some of) his ultrafilters.

Example Consider $ℝ$ with the standard topology $\mathcal T\leq \mathscr P(ℝ)$. There, for any point $x\in ℝ$, we have an ultrafilter $\mathcal O_x := \{O\subseteq \mathcal T \mid x\in O\}$ which is non-principal. However, it arises as the restriction of $\{O\subseteq \mathscr P(ℝ)\mid x\in O\}$, which is principal.

Question Given a lattice $L$ and an Ultrafilter $\mathcal U\subseteq L$, when (and only when) does there exist a lattice embedding $L\to M$ into an atomic lattice such that $U$ is the restriction of a principal ultrafilter? When does this work when we allow poset embeddings?

$\endgroup$
  • 3
    $\begingroup$ That's a very nice question! $\endgroup$ – Asaf Karagila Aug 19 at 14:16
5
$\begingroup$

Unless I'm missing something, every ultrafilter is the restriction of a principal ultrafilter.

Suppose I have a lattice $L$. Let $\mathcal{U}$ be the set of ultrafilters on $L$, and let $\mathfrak{L}$ be the lattice of subsets of $\mathcal{U}$. This lattice is atomic (think of the singletons). Moreover, the map $$i: L\rightarrow\mathfrak{L}: x\mapsto\{X\in\mathcal{U}: x\in X\}$$ is a lattice embedding.

Now let $X$ be any ultrafilter on $L$, and think about the principal ultrafilter $\hat{X}$ in $\mathfrak{L}$ generated by $\{X\}$. For $x\in L$ we have $$i(x)\in \hat{X}\iff X\in i(x)\iff x\in X.$$ So $$\hat{X}\cap im(i)=i[X],$$ as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.