#  Strongly rigid metrics in spaces of metrics

Yoshito Ishiki  Photonics Control Technology Team RIKEN Center for Advanced
Photonics 2-1 Hirasawa, Wako, Saitama 351-0198, Japan<EMAIL_ADDRESS>
###### Abstract.

A metric space is said to be strongly rigid if no positive distance is taken
twice by the metric. In 1972, Janos proved that a separable metrizable space
has a strongly rigid metric if and only if it is zero-dimensional. In this
paper, we shall develop this result for the theory of space of metrics. For a
strongly zero-dimensional metrizable space, we prove that the set of all
strongly rigid metrics is dense in the space of metics. Moreover, if the space
is the union of countable compact subspaces, then that set is comeager. As its
consequence, we show that for a strongly zero-dimensional metrizable space,
the set of all metrics possessing no nontrivial (bijective) self-isometry is
comeager in the space of metrics.

###### Key words and phrases:

Strong rigidity, Rigidity, Space of metrics, Comeagerness, Baire space

###### 2020 Mathematics Subject Classification:

Primary 54E35, Secondary 54E52, 51F99

##  1\. Introduction

###  1.1. Background

Let $X$ be a topological space, and $S$ a subset of $[0,\infty)$ with $0\in
S$. We denote by $\operatorname{Met}(X;S)$ the set of all metrics on $X$
taking values in $S$ and generating the same topology of $X$. We also denote
by $\operatorname{\mathcal{D}}_{X}$ the supremum metric on
$\operatorname{Met}(X;S)$; namely,
$\operatorname{\mathcal{D}}_{X}(d,e)=\sup_{x,y\in X}|d(x,y)-e(x,y)|$. We often
write $\operatorname{Met}(X)=\operatorname{Met}(X;[0,\infty))$. Remark that
$\operatorname{\mathcal{D}}_{X}$ is a metric taking values in $[0,\infty]$. As
is the case of ordinary metric spaces, we can introduce the topology on
$\operatorname{Met}(X)$ generated by open balls. In what follows, we consider
that $\operatorname{Met}(X)$ is equipped with this topology. The author [6, 7,
8, 10] proved the denseness and determined the Borel hierarchy of a subset of
$\operatorname{Met}(X)$ which can be represented as
$\\{\,d\in\operatorname{Met}(X)\mid\text{$d$ satisfies $P$}\,\\}$ for some
property $P$ on metric spaces under certain conditions. For example, in [8],
the author proved that the set of all doubling metrics in
$\operatorname{Met}(X)$ is dense and $F_{\sigma}$ for every compact metrizable
space $X$.

A metric $d$ on a set $X$ is said to be _strongly rigid_ if for all
$x,y,u,v\in X$, the relations $d(x,y)=d(u,v)$ and $d(x,y)\neq 0$ imply
$\\{x,y\\}=\\{u,v\\}$.

In 1972, Janos [11] proved that a separable metric space $X$ is strongly
$0$-dimensional if and only if there exists a strongly rigid metric
$d\in\operatorname{Met}(X)$ (see also [14]). The existence of strongly rigid
metrics affected research on a characterization of the dimension of metrizable
spaces using values of metrics (see, for example, [1], [12] and [4]).

A topological space $X$ is said to be _strongly $0$-dimensional_ if for all
pair $A,B$ of disjoint closed subsets of $X$, there exists a clopen subset $V$
of $X$ such that $A\subset V$ and $V\cap B=\emptyset$. Such a space is
sometimes said to be _ultranormal_.

In this paper, we develop the result on the existence of strongly rigid
metrics for the theory of spaces of metrics. For a strongly zero-dimensional
metrizable space, we prove that the set of all strongly rigid metrics is dense
in the space of metics. Moreover, if the space is $\sigma$-compact, then that
set is $G_{\delta}$. As its consequence, we show that for a strongly zero-
dimensional metrizable space, the set of all metrics possessing no nontrivial
(bijective) self-isometry is comeager in the space of metrics.

###  1.2. Main results

The symbol “$\mathfrak{c}$” stands for the cardinality of the continuum. For a
set $S$, we denote by $\operatorname{Card}(S)$ the cardinality of $S$. A
subset of a topological space is said to be $G_{\delta}$ if it is the
intersection of countable open subsets.

Let $X$ be a metrizable space. We denote by $\mathrm{LI}(X)$ the set of all
metrics $d$ such that if $x,y,u,v\in X$ satisfies $x\neq y$, $u\neq v$, and
$\\{x,y\\}\neq\\{u,v\\}$, then $d(x,y)$ and $d(u,v)$ are linearly independent
over $\mathbb{Q}$. The following is our first result:

######  Theorem 1.1.

Let $X$ be a strongly $0$-dimensional metrizable space with
$\operatorname{Card}(X)\leq\mathfrak{c}$. Let $\epsilon\in(0,\infty)$ and
$d\in\operatorname{Met}(X)$. Then there exists $e\in\mathrm{LI}(S)$ such that
$\operatorname{\mathcal{D}}_{X}(d,e)\leq\epsilon$. Namely, the set
$\mathrm{LI}(X)$ is dense in
$(\operatorname{Met}(X),\operatorname{\mathcal{D}}_{X})$. Moreover, if $X$ is
completely metrizable, we can choose $e$ as a complete metric.

We denote by $\mathrm{SR}(X)$ the set of all strongly rigid metrics in
$\operatorname{Met}(X)$. As a consequence of Theorem 1.1, we obtain our second
result:

######  Theorem 1.2.

Let $X$ be a strongly $0$-dimensional metrizable space with
$\operatorname{Card}(X)\leq\mathfrak{c}$. Then the set $\mathrm{SR}(X)$ is
dense in $\operatorname{Met}(X)$. Moreover, if $X$ is $\sigma$-compact, then
$\mathrm{SR}(X)$ is dense $G_{\delta}$ in
$(\operatorname{Met}(X),\operatorname{\mathcal{D}}_{X})$.

######  Remark 1.1.

Note that Theorem 1.2 is true even if $X$ is not locally compact. For example,
Theorem 1.2 is ture for $X=\mathbb{Q}$.

######  Remark 1.2.

Our Theorem 1.2 can be considered as an analogue of Rouyer’s result that
generic metric spaces in the Gromov–Hausdorff space are strongly rigid [20,
Theorem 2]. Rouyer uses the term “totally anisometric” instead of “strongly
rigid”.

We say that a metric $d$ on a set $X$ is said to be _rigid_ if every bijective
isometry $f\colon(X,d)\to(X,d)$ must be the identity map. We denote by
$\mathrm{R}(X)$ the set of all rigid metrics in $\operatorname{Met}(X)$.

Let $X$ be a topological space. A subset $S$ of $X$ is said to be _comeager_
if $S$ contains a dense $G_{\delta}$ subset of $X$. As an application of
Theorem 1.1, we obtain our third result:

######  Theorem 1.3.

Let $X$ be a strongly $0$-dimensional metrizable space. If $X$ is
$\sigma$-compact and satisfies $3\leq\operatorname{Card}(X)\leq\mathfrak{c}$,
then the set $\mathrm{R}(X)$ is comeager in
$(\operatorname{Met}(X),\operatorname{\mathcal{D}}_{X})$.

######  Remark 1.3.

Our Theorem 1.3 can be considered as a $0$-dimensional analogue of the result
that the set of all rigid Riemannian (or pseudo-Riemannian) metrics is open
dense in the space of Riemannian metrics with respect to the Whitney
$C^{\infty}$-topology (see [2] and [15]).

As a consequence of Theorem 1.2, we know the existence of strange metrics on
$\sigma$-compact locally compact Hausdorff spaces.

######  Theorem 1.4.

Let $X$ be a strongly zero-dimensional $\sigma$-compact locally compact
Hausdorff space. Then there exists $d\in\operatorname{Met}(X)$ such that for
every $\xi\in X$, the map $F_{\xi}\colon X\to[0,\infty)$ defined by
$F_{\xi}(x)=d(x,\xi)$ is a topological embedding.

The organization of this paper is as follows: In Section 2, we introduce a
ubiquitously dense subset of a metrizable space, and we construct a strongly
rigid discrete metric taking values in a given ubiquitously dense subset of
$[0,\infty)$. In Section 3, we give a system of linearly independent real
numbers over $\mathbb{Q}$ using a bijection between $\mathbb{Z}_{\geq 0}$ and
$\mathbb{Q}_{\geq 0}$. The argument in this section is devoted to show Lemma
4.18, which is a important part of the proof of Theorem 1.1. In Section 4, we
first construct a strongly rigid metric on the countable power $N(\Omega)$ of
the discrete space $\Omega$ with $\operatorname{Card}(\Omega)=\mathfrak{c}$.
Since every strongly $0$-dimensional metrizable space $X$ with
$\operatorname{Card}(X)\leq\mathfrak{c}$ can be topologically embedded into
$N(\Omega)$, we can obtain a strongly rigid metric on $X$. In Section 5, we
prove Theorems 1.1, 1.2, 1.3, and 1.4.

##  2\. A construction of strongly rigid discrete metrics

In this paper, we say that a metric is _discrete_ if it generates the discrete
topology. We first give a construction of strongly rigid metrics on discrete
spaces. We begin with the following basic proposition on the triangle
inequality.

######  Proposition 2.1.

Let $N_{1},N_{2},N_{3}\in\mathbb{Z}_{\geq 1}$. We assume that
$M_{1},M_{2},M_{3}\in(0,\infty)$ satisfy
$M_{i}\in(N_{i}+2^{-N_{i}-1},N_{i}+2^{-N_{i}})$ for all $i\in\\{1,2,3\\}$. If
$N_{1}\leq N_{2}+N_{3}$, then we have $M_{1}<M_{2}+M_{3}$.

###### Proof.

We may assume that $N_{2}\leq N_{3}$. If $N_{1}<N_{2}+N_{3}$, then we have
$M_{1}<N_{1}+1\leq N_{2}+N_{3}<M_{2}+M_{3}$.

In the case of $N_{1}=N_{2}+N_{3}$, we have

$\displaystyle M_{1}$
$\displaystyle<N_{1}+2^{-N_{1}}=N_{1}+2^{-(N_{2}+N_{3})}\leq N_{1}+2^{-N_{3}}$
$\displaystyle=N_{1}+2^{-N_{3}-1}+2^{-N_{3}-1}\leq
N_{2}+N_{3}+2^{-N_{2}-1}+2^{-N_{3}-1}$
$\displaystyle=(N_{2}+2^{-N_{2}-1})+(N_{3}+2^{-N_{3}-1})<M_{2}+M_{3}.$

Thus, we conclude that $M_{1}<M_{2}+M_{3}$. ∎

######  Remark 2.1.

A crucial point of Proposition 2.1 is that we can choose $M_{i}$ depending
only on $N_{i}$.

######  Definition 2.1.

Let $X$ be a metrizable space. We say that a subset $S$ of $X$ is
_ubiquitously dense_ if for every non-empty open subset $U$ of $X$, we have
$\operatorname{Card}(U\cap S)=\operatorname{Card}(S)$.

Note that if there exists a ubiquitously dense subset $S$ of a metrizable
space $X$ with $1<\operatorname{Card}(X)$, then $S$ should be infinite and the
space $X$ has no isolated points.

In this paper, we use the set-theoretic representation of cardinal. For
example, the relation $\alpha<\mathfrak{c}$ means $\alpha\neq\mathfrak{c}$ and
$\alpha\in\mathfrak{c}$, and we have
$\mathfrak{c}=\\{\,\alpha\mid\alpha<\mathfrak{c}\,\\}$. For more discussion,
we refer the readers to [13].

######  Theorem 2.2.

Let $X$ be a separable metrizable space with $1<\operatorname{Card}(X)$ and
$S$ a ubiquitously dense subset of $X$. Put $\kappa=\operatorname{Card}(S)$.
Then there exists a family $\\{A(\alpha)\\}_{\alpha<\kappa}$ of subsets of $X$
such that

  1. (1)

each $A(\alpha)$ is countable;

  2. (2)

each $A(\alpha)$ is dense in $X$;

  3. (3)

if $\alpha,\beta\in\kappa$ satisfy $\alpha\neq\beta$, then $A(\alpha)\cap
A(\beta)=\emptyset$;

  4. (4)

$S=\bigcup_{\alpha<\kappa}A(\alpha)$.

###### Proof.

Notice that $\aleph_{0}\leq\kappa$. By $\kappa\times\aleph_{0}=\kappa$, there
exists a (strictly) linear order $\prec$ such that
$(\kappa\times\mathbb{Z}_{\geq 0},\prec)$ and $(\kappa,<)$ are isomorphic to
each other as ordered sets. Since $X$ is separable, there exists a countable
open base $\\{U_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ of $X$. Using transfinite
induction, we shall define a sequence
$\\{q(\alpha,i)\\}_{(\alpha,i)\in\kappa\times\mathbb{Z}_{\geq 0}}$ such that

  1. (A)

if $(\alpha,i),(\beta,j)\in\kappa\times\mathbb{Z}_{\geq 0}$ satisfy
$(\alpha,i)\neq(\beta,j)$, then $q(\alpha,i)\neq q(\beta,j)$;

  2. (B)

for all $(\alpha,i)\in\kappa\times\mathbb{Z}_{\geq 0}$, we have
$q(\alpha,i)\in U_{i}\cap S$.

We assume that we have already obtained a sequence
$\\{q(\alpha,i)\\}_{(\alpha,i)\prec(\theta,n)}$ satisfying the conditions (A)
and (B) for the set $\\{\,(\alpha,i)\mid(\alpha,i)\prec(\theta,n)\,\\}$
instead of $\kappa\times\mathbb{Z}_{\geq 0}$. We shall construct
$q(\theta,n)\in X$. Put
$B=\\{\,q(\alpha,i)\mid(\alpha,i)\prec(\theta,n)\,\\}$. Since
$(\kappa\times\mathbb{Z}_{\geq 0},\prec)$ is isomorphic to $(\kappa,<)$, we
have $\operatorname{Card}(B)<\kappa$. Thus, from
$\operatorname{Card}(U_{n}\cap S)=\kappa$, it follows that $(U_{n}\cap
S)\setminus B\neq\emptyset$. Take $q(\theta,n)\in(U_{n}\cap S)\setminus B$.
Then $\\{q(\theta,n)\\}\cup B$ satisfies the conditions (A) and (B) for the
set $\\{(\theta,n)\\}\cup\\{\,(\alpha,i)\mid(\alpha,i)\prec(\theta,n)\,\\}$
instead of $\kappa\times\mathbb{Z}_{\geq 0}$. Therefore, by transfinite
induction, we obtain a sequence
$\\{q(\alpha,i)\\}_{(\alpha,i)\in\kappa\times\mathbb{Z}_{\geq 0}}$ satisfying
the condition (A) and (B).

For each $\alpha\in\kappa$, we define $B(\alpha)=\\{\,q(\alpha,i)\mid
i\in\mathbb{Z}_{\geq 0}\,\\}$. Then the conditions (A) and (B) imply that the
family $\\{B(\alpha)\\}_{\alpha<\mathfrak{c}}$ satisfies the following
conditions:

  1. (a)

each $B(\alpha)$ is a subset of $S$;

  2. (b)

if $\alpha,\beta\in\mathfrak{c}$ satisfy $\alpha\neq\beta$, then
$B(\alpha)\cap B(\beta)=\emptyset$;

  3. (c)

each $B(\alpha)$ is countable and dense in $X$.

We put $\tau=\operatorname{Card}(S\setminus\bigcup_{\alpha<\kappa}B(\alpha))$
and we take a bijection $\xi\colon\tau\to
S\setminus\bigcup_{\alpha<\kappa}B_{\alpha}$. Notice that $\tau\leq\kappa$.
For each $\alpha<\kappa$, we define $A(\alpha)$ by
$A(\alpha)=B(\alpha)\cup\\{\xi(\alpha)\\}$ if $\alpha<\tau$; otherwise,
$A(\alpha)=B(\alpha)$. Then the family $\\{A(\alpha)\\}_{\alpha<\kappa}$ is a
desired one. ∎

The following lemma is identical with [10, Proposition 2.5], which is related
to metric-preserving functions.

######  Lemma 2.3.

Let $X$ be a discrete topological space. Let $\eta\in(0,\infty)$ and
$d\in\operatorname{Met}(X)$. Then, there exists a metric
$e\in\operatorname{Met}(X;\eta\cdot\mathbb{Z})$ such that
$\operatorname{\mathcal{D}}_{X}(d,e)\leq\eta$ and $\eta\leq e(x,y)$ for all
distinct $x,y\in X$.

For a set $X$, we define $[X]^{2}$ by $[X]^{2}=\\{\,\\{x,y\\}\mid x,y\in
X,x\neq y\,\\}$. Note that if $X$ is infinite, we have
$\operatorname{Card}(X)=\operatorname{Card}([X]^{2})$. A metric $d$ on a set
$X$ is said to be _uniformly discrete_ if there exists $c\in(0,\infty)$ such
that $c<d(x,y)$ for all distinct $x,y\in X$.

######  Theorem 2.4.

Let $S$ be a ubiquitously dense subset of $[0,\infty)$ with $0\in S$ and
$\kappa=\operatorname{Card}(S)$. Let $X$ be a discrete space with
$\operatorname{Card}(X)\leq\kappa$ and $d\in\operatorname{Met}(X)$. If
$\epsilon\in(0,\infty)$, then there exists a metric
$e\in\operatorname{Met}(X;S)$ such that

  1. (1)

we have $\operatorname{\mathcal{D}}_{X}(d,e)\leq\epsilon$;

  2. (2)

the metric $e$ is uniformly discrete;

  3. (3)

we have $e(x,y)<e(x,z)+e(z,y)$ for all distinct $x,y,z\in X$;

  4. (4)

the metric $e$ is strongly rigid.

###### Proof.

We put $\eta=\epsilon/2$ and $T=\eta^{-1}\cdot S=\\{\,\eta^{-1}s\mid s\in
S\,\\}$. Note that $T$ is ubiquitously dense in $[0,\infty)$ and $0\in T$.

Lemma 2.3 guarantees the existence of a metric
$h\in\operatorname{Met}(X;\eta\cdot\mathbb{Z}_{\geq 0})$ such that
$\operatorname{\mathcal{D}}_{X}(h,e)\leq\eta$. Put $u=\eta^{-1}\cdot
h\in\operatorname{Met}(X;\mathbb{Z}_{\geq 0})$. Due to Theorem 2.2, we can
take a mutually disjoint dense decomposition $\\{A(\alpha)\\}_{\alpha<\kappa}$
of $T$. Put $\tau=\operatorname{Card}(X)$. Then $\tau\leq\kappa$. We take a
bijection $\varphi\colon\tau\to[X]^{2}$. We represent
$\varphi(\alpha)=(x_{\alpha},y_{\alpha})$ and
$\theta_{\\{x,y\\}}=\varphi^{-1}(\\{x,y\\})$. For each $\alpha<\tau$, we put
$N_{\alpha}=u(x_{\alpha},y_{\alpha})\in\mathbb{Z}_{\geq 1}$ and take
$w(\alpha)\in(N_{\alpha}+2^{-N_{\alpha}-1},N_{\alpha}+2^{-N_{\alpha}})\cap
A(\alpha)$. Since each $A(\alpha)$ is dense in $[0,\infty)$, the existence of
$w(\alpha)$ is always guaranteed.

We define a function $v\colon X^{2}\to[0,\infty)$ by
$v(x,y)=w(\theta_{\\{x,y\\}})$ if $x\neq y$; otherwise, $v(x,x)=0$. According
to Proposition 2.1, the function $v\colon X^{2}\to[0,\infty)$ satisfies the
triangle inequality. Since $1+2^{-2}\leq v(x,y)$ for all distinct $x,y\in X$,
the metric $v$ generates the discrete topology on $X$, namely,
$v\in\operatorname{Met}(X;T)$. Notice that
$\operatorname{\mathcal{D}}_{X}(u,v)\leq 1$.

Put $e=\eta\cdot v$. Since $v$ is in $\operatorname{Met}(X)$, so is $e$. By
$v\in\operatorname{Met}(X;T)$ and $T=\eta^{-1}\cdot S$, we have
$e\in\operatorname{Met}(X;S)$.

Using $h=\eta\cdot u$, $e=\eta\cdot v$,
$\operatorname{\mathcal{D}}_{X}(d,h)\leq\eta$, and
$\operatorname{\mathcal{D}}_{X}(u,v)\leq 1$, we obtain

$\displaystyle\operatorname{\mathcal{D}}_{X}(d,e)$
$\displaystyle\leq\operatorname{\mathcal{D}}_{X}(d,h)+\operatorname{\mathcal{D}}_{X}(h,e)=\operatorname{\mathcal{D}}_{X}(d,h)+\operatorname{\mathcal{D}}_{X}(\eta\cdot
u,\eta\cdot v)$
$\displaystyle\leq\eta+\eta\operatorname{\mathcal{D}}_{X}(u,v)\leq
2\eta=\epsilon.$

This implies the condition (1).

Since $1+2^{-2}\leq v(x,y)$ for all distinct $x,y\in X$, we observe that
$e(=\eta\cdot v)$ is uniformly discrete. This means that the condition (2) is
true.

From Proposition 2.1, it follows that $e(x,y)<e(x,z)+e(z,y)$ for all distinct
$x,y,z\in X$. This proves the condition (3).

We shall prove $e$ is strongly rigid. Take $\\{x,y\\},\\{a,b\\}\in[X]^{2}$
with $\\{x,y\\}\neq\\{a,b\\}$. Then $\theta_{\\{x,y\\}}\neq\theta_{\\{a,b\\}}$
and $A\left(\theta_{\\{x,y\\}}\right)\cap
A\left(\theta_{\\{a,b\\}}\right)=\emptyset$. In particular, we have
$w\left(\theta_{\\{x,y\\}}\right)\neq w\left(\theta_{\\{a,b\\}}\right)$, and
hence $e(x,y)\neq e(a,b)$. Namely, the metric $e$ is strongly rigid. This
implies that $e$ satisfies the condition (4). ∎

######  Remark 2.2.

In the case where $X$ is finite, Theorem 2.4 gives a new proof of [20, Lemma
3] stating that the set of finite metric spaces which are totally anisometric
(strongly rigid) and without collinear points is dense in the Gromov–Hausdorff
space.

##  3\. A system of linearly independent numbers over $\mathbb{Q}$

In this subsection, we give a system yielding real numbers which are linearly
independent over $\mathbb{Q}$.

######  Definition 3.1.

Let $\alpha=\\{a_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ be a summable sequence of
positive real numbers. Let $Q\colon\mathbb{Z}_{\geq 0}\to\mathbb{Q}_{\geq 0}$
be a bijection. For a non-empty subset $B$ of $\mathbb{Q}_{\geq 0}$, we define

$\Sigma_{Q}[\alpha,B]=\sum_{Q(i)\in B}a_{i}.$

If $B=\emptyset$, we also define $\Sigma_{Q}[\alpha,\emptyset]=0$.

######  Definition 3.2.

We denote by $\mathcal{Q}$ the set of all $F\colon\mathbb{Z}_{\geq
0}\to\mathbb{Z}_{\geq 0}$ such that $F$ is strictly increasing and satisfies

$\lim_{n\to\infty}(F(n+1)-F(n))=\infty,$

and

$\lim_{n\to\infty}\sum_{m=n+1}^{\infty}2^{F(n)-F(m)}=0.$

The author is inspired by [5] (see also [17] and [16]) with respect to a
construction of linearly independent real numbers over $\mathbb{Q}$ in the
following proposition:

######  Proposition 3.1.

Let $F\in\mathcal{Q}$. We define a sequence
$\lambda=\\{\lambda_{i}\\}_{i\in\mathbb{Z}_{\geq}}$ by
$\lambda_{i}=2^{-F(i)}$. Let $k\in\mathbb{Z}_{\geq 0}$ and
$\\{P_{i}\\}_{i=0}^{k}$ a family of subsets of $\mathbb{Q}_{\geq 0}$. Let $S$
be a subset of $\mathbb{Q}_{\geq 0}$. We assume that there exist $a$,
$b_{0},\dots,b_{k}$ in $[0,\infty)$ such that

  1. (I)

$a<b_{i}$ for all $i\in\\{0,\dots,k\\}$;

  2. (II)

if $i\neq j$, then $b_{i}\neq b_{j}$;

  3. (III)

we have $S\cap P_{i}=[a,b_{i})\cap\mathbb{Q}_{\geq 0}$ for all
$i\in\\{0,\dots,k\\}$.

Then the $(k+2)$-many numbers
$\Sigma_{Q}[\lambda,P_{0}],\dots,\Sigma_{Q}[\lambda,P_{k}]$, and $1$ are
linearly independent over $\mathbb{Q}$.

###### Proof.

We may assume that $b_{i}<b_{i+1}$ for all $i\in\\{0,\dots,k-1\\}$. For every
$i\in\\{0,\dots,k\\}$, we define $r(i)=\Sigma_{Q}[\lambda,P_{i}]$. To prove
the proposition, we assume that integers
$c_{0},\dots,c_{k},c_{k+1}\in\mathbb{Z}$ satisfy

(3.1) $\displaystyle c_{0}\cdot r(0)+\dots+c_{k}\cdot r(k)+c_{k+1}\cdot 1=0.$

We first prove $c_{k}=0$. Since $S\cap P_{k}=[a,b_{k})\cap\mathbb{Q}$ and
$b_{k-1}<b_{k}$, we observe that the set
$P_{k}\setminus\left(\bigcup_{i=0}^{k-1}P_{i}\right)$ is infinite. Thus, by
$F\in\mathcal{Q}$, we can take a sufficient large $n\in\mathbb{Z}_{\geq 0}$
such that

  1. (1)

$(|c_{0}|+\dots+|c_{k}|)\sum_{m=n+1}^{\infty}2^{F(n)-F(m)}<1$.

  2. (2)

$n\in P_{k}$

  3. (3)

we have $n\not\in\bigcup_{i=0}^{k-1}P_{i}$.

  4. (4)

$|c_{k}|<2^{F(n)-F(n-1)}$.

Put

$I=c_{0}\sum_{j\in P_{i}\cap[0,n]}2^{F(n)-F(j)}+\dots+c_{k}\sum_{j\in
P_{i}\cap[0,n]}2^{F(n)-F(j)}+c_{k+1}2^{F(n)}.$

and

$J=c_{0}\sum_{j\in P_{i}\cap(n,\infty)}2^{F(n)-F(j)}+\dots+c_{k}\sum_{j\in
P_{i}\cap(n,\infty)}2^{F(n)-F(j)}.$

Then the equality (3.1) implies

$I+J=2^{F(n)}\times(c_{0}r(0)+\dots+c_{k}r(k)+c_{k+1})=0.$

By (1), we have $|J|<1$. Note that $I\in\mathbb{Z}$ (if $j\leq n$, then
$F(n)-F(j)\geq 0$). By $J=-I$, we obtain $J\in\mathbb{Z}$. Combining
$J\in\mathbb{Z}$ and $|J|<1$, we conclude that $J=0$. Thus, we also have
$I=0$.

Since for all $j\leq n$, we have $2^{F(n)-F(j)}$ can be divided by
$2^{F(n)-F(n-1)}$. Thus, by $n\in P_{k}$, we have

(3.2) $\displaystyle\sum_{j\in P_{k}\cap[0,n]}2^{F(n)-F(j)}=1+L_{k}\cdot
2^{F(n)-F(n-1)}$

for some $L_{k}\in\mathbb{Z}$. Due to (3), for all $i$ with $i<k$, we have
$n\not\in P_{i}$. Hence

(3.3) $\displaystyle\sum_{j\in P_{i}\cap[0,n]}2^{F(n)-F(j)}=L_{i}\cdot
2^{F(n)-F(n-1)}$

for some $L_{i}\in\mathbb{Z}$. Since $c_{k+1}$ is an integer, and since
$F(n)-F(n-1)<F(n)$,

(3.4) $\displaystyle c_{k+1}2^{F(n)}=L_{k+1}\cdot 2^{F(n)-F(n-1)}$

for some $L_{k+1}\in\mathbb{Z}$ (of cause, $L_{k+1}=c_{k+1}2^{F(n-1)}$).

From $I=0$, and the equalities (3.2), (3.3), and (3.4), it follows that

$c_{k}=M\cdot 2^{F(n)-F(n-1)}$

for some $M\in\mathbb{Z}$. By $|c_{k}|<2^{F(n)-F(n-1)}$ (the condition (4)),
we conclude that $M=0$. Thus $c_{k}=0$. Using the same argument, by induction,
we obtain $c_{k}=c_{k-1}=\dots=c_{0}=0$. Hence $c_{k+1}=0$. This means that
the numbers $\Sigma_{Q}[\lambda,P_{0}],\dots,\Sigma_{Q}[\lambda,P_{k}]$ and
$1$ are linearly independent over $\mathbb{Q}$. ∎

##  4\. A construction of strongly rigid metrics

In this subsection, we construct a strongly rigid metric on a given strongly
$0$-dimensional metrizable space.

The following proposition is deduced from the existence and uniqueness of the
binary representation of a real number with infinitely many digits which are
$1$. We omit the proof.

######  Proposition 4.1.

Let $f\colon\mathbb{Z}_{\geq 0}\to\mathbb{Z}_{\geq 0}$ be an injective map and
define a sequence $\lambda=\\{\lambda_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ by
$\lambda_{i}=2^{-f(i)}$. Fix a bijection $Q\colon\mathbb{Z}_{\geq
0}\to\mathbb{Q}_{\geq 0}$. Let $S$ and $T$ be infinite or empty subsets of
$\mathbb{Q}_{\geq 0}$. If $\Sigma_{Q}[\lambda,S]=\Sigma_{Q}[\lambda,T]$, then
we have $S=T$.

######  Definition 4.1.

Let $k\in\mathbb{Z}_{0}$. We define $F_{k}\colon\mathbb{Z}_{\geq
0}\to\mathbb{Z}_{\geq 0}$ by $F_{k}(n)=2^{n}+k$. Remark that
$F_{k}\in\mathcal{Q}$. We also define a sequence
$\zeta_{(k)}=\\{\zeta_{(k),i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ by
$\zeta_{(k),i}=2^{-F_{k}(i)}$.

######  Definition 4.2.

In what follows, we fix the discrete space $\Omega$ with
$\operatorname{Card}(\Omega)=\mathfrak{c}$. We define
$N(\Omega)=\Omega^{\mathbb{Z}_{\geq 0}}$. We consider that the set $N(\Omega)$
is always equipped with the product topology.

Remark that the space $N(\Omega)$ is sometimes called the _Baire space of
weight $\mathfrak{c}$_ (see [19] or [21]).

Our first purpose is to construct strongly rigid metrics on $\Omega$ and
$N(\Omega)$. For this purpose, we utilize strongly rigid semi-metrics.

######  Definition 4.3.

Let $S$ be a subset of $[0,\infty)$. We say that a map $r\colon X\times
X\to[0,\infty)$ is an _$S$ -semi-metric on $X$_ if

  1. (1)

for all $x,y\in X$, we have $r(x,y)=r(y,x)$;

  2. (2)

for all $x,y\in X$, we have $r(x,y)=0$ if and only if $x=y$;

  3. (3)

for all distinct $x,y\in X$, we have $r(x,y)\in S$.

The _strong rigidity of an $S$-semi-metric_ is defined in the same way of
ordinary metrics.

Note that semi-metrics are not assumed to satisfy the triangle inequality.

######  Proposition 4.2.

Let $S$ be a subset $(0,\infty)$ with $\operatorname{Card}(S)=\mathfrak{c}$.
Then there exists a strongly rigid $S$-semi-metric $r$ on $\Omega$.

###### Proof.

Since $\operatorname{Card}([\Omega]^{2})=\mathfrak{c}$, we can take a
bijection $\phi\colon[\Omega]^{2}\to S$. We define $r(x,y)=\phi(\\{x,y\\})$ if
$x\neq y$, otherwise, $r(x,x)=0$. Then $r$ is a desired one. ∎

######  Definition 4.4.

Let $Q\colon\mathbb{Z}_{\geq 0}\to\mathbb{Q}_{\geq 0}$ be a bijection. We
define $\mu(m)=\min Q^{-1}([m,m+1)\cap\mathbb{Q})$. We say that the map $Q$
satisfies the _property $(\mathrm{M})$_ if

  1. (1)

for all $m\in\mathbb{Z}_{\geq 0}$, we have $Q(\mu(m))=m$;

  2. (2)

for all $m\in\mathbb{Z}_{\geq 0}$, we have $\mu(m)<\mu(m+1)$.

Notice that $\mu(0)=0$.

######  Lemma 4.3.

There exists a bijection $Q\colon\mathbb{Z}_{\geq 0}\to\mathbb{Q}_{\geq 0}$
satisfying the property $(\mathrm{M})$.

###### Proof.

Take a mutually disjoint family $\\{A_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ of
infinite subsets of $\mathbb{Z}_{\geq 0}$ satisfying that
$\bigcup_{i\in\mathbb{Z}_{\geq 0}}A_{i}=\mathbb{Z}_{\geq 0}$ and $\min
A_{i}<\min A_{i+1}$ for all $i\in\mathbb{Z}_{\geq 0}$. For each
$i\in\mathbb{Z}_{\geq 0}$, we take a bijection $\theta_{i}\colon
A_{i}\to[i,i+1)\cap\mathbb{Q}$ with $\theta_{i}(\min A_{i})=i$. Glueing them
together, we obtain a bijection $\mathbb{Z}_{\geq 0}\to\mathbb{Q}_{\geq 0}$
with the property $(\mathrm{M})$. ∎

######  Remark 4.1.

In what follows, based on Lemma 4.3, we fix the bijection
$Q\colon\mathbb{Z}_{\geq 0}\to\mathbb{Q}_{\geq 0}$ with the property
$(\mathrm{M})$.

######  Definition 4.5.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let $m\in\mathbb{Z}_{\geq 0}$ and $r$ an
$S$-semi-metric on $\Omega$. We define
$H_{m,r(x,y)}=[m,r(x,y))\cap\mathbb{Q}$. Note that if $x=y$, the set
$H_{m,r(x,y)}$ is always empty. We also define
$[r]_{k,m}\colon\Omega\times\Omega\to[0,\infty)$ by
$[r]_{k,m}(x,y)=\Sigma_{Q}[\zeta_{(k)},H_{m,r(x,y)}]$.

######  Remark 4.2.

Under the same assumptions in Definition 4.5, we notice that
$H_{0,r(x,x)}=[0,0)\cap\mathbb{Q}=\emptyset$ for all $x\in\Omega$.

######  Lemma 4.4.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let $m\in\mathbb{Z}_{\geq 0}$ and $S$ a subset
of $(m,m+1)$. Let $X$ be a discrete space with
$\operatorname{Card}(X)\leq\mathfrak{c}$. If $r$ is a strongly rigid $S$-semi-
metric on $\Omega$, then the following statements are true:

  1. (1)

We have $[r]_{k,m}(x,y)\in(\zeta_{(k),\mu(m)},2\zeta_{(k),\mu(m)})$ for all
distinct $x,y\in X$.

  2. (2)

The function $[r]_{k,m}$ is a metric on $\Omega$ and we have
$[r]_{k,m}(x,y)\in\operatorname{Met}(\Omega)$.

  3. (3)

The metric $[r]_{k,m}$ is strongly rigid.

###### Proof.

We first prove the statement (1). Take distinct $x,y\in\Omega$. By the
property $(\mathrm{M})$, we have $Q(\mu(m))=m$ and
$m\in[m,d(x,y))\cap\mathbb{Q}=H_{m,r(x,y)}$. Thus we obtain

$\zeta_{(k),\mu(m)}\leq\Sigma_{Q}[\zeta_{(k)},H_{m,r(x,y)}]=[r]_{k,m}(x,y).$

Since $\mu(m)$ is the minimal number of the set
$Q^{-1}([m,m+1)\cap\mathbb{Q})$ (see Definition 4.4), and since $F_{k}$ is
strictly increasing, we have

$\displaystyle\Sigma_{Q}[\zeta_{(k)},H_{m,r(x,y)}]$ $\displaystyle=\sum_{i\in
Q^{-1}([m,m+1)\cap\mathbb{Q})}2^{-F_{k}(i)}\leq\sum_{F_{k}(\mu(m))\leq
i}2^{-i}$ $\displaystyle=2\cdot 2^{-F_{k}(\mu(m))}=2\zeta_{(k),\mu(m)}.$

Hence $\Sigma_{Q}[\zeta_{(k)},H_{m,r(x,y)}]\leq 2\zeta_{(k),\mu(m)}$. This
implies the statement (1).

We next prove the statement (2). If $x,y\in\Omega$ satisfies $x=y$, we have
$H_{m,r(x,y)}=\emptyset$. Thus
$[r]_{k,m}(x,y)=\Sigma_{Q}[\zeta_{(k)},H_{m,r(x,y)}]=0$. To show that
$[r]_{k,m}$ satisfies the triangle inequality, we take distinct
$x,y,z\in\Omega$. According to the statement (1), we have

$[r]_{k,m}(x,y)\leq
2\zeta_{(k),\mu(m)}=\zeta_{(k),\mu(m)}+\zeta_{(k),\mu(m)}<[r]_{k,m}(x,z)+[r]_{k,m}(z,y).$

Thus, the function $[r]_{k,m}$ satisfies the triangle inequality. According to
the statement (1) again, the metric $[r]_{k,m}$ is uniformly discrete, and
hence $[r]_{k,m}\in\operatorname{Met}(\Omega)$. This finishes the proof of
(2).

We shall prove the statement (3). We assume that $x,y,u,v\in X$ satisfy
$0<[r]_{k,m}(x,y)$ and $[r]_{k,m}(x,y)=[r]_{k,m}(u,v)$. Then we have
$\Sigma_{Q}[\zeta_{(k)},H_{m,r(x,y)}]=\Sigma_{Q}[\zeta_{(k)},H_{m,r(u,v)}]$.
By Proposition 4.1, we obtain $H_{m,r(x,y)}=H_{m,r(u,v)}$. Namely, we have
$[m,r(x,y))\cap\mathbb{Q}=[m,r(u,v))\cap\mathbb{Q}$, and hence
$r(x,y)=r(u,v)$. Since $r$ is strongly rigid, we conclude that
$\\{x,y\\}=\\{u,v\\}$. This means that $d$ is strongly rigid. Then the
statement (3) is true. This completes the proof of the lemma. ∎

######  Definition 4.6.

Let $S$ be a dense subset of $[0,\infty)$. We say that a family
$\\{r_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ is an _$S$ -gauge system on $\Omega$_
if each $r_{i}$ is a strongly rigid $(S\cap(i,i+1))$-semi-metric on $\Omega$.

######  Proposition 4.5.

Let $S$ be a ubiquitously dense subset in $[0,\infty)$ with
$\operatorname{Card}(S)=\mathfrak{c}$. Then, there exists an $S$-gauge system
$\\{r_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ on $\Omega$.

###### Proof.

Since $S$ is ubiquitously dense, we have
$\operatorname{Card}(S\cap(i,i+1))=\mathfrak{c}$ for all $i\in\mathbb{Z}_{\geq
0}$. Then, by Proposition 4.2, there exists a strongly rigid
$(S\cap(i,i+1))$-semi-metric $r_{i}$ on $\Omega$. Thus, the sequence
$\\{r_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ is an $S$-gauge system on $\Omega$. ∎

######  Definition 4.7.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let $S$ be a ubiquitously dense subset of
$[0,\infty)$ and $R=\\{r_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ an $S$-gauge system
on $\Omega$. In this case, $H_{m,r_{m}(x_{m},y_{m})}\cap
H_{m^{\prime},r_{m^{\prime}}(x_{m^{\prime}},y_{m^{\prime}})}=\emptyset$ for
all distinct $m,m^{\prime}\in\mathbb{Z}_{\geq 0}$ and for all
$x=(x_{i})_{i\in\mathbb{Z}_{\geq 0}}$ and $y=(y_{i})_{i\in\mathbb{Z}_{\geq
0}}$ in $N(\Omega)$. We define

$I_{R,x,y}=\coprod_{m\in\mathbb{Z}_{\geq 0}}H_{m,r_{m}(x_{m},y_{m})}.$

We also define a function $[R]_{k}\colon N(\Omega)\times
N(\Omega)\to[0,\infty)$ by

$[R]_{k}(x,y)=\Sigma_{Q}[\zeta_{(k)},I_{R,x,y}].$

Under the same assumptions in Definition 4.7, notice that we have

$[R]_{k}(x,y)=\sum_{i=0}^{\infty}[r_{m}]_{k,m}(x_{m},y_{m}),$

where $x=(x_{i})_{i\in\mathbb{Z}_{\geq 0}}$ and
$y=(y_{i})_{i\in\mathbb{Z}_{\geq 0}}$. Since $\zeta_{(k)}$ is summable, we
have $[R]_{k}(x,y)<\infty$ for all $x,y\in N(\Omega)$.

######  Lemma 4.6.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let $S$ be a ubiquitously dense subset of
$[0,\infty)$ and $R=\\{r_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ an $S$-gauge system
on $\Omega$. Let $m\in\mathbb{Z}_{\geq 0}$ and
$x=(x_{i})_{i\in\mathbb{Z}_{\geq 0}},y=(y_{i})_{i\in\mathbb{Z}_{\geq 0}}\in
N(\Omega)$. Then the following statements hold:

  1. (A)

If $x,y\in X$ satisfies $[R]_{k}(x,y)\leq\zeta_{(k),\mu(m)}$, then we have
$x_{i}=y_{i}$ for all $i\in\\{0,\dots,m\\}$.

  2. (B)

If $x_{i}=y_{i}$ for all $i\in\\{0,\dots,m\\}$, then we have $[R]_{k}(x,y)\leq
4\zeta_{(k),\mu(m+1)}$.

###### Proof.

We first prove (A). For the sake of contradiction, we suppose that there
exists $i\in\\{0,\dots,m\\}$ such that $x_{i}\neq y_{i}$ and
$[R]_{k}(x,y)\leq\zeta_{(k),\mu(m)}$. Since $Q$ satisfies the property
$(\mathrm{M})$, we have $\mu(i)\leq\mu(m)$, and hence
$\zeta_{(k),\mu(m)}\leq\zeta_{(k),\mu(i)}$. By
$[R]_{k}(x,y)=\sum_{i=0}^{\infty}[r_{m}]_{k,m}(x_{m},y_{m})$ and by
$\zeta_{(k),\mu(i)}<[r_{i}]_{k,i}(x_{i},y_{i})$ (see the statement (1) in
Lemma 4.4), we have $\zeta_{(k),\mu(m)}<[R]_{k}(x,y)$. This is a
contradiction. Therefore the statement (A) is true.

We next prove (B). If $x_{i}=y_{i}$ for all $i\in\\{0,\dots,m\\}$, we have
$[R]_{k}(x,y)=\sum_{i=m+1}^{\infty}[r_{m}]_{k,m}(x_{m},y_{m})$. According to
the statement (1) in Lemma 4.4, for each $i\in\mathbb{Z}_{\geq m+1}$, we have
$[r_{m}]_{k,m}(x_{m},y_{m})<2\zeta_{(k),\mu(i)}$. Since
$[R]_{k}(x,y)=\sum_{i=0}^{\infty}[r_{m}]_{k,m}(x_{m},y_{m})$, and since
$\sum_{i=m+1}^{\infty}\zeta_{(k),\mu(i)}\leq 2\zeta_{(k),\mu(m+1)}$, we obtain
$[R]_{k}(x,y)\leq 4\zeta_{(k),\mu(m+1)}$. This proves the statement (B). ∎

######  Proposition 4.7.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let $S$ be a ubiquitously dense subset of
$[0,\infty)$ and $R=\\{r_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ an $S$-gauge system
on $\Omega$. Then the function $[R]_{k}\colon N(\Omega)\times
N(\Omega)\to[0,\infty)$ satisfies the following statements:

  1. (1)

We have $[R]_{k}\in\operatorname{Met}\left(N(\Omega)\right)$.

  2. (2)

We have $\operatorname{diam}_{[R]_{k}}\left(N(\Omega)\right)\leq 2\cdot
2^{-F_{k}(0)}(=2^{-k})$.

  3. (3)

The metric $[R]_{k}$ is complete.

###### Proof.

We first prove the statement (1). Since each $[r_{i}]_{k,i}$ generates the
same topology on $\Omega$, the statements (A) and (B) in Lemma 4.6 imply that
$[R]_{k}$ generates the same topology on $N(\Omega)$ (recall that $N(\Omega)$
is equipped with the product topology).

We next prove the statement (2). For all $x,y\in N(\Omega)$, we have

$[R]_{k}(x,y)\leq\Sigma_{Q}[\zeta_{(k)},\mathbb{Q}_{\geq
0}]=\sum_{i\in\mathbb{Z}_{\geq 0}}2^{-F_{k}(i)}\leq\sum_{F_{k}(0)\leq
j}2^{-j}=2\cdot 2^{-F_{k}(0)}.$

Since $F_{k}(0)=1+k$, and $[R]_{k}(x,y)\leq 2\cdot 2^{-F_{k}(0)}$, the
statement (2) is true.

To prove the statement (3), we take an arbitrary Cauchy sequence
$\\{a(n)\\}_{n\in\mathbb{Z}_{\geq 0}}$ in $N(\Omega)$. We put
$a(n)=\\{a(n,i)\\}_{i\in\mathbb{Z}_{\geq 0}}$ for all $i\in\mathbb{Z}_{\geq
0}$ and $n\in\mathbb{Z}_{\geq 0}$. Due to the statement (A) in Lemma 4.6, for
each $i\in\mathbb{Z}_{\geq 0}$, there exists $N_{i}$ such that
$a(n,i)=a(n+1,i)$ for all $n\in\mathbb{Z}_{\geq 0}$ with $N_{i}\leq n$. We
define $b_{i}=a(N_{i},i)$. Then the point $b=(b_{i})_{i\in\mathbb{Z}_{\geq
0}}\in N(\Omega)$ is a limit of $\\{a(n)\\}_{n\in\mathbb{Z}_{\geq 0}}$. Thus,
the metric $[R]_{k}$ is complete. This finishes the proof of the proposition.
∎

######  Remark 4.3.

The metric $[R]_{k}$ in Definition 4.7 is not strongly rigid. Indeed, if we
take $a,b\in\Omega$ with $a\neq b$, and define $x,y,u,v\in N(\Omega)$ by
$x=(a,a,a,\dots,)$, $y=(a,b,b,\dots,)$, $u=(b,a,a,\dots,)$, and
$v=(b,b,b,\dots,)$, then we have $\\{x,y\\}\neq\\{u,v\\}$, and
$[R]_{k}(x,y)=[R]_{k}(u,v)$ for a fixed integer $k\in\mathbb{Z}_{\geq 0}$, and
for every $S$-gauge system $R$ on $\Omega$.

To obtain a strongly rigid metric on $N(\Omega)$, we construct a topological
embedding $\Psi_{\mathcal{F}}\colon N(\Omega)\to N(\Omega)$ such that
$\Psi_{\mathcal{F}}(x)$ contains information of all finite prefixes of
$x=(x_{i})_{i\in\mathbb{Z}_{\geq 0}}$ in $N(\Omega)$.

######  Definition 4.8.

For $n\in\mathbb{Z}_{\geq 0}$, and for a point
$x=(x_{i})_{i\in\mathbb{Z}_{\geq 0}}\in N(\Omega)$, we denote by
$\pi_{n}(x)\in\Omega^{n+1}$ the point $(x_{0},x_{1},\dots,x_{n})$. For each
$i\in\mathbb{Z}_{\geq 0}$, we take an injective map
$f_{i}\colon\Omega^{i+1}\to\Omega$. Since
$\operatorname{Card}(\Omega)=\mathfrak{c}$ and
$\mathfrak{c}^{i+1}=\mathfrak{c}$, the injective map
$f_{i}\colon\Omega^{i+1}\to\Omega$ always exists. Put
$\mathcal{F}=\\{f_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$. We define a map
$\Psi_{\mathcal{F}}\colon N(\Omega)\to N(\Omega)$ as follows: The $i$-th entry
$y_{i}$ of $\Psi_{\mathcal{F}}(x)$ is defined by

$y_{i}=\begin{cases}x_{n}&\text{if $i=2n$};\\\ f_{n}(\pi_{n}(x))&\text{if
$i=2n+1$, }\end{cases}$

where $x=(x_{i})_{i\in\mathbb{Z}_{\geq 0}}$. In what follows, we fix the
family $\mathcal{F}=\\{f_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$.

######  Lemma 4.8.

The map $\Psi_{\mathcal{F}}\colon N(\Omega)\to N(\Omega)$ is a topological
embedding and the image of $\Psi_{\mathcal{F}}$ is closed in $N(\Omega)$.

###### Proof.

By the definition of $\Psi_{\mathcal{F}}$, we observe that
$\Psi_{\mathcal{F}}$ is a topological embedding. To prove that
$\Psi_{\mathcal{F}}(N(\Omega))$ is closed, we take a sequence
$\\{x_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ in $\Psi_{\mathcal{F}}(N(\Omega))$
such that $x_{i}\to a$ as $i\to\infty$ for some point
$a=(a_{i})_{i\in\mathbb{Z}_{\geq 0}}\in N(\Omega)$. We define
$b=(b_{i})_{i\in\mathbb{Z}_{\geq 0}}\in N(\Omega)$ by $b_{i}=a_{2i}$. Then
$\Psi_{\mathcal{F}}(b)=a$ and hence $a\in\Psi_{\mathcal{F}}(N(\Omega))$. Thus,
the set $\Psi_{\mathcal{F}}(N(\Omega))$ is closed in $N(\Omega)$. ∎

Since for all $x=(x_{i})_{i\in\mathbb{Z}_{\geq
0}},y=(y_{i})_{i\in\mathbb{Z}_{\geq 0}}\in N(\Omega)$ we have $x=y$ if and
only if $x_{i}=y_{i}$ for all $i\in\mathbb{Z}_{\geq 0}$, we obtain the
following lemma:

######  Lemma 4.9.

Let $l\in\mathbb{Z}_{\geq 0}$ and $p(0),\dots p(l)\in N(\Omega)$. If
$p(0),\dots p(l)$ are mutually distinct, then there exists
$N\in\mathbb{Z}_{\geq 0}$ such that for all $n\in\mathbb{Z}_{\geq 0}$ with
$N<n$, the points $\pi_{n}(p(0)),\dots\pi_{n}(p(l))$ are mutually distinct.

As a consequence of Lemma 4.9, we obtain:

######  Lemma 4.10.

Let $l\in\mathbb{Z}_{\geq 0}$ and $x(0),y(0),x(1),y(1),\dots,x(l),y(l)\in
N(\Omega)$. If $x(s)\neq y(s)$ for all $s\in\\{0,\dots,l\\}$, and if
$\\{x(s),y(s)\\}\neq\\{x(t),y(t)\\}$ for all distinct $s,t\in\\{0,\dots,l\\}$,
then there exists $N$ such that for all $n\in\mathbb{Z}_{\geq 0}$ with $N<n$,

  1. (1)

we have
$\\{\pi_{n}(x(s)),\pi_{n}(y(s))\\}\neq\\{\pi_{n}(x(t)),\pi_{n}(y(t))\\}$ for
all distinct $s,t\in\\{0,\dots,l\\}$.

  2. (2)

we have $\pi_{n}(x(s))\neq\pi_{n}(y(s))$ for all $s\in\\{0,\dots,l\\}$.

###### Proof.

Put $x(s)=(x_{i}(s))_{i\in\mathbb{Z}_{\geq 0}}$ and
$y(s)=(y_{i}(s))_{i\in\mathbb{Z}_{\geq 0}}$. For each $s\in\\{0,\dots,l\\}$,
we define $u(s)=(u_{i}(s))_{i\in\mathbb{Z}_{\geq 0}}$ and
$v(s)=(v_{i}(s))_{i\in\mathbb{Z}_{\geq 0}}$ by

$u_{i}(s)=\begin{cases}x_{j}(s)&\text{if $i=2j$;}\\\ y_{j}(s)&\text{if
$i=2j+1$,}\end{cases}$

and

$v_{i}(s)=\begin{cases}y_{j}(s)&\text{if $i=2j$;}\\\ x_{j}(s)&\text{if
$i=2j+1$.}\end{cases}$

The inequality $x(s)\neq y(s)$ implies $u(s)\neq v(s)$. For all distinct
$s,t\in\\{0,\dots,l\\}$, by the assumption that
$\\{x(s),y(s)\\}\neq\\{x(t),y(t)\\}$, the points $u(s),v(s),u(t),v(t)$ are
mutually distinct. Therefore the points $u(0),v(0),\dots,u(l),v(l)$ are
mutually distinct. According to Lemma 4.10, there exists $M\in\mathbb{Z}_{\geq
0}$ such that for all $m\in\mathbb{Z}_{\geq 0}$ with $M<m$, the points
$\pi_{m}(u(0)),\pi_{m}(v(0)),\dots,\pi_{m}(u(l)),\pi_{m}(v(l))$ are mutually
distinct. Take $N\in\mathbb{Z}_{\geq 0}$ with $M\leq 2N$. Then, by the
definition of $u(s)$ and $v(s)$, the integer $N$ satisfies the two conditions
stated in the proposition. ∎

Using the metric $[R]_{k}$ and the embedding $\Psi_{\mathcal{F}}$, we
construct a strongly rigid metric on $N(\Omega)$.

######  Definition 4.9.

Let $S$ be a subset of $[0,\infty)$ and let
$R=\\{r_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ be an $S$-gauge system on $\Omega$.
We define

$\llbracket
R\rrbracket_{k}(x,y)=[R]_{k}(\Psi_{\mathcal{F}}(x),\Psi_{\mathcal{F}}(y)).$

Note that we have

$\llbracket
R\rrbracket_{k}(x,y)=\Sigma_{Q}[\zeta_{(k)},I_{R,\Psi_{\mathcal{F}}(x),\Psi_{\mathcal{F}}(y)}].$

For a metric space $(X,d)$, and for a subset $A$ of $X$, we denote by
$\operatorname{diam}_{d}(A)$ the diameter of $A$ with respect to $d$. The
following theorem plays an important role to prove our main results.

######  Theorem 4.11.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let $S$ be a ubiquitously dense subset of
$[0,\infty)$, and $R=\\{r_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ an $S$-gauge
system on $\Omega$. Then the metric $\llbracket R\rrbracket_{k}\colon
N(\Omega)\times N(\Omega)\to[0,\infty)$ satisfies the following statements:

  1. (1)

We have $\llbracket
R\rrbracket_{k}\in\operatorname{Met}\left(N(\Omega)\right)$.

  2. (2)

We have $\operatorname{diam}_{\llbracket
R\rrbracket_{k}}\left(N(\Omega)\right)\leq 2\cdot 2^{-F_{k}(0)}(=2^{-k})$.

  3. (3)

The metric $\llbracket R\rrbracket_{k}$ is complete.

  4. (4)

The metric $\llbracket R\rrbracket_{k}$ is strongly rigid.

###### Proof.

Since $\Psi_{\mathcal{F}}\colon N(\Omega)\to N(\Omega)$ is a topological
embedding and its image is closed (see Lemma 4.8), the statements (1) and (3)
in Proposition 4.7 implies that (1) and (3) in the theorem are true.

By (2) in Proposition 4.7, the statement (2) is true.

We now prove the statement (4) in the theorem. Take $x,y,u,v\in N(\Omega)$
such that $x\neq y$, $u\neq v$, and $\\{x,y\\}\neq\\{u,v\\}$. Due to Lemma
4.10, we can take $n\in\mathbb{Z}_{\geq 0}$ such that
$\\{\pi_{n}(x),\pi_{n}(y)\\}\neq\\{\pi_{u}(n),\pi_{n}(v)\\}$. Thus, using the
injectivity of $f_{n}$, we have

$\\{f_{n}(\pi_{n}(x)),f_{n}(\pi_{n}(y))\\}\neq\\{f_{n}(\pi_{n}(u)),f_{n}(\pi_{n}(v))\\}.$

Thus, we have

$r_{2n+1}(f_{n}(\pi_{n}(x)),f_{n}(\pi_{n}(y)))\neq
r_{2n+1}(f_{n}(\pi_{n}(u)),f_{n}(\pi_{n}(v))).$

This inequality implies that

$H_{2n+1,r_{2n+1}(f_{n}(\pi_{n}(x)),f_{n}(\pi_{n}(y)))}\neq
H_{2n+1,r_{2n+1}(f_{n}(\pi_{n}(u)),f_{n}(\pi_{n}(v)))}.$

Therefore, we obtain

$I_{R,\Psi_{\mathcal{F}}(x),\Psi_{\mathcal{F}}(y)}\neq
I_{R,\Psi_{\mathcal{F}}(x),\Psi_{\mathcal{F}}(y)}.$

According to Proposition 4.1, we notice that $\llbracket
R\rrbracket_{k}(x,y)\neq\llbracket R\rrbracket_{k}(u,v)$. This means that the
metric $\llbracket R\rrbracket_{k}$ is strongly rigid, and the statement (4)
is true. This completes the proof of the theorem. ∎

######  Definition 4.10.

We denote by $\mathrm{Suc}(\mathfrak{c})$ the set
$\mathfrak{c}\cup\\{\mathfrak{c}\\}$. This is nothing but the successor of
$\mathfrak{c}$ as an ordinal.

######  Lemma 4.12.

There exists a family $\\{K(\alpha)\\}_{\alpha\in\mathrm{Suc}(\mathfrak{c})}$
satisfying that

  1. (1)

each $K(\alpha)$ is a subset of $[0,\infty)$;

  2. (2)

each $K(\alpha)$ is ubiquitously dense $in[0,\infty)$ and
$\operatorname{Card}(K(\alpha))=\mathfrak{c}$;

  3. (3)

if $\alpha,\beta\in\mathrm{Suc}(\mathfrak{c})$ satisfy $\alpha\neq\beta$, then
$K(\alpha)\cap K(\beta)=\emptyset$.

###### Proof.

The set $[0,\infty)$ is ubiquitously dense in $[0,\infty)$ and
$\operatorname{Card}([0,\infty))=\mathfrak{c}$. Thus, by Theorem 2.2, we
obtain a mutually disjoint decomposition
$\\{T(\alpha)\\}_{\alpha<\mathfrak{c}}$ of $[0,\infty)$ such that each
$T(\alpha)$ is countable and dense in $[0,\infty)$. Since
$\operatorname{Card}(\mathrm{Suc}(\mathfrak{c}))=\mathfrak{c}$, and since
$\mathfrak{c}\times\mathfrak{c}=\mathfrak{c}$ as cardinals, we can take a
bijection
$\phi\colon\mathrm{Suc}(\mathfrak{c})\to\mathrm{Suc}(\mathfrak{c})\times\mathfrak{c}$,
where $\mathrm{Suc}(\mathfrak{c})\times\mathfrak{c}$ stands for the product as
sets. Put $\phi(\alpha)=(\theta(\alpha),\lambda(\alpha))$. For each
$\alpha\in\mathrm{Suc}(\mathfrak{c})$, we define
$K(\alpha)=\bigcup_{\beta\in\mathrm{Suc}(\mathfrak{c}),\theta(\beta)=\alpha}T(\beta)$.
Then the family $\\{K(\alpha)\\}_{\alpha\in\mathrm{Suc}(\mathfrak{c})}$
satisfies the three conditions stated in the lemma. This finishes the proof. ∎

######  Definition 4.11.

Fix $k\in\mathbb{Z}_{\geq 0}$. In what follows, we fix a family
$\\{K(\alpha)\\}_{\alpha\in\mathrm{Suc}(\mathfrak{c})}$ stated in Lemma 4.12.
For each $\alpha\in\mathrm{Suc}(\mathfrak{c})$, we also fix a
$K(\alpha)$-gauge system $R(\alpha)=\\{r_{i,\alpha}\\}_{i\in\mathbb{Z}_{\geq
0}}$ (see Proposition 4.5).

For each $\alpha\in\mathrm{Suc}(\mathfrak{c})$, we define

$\mathbb{F}_{k,\alpha}=\\{\,\llbracket R(\alpha)\rrbracket_{k}(x,y)\mid x\neq
y,x,y\in X\,\\},$

and

$\mathbb{G}_{k,\alpha}=\\{0\\}\sqcup\mathbb{F}_{k,\alpha}.$

Namely, the set $\mathbb{G}_{k,\alpha}$ is the image of the metric $\llbracket
R(\alpha)\rrbracket_{k}$ on $N(\Omega)$.

In a similar way to the proof of (4) in Theorem 4.11, we obtain Lemmas 4.13
and 4.14.

######  Lemma 4.13.

Fix $k\in\mathbb{Z}_{\geq 0}$. If $\alpha,\beta\in\mathrm{Suc}(\mathfrak{c})$
satisfy $\alpha\neq\beta$, then
$\mathbb{F}_{k,\alpha}\cap\mathbb{F}_{k,\beta}=\emptyset$.

###### Proof.

For the sake of contradiction, we suppose that there exists
$\alpha,\beta\in\mathrm{Suc}(\mathfrak{c})$ with $\alpha\neq\beta$ and
$\mathbb{F}_{k,\alpha}\cap\mathbb{F}_{k,\beta}\neq\emptyset$. Take
$d\in\mathbb{F}_{k,\alpha}\cap\mathbb{F}_{k,\beta}$ and put $d=\llbracket
R(\alpha)\rrbracket_{k}(x,y)$ and $d=\llbracket R(\beta)\rrbracket_{k}(u,v)$,
where $x,y,u,v\in N(\Omega)$, and $x\neq y$ and $u\neq v$. Take
$i\in\mathbb{Z}_{\geq 0}$ such that $x_{i}\neq y_{i}$. Then
$r_{i,\alpha}(x_{i},y_{i})\in K(\alpha)\setminus\\{0\\}$ and
$r_{i,\beta}(u_{i},v_{i})\in K(\beta)\cup\\{0\\}$. From $K(\alpha)\cap
K(\beta)=\emptyset$, it follows that $r_{i,\alpha}(x_{i},y_{i})\neq
r_{i,\beta}(u_{i},v_{i})$. This implies that $H_{i,r_{i,\alpha}(x,y)}\neq
H_{i,r_{i,\beta}(u,v)}$ and hence
$I_{R(\alpha),\Psi_{\mathcal{F}}(x),\Psi_{\mathcal{F}}(y)}\neq
I_{R(\beta),\Psi_{\mathcal{F}}(u),\Psi_{\mathcal{F}}(v)}$. According to
Proposition 4.1, we obtain $\llbracket
R(\alpha)\rrbracket_{k}(x,y)\neq\llbracket R(\beta)\rrbracket_{k}(u,v)$. This
contradicts $d=\llbracket R(\alpha)\rrbracket_{k}(x,y)$ and $d=\llbracket
R(\beta)\rrbracket_{k}(u,v)$. Therefore we conclude that
$\mathbb{F}_{k,\alpha}\cap\mathbb{F}_{k,\beta}=\emptyset$ for all distinct
$\alpha,\beta\in\mathrm{Suc}(\mathfrak{c})$. ∎

A subset $S$ of $\mathbb{R}$ is said to be _linearly independent over
$\mathbb{Q}$_ if all finite distinct elements in $S$ are linearly independent
over $\mathbb{Q}$.

######  Lemma 4.14.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let $s,t\in\mathbb{Z}_{\geq 0}$. Let
$\alpha_{0},\dots,\alpha_{s}\in\mathrm{Suc}(\mathfrak{c})$. For each
$i\in\\{0,\dots,s\\}$, we take $(t+1)$-many arbitrary distinct elements
$d_{i,0},\dots,d_{i,t}\in\mathbb{F}_{k,\alpha_{i}}$. Then the set

$\\{1\\}\cup\\{d_{i,j}\mid i\in\\{0,\dots,s\\},j\in\\{0,\dots,t\\}\\}$

is linearly independent over $\mathbb{Q}$.

###### Proof.

Put $d_{i,j}=\llbracket R(\alpha_{i})\rrbracket_{k}(x(i,j),y(i,j))$, where
$x(i,j),y(i,j)\in N(\Omega)$. Note that $x(i,j)\neq y(i,j)$. According to
Lemma 4.10, we can take a sufficient large integer $n\in\mathbb{Z}_{\geq 0}$
such that for each $i\in\\{0,\dots,s\\}$, for all distinct
$j,j^{\prime}\in\\{0,\dots,t\\}$, we obtain the inequality

$\\{\pi_{n}(x(i,j)),\pi_{n}(y(i,j))\\}\neq\\{\pi_{n}(x(i,j^{\prime}),\pi_{n}(y(i,j^{\prime}))\\}.$

We put
$r(i,j)=r_{2n+1,\alpha_{i}}(f_{n}(\pi_{n}(x(i,j))),f_{n}(\pi_{n}(y(i,j))))$.
Then, for a fixed number $i\in\\{0,\dots,s\\}$, for all distinct
$j,j^{\prime}\in\\{0,\dots,t\\}$, we have $r(i,j)\neq r(i,j^{\prime})$. Since
$r(i,j)\in K(\alpha_{i})$ for all $j\in\\{0,\dots,t\\}$, and since
$K(\alpha_{i})\cap K(\alpha_{i^{\prime}})=\emptyset$ for all distinct
$i,i^{\prime}\in\\{0,\dots,s\\}$, we obtain $r(i,j)\neq
r(i^{\prime},j^{\prime})$ for all distinct $(i,j)$, $(i^{\prime},j^{\prime})$.

We put
$I_{i,j}=I_{R(\alpha_{i}),\Psi_{\mathcal{F}}(x(i,j)),\Psi_{\mathcal{F}}(y(i,j))}$,
and $S=[2n+1,2n+2)\cap\mathbb{Q}$. Then $S\cap
I_{i,j}=[2n+1,r(i,j))\cap\mathbb{Q}$. Thus, by
$d_{i,j}=\Sigma_{Q}[\zeta_{(k)},I_{i,j}]$, the numbers $d_{i,j}$
$(i\in\\{0,\dots,s\\},j\in\\{0,\dots,t\\})$ satisfy the assumptions of
Proposition 3.1. Therefore, due to Proposition 3.1, we conclude that the set

$\\{1\\}\cup\\{d_{i,j}\mid i\in\\{0,\dots,s\\},j\in\\{0,\dots,t\\}\\}$

is linearly independent over $\mathbb{Q}$. ∎

######  Corollary 4.15.

Fix $k\in\mathbb{Z}_{\geq 0}$. Then the set
$\\{1\\}\cup\bigcup_{\alpha\in\mathrm{Suc}(\mathfrak{c})}\mathbb{F}_{k,\alpha}$
is linearly independent over $\mathbb{Q}$.

######  Definition 4.12.

Fix $k\in\mathbb{Z}_{\geq 0}$. Recall that in Definition 4.11, we construct
the family
$\\{\mathbb{F}_{k,\alpha}\\}_{\alpha\in\mathrm{Suc}(\mathfrak{c})}$, where
$\mathrm{Suc}(\mathfrak{c})=\mathfrak{c}\cup\\{\mathfrak{c}\\}$. Using
$\mathfrak{c}\times\aleph_{0}=\mathfrak{c}$, we can represent

$\mathbb{F}_{k,\mathfrak{c}}=\left\\{s_{k,\alpha,i}\mid\alpha\in\mathfrak{c},i\in\mathbb{Z}_{\geq
0}\right\\}.$

We assume that if $s_{k,\alpha,i}=s_{k,\beta,j}$, then $(\alpha,i)=(\beta,j)$.
For each $(\alpha,i)\in\mathfrak{c}\times\mathbb{Z}_{\geq 0}$, we take
$q_{k,\alpha,i}\in\mathbb{Q}_{>0}$ such that $q_{k,\alpha,i}\cdot
s_{k,\alpha,i}\leq 2^{-i}$. We fix a bijection $P\colon\mathbb{Z}_{\geq
0}\to\mathbb{Q}_{\geq 0}$. We define set $\mathbb{A}_{k}$ and $\mathbb{X}_{k}$
by

$\mathbb{A}_{k}=\left\\{\,P(i)+q_{k,\alpha,i}s_{k,\alpha,i}\mid\alpha<\mathfrak{c},i\in\mathbb{Z}_{\geq
0}\,\right\\},$

and

$\mathbb{X}_{k}=\\{0\\}\sqcup\mathbb{A}_{k},$

respectively. In what follows, we no longer use the fact that the set
$\mathbb{F}_{k,\mathfrak{c}}$ is defined as a set of values of a metric on
$N(\Omega)$. We rather use the property that the union of
$\mathbb{F}_{k,\mathfrak{c}}$ and
$\\{1\\}\cup\bigcup_{\alpha<\mathfrak{c}}\mathbb{F}_{k,\alpha}$ is linearly
independent over $\mathbb{Q}$ (see Corollary 4.15).

######  Lemma 4.16.

Fix $k\in\mathbb{Z}_{\geq 0}$. Then the set $\mathbb{X}_{k}$ is ubiquitously
dense in $[0,\infty)$ and $\operatorname{Card}(\mathbb{X}_{k})=\mathfrak{c}$.

###### Proof.

It suffices to show that $\mathbb{A}_{k}$ is ubiquitously dense in
$[0,\infty)$. Take $x\in[0,\infty)$ and $\epsilon\in(0,\infty)$. Since
$\mathbb{Q}_{\geq 0}$ is dense in $[0,\infty)$, we can take
$n\in\mathbb{Z}_{\geq 0}$ such that $|P(n)-x|<\epsilon/2$ and
$2^{-n}\leq\epsilon/2$. Then, for all $\alpha<\mathfrak{c}$, we have

$\left|P(n)+q_{k,\alpha,n}s_{k,\alpha,n}-x\right|\leq|P(n)-x|+\left|q_{k,\alpha,n}s_{k,\alpha,n}\right|<\epsilon/2+2^{-n}\leq\epsilon.$

Thus, the set $\mathbb{A}_{k}$ is ubiquitously dense in $[0,\infty)$. ∎

By Lemma 4.13 and Corollary 4.15, and by the definitions of $\mathbb{A}_{k}$
and $\mathbb{F}_{k,\alpha}$, we obtain:

######  Proposition 4.17.

Fix $k\in\mathbb{Z}_{\geq 0}$. Then the following statements are true:

  1. (1)

For all $\alpha\in\mathrm{Suc}(\mathfrak{c})$, we have
$\mathbb{A}_{k}\cap\mathbb{F}_{k,\alpha}=\emptyset$;

  2. (2)

The set $\mathbb{A}_{k}\cup\bigcup_{\alpha<\mathfrak{c}}\mathbb{F}_{k,\alpha}$
is linearly independent over $\mathbb{Q}$.

For two sets $A$ and $B$, we denote by $A\ominus B=(A\setminus
B)\cup(B\setminus A)$. Namely, the set $A\ominus B$ is the symmetric
difference of $A$ and $B$.

######  Lemma 4.18.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let
$\alpha,\beta,\widetilde{\alpha},\widetilde{\beta}\in\mathfrak{c}$ with
$\alpha\neq\beta$ and $\widetilde{\alpha}\neq\widetilde{\beta}$. If
$x\in\mathbb{G}_{k,\alpha}$, $a\in\mathbb{G}_{k,\widetilde{\alpha}}$
$y\in\mathbb{G}_{k,\beta}$, $b\in\mathbb{G}_{k,\widetilde{\beta}}$ and
$z,c\in\mathbb{X}_{k}$. If $\\{x,y,z\\}\neq\\{a,b,c\\}$, $x+y+z\neq 0$, and
$a+b+c\neq 0$, then the numbers $x+y+z$ and $a+b+c$ are linearly independent
over $\mathbb{Q}$.

###### Proof.

We first prove the following claim:

  * •

There exists non-zero $r$ such that $r\in\\{x,y,z\\}\ominus\\{a,b,c\\}$.

By $\\{x,y,z\\}\neq\\{a,b,c\\}$, we observe that
$\\{x,y,z\\}\ominus\\{a,b,c\\}\neq\emptyset$. If
$0\not\in\\{x,y,z\\}\ominus\\{a,b,c\\}$, then any element in this set
satisfies the condition. If $0\in\\{x,y,z\\}\ominus\\{a,b,c\\}$, then
$0\in\\{x,y,z\\}$ or $0\in\\{a,b,c\\}$. We may assume that $0\in\\{a,b,c\\}$.
Thus, $0\not\in\\{x,y,z\\}$. Combining Lemma 4.13 and the statement (1) in
Proposition 4.17, we obtain
$\mathbb{F}_{k,\alpha}\cap\mathbb{F}_{k,\beta}=\mathbb{F}_{k,\alpha}\cap\mathbb{A}_{k}=\mathbb{F}_{k,\beta}\cap\mathbb{A}_{k}=\emptyset$.
Since $x\in\mathbb{F}_{k,\alpha}$, $y\in\mathbb{F}_{k,\beta}$, and
$z\in\mathbb{A}_{k}$, we conclude that $\operatorname{Card}(\\{x,y,z\\})=3$.
Since $\\{a,b,c\\}$ contains at most two non-zero numbers, and since all the
three numbers $x,y,z$ are non-zero, there exists non-zero
$r\in\\{x,y,z\\}\ominus\\{a,b,c\\}$. This finishes the proof of the claim.

To prove the linear independence of $x+y+z$ and $a+b+c$ over $\mathbb{Q}$, we
assume that integers $h_{0}$ and $h_{1}$ satisfy

(4.1) $\displaystyle h_{0}(x+y+z)+h_{1}(a+b+c)=0.$

We put $A=\\{x,y,z\\}\cap(0,\infty)$ and $B=\\{a,b,c\\}\cap(0,\infty)$. We
also put $u=\sum_{l\in A\setminus B}l$, $v=\sum_{l\in B\setminus A}l$, and
$w=\sum_{l\in A\cap B}l$. Then the equality (4.1) implies that

(4.2) $\displaystyle h_{0}u+h_{1}v+(h_{0}+h_{1})w=0.$

Remark that it can happen that some of $u,v,w$ are $0$. Using the claim
explained above, we have $u\neq 0$ or $v\neq 0$. We may assume that $u\neq 0$.
Put $C=\\{u,v,w\\}\cap(0,\infty)$. Then $u\in C$. According to Proposition
4.17, the set $A\cup B$ is linearly independent over $\mathbb{Q}$. Since
$A\setminus B$, $B\setminus A$, and $A\cap B$ are mutually disjoint subsets of
$A\cup B$, by the definitions of $u$, $v$, and $w$, the set $C$ is linearly
independent over $\mathbb{Q}$. Thus, from (4.2) and and $u\neq 0$, it follows
that $h_{0}=0$. The equality (4.1) implies $h_{1}(a+b+c)=0$. Since $a+b+c\neq
0$, we obtain $h_{1}=0$. Thus, we conclude that $h_{0}=h_{1}=0$. Hence $x+y+z$
and $a+b+c$ are linearly independent over $\mathbb{Q}$. ∎

######  Lemma 4.19.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let $X$ be a discrete space with
$\operatorname{Card}(X)\leq\mathfrak{c}$. Let $d\in\operatorname{Met}(X)$ and
$\epsilon\in(0,\infty)$. Then there exists a strongly rigid uniformly discrete
metric $e\in\operatorname{Met}(X;\mathbb{X}_{k})$ with
$\operatorname{\mathcal{D}}_{X}(d,e)\leq\epsilon$.

###### Proof.

Combining Lemma 4.16 and Theorem 2.4, we obtain the lemma. ∎

The following is deduced from [19, Theorem 3.1, Chapter 7]. The latter part is
deduced from [21, Theorem 2].

######  Theorem 4.20.

If $X$ is a strongly $0$-dimensional metrizable space with
$\operatorname{Card}(X)\leq\mathfrak{c}$, then $X$ can be topologically
embedded into $N(\Omega)$. Moreover, if $X$ is completely metrizable, the
space $X$ is homeomorphic to a closed subset of $N(\Omega)$.

######  Lemma 4.21.

Fix $k\in\mathbb{Z}_{\geq 0}$. Let $X$ be a strongly $0$-dimensional
metrizable space with $\operatorname{Card}(X)\leq\mathfrak{c}$. Let
$\alpha\in\mathfrak{c}$. Then there exists a strongly rigid metric
$e\in\operatorname{Met}(X;\mathbb{G}_{k,\alpha})$ such that
$\operatorname{diam}_{e}(X)\leq 2^{-k}$. Moreover, if $X$ is completely
metrizable, we can choose $e$ as a complete metric.

###### Proof.

Due to Theorem 4.20, we can take a topological embedding $h\colon X\to
N(\Omega)$. Put $e(x,y)=\llbracket R(\alpha)\rrbracket_{k}(h(x),h(y))$. Then,
by Theorem 4.11, the metric $e$ satisfies the desired properties. If $X$ is
completely metrizable, we can choose $h$ as a closed map. Since $\llbracket
R\rrbracket_{k}$ is complete, in this case, so is the metric $e$. ∎

The following was first stated in [14].

######  Corollary 4.22.

Let $X$ be a strongly $0$-dimensional metrizable space. Then the inequality
$\operatorname{Card}(X)\leq\mathfrak{c}$ holds if and only if there exists a
strongly rigid metric $d\in\operatorname{Met}(X)$.

##  5\. Proofs of main results

The following proposition can be found in [10, Proposition 2.1] (See also [8,
Proposition 3.1]).

######  Proposition 5.1.

Let $(X,d)$ be a metric space. Let $I$ be a set and $\\{B_{i}\\}_{i\in I}$ a
covering of $X$ consisting of mutually disjoint clopen subsets. Let
$P=\\{p_{i}\\}_{i\in I}$ be points with $p_{i}\in B_{i}$ and
$\\{e_{i}\\}_{i\in I}$ a set of metrics such that
$e_{i}\in\operatorname{Met}(B_{i})$. Let $h$ be a metric on $P$ generating the
discrete topology on $P$. We define a function $D:X^{2}\to[0,\infty)$ by

$D(x,y)=\begin{cases}e_{i}(x,y)&\text{ if $x,y\in B_{i}$;}\\\
e_{i}(x,p_{i})+h(p_{i},p_{j})+e_{j}(p_{j},y)&\text{if $x\in B_{i}$ and $y\in
B_{j}$.}\end{cases}$

Then $D\in\operatorname{Met}(X)$ and $D|_{B_{i}^{2}}=e_{i}$ for all $i\in I$.
Moreover, if for every $i\in I$ we have
$\operatorname{diam}_{d}(B_{i})\leq\epsilon$ and
$\operatorname{diam}_{e_{i}}(B_{i})\leq\epsilon$, then
$\operatorname{\mathcal{D}}_{X}(D,d)\leq
4\epsilon+\operatorname{\mathcal{D}}_{P}(d|_{P^{2}},h)$.

######  Proposition 5.2.

Under the same assumptions in Proposition 5.1, if $h$ is uniformly discrete
and each $e_{i}$ is complete, then the metric $D$ is complete.

###### Proof.

Take $c\in(0,\infty)$ such that $c<h(p_{i},p_{j})$ for all distinct $i,j\in
I$. By the definition of $D$, if $x\in B_{i}$ and $y\in B_{j}$ and $i\neq j$,
we have $c<D(x,y)$. Take a Cauchy sequence $\\{x_{n}\\}_{n\in\mathbb{Z}_{\geq
0}}$ in $(X,D)$ and take a sufficient large number $N\in\mathbb{Z}_{\geq 0}$
such that for all $n,m>N$, we have $D(x_{n},x_{m})<c$. Then we observe that
there exists $i\in I$ satisfying that $\\{\,x_{n}\mid N<n\,\\}\subset B_{i}$.
Since $D|_{B_{i}^{2}}=e_{i}$ and $e_{i}$ is complete, the sequence
$\\{x_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ has a limit point. Therefore we
conclude that $(X,D)$ is complete. ∎

We now prove Theorem 1.1.

###### Proof of Theorem 1.1.

Let $X$ be a strongly $0$-dimensional metrizable space with
$\operatorname{Card}(X)\leq\mathfrak{c}$. Let $d\in\operatorname{Met}(X)$ and
$\epsilon\in(0,\infty)$.

Put $\eta=\epsilon/5$. Take $k\in\mathbb{Z}_{\geq 0}$ such that
$2^{-k}\leq\eta$. Since $X$ is paracompact and strongly zero-dimensional, we
can take a mutually disjoint open cover $\\{O_{\alpha}\\}_{\alpha<\tau}$ of
$X$ with $\operatorname{diam}_{d}(O_{\alpha})\leq\epsilon$, where
$\tau\leq\mathfrak{c}$ (see [3, Proposition 1.2 and Corollary 1.4]).

For each $\alpha<\tau$, we take $p_{\alpha}\in O_{\alpha}$. Put
$P=\\{\,p_{\alpha}\mid\alpha\in\tau\,\\}$. Then $P$ is a discrete space and
$d|_{P^{2}}$ is a discrete metric on $P$. Lemma 4.19 guarantees the existence
of a strongly rigid uniformly discrete metric
$h\in\operatorname{Met}(P;\mathbb{X}_{k})$ with
$\operatorname{\mathcal{D}}_{P}(d_{P^{2}},h)\leq\eta$. Applying Lemma 4.21 to
$O_{i}$, we obtain a strongly rigid metric
$e_{\alpha}\in\operatorname{Met}(O_{\alpha};\mathbb{G}_{k,\alpha})$ with
$\operatorname{diam}_{e_{\alpha}}(O_{\alpha})\leq 2^{-k}\leq\eta$. We define a
metric $e$ by

$e(x,y)=\begin{cases}e_{\alpha}(x,y)&\text{ if $x,y\in B_{\alpha}$;}\\\
e_{\alpha}(x,p_{\alpha})+h(p_{\alpha},p_{\beta})+e_{\beta}(p_{\beta},y)&\text{if
$x\in B_{\alpha}$ and $y\in B_{\beta}$.}\end{cases}$

Applying Proposition 5.1 to $\\{O_{\alpha}\\}_{\alpha<\tau}$, $P$,
$\\{e_{\alpha}\\}_{\alpha<\tau}$, $h$, and $\eta$, we obtain
$e\in\operatorname{Met}(X)$ and $\operatorname{\mathcal{D}}(d,e)\leq
5\eta=\epsilon$.

We shall prove $e\in\mathrm{LI}(X)$. From the definition of $e$, and Lemma
4.18, it follows that for all $x,y,u,v\in X$ with $x\neq y$, $u\neq v$, and
$\\{x,y\\}\neq\\{u,v\\}$, the numbers $e(x,y)$ and $e(u,v)$ are linearly
independent over $\mathbb{Q}$. Hence we conclude that $e\in\mathrm{LI}(X)$.

To prove the latter part, we assume that $X$ is completely metrizable. Since
each $O_{\alpha}$ is clopen in $X$, the set $O_{\alpha}$ is completely
metrizable. By the latter part of Lemma 4.21, we can choose each $e_{\alpha}$
as a complete metric. Since $h$ is uniformly discrete, Proposition 5.2 implies
that $e$ is a complete metric. This finishes the proof of Theorem 1.1. ∎

The proof of the following proposition is analogous with that of [20, Theorem
2].

######  Proposition 5.3.

If $X$ is a strongly $0$-dimensional $\sigma$-compact metrizable space, then
the set $\mathrm{SR}(X)$ is $G_{\delta}$ in $\operatorname{Met}(X)$.

###### Proof.

Take a sequence $\\{K_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ of compact subsets of
$X$ such that $X=\bigcup_{i\in\mathbb{Z}_{\geq 0}}K_{i}$ and $K_{i}\subset
K_{i+1}$ for all $i\in\mathbb{Z}_{\geq 0}$. For $n,m\in\mathbb{Z}_{\geq 0}$,
we denote by $F_{n,m}$ the set of all $d\in\operatorname{Met}(X)$ such that
there exists $x,y,u,v\in K_{n}$ with

  1. (1)

$d(x,y)=d(u,v)\geq 2^{-m}$;

  2. (2)

$d(x,u)+d(y,v)\geq 2^{-m}$;

  3. (3)

$d(x,v)+d(u,y)\geq 2^{-m}$.

We now show that $L_{n,m}$ is a closed subset of $\operatorname{Met}(X)$. Take
a sequence $\\{e_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ in $L_{n,m}$ and take
$d\in\operatorname{Met}(X)$ such that $e_{i}\to d$ as $i\to\infty$. We shall
show $d\in L_{n,m}$. By extracting a subsequence using the compactness of
$K_{n}$ if necessary, we may assume that there exist sequences
$\\{x_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$, $\\{y_{i}\\}_{i\in\mathbb{Z}_{\geq
0}}$, $\\{u_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$, and
$\\{v_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ in $K_{n}$, and points $x,y,z,w\in
K_{n}$, such that

  1. (1)

$e_{i}(x_{i},y_{i})=e_{i}(u_{i},v_{i})\geq 2^{-m}$ for all
$i\in\mathbb{Z}_{\geq 0}$;

  2. (2)

$e_{i}(x_{i},u_{i})+e_{i}(y_{i},v_{i})\geq 2^{-m}$ for all
$i\in\mathbb{Z}_{\geq 0}$;

  3. (3)

$e_{i}(x_{i},v_{i})+e_{i}(y_{i},u_{i})\geq 2^{-m}$ for all
$i\in\mathbb{Z}_{\geq 0}$;

  4. (4)

$x_{i}\to x$, $y_{i}\to y$, $u_{i}\to u$, and $v_{i}\to v$ as $i\to\infty$.

Since $d$ and $e_{i}$ generate the same topology of $X$ and since $e_{i}\to d$
as $i\to\infty$, if $p\in\\{x,y,u,v\\}$ and $q\in\\{x,y,u,v\\}$, then we have
$e_{i}(p_{i},q_{i})\to d(p,q)$ as $i\to\infty$. Thus, we obtain
$d(x,y)=d(u,v)\geq 2^{-m}$ and $d(x,u)+d(y,v)\geq 2^{-m}$ and
$d(x,v)+d(y,u)\geq 2^{-m}$. Therefore $d\in L_{n,m}$, and hence $L_{n,m}$ is
closed. Put $G_{n,m}=\operatorname{Met}(X)\setminus L_{n,m}$. Then each
$G_{n,m}$ is open in $\operatorname{Met}(X)$, and we obtain

$\mathrm{SR}(X)=\bigcap_{n,m\in\mathbb{Z}_{\geq 0}}G_{n,m}$

This proves the proposition. ∎

###### Proof of Theorem 1.2.

Let $X$ be a strongly $0$-dimensional metrizable space with
$\operatorname{Card}(X)\leq\mathfrak{c}$.

Due to Proposition 5.3, we only need to prove that $\mathrm{SR}(X)$ is dense
in $\operatorname{Met}(X)$. We now show $\mathrm{LI}(X)\subset\mathrm{SR}(X)$.
Take $d\in\mathrm{LI}(X)$. Take $x,y,u,v\in X$ with $x\neq y$ and $u\neq v$,
and $\\{x,y\\}\neq\\{u,v\\}$. Then $d(x,y)$ and $d(u,v)$ are linearly
independent over $\mathbb{Q}$. In particular, we obtain $d(x,y)\neq d(u,v)$,
and hence $d\in\mathrm{SR}(X)$. Thus, we have
$\mathrm{LI}(X)\subset\mathrm{SR}(X)$. According to Theorem 1.1, we observe
that $\mathrm{SR}(X)$ is dense in $\operatorname{Met}(X)$. This complete the
proof of Theorem 1.2. ∎

###### Proof of Theorem 1.3.

Let $X$ be a strongly $0$-dimensional metrizable space. Assume that $X$ is
$\sigma$-compact and satisfies $3\leq\operatorname{Card}(X)\leq\mathfrak{c}$.

We only need to prove $\mathrm{SR}(X)\subset\mathrm{R}(X)$. Take
$d\in\mathrm{SR}(X)$, and let $f\colon(X,d)\to(X,d)$ be a bijective isometry.
Take arbitrary $x\in X$, and take two points $y,z$ in $X$ with
$\operatorname{Card}(\\{x,y,z\\})=3$. Since $d(x,y)=d(f(x),f(y))$ and
$d(x,z)=d(f(x),f(z))$, we have $\\{x,y\\}=\\{f(x),f(y)\\}$ and
$\\{x,z\\}=\\{f(x),f(z)\\}$. Then we obtain
$f(x)\in\\{x,y\\}\cap\\{x,z\\}=\\{x\\}$, and hence $f(x)=x$. Since $x$ is
arbitrary, we conclude that $f$ is the identity map, which implies that
$\mathrm{SR}(X)\subset\mathrm{R}(X)$. This finishes the proof of Theorem 1.3.
∎

Before proving Theorem 1.4, we introduce some notions. Let $X$ and $Y$ be
topological space. A continuous map $f\colon X\to Y$ is said to be proper if
for every compact subset of $Y$, the set $f^{-1}(K)$ is compact. For a metric
space $(X,d)$, the metric $d$ is _proper_ if all closed balls in $(X,d)$ are
compact. Note that $d$ is proper if and only if a map $x\mapsto d(x,p)$ is a
proper map for all $p\in X$. We denote by $B(x,r;d)$ the closed ball of
$(X,d)$ centered at $x$ with radius $r$.

###### Proof of Theorem 1.4.

Let $X$ be a strongly $0$-dimensional $\sigma$-compact locally compact
Hausdorff space. Remark that $X$ is metrizable.

Let $\mathrm{P}$ be the set of all proper metrics in $\operatorname{Met}(X)$.
Since $X$ is $\sigma$-compact locally compact, we have
$\mathrm{P}\neq\emptyset$ (see for example [9]). We now show that $\mathrm{P}$
is open in $\operatorname{Met}(X)$. Take $d\in\mathrm{P}$ and take
$e\in\operatorname{Met}(X)$ such that $\operatorname{\mathcal{D}}_{X}(d,e)\leq
1$. In this setting, we notice that $B(x,r;e)\subset B(x,r+1;d)$ for all $x\in
X$ and $r\in(0,\infty)$. Since $d$ is proper, so is $e$. Thus, the set
$\mathrm{P}$ is open in $\operatorname{Met}(X)$.

Due to Theorem 1.2, the set $\mathrm{SR}(X)$ is dense in
$\operatorname{Met}(X)$. Since $\mathrm{P}$ is non-empty and open, we obtain
$\mathrm{P}\cap\mathrm{SR}(X)\neq\emptyset$, and we can take a member from
this set, say $d$. Recall that for each $\xi\in X$, the map $F_{\xi}\colon
X\to[0,\infty)$ is defined as $F_{\xi}(x)=d(x,\xi)$. Fix $\xi\in X$. Next we
verify that $F_{\xi}$ is a topological embedding. Since $d$ is strongly rigid,
the map $F_{\xi}$ is injective. According to $d\in\mathrm{P}$, the map
$F_{\xi}$ is proper as a map. From the fact that every proper map into a
metrizable space is a closed map (see for instance [18]), it follows that
$F_{\xi}$ is a closed map. Therefore we conclude that $F_{\xi}$ is a
topological embedding. This finishes the proof of Theorem 1.4. ∎

## References

  * [1] K. A. Broughan, _A metric characterizing Čech dimension zero_ , Proc. Amer. Math. Soc. 39 (1973), 437–440.
  * [2] D. G. Ebin, _The manifold of Riemannian metrics_ , Global Analysis (Proc. Sympos. Pure Math., Vol. XV, Berkeley, Calif., 1968), Amer. Math. Soc., Providence, R.I., 1970, pp. 11–40.
  * [3] R. L. Ellis, _Extending continuous functions on zero-dimensional spaces_ , Math. Ann. 186 (1970), no. 2, 114–122.
  * [4] Y. Hattori, _Congruence and dimension of nonseparable metric spaces_ , Proc. Amer. Math. Soc. 108 (1990), no. 4, 1103–1105.
  * [5] F. G. Dorais (https://mathoverflow.net/users/2000/françois-g dorais), _explicit big linearly independent sets_ , MathOverflow, URL:https://mathoverflow.net/q/23206 (version: 2017-04-13).
  * [6] Y. Ishiki, _An interpolation of metrics and spaces of metrics_ , (2020), preprint, arXiv:2003.13277.
  * [7] by same author, _An embedding, an extension, and an interpolation of ultrametrics_ , $p$-Adic Numbers Ultrametric Anal. Appl. 13 (2021), no. 2, 117–147.
  * [8] by same author, _On dense subsets in spaces of metrics_ , (2021), preprint arXiv:2104.12450, to apper in Colloq. Math.
  * [9] by same author, _Extending proper metrics_ , (2022), preprint arXiv:2207.12905, to appear in Topology Appl.
  * [10] by same author, _On comeager sets of metrics whose ranges are disconnected_ , (2022), preprint arXiv:2207.12765, to appear in Topology Appl.
  * [11] L. Janos, _A metric characterization of zero-dimensional spaces_ , Proc. Amer. Math. Soc. 31 (1972), 268–270.
  * [12] L. Janos and H. Martin, _Metric characterizations of dimension for separable metric spaces_ , Proc. Amer. Math. Soc. 70 (1978), no. 2, 209–212.
  * [13] T. Jech, _Set theory_ , Springer Monographs in Mathematics, Springer-Verlag, Berlin, 2003, The third millennium edition, revised and expanded.
  * [14] H. W. Martin, _Strongly rigid metrics and zero dimensionality_ , Proc. Amer. Math. Soc. 67 (1977), no. 1, 157–161.
  * [15] P. Mounoud, _Metrics without isometries are generic_ , Monatsh. Math. 176 (2015), no. 4, 603–606.
  * [16] J. Mycielski, _Independent sets in topological algebras_ , Fund. Math. 55 (1964), 139–147.
  * [17] J. v. Neumann, _Ein System algebraisch unabhängiger Zahlen_ , Math. Ann. 99 (1928), no. 1, 134–141.
  * [18] R. S. Palais, _When proper maps are closed_ , Proc. Amer. Math. Soc. 24 (1970), 835–836.
  * [19] A. R. Pears, _Dimension theory of general spaces_ , Cambridge University Press, Cambridge, England-New York-Melbourne, 1975.
  * [20] J. Rouyer, _Generic properties of compact metric spaces_ , Topology Appl. 158 (2011), no. 16, 2140–2147.
  * [21] A. H. Stone, _Non-separable Borel sets_ , Rozprawy Mat. 28 (1962), 41.


# Photoacoustic effect from an oscillating source in a one-dimensional
resonator

Xingchi Yan<EMAIL_ADDRESS>Siyuan Song  Gerald J. Diebold  Department
of Chemistry, Brown University, Providence, RI, 02912, USA  School of
Engineering, Brown University, Providence, RI, 02912, USA

###### Abstract

Although the photoacoustic effect is most commonly generated by pulsed or
amplitude modulated continuous optical sources, it is possible to generate
acoustic waves by moving a constant amplitude, continuous light beam. If the
light beam moves at the speed of sound, an amplification effect takes place
which can be used in trace gas detection. Here, the properties of the
photoacoustic effect are investigated for a continuous optical beam moving in
a one-dimensional resonator. The solution shows the additive effects of
sweeping the optical beam the length of the cell and back.

##  I Introduction

Photoacoustic trace gas detection has a rich history dating from the 1960’s[1,
2, 3, 4]. The research on this subject[5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
has focused on optical excitation schemes and the design of the resonator with
the goal of providing the highest possible sensitivity. The photoacoustic
effect’s amplitude exhibits unbounded linear growth over time when an optical
source moves at the speed of sound in a one-dimensional geometry within the
linear acoustic regime [16, 17, 18]. Recently, the use of a moving infrared
optical grating has been employed to generate the photoacoustic effect in the
hundreds of kilohertz range in a cavity with a resonant piezoelectric crystal
used as the detector [17]. In this device a principle formulated by Gusev and
Karabutov [16] where the optical source is made to move at the sound speed
results in coherent addition of the instantaneously produced light induced
pressure wave with the traveling pressure wave generated at a previous time
was used to advantage. The reported detection limit for this device is in the
parts per quadrillion range for $\text{SF}_{6}$, however, the instrument uses
expensive technology and requires a crystal detector that is not at this point
commercially available. As a result, it was decided to investigate the
principles of operation of a longitudinal resonant cavity where a laser beam
is directed to move back and forth along the axis of the resonator. Such an
instrument would employ a microphone operating in the range of kHz instead of
the much higher frequency reported in ref. [17].

For optically thin bodies irradiated by short light pulses, solutions to the
photoacoustic wave equations [19, 20, 21, 22] have been previously given in a
list of references. These solutions, depending on the geometries such as
spheres [23, 24, 25], layers [26], and cylinders [27], show distinctive
temporal features that permit applications in sensing and imaging [28]. Here
we give solutions to the wave equation for the photoacoustic effect generated
by oscillating sources in a one-dimensional cell. Sequentially, the paper
gives the solutions to the wave equation for a single delta source, a moving
delta source, and a moving Gaussian light source in a one-dimensional
resonator in each respective section. The concluding section discusses the
results and outlines directions for future research.

##  II Single delta pulse in a 1D resonator

The motion of a constant amplitude, continuous laser beam along a longitudinal
resonator can be modeled as a summation of delta functions at various $z_{n}$
which are the coordinates of each moving delta source.

Consider a single moving delta heat source in a one-dimensional resonator that
extends from 0 to $L$ irradiated with a laser beam, the heating function can
be written as the form $H=\delta(z-ct)=\frac{I}{c}\delta(t-z/c)$ where $I$ is
the energy per unit area and time delivered to the gas and $z$ is the
coordinate along the symmetry axis of the resonator. The photoacoustic effect
in a inviscid fluid is described by [16, 20],

$\left(\nabla^{2}-\frac{1}{c^{2}}\frac{\partial^{2}}{\partial
t^{2}}\right)p=-\frac{\beta}{C_{P}}\frac{\partial H}{\partial t},$ (1)

where $p$ is the pressure, $c$ is the sound speed, $\beta$ is the thermal
expansion coefficient, $C_{p}$ is the specific heat capacity, $t$ is the time,
and $H$ is the heating function, which describes the energy per unit time and
volume deposited by the optical source. The one-dimensional photoacoustic wave
equation can be written as,

$\Bigg{(}\frac{\partial^{2}}{\partial
z^{2}}-\frac{1}{c^{2}}\frac{\partial^{2}}{\partial
t^{2}}\Bigg{)}p=-\frac{\beta I}{C_{p}c}\frac{\partial}{\partial
t}\delta(t-z/c).$ (2)

Fourier transform of eq. 2 followed by integration of the source term by parts
gives a Helmholtz equation for the frequency domain pressure $\tilde{p}$ as,

$\begin{split}\Bigg{(}\frac{\partial^{2}}{\partial
z^{2}}+\frac{\omega^{2}}{c^{2}}\Bigg{)}\tilde{p}=-\frac{\beta
I}{C_{p}c}\int_{-\infty}^{\infty}e^{-i\omega t}\frac{\partial}{\partial
t}\delta(t-z/c)dt\\\ =-\frac{i\omega\beta
I}{C_{p}c}\int_{-\infty}^{\infty}e^{-i\omega
t}\delta(t-z/c)dt=-\frac{i\omega\beta I}{C_{p}c}e^{-i\omega z/c}.\end{split}$
(3)

A Green’s function [29, 30] for the one-dimensional resonator is given as,

$G(z,z^{\prime})=\frac{2}{L}\sum_{m=1}^{\infty}\frac{\cos k_{m}z\cos
k_{m}z^{\prime}}{k^{2}-k_{m}^{2}},$ (4)

where $k=\omega/c$ and $k_{m}=\frac{m\pi}{L}$. Following an integration of eq.
3 over Green’s function, the frequency domain pressure can be obtained as,

$\tilde{p}=-\frac{2i\omega\beta
I}{C_{p}cL}\sum_{m=1}^{\infty}\frac{\cos(k_{m}z)}{k^{2}-k_{m}^{2}}\int_{0}^{L}e^{-i\omega
z^{\prime}/c}\cos(k_{m}z^{\prime})dz^{\prime}.$ (5)

The integral in eq. 5, denoted $K_{0}$, can be evaluated for the condition
where $\sin(k_{m}L)=0$ and where $r=\cos(k_{m}L)=(-1)^{m}$,

$K_{0}=\int_{0}^{L}e^{-i\omega
z^{\prime}/c}\cos(k_{m}z^{\prime})dz^{\prime}=\frac{-i\omega}{c(\frac{\omega^{2}}{c^{2}}-k_{m}^{2})}\Big{(}1-e^{-i\omega
L/c}r\Big{)}.$ (6)

Inverse Fourier transformation of eq. 5 gives the time domain photoacoustic
pressure as,

$p(t)=\frac{-\beta
Ic}{LC_{p}}\sqrt{\frac{\pi}{2}}\sum_{m=1}^{M}\frac{\cos(k_{m}z)}{k_{m}}e^{ick_{m}t}\Bigg{[}i-ck_{m}t+e^{iLk_{m}}(-1)^{m}(-i+k_{m}L+ck_{m}t)\Bigg{]}.$
(7)

For values of the parameters $z=L/2$, $c=100$ m/s, $L=0.1$ m, $\beta=1$,
$C_{p}=1$, $I=1$ and $M=200$, eq. 7 gives the pressures plotted in fig. 1 and
fig. 2. The real and imaginary parts of the solution for a single delta
function pulse source in the time interval of $(0,0.001\text{s})$ are shown in
fig. 1. For longer time intervals $(0,0.01\text{s})$ the response is shown in
fig. 2. Note that for $z=L/2$, the pressure peaks are observed at 0.5 ms, 1.5
ms, 2.5 ms, 3.5 ms, etc. as shown in fig. 2 (b). When $z=L$, the pressure
peaks are observed at 1 ms, 3 ms, 5 ms, 7 ms, etc. A similar mathematical
treatment can be applied for the other delta sources, for example, for a delta
source with $z_{1}=L-c(t-L/c)=2L-ct$. Results for a summation of single delta
sources are given immediately below.

Figure 1: Time domain pressure for single delta pulse in a one-dimensional
resonator in short time interval (0, 0.001 s) (a) pressure, (b) real part of
pressure. The parameters used are $z=\frac{L}{2}$, $c=100$ m/s, $L=0.1$ m,
$\beta=1$, $C_{p}=1$, $I=1$ and $M=200$. Figure 2: Evaluation of the time
domain pressure for single delta pulse in a one-dimensional resonator in a
long time interval (0, 0.01 s). The parameters taken are (a) $z=\frac{L}{4}$,
(b) $z=\frac{L}{2}$, (c) $z=\frac{3L}{4}$, $c=100$ m/s, $L=0.1$ m, $\beta=1$,
$C_{p}=1$, $I=1$ and $M=200$.

##  III Moving optical sources in a 1D resonator

Consider a one-dimensional resonator that extends from 0 to $L$ irradiated
with a laser beam that repeatedly moves the length of the resonator and then
reverses direction. For the present problem the heating function is taken to
be

$H(z)=I\sum_{n}\delta(z-z_{n}),$ (8)

where $I$ is the energy per unit area and time delivered to the gas, $z$ is
the coordinate along the symmetry axis of the resonator, and $z_{n}$ is the
coordinate of each of the moving delta function sources. The motion of the
laser beam along the $z$ axis is described by the arguments of the delta
functions containing the various $z_{n}$. Table 1 gives values of $z-z_{n}$
and the corresponding delta function (absent the factor $c$ in the
denominators in the right hand column) to be used in eq. 8 for the first few
values of $n$, where the identity $\delta(a-ct)=\delta(t-a/c)/c$ has been
used.

$z-z_{0}=z-ct$ | $\delta(t-\frac{z}{c})$
---|---
$z-z_{1}=z-2L+ct$ | $\delta(t-[\frac{2L}{c}-\frac{z}{c}])$
$z-z_{2}=z+2L-ct$ | $\delta(t-[\frac{2L}{c}+\frac{z}{c}])$
$z-z_{3}=z-4L+ct$ | $\delta(t-[\frac{4L}{c}-\frac{z}{c}])$
$z-z_{4}=z+4L-ct$ | $\delta(t-[\frac{4L}{c}+\frac{z}{c}])$
Table 1:  Arguments of the delta functions in Eq. 8

The first delta function originates at the left end of the resonator and moves
to the right along the $z$ axis. At the time $t=L/c$, the $z_{1}$ delta
function begins moving from the right end of the resonator traveling to the
origin. Thus all of the even numbered entries move from left to right while
the odd numbered ones move from the right end of the resonator to the left.

Fourier transformation of eq. 1 followed by integration of the source term by
parts gives a Helmholtz equation for the frequency domain pressure $\tilde{p}$
as

$\Big{(}\frac{\partial^{2}}{\partial
z^{2}}+\frac{\omega^{2}}{c^{2}}\Big{)}\tilde{p}=-\frac{i\omega\beta
I}{C_{p}c}\int_{-\infty}^{\infty}\sum_{n=0}\delta(t-t_{n})e^{-i\omega t}dt,$
(9)

where $t_{n}=(nL+z)/c$ for even $n$ and $t_{n}=((n+1)L-z)/c$ for odd $n$.

Following an integration of eq. 9 over the Green’s function, the frequency
domain pressure can be found as follows: for even $n$ the wave equation
becomes,

$\Big{(}\frac{\partial^{2}}{\partial
z^{2}}+\frac{\omega^{2}}{c^{2}}\Big{)}\tilde{p}=-\frac{i\omega\beta
I}{C_{p}c}e^{-i\frac{n\omega L}{c}}K_{\text{even}},$ (10)

whereas for odd it is,

$\Big{(}\frac{\partial^{2}}{\partial
z^{2}}+\frac{\omega^{2}}{c^{2}}\Big{)}\tilde{p}=-\frac{i\omega\beta
I}{C_{p}c}e^{-i\frac{(n+1)\omega L}{c}}K_{\text{odd}},$ (11)

where
$K_{\text{even}}=\displaystyle\int_{0}^{L}\cos(k_{m}z^{\prime})e^{-i\omega
z^{\prime}/c}dz^{\prime}$ as evaluated in eq. 6,
$K_{\text{odd}}=K_{\text{even}}^{*}$ and $r=(-1)^{m}$. The frequency domain
photoacoustic pressure for even $n$ becomes,

$\tilde{p}=-\frac{2\omega^{2}\beta
I}{C_{p}c^{2}L}\sum_{n,m}\frac{\cos(k_{m}z)}{\Big{(}\omega^{2}/c^{2}-k_{m}^{2}\Big{)}^{2}}e^{-in\omega
L/c}\Bigg{(}1-re^{-i\omega L/c}\Bigg{)},$ (12)

while for odd $n$ it is

$\tilde{p}=\frac{2\omega^{2}\beta
I}{C_{p}c^{2}L}\sum_{n,m}\frac{\cos(k_{m}z)}{\Big{(}\omega^{2}/c^{2}-k_{m}^{2}\Big{)}^{2}}e^{-i(n+1)\omega
L/c}\Bigg{(}1-re^{i\omega L/c}\Bigg{)};$ (13)

combining the two expression, it becomes,

$\tilde{p}=-\frac{2\omega^{2}\beta
I}{C_{p}c^{2}L}\sum_{n,m}\frac{x\cos(k_{m}z)}{(\omega^{2}/c^{2}-k_{m}^{2})^{2}}e^{-i\omega(\frac{Ln}{c}+\frac{L(x-1)}{2xc})}\Big{(}1-re^{-ix\omega
L/c}\Big{)}.$ (14)

where $x=(-1)^{n}$.

Inverse Fourier transformation of eq. 12 and eq. 13 can be obtained using
Mathematica, which gives,

$p=\frac{-\beta
Ic}{LC_{p}}\sqrt{\frac{\pi}{2}}\sum_{n,m}\frac{\cos(k_{m}z)}{k_{m}}\Theta(t-nL/c)(-1)^{n+1}\bar{p},$
(15)

where $\Theta$ is the Heaviside function and $\bar{p}$ in eq. 15 is defined
as,

$\bar{p}=ck_{m}r(a-b+t)\cos\Big{(}ck_{m}(a-b+t)\Big{)}-\sin\Big{(}ck_{m}(a+t)\Big{)}\\\
-ck_{m}(a+t)\cos\Big{(}ck_{m}(a+t)\Big{)}+r\sin\Big{(}ck_{m}(a-b+t)\Big{)},$
(16)

where $a$ and $b$ are defined as for even $n$: $a=nL/c$ and $b=-L/c$; and for
odd $n$: $a=(n+1)L/c$ and $b=L/c$. The summation in eq. 15 over $n$ refers the
number of sweeps of the optical beam, beginning at 0; the summation over $m$
starts at 1, and in principle extends to infinity.

For values of the parameters $c=100$ m/s, $L=0.1$ m, $\beta=1$, $C_{p}=1$,
$I=1$, $N=50$, $M=200$, eq. 15 gives the pressures as plotted in fig. 3. Note
that for $z=L/2$, the pressure peaks are observed at 0.5 ms, 1.5 ms, 2.5 ms,
3.5 ms, etc. as shown in fig. 3 (b). The amplitude of the pressure increases
linearly with the number of sweeps across the resonator.

Figure 3: Time domain pressure for an oscillating delta pulse in a one-
dimensional resonator. Magnitude of the photoacoustic pressure $p$ versus time
in milliseconds from eq. 15 evaluated at (a) $z=\frac{L}{4}$, (b)
$z=\frac{L}{2}$ and $z=\frac{3L}{4}$ of the resonator for $N=50$, $M=200$,
$L=0.1$ m and $c=100$ m/s.

##  IV Oscillating Gaussian

Consider an oscillating Gaussian source with speed $v$ along the longitudinal
axis. The heating function can be written as
$H=\frac{I_{0}}{\sigma\sqrt{\pi}}\sum_{n}e^{-(\frac{z-z_{n}}{\sigma})^{2}}$.
Fourier transformation of eq. 1 followed by integration of the source term by
parts gives the frequency domain pressure $\tilde{p}$ as,

$\Big{(}\frac{\partial^{2}}{\partial
z^{2}}+\frac{\omega^{2}}{c^{2}}\Big{)}\tilde{p}=-\frac{i\omega\beta
I}{C_{p}\sigma\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-i\omega
t}\sum_{n}e^{-(\frac{z-z_{n}}{\sigma})^{2}}dt,$ (17)

where the coordinate of each of the moving Gaussian source is
$z_{n}=\frac{1}{2}L(1-x)+xv(t-\frac{nL}{v})$. Through use of
$z-z_{n}=-xv(t-t_{0})$ and
$t_{0}:=\frac{z}{xv}+\frac{L}{2xv}(x-1)+\frac{nL}{v}$ (See Appendix), the
integral in eq. 17 becomes,

$\int_{-\infty}^{\infty}e^{-i\omega
t}\exp\Big{(}-\frac{x^{2}v^{2}}{\sigma^{2}}(t-t_{0})^{2}\Big{)}dt=\frac{\sqrt{\pi}\sigma}{v}\exp\Big{(}-\frac{\omega^{2}\sigma^{2}}{4v^{2}}-i\omega
t_{0}\Big{)}.$ (18)

Evaluation of eq. 17 at $v=c$ using the Green’s function in eq. 4 gives the
frequency domain pressure as,

$\tilde{p}=-\frac{2i\omega\beta
I}{C_{p}Lc}\sum_{n,m}\frac{\cos(k_{m}z)}{k^{2}-k_{m}^{2}}e^{-\frac{\omega^{2}\sigma^{2}}{4c^{2}}-i\omega(\frac{Ln}{c}+\frac{L(x-1)}{2xc})}\int_{0}^{L}\cos(k_{m}z^{\prime})e^{-i\omega\frac{z^{\prime}}{xc}}dz^{\prime}.$
(19)

Evaluation of the integral in eq. 19 gives,

$\int_{0}^{L}\cos(k_{m}z^{\prime})e^{-i\omega\frac{z^{\prime}}{xc}}dz^{\prime}=\frac{-i\omega
x}{c}\frac{[1-r\exp(-i\omega L/(xc))]}{(\omega/c)^{2}-k_{m}^{2}},$ (20)

where $x=(-1)^{n}$ and $r=(-1)^{m}$. Hence, the photoacoustic pressure
becomes,

$\tilde{p}=-\frac{2\beta
I}{C_{p}c^{2}L}\sum_{n,m}\frac{x\cos(k_{m}z)\omega^{2}}{(\omega^{2}/c^{2}-k_{m}^{2})^{2}}e^{-\frac{\omega^{2}\sigma^{2}}{4c^{2}}}e^{-i\omega(\frac{Ln}{c}+\frac{L(x-1)}{2xc})}\Big{(}1-re^{-ix\omega
L/c}\Big{)}.$ (21)

Inverse Fourier transform of
$F(\omega)=\frac{\omega^{2}}{(\omega^{2}/c^{2}-k_{m}^{2})^{2}}e^{-\frac{\omega^{2}\sigma^{2}}{4c^{2}}}$
(see Appendix) gives,

$\begin{split}f(t)=\frac{c^{2}}{2k_{m}^{2}}\Bigg{(}e^{-ick_{m}t}g^{[2]}(t;ck_{m})+e^{ick_{m}t}g^{[2]}(t;-ck_{m})\\\
-\frac{1}{ck_{m}}\Big{(}e^{-ick_{m}t}g^{[1]}(t;ck_{m})-e^{ick_{m}t}g^{[1]}(t;-ck_{m})\Big{)}\Bigg{)},\end{split}$
(22)

where the functions $g^{[1]}(t;ck_{m})$ and $g^{[2]}(t;ck_{m})$ are given in
the Appendix.

The other terms contributes as a phase shift thus the time domain
photoacoustic pressure becomes,

$\displaystyle p(t)$ $\displaystyle=\frac{-2\beta
I}{C_{p}c^{2}L}\sum_{n,m}x\cos(k_{m}z)\Theta(t-nL/c)$ (23)
$\displaystyle\times\Bigg{(}f\Big{(}t+\frac{Ln}{c}+\frac{L(x-1)}{2xc}\Big{)}-rf\Big{(}t+\frac{Ln}{c}+\frac{L(x-1)}{2xc}+\frac{xL}{c}\Big{)}\Bigg{)}$

In the limit $\sigma\rightarrow 0$, the solution for oscillating Gaussian eq.
23 reduces to the solution for oscillating delta eq. 15.

For values of the parameters $c=100$ m/s, $L=0.1$ m, $\beta=1$, $C_{p}=1$,
$I=1$, $N=50$, $M=150$, $\sigma=0.01$, and $z=L/2$, eq. 23 gives the pressures
plotted in fig. 4. The pressure peaks are observed at 0.5 ms, 1.5 ms, 2.5 ms,
3.5 ms, etc. as shown in Fig. 4.

Figure 4: Time domain pressure for an oscillating Gaussian source in a one-
dimensional resonator. Magnitude of the photoacoustic pressure $p$ versus time
in milliseconds from eq. 23 at $z=\frac{L}{2}$ of the resonator for $N=50$,
$M=150$, $L=0.1$ m, and $c=100$ m/s.

##  V Discussion

The results of the calculations presented here give an expression for the
photoacoustic pressure generated by a moving light beam in an optically thin
absorbing fluid in a one-dimensional resonator. It can be seen that the
amplitude of the pressure increases linearly with the number of sweeps across
the resonator. This addition is reminiscent of pulsed excitation of a
conventional resonator when the time between pulses corresponds to a resonance
of the photoacoustic cell. Such pulse addition has been used to advantage in
ref. [10] where a cylindrical resonator containing infrared absorbing gas was
excited by a pulsed laser. Both the pulsed excitation and moving source have
in common synchronization of excitation, however, only the latter gives the
amplification that appears to be unique in that the excitation takes place
additively at all times.

Of note is that only the case of a source that instantaneously switches
direction as it approaches the ends of the cavity has been considered, which,
in practice may present a difficulty requiring a large bandwidth for the
optical modulator that provides the directional modulation. A simpler
experimental arrangement might employ sinusoidal motion in space, where the
speed of the optical source varies and does not match the sound speed at all
times. Such a configuration was not treated here, but it is clear that rapid
switching of the direction is optimal, which simplifies the mathematics and
shows the major features of a moving source photoacoustic effect in a one-
dimensional resonator.

As the above calculations show, the amplification effect for a light source
comoving with the sound speed as described in ref. [16] is extant in a cavity
for a source whose position instantaneously changes direction at the end
points of the resonator. It is clear from inspection of eq. 17 and eq. 23 that
the pressure in the cavity is a summation of wavelets, with the final pressure
coming as a result of acoustic interference. The results given here suggest a
straightforward implementation of the amplification effect is possible for
trace gas detection without the complicated moving grating design and
sophisticated detector as reported in ref. [17]. For other possible further
investigation, it might be interesting to detect and engineer the spatial-
temporal characteristics of the emitted acoustic waves [16, 24, 21, 31]
through multiscale modeling [32] , machine learning [33, 34] and experiments,
given extensive literature in model-based and data-driven approaches for
inverse design of mechanical systems [35, 36].

Regarding experimental implementation of the moving source concept developed
here, construction of a resonator with an audio frequency microphone is
straightforward. A moving source can be implemented using an electro-optic
oscillating mirror, which would yield sinusoidal or some variant of sinusoidal
motion in the resonator depending on the design of the modulator, or,
alternately, through use of a wide bandwidth acoustic-optic modulator which
could provide near perfect matching of the motion of the optical beam to the
motion of the acoustic wave at all times.

##  VI Appendix

The following identity has been used in the evaluation of eq. 17,

$z-z_{n}=z-\frac{1}{2}L(1-x)-xvt+\frac{xvnL}{v}=-xv\Big{[}t-\Big{(}\frac{z}{xv}+\frac{L}{2xv}(x-1)+\frac{nL}{v}\Big{)}\Big{]}=-xv(t-t_{0}),$
(A1)

where $t_{0}=\frac{z}{xv}+\frac{L}{2xv}(x-1)+\frac{nL}{v}$.

###  VI.1 Derivation Eq. 22: the forms of $g^{[1]}$ and $g^{[2]}$

Rewrite $F(\omega)$ as,

$\displaystyle F(\omega)$
$\displaystyle=\frac{\omega^{2}}{(\omega^{2}/c^{2}-k_{m}^{2})^{2}}e^{-\frac{\omega^{2}\sigma^{2}}{4c^{2}}}$
(A2)
$\displaystyle=\frac{c^{2}\omega^{2}}{2k_{m}^{2}}e^{-\frac{\omega^{2}\sigma^{2}}{4c^{2}}}\Bigg{(}\frac{1}{(\omega-
ck_{m})^{2}}+\frac{1}{(\omega+ck_{m})^{2}}-\frac{1}{ck_{m}(\omega-
ck_{m})}+\frac{1}{ck_{m}(\omega+ck_{m})}\Bigg{)}.$

The Fourier transform of each term can be evaluated separately, which gives
Eq. 22 where

$g^{[1]}(t;ck_{m})=\frac{2\sqrt{2}c^{3}}{\sigma^{3}}\Big{(}-it+\frac{k_{m}\sigma^{2}}{2c}\Big{)}e^{ick_{m}t-t^{2}c^{2}/\sigma^{2}}+\frac{c^{2}k_{m}^{2}i\sqrt{\pi}}{\sqrt{2}}e^{-\frac{k_{m}^{2}\sigma^{2}}{4}}\operatorname{erf}\Bigg{(}{\frac{-ct+ik_{m}\sigma^{2}/2}{\sigma}}\Bigg{)}$
(A3) $\displaystyle g^{[2]}(t;ck_{m})$
$\displaystyle=\frac{\sqrt{2}}{\sigma/c}e^{ick_{m}t-t^{2}c^{2}/\sigma^{2}}$
(A4)
$\displaystyle\quad+ck_{m}e^{-k_{m}^{2}\sigma^{2}/4}\sqrt{\frac{\pi}{2}}\Bigg{(}-ck_{m}t+\frac{ik_{m}^{2}\sigma^{2}}{2}+2i\operatorname{erf}\Bigg{(}\frac{-ct+ik_{m}\sigma^{2}/2}{\sigma}\Bigg{)}\Bigg{)}$

where erf is the error function.

###  VI.2 Numerical evaluation of the theoretical solution

Eq. 23 is the general solution to the Gaussian case. For any fixed $n$ and
$m$, the Fourier transform of Eq. 19 can be compared to the theoretical
solution in Eq. 23. The singularity term in the denominator will lead to
instability in evaluating of fast Fourier transform (FFT). The singular term
and the non-singular term can be sepatated into
$\tilde{p}=\tilde{p}_{\text{singular}}+\tilde{p}_{\text{non-singular}}$. For
the singular term, there is an analytical Fourier transform formula.
Evaluation of the Fourier transform of the singular term gives,

$F\Bigg{[}\frac{\omega^{2}}{(\omega^{2}/c^{2}-k_{m}^{2})^{2}}\Bigg{]}=\frac{-c^{3}\sqrt{\pi}}{2\sqrt{2}k_{m}}\Bigg{(}ck_{m}t\cos(ck_{m}t)+\sin(ck_{m}t)\Bigg{)}$
(A5)

Figure 5: Numerical Evaluation of the term in eq. 23 with $n=5$ and $m=10$ via
both the theoretical expression and FFT evaluation of the term in eq. 19.

Evaluation of the theoretical expression in Eq. 23 with $n=5$ and $m=10$ and
numerical FFT of the non-singular term, adding the theoretical evaluation of
the singular term eq. A5, gives the pressure as shown in Fig. 5. The FFT can
be seen to align precisely with the theoretical solution described in Eq. 23.

## References

  * Kreuzer [1971] L. Kreuzer, Ultralow gas concentration infrared absorption spectroscopy, Journal of Applied Physics 42, 2934 (1971).
  * Dewey Jr  _et al._ [1973] C. Dewey Jr, R. Kamm, and C. Hackett, Acoustic amplifier for detection of atmospheric pollutants, Applied Physics Letters 23, 633 (1973).
  * Lyamshev [2004] L. M. Lyamshev,  _Radiation Acoustics_ (CRC Press, 2004).
  * Bunkin  _et al._ [1991] F. V. Bunkin, A. A. Kolomensky, and V. G. Mikhalevich,  _Lasers in Acoustics_ (Harwood Academic, Reading, MA, 1991).
  * Hodgkinson and Tatam [2012] J. Hodgkinson and R. P. Tatam, Optical gas sensing: a review, Measurement Science and Technology 24, 012004 (2012).
  * Hess [1983] P. Hess,  _Resonant photoacoustic spectroscopy_ (Springer Berlin Heidelberg, 1983) pp. 1–32.
  * Sigrist [1994] M. W. Sigrist,  _Air monitoring by laser photoacoustic spectroscopy_ , Vol. 127 (Chemical Analysis (Wiley New York), 1994) pp. 163–238.
  * Sigrist [1992] M. W. Sigrist,  _Environmental and chemical trace gas analysis by photoacoustic methods_ , Vol. 1 (Elsevier, New York, 1992) pp. 369–427.
  * Sigrist [1998] M. W. Sigrist,  _Air monitoring, optical spectroscopic methods_ (Wiley, New York, 1998) pp. 84–117.
  * Brand  _et al._ [1995] C. Brand, A. Winkler, P. Hess, A. Miklós, Z. Bozóki, and J. Sneider, Pulsed-laser excitation of acoustic modes in open high-q photoacoustic resonators for trace gas monitoring: results for ethylene, Applied Optics 34, 3257 (1995).
  * Zharov and Letokhov [2013] V. P. Zharov and V. S. Letokhov,  _Laser optoacoustic spectroscopy_ , Vol. 37 (Springer, 2013).
  * Claspy [1977] P. Claspy,  _Infrared optoacoustic spectroscopy and detection_ (Academic, New York, 1977) pp. 133–166.
  * Elia  _et al._ [2009] A. Elia, P. M. Lugarà, C. Di Franco, and V. Spagnolo, Photoacoustic techniques for trace gas sensing based on semiconductor laser sources, Sensors 9, 9616 (2009).
  * Karbach and Hess [1985] A. Karbach and P. Hess, High precision acoustic spectroscopy by laser excitation of resonator modes, The Journal of chemical physics 83, 1075 (1985).
  * Karbach and Hess [1986] A. Karbach and P. Hess, Photoacoustic signal in a cylindrical resonator: theory and laser experiments for ch4 and c2h6, The Journal of chemical physics 84, 2945 (1986).
  * Gusev and Karabutov [1993] V. E. Gusev and A. A. Karabutov,  _Laser Optoacoustics_ (American Institute of Physics, New York, 1993).
  * Xiong  _et al._ [2017] L. Xiong, W. Bai, F. Chen, X. Zhao, F. Yu, and G. J. Diebold, Photoacoustic trace detection of gases at the parts-per-quadrillion level with a moving optical grating, Proceedings of the National Academy of Sciences 114, 7246 (2017).
  * Bai and Diebold [2019] W. Bai and G. J. Diebold, Moving photoacoustic sources: Acoustic waveforms in one, two, and three dimensions and application to trace gas detection, Journal of Applied Physics 125 (2019).
  * Morse and Ingard [1986] P. M. Morse and K. U. Ingard,  _Theoretical acoustics_ (Princeton university press, 1986).
  * Westervelt and Larson [1973] P. J. Westervelt and R. S. Larson, Laser-excited broadside array, The Journal of the Acoustical Society of America 54, 121 (1973).
  * Diebold  _et al._ [1991] G. Diebold, T. Sun, and M. Khan, Photoacoustic monopole radiation in one, two, and three dimensions, Physical Review Letters 67, 3384 (1991).
  * Calasso  _et al._ [2001] I. G. Calasso, W. Craig, and G. J. Diebold, Photoacoustic point source, Physical Review Letters 86, 3550 (2001).
  * Diebold and Westervelt [1988] G. Diebold and P. Westervelt, The photoacoustic effect generated by a spherical droplet in a fluid, The Journal of the Acoustical Society of America 84, 2245 (1988).
  * Diebold  _et al._ [1990] G. Diebold, M. Khan, and S. Park, Photoacoustic” signatures” of particulate matter: optical production of acoustic monopole radiation, Science 250, 101 (1990).
  * Yan and Diebold [2021a] X. Yan and G. J. Diebold, Generation of high amplitude compressions and rarefactions in a photoacoustically excited droplet, Photoacoustics 23, 100289 (2021a).
  * Khan  _et al._ [1993] M. Khan, T. Sun, and G. Diebold, Photoacoustic waves generated by absorption of laser radiation in optically thin layers, The Journal of the Acoustical Society of America 93, 1417 (1993).
  * Khan and Diebold [1996] M. Khan and G. Diebold, The photoacoustic effect generated by laser irradiation of an isotropic solid cylinder, Ultrasonics 34, 19 (1996).
  * Wang [2017] L. Wang,  _Photoacoustic imaging and spectroscopy_ (CRC press, 2017).
  * Mathews and Walker [1970] J. Mathews and R. L. Walker,  _Mathematical methods of physics_ , Vol. 501 (WA Benjamin New York, 1970).
  * Arfken  _et al._ [2005] G. B. Arfken, H. J. Weber, and F. E. Harris,  _Mathematical Methods for Physicists: A Comprehensive Guide_ , 7th ed. (Academic Press, 2005).
  * Yan and Diebold [2021b] X. Yan and G. J. Diebold, Abel inversion of optical beam deflection signals from photoacoustic waves with symmetry in one, two, and three dimensions, Journal of Applied Physics 130 (2021b).
  * Weinan [2011] E. Weinan,  _Principles of multiscale modeling_ (Cambridge University Press, 2011).
  * Jin  _et al._ [2023] H. Jin, E. Zhang, and H. D. Espinosa, Recent advances and applications of machine learning in experimental solid mechanics: A review, Applied Mechanics Reviews 75, 061001 (2023).
  * Molinaro  _et al._ [2023] R. Molinaro, Y. Yang, B. Engquist, and S. Mishra, Neural inverse operators for solving PDE inverse problems, in  _Proceedings of the 40th International Conference on Machine Learning_ , Proceedings of Machine Learning Research, Vol. 202 (PMLR, 2023) pp. 25105–25139.
  * Jin  _et al._ [2022] H. Jin, T. Jiao, R. J. Clifton, and K.-S. Kim, Dynamic fracture of a bicontinuously nanostructured copolymer: A deep-learning analysis of big-data-generating experiment, Journal of the Mechanics and Physics of Solids 164, 104898 (2022).
  * Jin  _et al._ [2021] H. Jin, A. K. Landauer, and K.-S. Kim, Ruga mechanics of soft-orifice closure under external pressure, Proceedings of the Royal Society A 477, 20210238 (2021).