Existence of strongly proper dyadic subbases

TL;DR

This paper proves that every locally compact separable metric space admits a strongly proper dyadic subbase, enabling a domain representation that is robust and independent of enumeration.

Contribution

It establishes the existence of strongly proper dyadic subbases for all locally compact separable metric spaces, advancing the theory of domain representations.

Findings

Every locally compact separable metric space has a strongly proper dyadic subbase.

Strongly proper dyadic subbases induce admissible domain representations.

The result is independent of the enumeration of the subbase.

Abstract

We consider a topological space with its subbase which induces a coding for each point. Every second-countable Hausdorff space has a subbase that is the union of countably many pairs of disjoint open subsets. A dyadic subbase is such a subbase with a fixed enumeration. If a dyadic subbase is given, then we obtain a domain representation of the given space. The properness and the strong properness of dyadic subbases have been studied, and it is known that every strongly proper dyadic subbase induces an admissible domain representation regardless of its enumeration. We show that every locally compact separable metric space has a strongly proper dyadic subbase.

Equations

$$\sigma(k):=\left\{\begin{array}[]{ll}\sigma(k)&(k<n)\\ \bot&(n\leq k<\omega).\end{array}\right.$$

$$\sigma\tau(k):=\left\{\begin{array}[]{ll}\sigma(k)&(k<\operatorname{len}(\sigma))\\ \tau(k-\operatorname{len}(\sigma))&(\operatorname{len}(\sigma)\leq k<\omega).\end{array}\right.$$

$$S(\sigma )$$

$$\bar{S}(\sigma )$$

$$\bar{S}(\sigma )=\bigcap _{\begin{array}{c}k\in \mathrm{dom}(\sigma )\end{array}}(S_k^{\sigma (k)}\cup S_k^{\partial })=\bigcap _{\begin{array}{c}k\in \mathrm{dom}(\sigma )\end{array}}X\setminus S_k^{1-\sigma (k)}=X\setminus \bigcup _{\begin{array}{c}k\in \mathrm{dom}(\sigma )\end{array}}{S}_k^{1-\sigma (k)},$$

$$S_n^a=X\setminus \mathrm{cl}{S}_n^{1-a}=\mathrm{int}(X\setminus S_n^{1-a})=\mathrm{int}(S_n^a\cup S_n^{\partial }).$$

$$\mathrm{int}\bar{S}(\sigma )=\mathrm{int}\bigcap _{\begin{array}{c}k\in \mathrm{dom}(\sigma )\end{array}}(S_k^{\sigma (k)}\cup S_k^{\partial })=\bigcap _{\begin{array}{c}k\in \mathrm{dom}(\sigma )\end{array}}\mathrm{int}(S_k^{\sigma (k)}\cup S_k^{\partial })=\bigcap _{\begin{array}{c}k\in \mathrm{dom}(\sigma )\end{array}}{S}_k^{\sigma (k)}=S(\sigma ).$$

$${\sigma }_{0,1}(n)$$

$${\sigma }_{\partial }(n)$$

$$\varphi_{S}(x)(n):=\left\{\begin{array}[]{ll}0&(x\in S_{n}^{0})\\ 1&(x\in S_{n}^{1})\\ \bot&(\text{otherwise})\end{array}\right.$$

$$K_S$$

$$D_S$$

$$\varphi_{S}(x):=\left\{\begin{array}[]{ll}\bot\bot 10^{\omega}&(\text{if }x=p)\\ \varphi_{G}(x)&(\text{otherwise}).\end{array}\right.$$

$$\begin{array}{rl}{U}^0(f,c)& :=\{x\in X:f(x)<c\},\\ U^1(f,c)& :=\{x\in X:f(x)>c\},\\ U^{\partial }(f,c)& :=f^{-1}(c).\end{array}$$

$$S_n^a:=U^a(f_n,c_n)$$

$$\bar{S}(\sigma [n\mapsto a])=\bar{S}(\sigma )\setminus S_n^{1-a}=\mathrm{cl}S(\sigma )\setminus S_n^{1-a}$$

$$\mathrm{cl}S(\sigma )\setminus S_n^{1-a}=\mathrm{cl}(S(\sigma )\cap S_n^a)=\mathrm{cl}S(\sigma [n\mapsto a]).$$

$$A\text{ is codense }\iff A\text{ is nowhere dense }\iff A\text{ is meagre }\iff A=\varnothing .$$

$$g(K\cap U^{\partial }(f,c))\setminus g(B\cap U^a(f,c))\subseteq g(K\cap U^{\partial }(f,c))\setminus \mathrm{int}g(B\cap U^a(f,c)).$$

$$\varnothing \ne {\mathrm{fat}}_g(K\cap U^{\partial }(f,c)\cap g^{-1}(V))$$

$$\subseteq {\mathrm{fat}}_{g}X\cap B\cap U^{\partial }(f,c)\cap g^{-1}(V).$$

$$\bigcup _{n<\omega }g(K_n\cap U^{\partial }(f,c))\setminus g(B\cap U^a(f,c))=g(B\cap U^{\partial }(f,c))\setminus g(B\cap U^a(f,c))$$

$$M:=\bigcup _{\begin{array}{c}n<\omega \\ a\in \{0,1\}\end{array}}g(B_n\cap U^{\partial }(f,c))\setminus g(B_n\cap U^a(f,c))$$

$$\{p\in {\mathbb{R}}^r:c\text{ is a local extremum of }f{|}_{g^{-1}(p)}\}\subseteq M.$$

$$\bigcup _{n<\omega }\{c\in \mathbb{R}:B_n\times \{c\}\subseteq K\}=\{c\in \mathbb{R}:K\cap H_c\text{ is not codense in }{H}_c\}$$

$$C:=\bigcap _{n<\omega }\{c\in \mathbb{R}:K_n\cap H_c\text{ is nowhere dense in }{H}_c\}$$

$$M(k,\tau ,\upsilon ):=\{p\in {\mathbb{R}}^{\mathrm{dom}(\upsilon )}:c_k\text{ is a local extremum of }{f}_k{|}_{S(\tau )\cap f_{\upsilon }^{-1}(p)}\}$$

$$\forall \tau \in \{\partial ,\perp {\}}^n.\forall \upsilon ={\perp }^n{\upsilon }^{\prime }\in \{\top ,\perp {\}}^{\ast }.\forall k\in n\setminus \mathrm{dom}(\tau ).M(k,\tau ,\upsilon )\text{ is meagre.}$$

$$\{c\in \mathbb{R}:M(k,\tau ,\upsilon )\cap H_c(n,\upsilon )\text{ is meagre in }{H}_c(n,\upsilon )\}$$

$$\begin{gathered} \forall \tau \in \{\partial ,\perp {\}}^n.\forall \upsilon ={\perp }^n\top {\upsilon }^{\prime }\in \{\top ,\perp {\}}^{\ast }.\forall k\in n\setminus \mathrm{dom}(\tau ). \\ M(k,\tau [n\mapsto \partial ],\upsilon [n\mapsto \perp ])\text{ is meagre.} \end{gathered}$$

$$W:=V\cap \bigcap _{\begin{array}{c}k\in \mathrm{dom}(\sigma )\\ \cap \mathrm{dom}({\phi }_S(x))\end{array}}{S}_k^{{\phi }_S(x)(k)}.$$

$$W\cap S(\tau )=V\cap \bigcap _{\begin{array}{c}k\in \mathrm{dom}(\sigma )\\ \cap \mathrm{dom}({\phi }_S(x))\end{array}}{S}_k^{{\phi }_S(x)(k)}\cap \bigcap _{\begin{array}{c}\begin{array}{rl}k\in & \mathrm{dom}(\sigma )\\ & \setminus \mathrm{dom}({\phi }_S(x))\end{array}\end{array}}{S}_k^{\partial }\subseteq V\cap \bigcap _{\begin{array}{c}k\in \mathrm{dom}(\sigma )\end{array}}(S_k^{\sigma (k)}\cup S_k^{\partial })=V\cap \bar{S}(\sigma ).$$

Full text

1. Introduction

Let $X$ be a second-countable Hausdorff space. We consider a subbase of $X$ which induces a coding for each point of $X$. Since $X$ is second-countable, it has a subbase that is the union of countably many pairs of disjoint open subsets. A dyadic subbase $S$ is such a subbase with a fixed enumeration. Let $(S_{n}^{0},S_{n}^{1})$, $n<\omega$, be the $n$-th pair of $S$. For each point $x\in X$, the $n$-th digit of the coding of $x$ will be decided between [math] and $1$ depending on which of $S_{n}^{0}$ and $S_{n}^{1}$ contains $x$. Since $x$ may belong to none of the two, we allow undefinedness in the coding, and the bottom character $\bot$ is used in the sequence unless it is decided. Every sequence that contains the bottom character is called a bottomed sequence.

When a dyadic subbase $S$ is given, each point $x$ of $X$ is represented by a bottomed sequence $\varphi_{S}(x)\in\{0,1,\bot\}^{\omega}$. Every finite prefix of the bottomed sequence $\varphi_{S}(x)$ can be considered as a finite time state of the output of computation, and the properties of the set $K_{S}:=\{\varphi_{S}(x)|_{n}:x\in X,\,n<\omega\}$ of these sequences have been studied. If $S$ is proper and $K_{S}$ is a conditional upper semilattice with least element (cusl), then we obtain an admissible domain representation of $X$ [6]. Whether $K_{S}$ is a cusl depends not only on $S$ itself but also on the enumeration of $S$. It has been proved that $K_{S}$ is a cusl regardless of the enumeration of $S$ if and only if $S$ is strongly proper [7]. The definitions of proper and strongly proper dyadic subbases will be recalled in the next section.

We study a condition which ensures the existence of strongly proper dyadic subbases. It has been proved that every second-countable regular Hausdorff space has a proper dyadic subbase [2, 3]. First we give another proof of this fact that uses the metric induced by Urysohn’s metrisation theorem. Then we show that every locally compact separable metric space has a strongly proper dyadic subbase.

2. Preliminaries

2.1. Bottomed sequences

Let $\Sigma$ be a finite set containing the bottom character $\bot$, and $\omega$ the first infinite ordinal. As usual, an element of $\omega$ is identified with the set of its predecessors.

For an ordinal number $n\leq\omega$, $\Sigma^{n}$ denotes the set of sequences of elements of $\Sigma$ of length $n$, i.e., the maps from $n$ to $\Sigma$. We identify a sequence $\sigma\in\Sigma^{n}$ with its infinite extension

[TABLE]

By this identification, we have $\Sigma^{n}\subseteq\Sigma^{m}$ if $n\leq m$. For a sequence $\sigma\in\Sigma^{\omega}$, its domain is defined as $\operatorname{dom}(\sigma):=\{k:\sigma(k)\neq\bot\}$ and its length as $\operatorname{len}(\sigma):=\min\{n\leq\omega:\operatorname{dom}(\sigma)\subseteq n\}$. A sequence $\sigma\in\Sigma^{\omega}$ is compact if $\operatorname{len}(\sigma)$ is finite. The set of compact sequences is denoted by $\Sigma^{*}:=\bigcup_{n<\omega}\Sigma^{n}$.

Let $\sigma$ and $\tau$ be sequences of $\Sigma$, $a\in\Sigma$ an element and $n<\omega$ a finite ordinal. $\sigma[n\mapsto a]$ is the sequence which maps $n$ to $a$ and equals $\sigma$ elsewhere. If $\sigma$ is compact, then a concatenation $\sigma\tau\in\Sigma^{\omega}$ is defined as

[TABLE]

The $n$-fold concatenation of $\sigma$ with itself is denoted by $\sigma^{n}$. Elements $a\in\Sigma$ are identified with sequences of length one.

2.2. The domain $\mathbb{T}^{\omega}$

We set $\mathbb{T}:=\{0,1,\bot\}$ ordered by $\bot\sqsubseteq 0$ and $\bot\sqsubseteq 1$. A $T_{0}$ topology of $\mathbb{T}$ is defined as $\{\emptyset,\{0\},\{1\},\{0,1\},\mathbb{T}\}$. We equip $\mathbb{T}^{\omega}$ with the product order and the product topology. For all $\sigma\in\mathbb{T}^{\omega}$, we set $\operatorname{\uparrow}\sigma:=\{\tau\in\mathbb{T}^{\omega}:\tau\sqsupseteq\sigma\}$, and the family $\{\operatorname{\uparrow}\sigma:\sigma\in\mathbb{T}^{*}\}$ forms a base of $\mathbb{T}^{\omega}$ [4].

2.3. Dyadic subbases

We review proper and strongly proper dyadic subbases [5, 6, 7]. In this section, $X=(X,\mathfrak{O})$ is a second-countable Hausdorff space. For a subset $A$ of $X$, $\operatorname{cl}A$ denotes the closure of $A$, $\operatorname{int}A$ the interior of $A$. The exterior of a subset $A\subseteq X$ is the set $\operatorname{int}(X\setminus A)$ or equivalently $X\setminus\operatorname{cl}A$.

{defi}

A dyadic subbase of $X$ is a map $S:\omega\times\{0,1\}\rightarrow\mathfrak{O}$ such that

1. (1)

$\{S(n,a):n<\omega,\,a\in\{0,1\}\}$ is a subbase of $X$, 2. (2)

$S(n,0)\cap S(n,1)=\emptyset$ for all $n<\omega.$

For readability, $S(n,a)$ is denoted by $S_{n}^{a}$. We set $S_{n}^{\partial}:=X\setminus(S_{n}^{0}\cup S_{n}^{1})$ for $n<\omega$. If $S_{n}^{0}$ and $S_{n}^{1}$ are the exteriors of each other, then $S_{n}^{\partial}$ is their common boundary, and we have $\operatorname{cl}S_{n}^{a}=S_{n}^{a}\cup S_{n}^{\partial}$ for $a\in\{0,1,\partial\}$. We use the notations

[TABLE]

[TABLE]

for all $\sigma\in\{0,1,\partial,\bot\}^{\omega}$. {defi} A dyadic subbase $S$ of $X$ is

1. (1)

proper if $\bar{S}(\sigma)=\operatorname{cl}S(\sigma)\text{ for all }\sigma\in\mathbb{T}^{*}$. 2. (2)

strongly proper if $\bar{S}(\sigma)=\operatorname{cl}S(\sigma)$ for all $\sigma\in\{0,1,\partial,\bot\}^{*}$.

For all $\sigma\in\mathbb{T}^{*}$, by De Morgan’s law, we obtain

[TABLE]

and therefore, we have $X\setminus\bar{S}(\sigma)=\bigcup_{k\in\operatorname{dom}(\sigma)}S_{k}^{1-\sigma(k)}$. We have the following characterization of proper dyadic subbases.

Proposition 1**.**

A dyadic subbase $S$ of $X$ is proper if and only if $S(\sigma)$ and $X\setminus\bar{S}(\sigma)$ are the exteriors of each other for all $\sigma\in\mathbb{T}^{*}$.

Proof 2.1**.**

For all $\sigma\in\mathbb{T}^{*}$, we have $\bar{S}(\sigma)=\operatorname{cl}S(\sigma)$ if and only if $X\setminus\bar{S}(\sigma)=X\setminus\operatorname{cl}S(\sigma)$ if and only if $X\setminus\bar{S}(\sigma)$ is the exterior of $S(\sigma)$.

We show that if $S$ is proper, then for all $\sigma\in\mathbb{T}^{*}$, $S(\sigma)$ is the exterior of $X\setminus\bar{S}(\sigma)$, i.e., $S(\sigma)$ is the interior of $\bar{S}(\sigma)$. Suppose that $S$ is proper. For all $n<\omega$ and $a\in\{0,1\}$, we have $\operatorname{cl}S_{n}^{1-a}=S_{n}^{1-a}\cup S_{n}^{\partial}=X\setminus S_{n}^{a}$, and therefore,

[TABLE]

Hence, for all $\sigma\in\mathbb{T}^{*}$, we obtain

[TABLE]

Every sequence $\sigma\in\{0,1,\partial,\bot\}^{*}$ is decomposed into two sequences

[TABLE]

We have $S(\sigma)=S(\sigma_{0,1})\cap S(\sigma_{\partial})$, and $S(\sigma)$ is an open subset of $S(\sigma_{\partial})$. Since $S(\sigma_{\partial})$ is closed, we have $\bar{S}(\sigma)=\bar{S}(\sigma_{0,1})\cap S(\sigma_{\partial})$. Similarly to Proposition 1, if $S$ is strongly proper, then $S(\sigma)$ and $S(\sigma_{\partial})\setminus\bar{S}(\sigma)$ are the exteriors of each other in the space $S(\sigma_{\partial})$.

Proposition 2**.**

A dyadic subbase $S$ of $X$ is strongly proper if and only if for all $\sigma\in\{0,1,\partial,\bot\}^{*}$, $S(\sigma)$ and $S(\sigma_{\partial})\setminus\bar{S}(\sigma)$ are the exteriors of each other in the space $S(\sigma_{\partial})$.

Proof 2.2**.**

Since $S(\sigma_{\partial})$ is closed, $\operatorname{cl}(S(\sigma_{0,1})\cap S(\sigma_{\partial}))$ equals the closure of $S(\sigma_{0,1})\cap S(\sigma_{\partial})$ in the space $S(\sigma_{\partial})$. Therefore, we have $\operatorname{cl}S(\sigma)=\bar{S}(\sigma)$ if and only if $S(\sigma_{\partial})\setminus\bar{S}(\sigma)=S(\sigma_{\partial})\setminus\operatorname{cl}S(\sigma)$ if and only if $S(\sigma_{\partial})\setminus\bar{S}(\sigma)$ is the exterior of $S(\sigma)$ in the space $S(\sigma_{\partial})$. Similarly to the proof of Proposition 1, we can see that if $S$ is strongly proper, then $S(\sigma)$ is the interior of $\bar{S}(\sigma)$ in the space $S(\sigma_{\partial})$.

Let $S$ be a dyadic subbase of $X$. We define a map $\varphi_{S}:X\rightarrow\mathbb{T}^{\omega}$ as

[TABLE]

for $x\in X$ and $n<\omega$. We set

[TABLE]

The set $D_{S}$ is an algebraic pointed dcpo that is the ideal completion of $K_{S}$. We quote some results known from [6, 7]: If $S$ is a proper dyadic subbase of a regular Hausdorff space $X$, then $X$ is embedded in the space of minimal elements of $D_{S}\setminus K_{S}$. Moreover, if $X$ is compact in addition, then we have a quotient map $\rho_{S}:D_{S}\setminus K_{S}\rightarrow X$ and $X$ is homeomorphic to the space of minimal elements of $D_{S}\setminus K_{S}$. The triple $(D_{S},D_{S}\setminus K_{S},\rho_{S})$ is a domain representation of $X$. If $K_{S}$ is a conditional upper semilattice with least element (cusl), then the domain representation is admissible ([6]). Whether $K_{S}$ is a cusl depends not only on the subbase $\{S_{n}^{a}:n<\omega,a\in\{0,1\}\}$ itself but also on the enumeration of $S$. It has been proved that $K_{S}$ is a cusl regardless of the enumeration of $S$ if and only if $S$ is strongly proper ([7]). We refer the reader to [1] for the notion of domain representations of topological spaces.

We give an example of a strongly proper dyadic subbase and show some modifications. {exa}[The Gray subbase of the unit interval] Let $I$ be the unit interval $[0,1]\subseteq\mathbb{R}$. For every non-negative integer $n$, we define a function $f_{n}:I\rightarrow\mathbb{R}$ as $f_{n}(x):=-\cos(2^{n}\pi x)$. We set $G_{n}^{0}:=f_{n}^{-1}((-\infty,0))$ and $G_{n}^{1}:=f_{n}^{-1}((0,\infty))$ for all $n<\omega$. The family $\{G_{n}^{a}:n<\omega,a\in\{0,1\}\}$ forms a subbase of $I$, the Gray subbase.

We show that the Gray subbase is a strongly proper dyadic subbase. For every $n<\omega$, we have $f_{n}^{-1}(0)=\{(2k+1)/2^{n+1}:k=0,1,\dots,2^{n}-1\}$, and the set $G_{n}^{\partial}=f_{n}^{-1}(0)$ is the common boundary of $G_{n}^{0}$ and $G_{n}^{1}$. Note that $G_{m}^{\partial}$ and $G_{n}^{\partial}$ are disjoint if $m\neq n$. Let $\sigma\in\{0,1,\partial,\bot\}^{*}$ be a bottomed sequence. We have $\bar{G}(\sigma)\supseteq\operatorname{cl}G(\sigma)$ because $\bar{G}(\sigma)$ is closed and contains $G(\sigma)$. Suppose that $x$ belongs to $\bar{G}(\sigma)$, and we assume $x\in G_{m}^{\partial}$ for an $m\in\operatorname{dom}(\sigma)$, otherwise $x$ belongs to $G(\sigma)$. Since $G_{m}^{\partial}$ and $G_{k}^{\partial}$ are disjoint for all $k\in\operatorname{dom}(\sigma)\setminus\{m\}$, we get $x\in\operatorname{cl}G_{m}^{\sigma(m)}\cap G(\sigma[m\mapsto\bot])$ and $\sigma[m\mapsto\bot]\in\mathbb{T}^{*}$. For every open neighbourhood $V$ of $x$, we have $V\cap\operatorname{cl}G_{m}^{\sigma(m)}\cap G(\sigma[m\mapsto\bot])\neq\emptyset$. Since $V\cap G(\sigma[m\mapsto\bot])$ is open, we get $V\cap G_{m}^{\sigma(m)}\cap G(\sigma[m\mapsto\bot])\neq\emptyset$. Therefore, every open neighbourhood $V$ of $x$ intersects with $G_{m}^{\sigma(m)}\cap G(\sigma[m\mapsto\bot])=G(\sigma)$ non-trivially. Hence, we obtain $x\in\operatorname{cl}G(\sigma)$.

Remark 3**.**

We allow duplications in dyadic subbases, and a one point set $\{x\}$ has a strongly proper dyadic subbase given by $S_{n}^{0}=\{x\}$ and $S_{n}^{1}=\emptyset$ for all $n<\omega$. However, such a duplication induces non-properness in general. Let $G$ be the Gray subbase of the unit interval. We have $\operatorname{cl}(G_{n}^{0}\cap G_{n}^{1})=\emptyset$ and $\operatorname{cl}G_{n}^{0}\cap\operatorname{cl}G_{n}^{1}=G_{n}^{\partial}\neq\emptyset$. Therefore, if we enumerate a pair in $G$ twice, then we obtain a non-proper dyadic subbase.

{exa}

Let $I$ be the unit interval with the Gray subbase $G$. We define $X$ as the one point compactification of $I\setminus\{1/4,3/4\}$, and let $p$ be the added point. Since we have $\varphi_{G}(1/4)=0\bot 10^{\omega}$ and $\varphi_{G}(3/4)=1\bot 10^{\omega}$, $X$ has a dyadic subbase $S$ given by

[TABLE]

We have $p\in\operatorname{cl}(S_{0}^{a}\cap S_{1}^{b})$ for all $a,b\in\{0,1\}$, and the properness of $S$ can be proved similarly. However, $S$ is not strongly proper because we have $p\in S_{0}^{\partial}\cap\operatorname{cl}S_{1}^{0}$ whereas $p\not\in\operatorname{cl}(S_{0}^{\partial}\cap S_{1}^{0})$.

3. Existence of strongly proper dyadic subbases

We will show the following.

Theorem 4**.**

Every locally compact separable metric space has a strongly proper dyadic subbase.

Every separable metric space is second-countable and regular Hausdorff. Urysohn’s metrization theorem states that every second-countable regular Hausdorff space is metrisable. Therefore, Theorem 4 states that every locally compact second-countable Hausdorff space has a strongly proper dyadic subbase. It is still an open problem whether every separable metric space has a strongly proper dyadic subbase.

3.1. Existence of proper dyadic subbases

First, we show the following.

Proposition 5**.**

Every separable metric space $X=(X,d)$ has a proper dyadic subbase.

Proposition 5 has been proved already in [2, 3]. Using the metric directly, we give another proof of this fact. The proof is simpler than the previous proofs, and the idea is useful for proving Theorem 4. Let $f:X\rightarrow\mathbb{R}$ be a function, $c$ a real number. We use the notations

[TABLE]

We will construct a dyadic subbase $S:\omega\times\{0,1\}\rightarrow\mathfrak{O}$ of the form

[TABLE]

for every $n<\omega$ and $a\in\{0,1\}$, where $f_{n}:X\rightarrow\mathbb{R}$ is a continuous function and $c_{n}$ is a real number for every $n<\omega$.

We say that $c$ is a local maximum (resp. local minimum) of $f$ if $c$ is a maximum (resp. minimum) value of $f|_{V}$ for some open subset $V$. Local maxima and local minima are collectively called local extrema. If $c$ is a local maximum of $f:X\rightarrow\mathbb{R}$, then there exists a point $x\in U^{\partial}(f,c)$ with its neighbourhood $V$ such that $V\cap U^{1}(f,c)$ is empty. The point $x$ belongs to neither $\operatorname{cl}U^{1}(f,c)$ nor $U^{0}(f,c)$. Therefore, $U^{0}(f,c)$ is not the exterior of $U^{1}(f,c)$. Similarly, if $c$ a local minimum, then there exists $y\not\in\operatorname{cl}U^{0}(f,c)\cup U^{1}(f,c)$. Hence, local extrema should be avoided in order to obtain a proper dyadic subbase. We do not fix real numbers $c_{n}$ first, but give open intervals $I_{n}$ from which $c_{n}$ will be taken.

Lemma 6**.**

There exist a sequence $(f_{n})_{n<\omega}$ of continuous functions and a sequence $(I_{n})_{n<\omega}$ of open intervals such that the family $\{U^{a}(f_{n},c_{n}):n<\omega,\,a\in\{0,1\}\}$ is a subbase of $X$ if $c_{n}\in I_{n}$ for $n<\omega$.

Proof 3.1**.**

Since $X$ is separable, there exists a dense countable set $\{x_{n}\in X:n<\omega\}$. Suppose that $\{B_{n}:n<\omega\}$ is a family consisting of open intervals that forms a base of the space $\mathbb{R}_{>0}$ of positive real numbers. Note that if $b_{n}\in B_{n}$ for all $n<\omega$, then the set $\{b_{n}:n<\omega\}$ is dense in $\mathbb{R}_{>0}$. Let $n\mapsto(n_{0},n_{1})$ be a map from $\omega$ onto $\omega\times\omega$. We define $f_{n}(x):=d(x_{n_{0}},x)$ and $I_{n}:=B_{n_{1}}$. If $c_{n}\in I_{n}$ for all $n<\omega$, then the family $\{U^{a}(f_{n},c_{n}):n<\omega,\,a\in\{0,1\}\}$ is a subbase of $X$ because $\{U^{0}(f_{n},c_{n}):n<\omega\}$ forms a base. By definition, $U^{0}(f_{n},c_{n})$ and $U^{1}(f_{n},c_{n})$ are disjoint for all $n<\omega$.

As the following lemma shows, the set of local extrema is countable.

Lemma 7**.**

Every function $f:X\rightarrow\mathbb{R}$ has at most countably many local extrema.

Proof 3.2**.**

Let $\mathfrak{B}$ be a countable base of $X$. Since each local extremum is an extremum of $f|_{B}$ for some $B\in\mathfrak{B}$, the number of local extrema of $f$ is countable.

In order to obtain the properness property, we have only to avoid local extrema of finitely many functions. Therefore, we can avoid them.

Lemma 8**.**

Let $A$ be a subset of $X$, $f:X\rightarrow\mathbb{R}$ a continuous function, $c$ a real number. If $c\in\mathbb{R}$ is not a local extremum of $f|_{\operatorname{cl}A}$, then we have $\operatorname{cl}A\setminus U^{1-a}(f,c)=\operatorname{cl}(A\cap U^{a}(f,c))$ for $a\in\{0,1\}$.

Proof 3.3**.**

We can see $\operatorname{cl}A\setminus U^{1-a}(f,c)\supseteq\operatorname{cl}A\cap\operatorname{cl}U^{a}(f,c)\supseteq\operatorname{cl}(A\cap U^{a}(f,c))$. Suppose that $x\in\operatorname{cl}A\setminus U^{1-a}(f,c)$ and $V$ is an open neighbourhood of $x$. Since $c$ is not a local extremum of $f|_{\operatorname{cl}A}$, there exists $y\in V\cap\operatorname{cl}A$ such that $y\in U^{a}(f,c)$, i.e., $V\cap\operatorname{cl}A\cap U^{a}(f,c)\neq\emptyset$. Since $U^{a}(f,c)$ and $V$ are open, we have $V\cap A\cap U^{a}(f,c)\neq\emptyset$. Since this holds for every open neighbourhood $V$ of $x$, we obtain $x\in\operatorname{cl}(A\cap U^{a}(f,c))$.

Proof 3.4** (Proof of Proposition 5).**

By Lemma 6 we can take a sequence $(f_{n})_{n<\omega}$ of continuous functions and a sequence $(I_{n})_{n<\omega}$ of open intervals, such that the family $\{U^{a}(f_{n},c_{n}):n<\omega,\,a\in\{0,1\}\}$ is a subbase of $X$ if $c_{n}\in I_{n}$ for $n<\omega$.

First, we take $c_{0}\in I_{0}$ which is not a local extremum of $f_{0}$, and set $S_{0}^{a}:=U^{a}(f_{0},c_{0})$ for $a\in\{0,1\}$. By Lemma 8, $S_{0}^{0}$ and $S_{0}^{1}$ are the exteriors of each other.

Let $n$ be a finite ordinal. Suppose that we have obtained a family $\{S_{k}^{a}:k<n,a\in\{0,1\}\}$ such that $\bar{S}(\sigma)=\operatorname{cl}S(\sigma)$ for all $\sigma\in\mathbb{T}^{n}$. We take a real number $c_{n}\in I_{n}$ which is not a local extremum of $f_{n}|_{\bar{S}(\sigma)}$ for all $\sigma\in\mathbb{T}^{n}$. We set $S_{n}^{a}:=U^{a}(f_{n},c_{n})$ for $a\in\{0,1\}$. For all $\sigma\in\mathbb{T}^{n}$ and $a\in\{0,1\}$, we have

[TABLE]

by the assumption. Since $c_{n}$ is not a local extremum of $f_{n}|_{\operatorname{cl}S(\sigma)}$, by Lemma 8, we obtain

[TABLE]

Therefore, $\bar{S}(\sigma)=\operatorname{cl}S(\sigma)$ holds for all $\sigma\in\mathbb{T}^{n+1}$. Hence, we obtain a proper dyadic subbase inductively.

3.2. Fat points

In the proof of Proposition 5, $c_{n}$ could be a local extremum of $f_{n}|_{f_{m}^{-1}(c_{m})}$ for some $m,n<\omega$. If $c_{n}$ is a local extremum of $f_{n}|_{f_{m}^{-1}(c_{m})}$, then $U^{0}(f_{n},c_{n})\cap U^{\partial}(f_{m},c_{m})$ and $U^{1}(f_{n},c_{n})\cap U^{\partial}(f_{m},c_{m})$ are not the exteriors of each other in the space $U^{\partial}(f_{m},c_{m})$, and we fail to obtain the strong properness. We have to avoid $c_{m}$ such that an already chosen $c_{n}$ will be a local extremum of $f_{n}|_{f_{m}^{-1}(c_{m})}$. In Section 3.3, we show that such real numbers can be avoided if the space $X$ is locally compact.

In the rest of this section, $X$ is a locally compact separable metric space. {defi} A subset $A\subseteq X$ is

1. (i)

codense if $\operatorname{int}A=\emptyset$. 2. (ii)

nowhere dense if $\operatorname{int}\operatorname{cl}A=\emptyset$. 3. (iii)

meagre if $A$ is a countable union of nowhere dense subsets.

Let $r$ be a non-negative integer, $f:X\rightarrow\mathbb{R}^{r}$ a continuous map. We say that a point $x\in X$ is fat with respect to $f$ if $f(V)$ has an interior point for every neighbourhood $V$ of $x$. The set of all fat points with respect to $f$ is denoted by $\operatorname{fat}_{f}X$. For every subset $A\subseteq X$, $\operatorname{fat}_{f}A$ denotes the set of fat points of $A$ with respect to $f|_{A}$.

Lemma 9**.**

Let $f:X\rightarrow\mathbb{R}^{r}$ be a continuous map. For every subset $A\subseteq X$, we have $\operatorname{fat}_{f}A\subseteq\operatorname{fat}_{f}X\cap A$.

Proof 3.5**.**

Let $x\in\operatorname{fat}_{f}A$ be a point, $V\subseteq X$ its open neighbourhood. We can see $x\in A$. Since $x$ belongs to $\operatorname{fat}_{f}A$, $f(A\cap V)$ has an interior point. Therefore, $f(V)\supseteq f(A\cap V)$ also has an interior point, and hence $x\in\operatorname{fat}_{f}X$.

By definition, $X$ has no fat point with respect to $f$ if $f(X)$ is codense. We show its converse.

Proposition 10**.**

Let $r$ be a non-negative integer, $f:X\rightarrow\mathbb{R}^{r}$ a continuous map. If $\operatorname{fat}_{f}X$ is empty, then $f(X)$ is codense.

We make a remark about the case in which $r$ is zero. $\mathbb{R}^{0}$ is a one point set and every point $x\in X$ is mapped to the same point by $f$. Therefore, we have $\operatorname{fat}_{f}X=X$. Note that for every subset $A$ of a one point set, we have

[TABLE]

Therefore, Proposition 10 holds in this case.

Baire category theorem states that every meagre subset of a complete metric space is codense. Since $\mathbb{R}^{r}$ is a complete metric space, we have the following.

Lemma 11** (Baire category theorem).**

Every meagre subset of $\mathbb{R}^{r}$ is codense.∎

Proof 3.6** (Proof of Proposition 10).**

Suppose that $X$ has no fat point and $\{B_{n}:n<\omega\}$ is a base of $X$. Since $X$ is locally compact, all $B_{n}$ can be chosen to be relatively compact. For all $x\in X$, there exists a compact neighbourhood $\operatorname{cl}B_{n}$ of $x$ such that $f(\operatorname{cl}B_{n})$ is codense. Note that the image of a compact set by a continuous map is always compact and every compact codense subset is nowhere dense. Therefore, we have $X=\bigcup\{\operatorname{cl}B_{n}:f(\operatorname{cl}B_{n})\text{ is nowhere dense}\}.$ We can see that $f(X)=\bigcup\{f(\operatorname{cl}B_{n}):f(\operatorname{cl}B_{n})\text{ is nowhere dense}\}$ is meagre, and hence $f(X)$ is codense by Lemma 11.

Let $f,g:X\rightarrow\mathbb{R}$ be continuous functions, $c$ a real number. We consider whether there exists $b\in\mathbb{R}$ such that $b$ is a local extremum of neither $g$ nor $g|_{f^{-1}(c)}$ and $c$ is not a local extremum of $f|_{g^{-1}(b)}$. As the next proposition shows, if $c$ is not a local extremum of $f|_{\operatorname{fat}_{g}X}$, then the set $\{p\in\mathbb{R}:c\text{ is a local extremum of }f|_{g^{-1}(p)}\}$ is meagre, and such a $b$ exists. Since we will deal with countably many functions, we prove this in an extended form.

Proposition 12**.**

Let $f:X\rightarrow\mathbb{R}$ be a continuous function, $r$ a non-negative integer, $g:X\rightarrow\mathbb{R}^{r}$ a continuous map, $c$ a real number which is not a local extremum of $f|_{\operatorname{fat}_{g}X}$. The set $\{p\in\mathbb{R}^{r}:c\text{ is a local extremum of }f|_{g^{-1}(p)}\}$ is meagre.

Before proving Proposition 12, we provide two lemmas.

Lemma 13**.**

Suppose that $f,g$ and $c$ are as above. For any open subset $B$ of $X$ and any compact subset $K$ of $B$, $g(K\cap U^{\partial}(f,c))\setminus g(B\cap U^{a}(f,c))$ is nowhere dense for $a\in\{0,1\}$.

Proof 3.7**.**

For each $a\in\{0,1\}$, we have

[TABLE]

The right hand side is closed, and we will show that it is codense. Let $V\subseteq g(K\cap U^{\partial}(f,c))$ be a non-empty open subset. We can see that $K\cap U^{\partial}(f,c)\cap g^{-1}(V)$ is a locally compact separable metric space. Therefore, by Proposition 10, the set $\operatorname{fat}_{g}(K\cap U^{\partial}(f,c)\cap g^{-1}(V))$ is not empty. By Lemma 9, we obtain

[TABLE]

Since $c$ is not a local extremum of $f|_{\operatorname{fat}_{g}X}$, we obtain $\operatorname{fat}_{g}X\cap B\cap U^{a}(f,c)\cap g^{-1}(V)\neq\emptyset$ for $a\in\{0,1\}$, and thus $g(B\cap U^{a}(f,c))$ has an interior point in $V$. Thus, every non-empty open subset $V$ of $g(K\cap U^{\partial}(f,c))$ intersects with $\operatorname{int}g(B\cap U^{a}(f,c))$ non-trivially and, consequently, $g(K\cap U^{\partial}(f,c))\setminus\operatorname{int}g(B\cap U^{a}(f,c))$ is codense for $a\in\{0,1\}$.

Lemma 14**.**

Suppose that $f,g$ and $c$ are as above. For any open subset $B$ of $X$, $g(B\cap U^{\partial}(f,c))\setminus g(B\cap U^{a}(f,c))$ is meagre for $a\in\{0,1\}$.

Proof 3.8**.**

Since $B$ is a locally compact separable metric space, there are compact sets $K_{n}$, $n<\omega$, such that $B=\bigcup_{n<\omega}K_{n}$. By Lemma 13, $g(K_{n}\cap U^{\partial}(f,c))\setminus g(B\cap U^{a}(f,c))$ is nowhere dense for all $n<\omega$ and $a\in\{0,1\}$. Therefore, for each $a\in\{0,1\}$, their union

[TABLE]

is meagre.

Proof 3.9** (Proof of Proposition 12).**

Let $\{B_{n}:n<\omega\}$ be a countable base of $X$. By Lemma 14, $g(B_{n}\cap U^{\partial}(f,c))\setminus g(B_{n}\cap U^{a}(f,c))$ is meagre for all $n<\omega$ and $a\in\{0,1\}$. Therefore, their union

[TABLE]

is meagre.

Suppose that $c$ is a local extremum of $f|_{g^{-1}(p)}$ for a point $p\in\mathbb{R}^{r}$. There exists a base element $B_{n}$ such that $B_{n}\cap g^{-1}(p)\cap U^{\partial}(f,c)$ is non-empty, and either $B_{n}\cap g^{-1}(p)\cap U^{0}(f,c)$ or $B_{n}\cap g^{-1}(p)\cap U^{1}(f,c)$ is empty. That is, $p$ belongs to $g(B_{n}\cap U^{\partial}(f,c))\setminus g(B_{n}\cap U^{a}(f,c))$ for an $a\in\{0,1\}$. Therefore, we obtain

[TABLE]

We show that if the set of $(r+1)$-tuples that we should avoid is meagre, then we have only to avoid a meagre subset at each step. For a real number $c$, a hyperplane $H_{c}\subseteq\mathbb{R}^{r+1}$ is defined as $\{(x_{0},\dots,x_{r}):x_{r}=c\}$.

Proposition 15**.**

If $M\subseteq\mathbb{R}^{r+1}$ is meagre, then the set $\{c\in\mathbb{R}:M\cap H_{c}\text{ is meagre in }H_{c}\}$ is comeagre, i.e., its complement is meagre.

For the proof we need a lemma.

Lemma 16**.**

If $K\subseteq\mathbb{R}^{r+1}$ is closed nowhere dense, then $\{c\in\mathbb{R}:K\cap H_{c}$ is nowhere dense in $H_{c}\}$ is comeagre.

Proof 3.10**.**

Let $\{B_{n}:n<\omega\}$ be a countable base of $\mathbb{R}^{r}$. For a real number $c$, $B_{n}\times\{c\}$ denotes the set $\{(x_{0},\dots,x_{r-1},c):(x_{0},\dots,x_{r-1})\in B_{n}\}$. Note that if $K\cap H_{c}$ is not codense in $H_{c}$, then an open subset of $H_{c}$ is contained in $K$, and therefore, $B_{n}\times\{c\}\subseteq K$ for some $n<\omega$. For each $n<\omega$, the set $\{c\in\mathbb{R}:B_{n}\times\{c\}\subseteq K\}$ is nowhere dense because $K$ is nowhere dense. We can see that their union

[TABLE]

is meagre. Therefore, its complement $\{c\in\mathbb{R}:K\cap H_{c}\text{ is codense in }H_{c}\}$ is comeagre. Since $K$ is closed, $K\cap H_{c}$ is closed in $H_{c}$. Hence, $\{c\in\mathbb{R}:K\cap H_{c}\text{ is nowhere dense in }H_{c}\}$ is comeagre.

Proof 3.11** (Proof of Proposition 15).**

Suppose that $M\subseteq\mathbb{R}^{r+1}$ is meagre. There exists a countable covering $M\subseteq\bigcup_{n<\omega}K_{n}$, where $K_{n}$ is closed nowhere dense for $n<\omega$. By Lemma 16, $\{c\in\mathbb{R}:K_{n}\cap H_{c}\text{ is nowhere dense in }H_{c}\}$ is comeagre for all $n<\omega$. By De Morgan’s law, the intersection of countably many comeagre subsets is comeagre. Therefore, their intersection

[TABLE]

is comeagre. If $c\in C$, then $K_{n}\cap H_{c}$ is nowhere dense in $H_{c}$ for all $n<\omega$, and therefore, $\bigcup_{n<\omega}K_{n}\cap H_{c}\supseteq M\cap H_{c}$ is meagre in $H_{c}$. Hence, the set $\{c\in\mathbb{R}:M\cap H_{c}\text{ is meagre in }H_{c}\}\supseteq C$ is comeagre.

3.3. Proof of Theorem 4

3.3.1. Construction of a strongly proper dyadic subbase

By Lemma 6, we can take a sequence $(f_{n})_{n<\omega}$ of real-valued continuous functions and a sequence $(I_{n})_{n<\omega}$ of open intervals in $\mathbb{R}$, such that the family $\{U^{a}(f_{n},c_{n}):n<\omega,\,a\in\{0,1\}\}$ is a subbase of $X$ if $c_{n}\in I_{n}$ for $n<\omega$.

Every subset of $\omega$ is represented by the domain of a sequence of $\{\top,\bot\}$, where the top character $\top$ means non-bottom. For a sequence $\upsilon\in\{\top,\bot\}^{*}$, $f_{\upsilon}$ denotes the map $f_{\upsilon}:X\rightarrow\mathbb{R}^{\operatorname{dom}(\upsilon)}$ given by $f_{\upsilon}(x)=(f_{k}(x))_{k\in\operatorname{dom}(\upsilon)}$.

Let $n>0$ be a finite ordinal. If we have a sequence $(c_{i})_{i<n}$ and set $S_{i}^{a}=U^{a}(f_{i},c_{i})$ for all $i<n$ and $a\in\{0,1,\partial\}$, then we define $M(k,\tau,\upsilon)\subseteq\mathbb{R}^{\operatorname{dom}(\upsilon)}$ as

[TABLE]

for all $\tau\in\{\partial,\bot\}^{n}$, $\upsilon=\bot^{n}\upsilon^{\prime}\in\{\top,\bot\}^{*}$ and $k\in n\setminus\operatorname{dom}(\tau)$. For each $n$, we now define inductively a sequence $(c_{i})_{i<n}$ such that

[TABLE]

First we define $c_{0}\in\mathbb{R}$ as follows. By Lemma 7, the set of local extrema of $f_{0}|_{\operatorname{fat}_{f_{\upsilon}}X}$ is countable for all $\upsilon\in\{\top,\bot\}^{*}$ with $\upsilon(0)=\bot$. Their union $E_{0}$ is countable. Therefore, we can take $c_{0}\in I_{0}\setminus E_{0}$. We set $S_{0}^{a}:=U^{a}(f_{0},c_{0})$ for $a\in\{0,1,\partial\}$. Since $c_{0}\not\in E_{0}$, by Proposition 12, the set $M(0,\bot,\upsilon)$ is meagre for all $\upsilon=\bot\upsilon^{\prime}\in\{\top,\bot\}^{*}$. Therefore, (11) holds for $n=1$.

Let $n>0$ be a finite ordinal. Suppose that we have obtained a sequence $(c_{i})_{i<n}$ such that (11) holds. For $c\in\mathbb{R}$ and $\upsilon\in\{\top,\bot\}^{*}$ with $n\in\operatorname{dom}(\upsilon)$, we define a hyperplane $H_{c}(n,\upsilon):=\{(x_{k})_{k\in\operatorname{dom}(\upsilon)}\in\mathbb{R}^{\operatorname{dom}(\upsilon)}:x_{n}=c\}$. By Proposition 15, if $n\in\operatorname{dom}(\upsilon)$, then the set

[TABLE]

is comeagre for all $\tau\in\{\partial,\bot\}^{n}$, $\upsilon=\bot^{n}\upsilon^{\prime}\in\{\top,\bot\}^{*}$ and $k\in n\setminus\operatorname{dom}(\tau)$. Therefore, their intersection $C_{n}$ is comeagre. By Lemma 7, the set of local extrema of $f_{n}|_{\operatorname{fat}_{f_{\upsilon}}S(\tau)}$ is countable for all $\tau\in\{\partial,\bot\}^{n}$ and $\upsilon=\bot^{n+1}\upsilon^{\prime}\in\{\top,\bot\}^{*}$. Their union $E_{n}$ is countable. We can take $c_{n}$ from $(I_{n}\cap C_{n})\setminus E_{n}$, and we set $S_{n}^{a}:=U^{a}(f_{n},c_{n})$ for $a\in\{0,1,\partial\}$. Since $c_{n}\in C_{n}$, we obtain

[TABLE]

Note that $S(\tau)$ is a locally compact separable metric space for all $\tau\in\{\partial,\bot\}^{n}$. Since $c_{n}\not\in E_{n}$, by Proposition 12, $M(n,\tau,\upsilon)$ is meagre for all $\tau\in\{\partial,\bot\}^{n}$ and $\upsilon=\bot^{n+1}\upsilon^{\prime}\in\{\top,\bot\}^{*}$. Therefore, we can obtain a sequence $(c_{n})_{n<\omega}$ which satisfies (11) for all $n<\omega$ inductively.

3.3.2. Proof of strong properness

Suppose that $(f_{n})_{n<\omega}$ is a sequence of functions, $(c_{n})_{n<\omega}$ is a sequence of real numbers, (11) holds for all $n<\omega$ and $\{S_{n}^{a}:n<\omega,\,a\in\{0,1\}\}$ forms a subbase of $X$, where $S_{n}^{a}=U^{a}(f_{n},c_{n})$ for $n<\omega$ and $a\in\{0,1\}$. We can easily see that $S$ is a dyadic subbase of $X$.

Lemma 17**.**

For all $n<\omega$, $c_{n}$ is not a local extremum of $f_{n}|_{\bar{S}(\sigma)}$ for all $\sigma\in\{0,1,\partial,\bot\}^{*}$ with $\sigma(n)=\bot$.

Proof 3.12**.**

In (11), setting $\upsilon:=\bot$, we can see that $c_{k}$ is not a local extremum of $f_{k}|_{S(\tau)}$ for all $\tau\in\{\partial,\bot\}^{n}$ and $k\in n\setminus\operatorname{dom}(\tau)$. Since (11) holds for all $n<\omega$, $c_{n}$ is not a local extremum of $f_{n}|_{S(\tau)}$ for all $\tau\in\{\partial,\bot\}^{*}$ and $n\in\omega\setminus\operatorname{dom}(\tau)$.

Suppose that $c_{n}$ is a local extremum of $f_{n}|_{\bar{S}(\sigma)}$ for a sequence $\sigma\in\{0,1,\partial,\bot\}^{*}$ with $\sigma(n)=\bot$. There exists an open subset $V\subseteq X$ such that $c_{n}$ is an extremum of $f_{n}|_{V\cap\bar{S}(\sigma)}$. Take a point $x\in V\cap\bar{S}(\sigma)\cap f_{n}^{-1}(c_{n})$ and set

[TABLE]

*Let $\tau\in\{\partial,\bot\}^{*}$ be a sequence whose domain is the set $\operatorname{dom}(\sigma)\setminus\operatorname{dom}(\varphi_{S}(x))$. For all $k\in\operatorname{dom}(\sigma)$, either $S_{k}^{\varphi_{S}(x)(k)}$ or $S_{k}^{\tau(k)}$ is defined, and the defined one is contained in $S_{k}^{\sigma(k)}\cup S_{k}^{\partial}$. Therefore, we obtain *

[TABLE]

We can see that $c_{n}$ is an extremum value of $f|_{W\cap S(\tau)}$ because $W\cap S(\tau)$ is a subset of $V\cap\bar{S}(\sigma)$ and contains the point $x\in f_{n}^{-1}(c_{n})$. Since $W$ is open, $c_{n}$ is a local extremum of $f|_{S(\tau)}$, a contradiction.

Proposition 18**.**

$S$* is a strongly proper dyadic subbase.*

Proof 3.13**.**

Let $\sigma\in\{0,1,\partial,\bot\}^{*}$ be a sequence,

[TABLE]

its restriction. Since the cardinality of $\sigma^{-1}(\{0,1\})$ is finite, we can take a finite sequence $(\tau_{k})_{k<m}$ such that $\tau_{0}=\sigma_{\partial}$, $\tau_{k+1}=\tau_{k}[n\mapsto\sigma(n)]$ for some $n\in\operatorname{dom}(\sigma)\setminus\operatorname{dom}(\tau_{k})$ for all $k<m$, and $\tau_{m-1}=\sigma$. We show that $\bar{S}(\tau_{k})=\operatorname{cl}S(\tau_{k})$ holds for all $k<m$ by induction. By definition, we have $\bar{S}(\tau_{0})=S(\tau_{0})=\operatorname{cl}S(\tau_{0})$. Assume that we have $\bar{S}(\tau_{k})=\operatorname{cl}S(\tau_{k})$ and $\tau_{k+1}=\tau_{k}[n\mapsto a]$ for some $k<m$, $n\in\operatorname{dom}(\sigma)\setminus\operatorname{dom}(\tau_{k})$ and $a=\sigma(n)\in\{0,1\}$. By the assumption, we have $\bar{S}(\tau_{k}[n\mapsto a])=\bar{S}(\tau_{k})\cap\operatorname{cl}S_{n}^{a}=\operatorname{cl}S(\tau_{k})\cap\operatorname{cl}S_{n}^{a}$. By Lemma 17, $c_{n}$ is not a local extremum of $f_{n}|_{\bar{S}(\tau_{k})}$. By Lemma 8, we obtain $\operatorname{cl}S(\tau_{k})\cap\operatorname{cl}S_{n}^{a}=\operatorname{cl}(S(\tau_{k})\cap S_{n}^{a})=\operatorname{cl}S(\tau_{k+1})$.

Acknowledgement

I would like to thank Hideki Tsuiki for his expert guidance and valuable discussions. I am also grateful to Haruto Ohta for technical advices. Finally, I thank the referees for valuable comments.