Sublattice of a power set, where elements are either empty or countably infinite

Question

I was developing a new logic system that deals with undecidability (like Continuum Hypothesis is to ZFC), and this question has emerged.

Let $X$ be a countably infinite set. Endow $\mathcal{P}(X)$ with the usual lattice, i.e. Order as $\subset$, join as $\cup$, meet as $\cap$. I need a sublattice $\mathcal{L} \subset \mathcal{P}(X)$ such that:

$\emptyset \in \mathcal{L}$ (Empty set is a member)
$\forall (Y \in \mathcal{L} \setminus \{\emptyset\}) \quad |Y| = \aleph_0$ (Nonempty members are countably infinite)
$\forall (Y \in \mathcal{L}) \quad X \setminus Y \in \mathcal{L}$ (Closure under complement)
$\forall (\mathcal{M} \subset \mathcal{L}) \quad \bigcup\mathcal{M} \in \mathcal{L} $ (Closure under arbitrary join)

Note that closure under arbitrary meet is not a requirement, though closure under finite meet is.

The trivial example is $\mathcal{L} = \{\emptyset, X\}$. But the presence of undecidable statements means that this is not the case.

Up to isomorphism, I need $\mathcal{L}$ to be the biggest. Is such $\mathcal{L}$ unique? In that case, I can call $X$ a house, and can call its members residents.

Edit

I thought about it, and found some examples of $\mathcal{L}$:

For some fixed $n \in \mathbb{Z}^+$, let $X = \mathbb{Z}$, and let $\mathcal{L_n}$ be the topology generated by a subbasis $\{\langle n \rangle + m : m \in \mathbb{Z} / n \mathbb{Z} \}$.

But $\bigcup_{n \in \mathbb{Z}^+} \mathcal{L}_n$ isn't closed under complement. Does this mean there is no biggest $\mathcal{L}$?

Edit 2

It turns out that $X$ needs another requirement to be a house:

$\forall (Y \in \mathcal{L} \setminus \{\emptyset\}) \quad |\{Y \cap Z : Z \in \mathcal{L}\}| = 2^{\aleph_0}$

Or in other words, a sublattice of $\mathcal{L}$ bounded above by a countably infinite member $Y$ must be uncountable. Note that it also must satisfy if $Y = X$.

Since $\bigcap M=\Bbb N\setminus(\bigcup \Bbb N\setminus M)$ you get meets as well. Also what you want is pretty much a $\sigma$ algebra containing no finite sets — Alessandro Codenotti, Dec 02 '20 at 23:47
@AlessandroCodenotti Note that there's a slight subtlety there: you're using the fact that a $\sigma$-algebra of subsets of $\mathbb{N}$ is in fact a complete Boolean algebra, but since there are uncountable $\subseteq$-chains of sets of naturals this isn't quite trivial. — Noah Schweber, Dec 03 '20 at 00:40
@NoahSchweber My new logic system is compatible with Boolean algebra, so that's no problem. — Dannyu NDos, Dec 03 '20 at 01:17

Dannyu NDos · Accepted Answer · 2020-12-09T01:29:56.780

Self-Answered

I think I found that such $\mathcal{L}$ doesn't exist.

Pick a nonempty $Y_0 \in \mathcal{L}$. Pick $x \in Y_0$. Pick a nonempty $Y_1 \in \mathcal{L}$ such that $x \in Y_1 \subsetneq Y_0$. Such $Y_1$ exists due to the requirement in Edit 2. Repeat.

Let $Y_\omega = \bigcap_{n=0}^\infty Y_n$. Then $x \in Y_\omega$, so $Y_\omega$ is countably infinite. Now we can continue this to arbitrary countable ordinals. We now have an uncountably long chain between $X$ and $\emptyset$. This contradicts against that $X$ is countable.

This arises a new question, but that should be another post.

Sublattice of a power set, where elements are either empty or countably infinite

Edit

Edit 2

1 Answers1

Self-Answered