Constructing a partition of a nested family of sets that is closed under unions and nonempty intersections.

Question

I have some set theory background, but neither topology nor functional analysis are fields where I feel comfortable, so this may be a very simple question to people familiar with those fields, but I'm stumped. Those considerations have popped up in my thesis, which covers topics from philosophical logic.

I have simplified the question since first posting it.

Let $U_x$ be a subset of $\mathcal P(S)$ such that $x \in \cap U_x$: with the following properties:

$(1)\,\,\,U_x$ is nested: for all $X,Y \in U_x$ either $X \subseteq Y$ or $Y \subseteq X$.

$(2)\,\,\,U_x$ is closed under unions

$(3)\,\,\,U_x$ is closed under non-empty intersections

Question: (I have refined my question, after receiving feedback that it's not clear enough.)

My question is whether from the properties of $U_x$ alone we can easily derive a particular type of partition of $S$?

The kind of partition I have in mind can be exemplified on the assumption that $S$ is finite. If $S$ is finite then we can order its elements by natural numbers. So there exists an $n \in \mathbb N$ such that $U_x=\{X^0, X^1,...,X^n\}$ has the following form: $X^0 \subseteq X^1 \subseteq... \subseteq X^{n-1} \subseteq X^n$. This allows for the definition of the following equivalence relation on $S$: $\; y \sim_x z \;$ iff $\; \{y,z\}\subseteq X^{k-1}\setminus X^k$ for some $ 0\lt k\le n$. And the set $S/\sim_x$ is a partition of $S$.

Can this kind of partition always be achieved for any $S$ and $U_x$ as defined, or would some additional assumptions need to be made? By this kind I mean all elements of $U_x$ contributing in an orderly manner as in the given example.

What if we additionally say that:

$(4)\,\,\,$ There are no infinitely descending $\subseteq$-sequences, $...\subseteq Y\subseteq X$ of smaller and smaller $\qquad X,Y,... \in U_x$? That is, the ordering of all $X \in U_x$ by set inclusion is a well-ordering.

Any help will be appreciated.

Answer 1

The problem with this question is that the answer is "yes", but stupidly so:

"My question is whether from the properties of $U_x$ alone we can easily derive a partition of $S$?"

Yes. In many ways:

1. Partition $S$ into singletons.
2. Partition $S$ into a single subset, $S$.
3. Partition $S$ into $S-\{x\}$ and $\{x\}$.

(any of these partitions might be identical, depending on the set $S$).

The first two use properties of $U_x$ vacuously; the third uses the property that $S$ is nonempty, because $x$ must be an element of $S$.

I think that you really need to try harder to formulate the question you want to ask (as both my and @Not Mike's comments suggested). The part where you can't quite say what exactly you mean? That's the hard one, and we can't help, because we don't know what you mean either. :)