Set closed under Boolean operations and the interior operation

Let $(\preceq, P)$ be a well-founded preorder: there is no infinite sequence $... a_3 \preceq a_2 \preceq a_1$ where $a_1, a_2 ...$ are all distinct.

Consider the topology on $P$ generated by letting the upwards closed sets be open. Given a set $X\subseteq P$ I'm interested in the smallest collection of subsets of $P$ that contains $X$ and is closed under the Boolean operations and the interior operation. We can build this up in stages as follows:

• $S_0 = \{X\}$
• $S_{n+1} = \{int(Y), Y\cap Z, P\setminus Y \mid Y, Z\in S_n\}$

The union of these sets will be closed under the required operations. My suspicion is that, since $\preceq$ is well-founded, $S_n$ should already be closed under those operations for some finite $n$. Is this true?

• Won't this end up just with $\{X, ∅\}$? – user87690 Dec 27 '17 at 11:59
• No. You will at least get $Int(X)$ in there, which is not guaranteed to be identical to $X$ or $\emptyset$. – Andrew Bacon Dec 29 '17 at 20:49
• What do you mean? The whole space is open in any topology, and also is trivially upwards closed. – user87690 Dec 30 '17 at 9:32
• Oh, sorry. In topology, $X$ very often denotes the whole space, it is so hardcoded in my brain that I missed the notation here. – user87690 Dec 30 '17 at 9:34
• @user87690 Yes. Consider the naturals with the opposite ordering (so there's a largest element and no smallest element). Then let $X$ be the evens, $Y_0 = int(X)$ and let $Y_{n+1} = int(X \cup Y_n)$ if $n$ is odd, and $int(\overline{X} \cup Y_n)$ if $n$ is even. $Y_n = \{0,...,n\}$ and first appears in $S_n$. (Thanks for your answer, I will think about it shortly.) – Andrew Bacon Dec 31 '17 at 8:43

Consider the set $n = \{0, 1, …, n - 1\} = [0, n)$ with the left order topology (corrseponding to the reversed ordering), and the starting set $X$ will be the odd numbers. We have $\overline{X} = [1, n)$, so $\overline{X} ∩ n \setminus X = [2, n) ∩ X$ and $\overline{[2, n) ∩ X} = [2, n)$. And so on. This way we build all the intervals and so whole $\mathcal{P}(n)$.
The point is we can consider disjoint union $P := \coprod_{n ∈ ω} n$, and start with the disjoint union of all copies of corresponing even numbers. The process above works in paralel, and builds infinitely many sets. On the other hand, since every $n$ is well-ordered, $P$ is also well-ordered.