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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?

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  • $\begingroup$ Won't this end up just with $\{X, ∅\}$? $\endgroup$ – user87690 Dec 27 '17 at 11:59
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    $\begingroup$ No. You will at least get $Int(X)$ in there, which is not guaranteed to be identical to $X$ or $\emptyset$. $\endgroup$ – Andrew Bacon Dec 29 '17 at 20:49
  • $\begingroup$ What do you mean? The whole space is open in any topology, and also is trivially upwards closed. $\endgroup$ – user87690 Dec 30 '17 at 9:32
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    $\begingroup$ Oh, sorry. In topology, $X$ very often denotes the whole space, it is so hardcoded in my brain that I missed the notation here. $\endgroup$ – user87690 Dec 30 '17 at 9:34
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    $\begingroup$ @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.) $\endgroup$ – Andrew Bacon Dec 31 '17 at 8:43
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It seems that the claim is not true.

In the following paper the author considers variations of the classical Kuratowski closure-complement problem. Considering all topological operators and all Boolean operators is one of the cases considered. The example from the paper (the last paragraph of section 4) showing that infinitely many sets can be generated can be modified to give a counterexample.

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.

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  • $\begingroup$ Thanks. And thanks for the link to the paper -- I hadn't heard of Kuratowski's problem when I asked the question, but it seems extremely relevant! $\endgroup$ – Andrew Bacon Jan 1 '18 at 8:33

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