Does the Rieger-Nishimura lattice over a subset of $\mathbb{R}^k$ stabilize?

Question

Notation: If $U,V$ are open subsets of a topological space $X$, let us write $U\Rrightarrow V$ for the Heyting operation: the largest open subset $W$ of $X$ such that $U\cap W \subseteq V$ (i.e., the set of points $x\in X$ in a neighborhood of which $U$ is included in $V$; or $\newcommand\interior{\mathop{\mathrm{int}}}\interior(V \cup (X\setminus U))$ if you will), and $\newcommand\mytilde{\mathop{\sim}}\mytilde U$ for $U\Rrightarrow \varnothing$ (the largest open set disjoint with $U$, that is, $\interior(X\setminus U)$), the pseudocomplement of $U$.

Given an open set $P$ of a topological space $X$, let us now define the Rieger-Nishimura lattice over $P$, a sequence $(P_m)$ of open sets of $X$, by induction on $m$, as follows:

• $P_{-1} = \varnothing$

• $P_1 = P$

• $P_{2n+2} = (P_{2n+1} \Rrightarrow P_{2n-1})$ for $n\geq 0$, which also happens to be $P_{2n} \Rrightarrow P_{2n-3}$ when $n\geq 1$

• $P_{2n+3} = (P_{2n+1} \cup P_{2n+2})$ for $n\geq 0$, which also happens to be $P_{2n+1} \cup P_{2n+2}$ when $n\geq 1$

so this starts as follows

• $P_2 = \mytilde P$

• $P_3 = (P \cup \mytilde P)$ is the complement of the boundary $\partial P$ of $P$

• $P_4 = \mytilde\mytilde P$ is the smallest regular open set $\newcommand{\regularopen}{\mathop{\mathrm{ro}}} \regularopen P = \newcommand{\closure}{\mathop{\mathrm{cl}}} \interior\closure P$ containing $P$ (viꝫ. the set of points in whose neighborhood $P$ is dense)

• $P_5 = (\mytilde P \cup \mytilde\mytilde P)$ is the complement of $\partial\regularopen P$

• $P_6 = (\mytilde\mytilde P \Rrightarrow P)$ is the set of points in whose neighborhood $P$ is regular open

…after which I start to lose intuitive sense of what the $P_m$ mean. But the set of all $P_m$ together with $P_\infty := X$ is the Heyting algebra generated by $P$, i.e., the smallest set of open subsets of $X$ containing $P$ and stable under $\cap$, $\cup$ and $\Rrightarrow$.

We have $P_m \subseteq P_{m'}$ when $m\leq m'$ except possibly when $m=2n+1$ and $m'=2n+2$ or when $m=2n$ and $m'=2n+2$ for some $n \geq 0$. In particular, if $P_m = X$ then all $P_{m'}$ are also $X$ eventually (for $m' \geq m+3$). For example, starting from $P := \mathopen]-1,0\mathclose[ \cup \mathopen]0,1\mathclose[ \subseteq \mathbb{R}$, we have $P_7 = \mathbb{R}$ and $P_m = \mathbb{R}$ for $m\geq 9$.

Now in general it is possible for the $P_m$ for $m<\infty$ to be all distinct from $X$, but this will be for some specially fabricated topological space $X$. What I don't know, and this is my question, is whether this can occur for $X = \mathbb{R}^k$:

Question: Is there $P \subseteq X$ open with $X := \mathbb{R}^k$ such that the $P_m$ are all distinct from $X$? Or to ask an even more specific question, if we call $m_0(P)$ the smallest $m \in \mathbb{N}\cup\{\infty\}$ such that $P_m = X$ for $m$ (which we might call the “Rieger-Nishimura rank” of $P$), what are the possible values of $m_0(P)$ for $P \subseteq \mathbb{R}^k$? (The answer could depend on $k$ for all I know.)

(My first intuition was to try to connect this “Rieger-Nishimura rank” to the Cantor-Bendixson rank of the complement of $P$, but I don't think they're related at all.)

Answer 1

The notation in the question prevents me from discussing a sequence of sets or applying the Rieger–Nishimura operations in several spaces at once, hence I will write $R(P,X,m)$ for what you denote $P_m$.

There is such a $P\subseteq\def\R{\mathbb R}\R^k$ for each $k\ge1$. First, intuitionistic propositional logic is complete with respect to the topological semantics when the topological space is fixed as $\R^k$ (this is true for every dense-in-itself separable metrizable space, see A. Tarski, Der Aussagenkalkül und die Topologie, Fundamenta Mathematicae 31 (1938), no. 1, pp. 103–134). This by itself implies that for each $m$, there exists $P$ such that $R(P,\R^k,m)\ne\R^k$. Better yet, we can fix sets $P_m\subseteq(m,m+1)^k$ such that $R(P_m,(m,m+1)^k,m)\ne(m,m+1)^k$. It is easy to check that $R(P,X,m)\cap U=R(P\cap U,U,m)$ for any open $U\subseteq X$, thus taking $P=\bigcup_mP_m$, we have $R(P,\R^k,m)\ne\R^k$ for all $m$.