text:
\begin{document}

\begin{abstract}
Let $v$ be a submultiplicative weight. Then we prove that
$v$ satisfies GRS-condition, if and only if $v\cdot e^{-\ep |\cdo |}$ is bounded
for every positive $\ep$. We use this equivalence to establish identification
properties between weighted Lebesgue spaces, and between
certain modulation spaces and Gelfand-Shilov spaces.
\end{abstract}

\maketitle

\section{Introduction}\label{sec0}

\par

In \cite{GRS}, Gel'fand, Raikov and Shilov considered a family of matrix
classes in $\zz d\times \zz d$, where each matrix class, $\mathcal A_v$,
depends on a weight function $v$ on $\rr d$, and consists of all matrices
$A=\big ( a(j,k) \big )_{j,k\in \zz d}$ such that
$$
\sup _{j\in \zz d}a(j,j-k)v(k) \in \ell ^1.
$$
Here the weight $v$ is positive and submultiplicative, i.{\,}e. $v$ is
even and fulfills $v(x+y)\le v(x)v(y)$.

\par

Since \cite{baskakov,GRS}, several investigations have been performed which confirm
the importance of the class $\mathcal A_v$. For example if $v \ge c$ for
some positive constant $c$, then $\mathcal A_v \subseteq \mathcal
B(\ell ^2(\zz d))$, the set of all matrices which are bounded on $\ell ^2(\zz d)$.
Furthermore, in \cite{gr-book, gr_sjostrand}, strong links
between $\mathcal A_v$ and certain modulation spaces are described,
which appear in natural ways when considering non-stationary filters
in signal analysis (cf. \cite{strohmer} and the references therein).

\par

In \cite{baskakov,GRS}, the condition

\begin{equation}\label{GRScondition}
\lim _{\ell \to \infty}v(\ell x)^{1/\ell } =1,\quad x\in \rr d,
\end{equation}

is introduced for weights $v$ on $\rr d$, and link this condition to
fundamental questions on inverse closed properties for matrix algebras.
More precisely, for any submultiplicative weight $v$ which fulfills
\eqref{GRScondition}, it is proved in \cite{GRS} that the matrix
class $\mathcal A_v$ is so-called \emph{inverse closed} on $\ell ^2(\zz d)$.
That is, $\mathcal A_v$ is contained in $\mathcal B(\ell ^2(\zz d))$, and if
$A\in A_v$ is invertible on $\ell ^2(\zz d)$, then its inverse $A^{-1}$, does
also
belong to $A_v$. Furthermore, if $v$ is submultiplicative and inverse closed
on $\ell ^2(\zz d)$, then it is necessary that $v$ satisfies
\eqref{GRScondition}
(cf. \cite[Corollary 5.31]{gr-survey}).

\par

The condition \eqref{GRScondition} (the so-called
\emph{GRS-condition} or \emph{Gel'fand-Raikov-Shilov condition})
is therefore strongly linked to inverse closed properties for matrix algebras,
and a submultiplicative weight which satisfies the GRS-conditions is called
a \emph{GRS weight}.

\par

Later on, similar links between the GRS-conditions and inverse closed
properties for pseudo-differential operators with symbols in certain
modulation spaces were established (see e.{\,}g. \cite{gr_sjostrand}).

\par

In \cite{gr_fieldspaper}, it is proved that if $v$ is submultiplicative, then
there are positive constants $c$ and $C$ such that
$$
v(x)\le Ce^{c|x|},
$$
and it is easily seen that the right-hand side is a submultiplicative weight.
We note however that the weight

\begin{equation}\label{ExpWeight}
v_c(x) =v_{c,1}(x)\equiv e^{c|x|}
\end{equation}

does not satisfy the GRS-condition. Consequently, the matrix algebra
$\mathcal A_v$ is not inverse closed when $v=v_c$ and $c>0$.
On the other hand, it is easily seen that the set of submultiplicative
weights satisfying the GRS-condition is a monoid and contains
the weights
$$
x\mapsto (1+|x|)^r \quad \text{and}\quad x\mapsto
v_{c,s}(x)\equiv e^{c|x|^s}
$$
when $r\ge 0$,  $c\ge 0$ and $0\le s<1$.

\par

Since the (non-GRS) weight $v_{c,1}$ is rather similar to the GRS
weight $v_{c,s}$ when $s<1$ is close to $1$, it might be anticipated
that the the algebra $\mathcal A_{v_{c,1}}$ should possess a weaker
form of inverse closed property. Such property is established in
\cite{jaffard}, where it is proved that if $A\in \mathcal A_{v_{c_1,1}}$ for some
$c_1>0$ is invertible on $\ell ^2(\zz d)$, then its inverse
$A^{-1}$ belongs to $\in \mathcal A_{v_{c_2,1}}$, for some $c_2$.

\par

In this context we remark that in \cite{FGT}, a condition on sequences
of weights, which extends the notion of GRS-condition and includes exponential
type weights is considered. Furthermore, certain type of
inverse closed property is established for matrix algebras which are
parameterized with such sequences of weights. In particular, the classical
equivalence between the inverse closed property of $\mathcal A_v$
and the fact that $v$ should be a GRS weight becomes a special
case of \cite[Theorem 2.1]{FGT}.

\medspace

In this paper we establish a more narrow link between the GRS-condition
and weights bounded by exponentials. More precisely, if $v$ is
submultiplicative,
then we prove that $v$ fulfills the GRS-conditions, if and only if for every
$\ep >0$, there is a constant $C_\ep$ such that $v\le C_\ep v_{\ep ,1}$.

\par

We also combine this result with Theorem 3.9 in \cite{toft} to establish
relations with modulation spaces parameterized with weights satisfying
GRS-conditions, and Gelfand-Shilov spaces.

\par

\section{Identifications of the GRS-condition in terms of
exponentials}\label{sec1}

\par

We recall that a weight on $\rr d$ is a positive function on $\rr d$
which belongs to $L^\infty _{loc}(\rr d)$. The weight $v$ on $\rr d$
is submultiplicative, if $v$ is positive and even, and satisfies
$$
v(x+y)\leq v(x)v(y),\quad \mbox{for}\quad x,y \in \rr d.
$$
It is well-known that
$$
v(x)\lesssim e^{c|x|}
$$
for some $c>0$, when $v$ is submultiplicative. Furthermore, the
weight $v$ satisfies the \emph{GRS-condition} whenever

\begin{equation}\label{eq:grslimite0}
\lim _{\ell \to \infty}v(\ell x)^{1/\ell }=1,\quad \text{when}\quad x\in \rr d,
\end{equation}

holds.

\par

We have now the following observation concerning submultiplicative weights.

\par

\begin{prop}\label{prop:main}
Let $v$ be a submultiplicative weight on $\rr d$. Then the following conditions
are equivalent:
\begin{enumerate}
\item $v$ satisfies the GRS-condition;

\vrum

\item $v$ satisfies $v(x)\lesssim e^{\ep |x|}$, for every $\ep >0$.
\end{enumerate}
\end{prop}

\par

\begin{proof}
First assume that (2) holds. Let $\ep >0$ and $x\in \rr d$ be arbitrary and
fixed. As
\begin{equation*}
v(nx) \le C e^{\ep n|x|},
\end{equation*}

we have $$
v(nx)^{1/n} \le  C^{1/n}e^{\ep |x|},
$$
and since $\ep $ is arbitrary we have
$$
\limsup _{n\to \infty} v(nx)^{1/n} \le 1.
$$

\par

On the other hand, since $v(x)\ge c$ for some $c>0$, it follows that
$$
\liminf _{n\to \infty} v(nx)^{1/n} \ge 1.
$$
A combination of these estimates gives that $\lim _{n\to \infty} v(nx)^{1/n}$ exists
and is equal to $1$, i.{\,}e. (1) holds.

\par

Next we assume that (2) does not hold. Let $\{e_1,\dots,e_d\}$
be the canonical  basis  in $\mathbb{R}^d.$ Since $v$ is
submultiplicative,
$$
v(x_1,\dots,x_d)\leq  \prod_{j=1}^{d}v(x_je_j).
$$
Hence, if (2) is not fulfilled, then there are $j\in \{1,
\dots, d\}$ and $\ep >0$ such that $v(te_j)e^{-\ep |t|}$ is
unbounded. Without loss of generality we may assume that $j=1$. The weight on $\mathbb{R}$ defined by
$$
u(t):=v(te_1)
$$
is submultiplicative. Since $u(t)e^{-\ep t}$ is
unbounded in $[0,\infty[$, we may find a sequence $\left(t_n\right)$
such that  $0<t_n+1<t_{n+1}\to \infty$ and
$$
u(t_n) > e^{\ep t_n}.
$$
Let $k_n$ be the
integer part of $t_n.$ As $t_n<k_n+1$ and $u$ is submultiplicative,
$$
u(t_n)\le  C u(k_n+1)\leq C u(1)u(k_n),
$$
where $C = \max_{|\xi|\leq 1}u(\xi)$. Hence
$$
u(k_n)\geq \frac{u(t_n)}{C u(1)}\geq \frac{e^{\ep t_n}}{C u(1)}
$$
and
$$
u(k_n)^{1/k_n}\geq \left(\frac{u(t_n)}{C u(1)}\right)^{1/t_n}\geq \frac{e^{\ep
}}{(C u(1))^{1/t_n}}.
$$
Consequently
$$
\liminf_{n\to \infty} u(k_n)^{1/k_n} \geq
e^{\ep }.
$$
This implies that $u,$ and therefore $v,$ does not satisfy the
GRS-condition. The proof is complete.
\end{proof}

\par\medskip\
The previous result means that ${\mathcal A}_v$ is inverse closed in $\mathcal B(\ell ^2(\zz d))$ if, and only if, it contains the Jaffard algebra of all matrices with exponential off-diagonal decay.

\begin{cor}\label{CorProp}
Let $\Omega$ be the set of all weights on $\rr {2d}$ which satisty
the GRS-condition. Then

\begin{equation}\label{ellAidentities}
\bigcup_{\ep >0}\ell^1_{(v_\ep )} =\bigcap_{v\in \Omega}
\ell^1_{(v)}
\quad \text{and} \quad
\bigcup_{\ep >0}\mathcal A_{v_\ep}
=\bigcap_{v\in \Omega} \mathcal A_v.
\end{equation}

\end{cor}

\par

\begin{proof}
This is a consequence of Proposition \ref{prop:main} and
\cite[Theorem 2.1]{FGT}.
\end{proof}

\par

Next we analyze a condition on sequences of weights
considered in  \cite{FGT}.

\par

\begin{prop}
Let ${\mathcal W} = \{ w_n\} _{n\in \mathbb N}$ be a decreasing sequence of
submultiplicative weights. Then the following conditions are
equivalent:
\begin{itemize}
\item[(1)] $\begin{displaystyle}\inf_n \lim_{\ell \to \infty}w_n(\ell
x)^{{1}/{\ell}} = 1\end{displaystyle}$ for every $x\in {\mathbb R}^d$.

\vrum

\item[(2)] for every $\ep > 0$, there exists an $n\in {\mathbb N}$ such
that $w_n(x)e^{-\ep |x|}$ is bounded.
\end{itemize}
\end{prop}

\par

\begin{proof}
$(2)\Rightarrow (1)$ is clear. Assume that (1) holds, and let
$V$ be the family of all submultiplicative
and weights $v\, :\,  \zz d\to \mathbb R_+$ such that
$$
\sup_{{k}}\frac{v(k)}{w_n(k)} < \infty\quad \forall \ n\in
{\mathbb N}.
$$
By \cite[Theorem 2.3]{bms} we have
$$
\bigcap_{v\in V}\ell^1_{(v)} = \bigcup_n \ell^1_{(w_n)}
$$
algebraic and topologically (see also the proof of \cite[Theorem 2.1]{FGT}).
Since every weight $v\in V$ satisfies the GRS-condition, we have $v =
O(v_{\ep})$ for every $\ep > 0.$
Consequently,
$$
\ell^1 _{(v_{\ep})} \subset \bigcup_n \ell^1_{(w_n)}
$$
with continuous inclusion. The result now follows from
Gr\"othendieck's  factorization theorem (see \cite[1.2.20]{bp}).
\end{proof}

\par

We finish the paper by applying our previous results to establish certain
links between modulation spaces and Gelfand-Shilov spaces.
We recall the definition of these spaces.

\par

Let $0<h,s\in
\mathbb R$ be fixed, and let $\mathcal S_{s,h}(\rr d)$ be the set of
all $f\in C^\infty (\rr d)$ such that

\begin{equation*}
\nm f{\mathcal S_{s,h}}\equiv \sup \frac {|x^\beta \partial ^\alpha
f(x)|}{h^{|\alpha | + |\beta |}\alpha !^s\, \beta !^s}
\end{equation*}

is finite. Here the supremum should be taken over all $\alpha ,\beta \in
\mathbb N_0^d$ and $x\in \rr d$. We note that $\mathcal S_{s,h}(\rr d)
\subseteq \mathscr S(\rr d)$, increases with $h$ and $s$, where
$\mathscr S(\rr d)$ is the set of all Schwartz functions on $\rr d$.
The \emph{Gelfand-Shilov space} $\mathcal S_{s}(\rr d)$ is the inductive limit
of $\mathcal S_{s,h}(\rr d)$ and the
\emph{Gelfand-Shilov distribution space} $\mathcal S_{s}'(\rr d)$
is the projective limit of $\mathcal S_{s,h}'(\rr d)$.
In particular,

\begin{equation}\label{GSspacecond1}
\mathcal S_{s}(\rr d) = \bigcup _{h>0}\mathcal S_{s,h}(\rr d),
\quad \text{and}\quad
\mathcal S_s'(\rr d) = \bigcap _{h>0}\mathcal S_{s,h}'(\rr d).
\end{equation}

Note that $\mathcal S_s'(\rr d)$
is the dual of $\mathcal S_s(\rr d)$ (\cite{GS, Ko, Pil}).

\par

Evidently, $\mathscr S(\rr d)\subset \mathcal S_s(\rr d)$ increases with
$s$, and $\mathscr S'(\rr d)\subseteq \mathcal S_s'(\rr d)$
decreases with $s$. If $s<1/2$, then $\mathcal S_s(\rr d)$ is trivial.

\par

The Fourier transform on $\mathscr S'(\rr d)$ is the linear and
continuous map which takes the form
$$
(\mathscr Ff)(\xi )= \widehat f(\xi ) \equiv (2\pi )^{-d/2}\int _{\rr
{d}} f(x)e^{-i\scal  x\xi }\, dx
$$
when $f\in L^1(\rr d)$. For every $s\ge 1/2$, the Fourier transform
is continuous and bijective on $\mathcal S_s(\rr d)$, and extends
uniquely to a continuous and bijective map on $\mathcal S_s'(\rr d)$.

\par

Next we recall the definition of modulation spaces. Let $1/2\le s_0<s$,
and let $\phi \in \maclS _{s_0}(\rr d)\setminus 0$ be fixed.
Then the \emph{short-time Fourier transform} $V_\phi f$ of
$f\in \maclS _s'(\rr d)$ is the element in $\maclS _s'(\rr {2d})
\cap C^\infty (\rr {2d})$, defined by the formula
$$
(V_\phi f)(x,\xi ) := \mathscr F(f\cdot \phi (\cdo -x))(\xi).
$$
If in addition $f$ is in the Schwartz class, then
$V_\phi f$ is given by
$$
(V_\phi f)(x,\xi ) = (2\pi )^{-d/2}\int _{\rr d} f(y)\overline{\phi (y-x)}
e^{-i\scal y\xi}\, dy.
$$

\par

Let $v$ be a submultiplicative weight on $\rr d$. Then the weight
$\omega$ on $\rr d$ is called $v$-moderate if
$$
\omega (x+y)\le C\omega (x)v(y),
$$
for some constant $C$ which is independent of $x\in \rr d$ and
$y\in \rr d$.

\par

Let $\phi \in \maclS _{1/2}(\rr d)$, $1/2<s<1$, $p,q\in (0,\infty]$, $v$ be
submultiplicative on $\rr {2d}$, and let $\omega$ be a $v$-moderate
weight on $\rr {2d}$. Then the \emph{modulation space}
$M^{p,q}_{(\omega )}(\rr d)$ is the quasi-Banach space which consists of
all $f\in \maclS _{1/2} '(\rr d)$ such that

\begin{equation}\label{modnorm}
\nm f{M^{p,q}_{(\omega )}} := \left ( \int _{\rr d} \left ( \int _{\rr d}
|V_\phi f (x,\xi )\omega (x,\xi ) |^p\, dx \right )^{q/p}\, d\xi \right )^{1/q}
\end{equation}

is finite (with obvious modifications when $p=\infty$ or $q=\infty$).
The definition of $M^{p,q}_{(\omega )}(\rr d)$ is
independent of the choice of $\phi \in \maclS _{1/2} (\rr d)\setminus 0$,
and different $\phi$ gives rise to equivalent norms. (See e.{\,}g.
\cite{gr-book}).

\medspace

According to \cite[Prop. 11.3.1]{gr-book},
$$
\bigcap_{s\geq 0}M^{\infty}_{(w_s)}({\mathbb R}^d) =
{\mathscr S}({\mathbb R}^d),
$$
where $w_s(x) = (1+|x|)^s$. However, when considering all the weights
satisfying the GRS-condition, the intersection is no longer a Fr\'echet
space but an inductive limit of Banach spaces.

\par

\begin{prop}
Let $p,q\in (0,\infty ]$, and let $\Omega$ be the set of all submultiplicative
weights on $\rr {2d}$ which satisty the GRS-condition.. Then

\begin{equation}\label{capcupidentities}
\bigcap_{v\in \Omega} M^{p,q}_{(v)}(\rr d) = \mathcal S_1(\rr d)
\quad \text{and}\quad
\bigcup_{v\in \Omega} M^{p,q}_{(1/v)}(\rr d) = \mathcal S_1'(\rr d).
\end{equation}

\end{prop}

\par

For the proof we note that if $\omega$ is a moderate weight on
$\rr {2d}$, $p\in (0,\infty ]$ and $f\in \maclS '_1(\rr d)$, then $f\in
M^{p,p}_{(\omega )}(\rr d)$, if and only if

\begin{equation}\label{CubeCoeff}
\left \{   \left ( \iint_{n+Q} |V_\phi f(x,\xi )|^p\, dxd\xi
\right )^{1/p} \right \} _{n\in \zz {2d}} \in
\ell ^{p}_{(\omega )}(\zz {2d}).
\end{equation}

Here $Q$ is the cube $[0,1]^{2d}$.

\par

\begin{proof}
By well-known embedding properties for modulation spaces, it
suffices to prove the result in the case $p=q=1$ or $p=q=\infty$
(cf. \cite{gr-book}). Let $\phi \in
{\mathcal S}_{1/2}({\mathbb  R}^d)\setminus \left\{0\right\}$ be fixed
and let $Q:=[0,1]^{2d}$ be the unit  cube. For any
submultiplicative weight $v$ on $\rr {2d}$ and $f\in
{\mathcal S}^\prime _1({\mathbb R}^d)$ it turns out that
$f\in M^1_{(v)}({\mathbb R}^d)$, if and only if \eqref{CubeCoeff}
holds for $p=1$. Consequently, the first identity in \eqref{ellAidentities}

gives
$$
\bigcap_{v\in \Omega} M^{1}_{(v)}(\rr d) = \bigcup_{\epsilon > 0} M^{1}
_{(v_{\epsilon})}(\rr d) = \mathcal S_1(\rr d),
$$
where the last identity follows from Theorem 3.9 in
\cite{toft}. The second statement
of the proposition follows from the fact that the dual of $M^1_{(v)}$
is given by $M^\infty _{(1/v)}$ (see e.{\,}g.  \cite{gr-book}).
\end{proof}

\vskip 0.5cm
\textbf{Acknowledgement.} The research of the first two authors was partially
supported by MEC and FEDER Project MTM2010-15200 and GVA
Prometeo II/2013/013.


TITLE: Circular motion of proton in special relativity
QUESTION [0 upvotes]: Consider a proton in uniform circular motion in a plane perpendicular to a magnetic field. The Lorentz force acts as the required centripetal force.
At a low speed, we have $\frac{m\cdot v^2}{r}=B\cdot q\cdot v\to$
$$v=\frac{B\cdot q\cdot r}{m}\qquad(1)$$
I wonder how this changes at high speeds when the rest energy is not yet negligible compared to the kinetic energy.
It is tempting to add a factor of gamma to the expression for the centripetal force and leave the expresssion for the Lorentz force unchanged:
$$v=\frac{B\cdot q\cdot r}{m_0}\cdot\sqrt{1-\frac{v^2}{c^2}}$$
$$v=\frac{1}{\sqrt{(\frac{1}{c})^2+(\frac{m_0}{B\cdot q\cdot r})^2}}\qquad(2)$$
Is this correct for a proton moving at a speed of the order of $c/2$, when the rest energy is not yet negligible compared to the kinetic energy?

REPLY [1 votes]: This is precisely what is seen in a relativistic particle accelerator. https://en.wikipedia.org/wiki/Cyclotron#Relativistic_considerations


TITLE: Compact Connected Set in $R^2$ and Connected Component
QUESTION [0 upvotes]: Let X = compact connected set in $R^2$. Let $X^c$ be its complement.
Am I right to say this:
The number of connected components of $X^c$ roughly refers to the number of "holes" in $X$. So, I can make $X^c$ contain any arbitrary number of connected components by starting with a closed and bounded set, then deleting the same number of open sets from its interior and defining the resulting set as $X$?

REPLY [0 votes]: The answer is yes. The explanation is in the comments.


text:
\begin{document}

\begin{abstract}
A set $A$ of nonnegative integers is an asymptotic basis of order $h$ if every sufficiently
large integer can be represented as the sum of $h$ not necessarily distinct elements of $A$.
An asymptotic basis $A$ is minimal if
removing any element of $A$ destroys every representation of infinitely many integers.
In this paper,  a new class of minimal asymptotic bases is constructed.
\end{abstract}

\maketitle

\section{\mcg-adic asymptotic bases}

Let $\N_0 = \{0,1,2,3,\ldots\}$ be the set of nonnegative integers and
let $h$ be a positive integer.
Let $A, A_1,\ldots, A_h$ be subsets of $\N_0$.
We define the \emph{sumset}
\[
A_1 + \cdots + A_h = \{a_1 + \cdots + a_h: a_i \in A_i \text{ for all } i  = 1,2,\ldots, h \}
\]
and the \emph{$h$-fold sumset}
\[
hA = \underbrace{A + \cdots + A}_{\text{$h$ summands}} = \{a_1 + \cdots + a_h:
a_i \in A \text{ for all } i =1,2,\ldots, h\}.
\]
The set $A$ is a \emph{basis of order $h$} if every nonnegative integer can be represented
as the sum of $h$ not necessarily distinct elements of $A$, that is, if $hA = \N_0$.
The set $A$ is an \emph{asymptotic basis of order $h$} if $hA$ contains all sufficiently large integers.
An asymptotic basis of order $h$ is \emph{minimal} if no proper subset of $A$
is an asymptotic basis of order $h$.
Thus, if $A$ is a minimal asymptotic basis of order $h$, then,
for all $a \in A$, there are infinitely many integers $n$ such that removing $a$ from $A$ destroys
every representation of $n$ as a sum of $h$ elements of $A$.

Minimal asymptotic bases are extremal objects in additive number theory, and are related to
the conjecture of Erd\H os and Tur\' an~\cite{erdo-tura41} that the representation function
of an asymptotic basis of order $h$ must be unbounded.

At this time there are few explicit constructions of minimal asymptotic bases.
Nathanson~\cite{nath1974-11,nath1988-66} used a 2-adic construction to produce
the first examples of minimal asymptotic bases.
This method was extended to $g$-adically defined sets by
Chen~\cite{chen12},
Chen and Chen~\cite{chen-chen11}, Chen and Tang~\cite{chen-tang18},
Jia~\cite{jia96}, Jia and Nathanson~\cite{nath1989-68},
Lee~\cite{lee93}, Li and Li~\cite{li-li16},
Ling and Tang~\cite{ling-tang15,ling-tang18},
Sun~\cite{sun19,sun21}, and
Sun and Tao~\cite{sun-tao19}.
This paper constructs a new class of minimal asymptotic bases.

An \emph{interval of integers of length $t$} is a set of $t$ consecutive integers.
For $u,v \in \N_0$ with $u \leq v$, the set
\[
[u,v] = \{x\in \N_0: u \leq x \leq v\}
\]
is an interval of integers of length $v - u + 1$.
Thus,  $[1,h] = \{1,2,\ldots, h\}$.

A \emph{\mcg-adic sequence} is a strictly increasing sequence of positive integers
$\mcg = (g_i)_{i=0}^{\infty}$
such that $g_ 0 = 1$ and $g_{i-1}$ divides $g_i$ for all $ i \geq 1$.
Let $(d_i)_{i=1}^{\infty}$ be the sequence  of  positive integers defined by
\[
d_i = \frac{g_i}{g_{i-1}}.
\]
For all $i \geq 1$, we have
\beq                                         \label{MinBasis:g1}
d_i \geq 2
\eeq
and
\beq                                         \label{MinBasis:g2}
g_i = d_1 d_2 \cdots d_i.
\eeq
For all $i$ and $j$ with $0 \leq i <  j$, we have
\beq                                        \label{MinBasis:g3}
\frac{g_{i+j}}{g_i} = d_{i+1}  d_{i+2} \cdots d_{i+j}.
\eeq
Every positive integer $n$ has a unique \textit{$\mcg$-adic representation}
\[
n = \sum_{j=0}^{\infty} x_j g_j
\]
where
\[
x_j \in [0, d_{j+1}-1]
\]
 for all $j \in \N_0$ and $x_j = 0$ for all sufficiently large $j$
(Nathanson~\cite{nath2014-150,nath2017-172}).
Equivalently, for every positive integer $n$, there is a unique nonempty finite set $F \subseteq \N_0$
and a unique set $\{x_j : j  \in F \}$ such that
\beq                             \label{MinBasis:g-adic}
n = \sum_{j \in F} x_j g_j
\eeq
where $x_j \in  [1, d_{j+1}-1]$ for all $j \in F$.

For every integer $g \geq 2$, the usual $g$-adic representation uses the
$\mcg$-adic sequence $\mcg = \left(g^i\right)_{i=0}^{\infty}$ with
quotients $d_i = g$ for all $i \geq 1$.

\bl                                             \label{MinBasis:lemma:Gbound}
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic system.
If $n =  \sum_{j \in F} x_j g_j $ is a positive integer
with $x_j \in  [1, d_{j+1}-1]$ for all $j \in F$,
then
\[
g_M \leq n < g_{M+1}
\]
if and only if
\[
\max(F) = M.
\]
\el

\begin{proof}
If $\max(F) = M$, then $F \subseteq [0,M]$ and
\begin{align*}
g_M & \leq x_M g_M \leq n \leq  \sum_{j \in F} (d_{j+1}-1) g_j \\
&  \leq  \sum_{j =0}^M (d_{j+1}-1) g_j =  \sum_{j =0}^M (g_{j+1}- g_j) \\
& = g_{M+1}-1 < g_{M+1}.
\end{align*}

Conversely, if $g_M \leq n < g_{M+1}$ and $\max(F) = M'$,
then the inequality
\[
g_{M'} \leq x_{M'} g_{M'} \leq n < g_{M+1}
\]
implies $M' \leq M$.  If $M' \leq M-1$, then
\[
n \leq \sum_{j =0}^{M'} (d_{j+1}-1) g_j < g_{M'+1} \leq g_M
\]
which is absurd.  Therefore, $g_M \leq n < g_{M+1}$ implies
$\max(F) = M$.
This completes the proof.
\end{proof}

Let $W$ be a nonempty set of nonnegative integers,
and let $\mcf^*(W)$ be the set of all nonempty finite subsets of $W$.
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence.
We define the set of positive integers
\[
A_{\mcg}(W) = \left\{ \sum_{j\in F} x_j g_j : F \in \mcf^*(W) \text{ and }  x_j \in [1,d_{j+1} -1] \right\}.
\]
Note that $0 \notin A_{\mcg}(W)$ because $\emptyset \notin  \mcf^*(W)$.

Let $h \geq 2$.
A \emph{partition} of $\N_0$ is a sequence $\mcw = (W_i)_{i=0}^{h-1}$
of nonempty pairwise disjoint sets such that
\[
\N_0 = W_0 \cup W_1 \cup \cdots \cup W_{h-1}.
\]

\bt                  \label{MinBasis:theorem:basis}
Let $h \geq 2$ and let $\mcw = (W_i)_{i=0}^{h-1}$ be a partition of $\N_0$.
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence.
The set
\[
A_{\mcg}(\mcw) = \bigcup_{i=0}^{h-1} A_{\mcg}(W_i)
\]
is an asymptotic basis of order $h$ with $h$-fold sumset
\[
hA_{\mcg}(\mcw) = \{n \in \N_0: n \geq h \}.
\]
\et

\begin{proof}
The smallest integer in the set  $A_{\mcg}(\mcw)$ is $1 = 1 \cdot g_0$.
It follows that $h \in hA_{\mcg}(\mcw)$ but
$[0,h-1] \cap hA_{\mcg}(\mcw) = \emptyset$.

Every positive integer $n$ has a unique \mcg-adic representation $n = \sum_{j \in F} x_j g_j$,
where $F$ is a nonempty finite set of nonnegative integers and $x_j \in [0, d_{j+1}-1] $.
For $i \in [0,h-1]$, let
\[
F_i = F  \cap W_i
\]
and
\[
n_i = \sum_{j\in F_i} x_j g_j.
\]
If   $F_i = \emptyset$, then $n_i = 0 \notin  A_{\mcg}(\mcw)$.  If $F_i \neq \emptyset$, then
\[
n_i = \sum_{j\in F_i} x_j g_j \in A_{\mcg}(W_i) \subseteq A_{\mcg}(\mcw).
\]
Let
\[
L = \left\{ i\in [0,h-1] :  F_i \neq \emptyset   \right\} = \left\{ i\in [0,h-1] :  n_i \geq 1  \right\}
\]
and
\[
|L| = \ell_0.
\]
We have  $\ell_0 \in [1,h]$ and
\[
n = \sum_{i\in L} n_i \in \ell_0  A_{\mcg}(\mcw).
\]
Let $\ell$ be the largest integer such that  $\ell \leq h$ and
$n \in \ell  A_{\mcg}(\mcw)$.   We must prove that $\ell = h$.

If $n \in \ell  A_{\mcg}(\mcw)$, then there exist integers
$n_1,\ldots, n_{\ell} \in A_{\mcg}(\mcw)$
such that
\[
n = n_1 + \cdots  + n_{k-1} +  n_k +  n_{k+1} + \cdots + n_{\ell}.
\]
For each $i \in [1,\ell]$ there is an integer $s_i \in [0,h-1]$ and  a set $F_{s_i} \in \mcf^*(W_{s_i})$
such that $n_i$ has the \mcg-adic representation
\[
n_i = \sum_{j \in F_{s_i}} x_j g_j \in A_{\mcg}(W_{s_i}).
\]

Suppose that $\ell < h$. If $|F_{s_k}| \geq 2$ for some $k \in [1, \ell]$,
then there are nonempty sets $F'_{s_i}$ and $F''_{s_i}$ such that
\[
F_{s_k} = F'_{s_k} \cup F''_{s_k}  \qqand F'_{s_k} \cap F''_{s_k} = \emptyset.
\]
The  integers
\[
n'_k = \sum_{j \in F'_{s_k}} x_j g_j  \in A_{\mcg}(W_{s_k})
 \qqand n''_k = \sum_{j \in F''_{s_k}} x_j g_j  \in A_{\mcg}(W_{s_k})
\]
satisfy
\[
n_k = n'_k + n''_k
\]
and so
\[
n = n_1 + \cdots n_{k-1} +  n'_k + n''_k +  n_{k+1} + \cdots + n_{\ell} \in (\ell+1) A_{\mcg}(\mcw).
\]
This contradicts the maximality of $\ell$, and so $|F_{s_i}| = 1$ for all $i \in [1,\ell]$ and
\[
n_i = x_{j_i} g_{j_i}
\]
for some $g_{j_i} \in W_{s_i}$ and $x_{j_i} \in [0, d_{j_i+1}-1]$.

If $x_{j_k} \geq 2$ for some $k \in [1,\ell]$, then
\[
g_{j_k} \in  A_{\mcg}(W_{s_k}) \qqand  (x_{j_k} -1) g_{j_k} \in  A_{\mcg}(W_{s_k})
\]
and
\[
n_k = g_{j_k} + (x_{j_k} -1) g_{j_k}.
\]
It follows that
\[
n = n_1 + \cdots n_{k-1} +  g_{j_k} + (x_{j_k} -1) g_{j_k} +  n_{k+1} + \cdots + n_{\ell} \in (\ell+1) A_{\mcg}(\mcw),
\]
which again contradicts the maximality of $\ell$.  Therefore, $x_{j_i} = 1$ for all $i \in [1,\ell]$, and
\[
n = g_{j_1}  + \cdots + g_{j_k}   + \cdots + g_{j_{\ell}}.
\]

If $j_k \geq 1$ for some $k \in [1,\ell]$, then
\[
g_{j_k} = g_{j_k-1} + (d_{j_k}-1) g_{j_k-1}
\]
and
\[
n = g_{j_1}  + \cdots + g_{j_k-1} + (d_{j_k}-1) g_{j_k-1} + \cdots + g_{j_{\ell}}
\in (\ell+1) A_{\mcg}(\mcw),
\]
which also contradicts the maximality of $\ell$.
Therefore, $j_k = 0$  and $g_{j_k} = g_0 = 1$ for all $k \in [1,\ell]$, and so
\[
n = \underbrace{g_0  + \cdots + g_0}_{\text{$\ell$ summands}}
= \underbrace{1+ \cdots + 1 + \cdots + 1}_{\text{$\ell$ summands}}
= \ell \leq h-1,
\]
which is absurd.
This completes the proof.
\end{proof}

\bt                         \label{MinBasis:theorem:Abasis}
Let $h \geq 2$ and let $\mcw = (W_i)_{i=0}^{h-1}$ be a partition of $\N_0$.
The set
\[
A = \{0\} \cup A_{\mcg}(\mcw)
\]
is a basis of order $h$ but not a minimal asymptotic basis of order $h$.
\et

\begin{proof}
Because $0 \in A$ and $1 \in A_{\mcg}(\mcw) \subseteq A$, we have
\[
\ell = (h-\ell)\cdot 0 + \ell \cdot 1 \in hA
\]
 for all $\ell \in [0,h-1]$.
By Theorem~\ref{MinBasis:theorem:basis},
\[
 \{n \in \N_0: n \geq h \} = hA_{\mcg}(\mcw) \subseteq hA
 \]
 and so $A$ is a basis of order $h$.

The set $A$ is not a minimal asymptotic basis of order $h$
because $0 \in A$ and the removal of 0 from $A$ gives the set $A \setminus \{0 \} = A_{\mcg}(\mcw)$,
which is still an asymptotic basis of order $h$.
This completes the proof.
\end{proof}

\section{Minimal asymptotic bases}

The following lemma generalizes a result of Jia~\cite{jia96}.

\bl                   \label{MinBasis:lemma:ineqM}
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence.
Let $(u_i)_{i=1}^p$ be a strictly increasing finite sequence of nonnegative integers,
and let $(v_j)_{j=1}^q$ be a finite sequence of not necessarily distinct nonnegative integers.
Let
\[
x_i \in [1, d_{u_i+1}-1]
\]
 for all $i \in [1,p]$ and
 \[
 y_j \in [1, d_{v_j+1}-1]
 \]
 for  all $j \in [1,q]$.
If
\beq                                 \label{MinBasis:ineqM-n}
n = \sum_{i=1}^p x_i g_{u_i} = \sum_{j=1}^q y_j g_{v_j}
\eeq
then
\beq                                 \label{MinBasis:ineqM}
\sum_{u_i\leq u_k} x_i g_{u_i}   \leq  \sum_{v_j \leq u_k} y_j g_{v_j}
\eeq
for all $k \in [1, p]$.
\el

\begin{proof}
Because the sequence  $(u_i)_{i=1}^p$is  strictly increasing,
Lemma~\ref{MinBasis:lemma:Gbound}  implies
\begin{align*}
\sum_{u_i\leq u_k} x_i g_{u_i}  =  \sum_{i=1}^k x_i g_{u_i}  < g_{u_{k+1}}
\end{align*}
for all $k \in [1,p]$.
Choosing $k = p$ gives
\[
n  = \sum_{i=1}^p x_i g_{u_i} < g_{u_{p+1}}.
\]
Relation~\eqref{MinBasis:ineqM-n} implies
\[
g_{v_j} \leq n < g_{u_p + 1}
\]
an so  $v_j \leq u_p$ for all $j \in [1,q]$.
This implies~\eqref{MinBasis:ineqM} for $k=p$.

In the sequence $\mcg = (g_i)_{i=0}^{\infty}$, the integer  $g_i$ divides $g_j$
for all $i \leq j$.
Let $k \in [1,p-1]$.
Because the  sequence $(u_i)_{i=1}^p$ is strictly increasing, for all $i \in [k+1,p]$ we have
\[
u_k < u_k+1 \leq u_{k+1} \leq u_i
\]
and $g_{u_k+1}$ divides $g_{u_i}$, that is,
\[
g_{u_i} \equiv 0 \pmod{g_{u_k+1}}.
\]
If $v_j \geq u_k +1$, then
\[
g_{v_j} \equiv 0 \pmod{g_{u_k+1}}.
\]
Rearranging~\eqref{MinBasis:ineqM-n}, we obtain
\begin{align}                                   \label{MinBasis:ineqM-cong}
\sum_{u_i\leq u_k} x_i g_{u_i} -  \sum_{v_j \leq u_k} y_j g_{v_j}
&  =   \sum_{v_j \geq u_k +1} y_j g_{v_j} - \sum_{u_i > u_k} x_i g_{u_i}   \\
&  \equiv 0 \pmod{g_{u_k+1}}.    \nonumber
 \end{align}
If
\[
\sum_{u_i\leq u_k} x_i g_{u_i}  >  \sum_{v_j \leq u_k} y_j g_{v_j}
\]
then Lemma~\ref{MinBasis:lemma:Gbound} gives
\[
0 <   \sum_{i=1}^k x_i g_{u_i}  - \sum_{v_j \leq u_k} y_j g_{v_j}
\leq  \sum_{i=1}^k x_i g_{u_i} < g_{u_{k+1}}.
\]
This inequality contradicts congruence~\eqref{MinBasis:ineqM-cong}.
This completes the proof.
\end{proof}

\bt                              \label{MinBasis:theorem:NewMinBasis}
Let $h \geq 2$ and let  $t$ be an integer such that
\beq             \label{MinBase:t}
t \geq 1 +  \frac{\log h}{\log 2}.
\eeq
Let $\mcw = (W_i)_{i=0}^{h-1}$ be a partition of $\N_0$  such that,
for all $i \in [0 ,h-1]$, there is an infinite set $\mcm_i$ of positive integers  such that
\beq             \label{MinBase:t-interval}
[M_i -t+ 1 , M_i] \subseteq W_i
\eeq
for all $M_i \in \mcm_i$.
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence.
The set
\[
A_{\mcg}(\mcw) = \bigcup_{i=0}^{h-1} A_{\mcg}(W_i)
\]
is a minimal asymptotic basis of order $h$.
\et

\begin{proof}
By Theorem~\ref{MinBasis:theorem:basis}, the set $ A_{\mcg}(\mcw) $
is an asymptotic basis of order $h$.

Let $a \in A_{\mcg}(\mcw)$.
Without loss of generality, we can assume that $a = a_0 \in A_{\mcg}(W_0)$
and
\beq             \label{MinBase:a0}
a_0 = \sum_{j \in F_0} x_{0,j} g_j
\eeq
where $F_0 \in \mcf^*(W_0)$ and $x_{0,j} \in [1, d_{j+1}-1]$ for all $j \in F_0$.
Let $M_0 = \max(F_0)$.  By Lemma~\ref{MinBasis:lemma:Gbound},
\[
g_{M_0} \leq a_0 < g_{M_0+1}.
\]

For all $i \in [1,h-1]$, choose an integer $M_i$ in the infinite set $\mcm_i$
such that
\beq             \label{MinBase:MiMot}
M_i \geq M_0 + t
\eeq
and let
\beq             \label{MinBase:ai}
a_i = \sum_{\substack{j \in W_i \\ j < M_0 }} (d_{j+1} - 1) g_j + g_{M_i} \in A_{\mcg}(W_i).
\eeq
This is the \mcg-adic representation of $a_i$.
Let
\beq             \label{MinBase:nn}
n = \sum_{i=0}^{h-1} a_i
= a_0 + \sum_{i=1}^{h-1} \sum_{\substack{j \in W_i \\ j < M_0 }} (d_{j+1} - 1) g_j
+ \sum_{i=1}^{h-1} g_{M_i}.
\eeq
This is the \mcg-adic representation of $n$.

Let
\[
n = \sum_{i=0}^{h-1} b_{k_i}
\]
be any representation of $n$ as the sum of $h$ elements of $A_{\mcg}(\mcw)$,
where $k_i \in  [0,h-1]$ for all $i\in  [0,h-1]$
and $b_{k_i} \in A_{\mcg}(W_{k_i})$.
We must prove that $b_{k_i} = a_0$ for some $i \in [0,h-1]$.

Each integer $b_{k_i}$  is of the form
\[
b_{k_i} = \sum_{j \in E_{k_i}} y_{i,j} g_j \in  A_{\mcg}(W_{k_i})
\]
where $E_{k_i} \in \mcf^*(W_{k_i})$
and $y_{i,j} \in [1,g_{j+1} - 1]$ for all $j \in E_{k_i}$.
The uniqueness of the \mcg-adic representation implies that if $\{k_0,k_1,\ldots, k_{h-1}\} = [0,h-1]$,
then, after rearrangement, $k_i = i$ and $a_i = b_i$ for all $i \in [0,h-1]$.

If $\{k_0, k_1,\ldots, k_{h-1}\} \neq [0,h-1]$, then there exists $s \in [0,h-1]$
such that $s \notin \{k_0,k_1,\ldots, k_{h-1}\}$.
Suppose that $s \neq 0$.
Recall that
\[
M_s \geq M_0 + t
\]
and
\[
[M_s -t+1,M_s] \subseteq W_s.
\]
 Because $k_i \neq s$ for all $i \in [0,h-1]$, we have
 \[
[M_s - t + 1, M_s] \cap E_{k_i}

\subseteq    W_s\cap W_{k_i}  = \emptyset.
\]
We construct the partition
\[
E_{k_i} = E'_{k_i} \cup E''_{k_i}
\]
with
\[
E_{k_i}' = \{j \in E_{k_i}: j \leq M_s - t\}
\]
and
\[
E_{k_i}'' = \{j \in E_{k_i}: j \geq M_s+1 \}.
\]
The sets $E_{k_i}' $ and $E_{k_i}''$ are not necessarily nonempty.
Let
\[
b_{k_i} = b'_{k_i} + b''_{k_i}
\]
where
\[
b'_{k_i} = \sum_{j \in E'_{k_i}} y_{i,j} g_j   \qqand   b''_{k_i} = \sum_{j \in E''_{k_i}} y_{i,j} g_j.
\]
Note that $b'_{k_i} = 0$ if  $E'_{k_i} = \emptyset$ and $b''_{k_i} = 0$ if  $E''_{k_i} = \emptyset$.

By Lemma~\ref{MinBasis:lemma:Gbound},
\beq                              \label{MinBase:g5}
b'_{k_i} < g_{M_s -t +1}
\eeq
and
\[
b''_{k_i} = 0  \qquad\text{or}\qquad b''_{k_i} \geq g_{M_s+1}.
\]
Let
\beq                             \label{MinBase:g6}
n' = \sum_{i=0}^{h-1} b'_{k_i}  \qqand n'' = \sum_{i=0}^{h-1} b''_{k_i}.
\eeq
Recall inequalities~\eqref{MinBasis:g1}  and~\eqref{MinBase:t}:
\[
d_i \geq 2 \qqand h \leq 2^{t-1}.
\]
From~\eqref{MinBase:g6},~\eqref{MinBase:g5}, and~\eqref{MinBasis:g3},  we obtain
\[
n' < hg_{M_s -t+1} \leq 2^{t-1} g_{M_s -t+1} \leq g_{M_s -t+1} \prod_{i=1}^{t-1}d_{M_s -t+i} =g_{M_s}.
\]
Therefore, the \mcg-adic representation of $n'$ is of the form
\[
n' = \sum_{j =0}^{M_s -1} z_ j g_j
\]
with $z_j \in [0,d_{j+1}-1]$.  Because
\[
n'' = 0 \qquad\text{or}\qquad  n'' \geq g_{M_s+1},
\]
the \mcg-adic representation of $n''$ is of the form
\[
n'' = \sum_{j= M_s+1}^{\infty}  z_j g_j
\]
with $z_j \in [0,d_{j+1}-1]$ and $z_j \geq 1$ for only finitely many $j$.
Therefore,
\[
n = n' + n'' = \sum_{i=0}^{ M_s -1 } z_i g_i + \sum_{i= M_s+1}^{\infty}  z_i g_i
\]
is the \mcg-adic representation of $n$.
In this representation,  the coefficient of $g_{M_s}$ is 0, which contradicts the construction of $n$.
It follows that
\[
[1,h-1] \subseteq \{k_0,k_1,\ldots, k_{h-1}\}.
\]
Renumbering the integers $b_i$, we can assume that $k_i = i$
and $b_i \in A_{\mcg}(W_i)$ for all $i \in [1,h-1]$.

We must prove that $k_0 = 0$, or, equivalently, that $b_0 \in A_{\mcg}(W_0)$.
If not, then $b_0 \in A_{\mcg}(W_r)$ for some $r \in [1,h-1]$.
Because
\[
M_0 = \max(F_0)  \in F_0 \subseteq W_0
\]
we have
\[
M_0 \notin \bigcup_{i=0}^{h-1} E_{k_i} \subseteq \bigcup_{i=1}^{h-1} W_i.
\]
Identity~\eqref{MinBase:a0} and Lemma~\ref{MinBasis:lemma:Gbound} give
\[
g_{M_0} \leq a_0 = \sum_{j \in F_0} x_{0,j} g_j < g_{M_0 +1}.
\]
From~\eqref{MinBase:ai} we have
\begin{align*}
n & =
 a_0 + \sum_{i=1}^{h-1} \sum_{\substack{j \in W_i \\ j < M_0 }} (d_{j+1} - 1) g_j
+ \sum_{i=1}^{h-1} g_{M_i} \\
& = \sum_{i=0}^{h-1} \sum_{ j \in E_i } y_{i,j} g_j.
\end{align*}
Summing only over terms $g_j$ with $j \leq M_0$ and
applying Lemma~\ref{MinBasis:lemma:ineqM}, we obtain
\begin{align*}
a_0 +  \sum_{i=1}^{h-1} \sum_{ \substack{j \in W_i \\ j < M_0 } } (d_{j+1}-1)  g_j
& \leq \sum_{i=0}^{h-1} \sum_{\substack{j \in E_i \\ j \leq M_0}} y_{i,j} g_j
=  \sum_{i=0}^{h-1} \sum_{\substack{j \in E_i \\ j < M_0}} y_{i,j} g_j \\
& =  \sum_{\substack{j \in E_0 \\ j < M_0}} y_{0,j} g_j
 + \sum_{i=1}^{h-1} \sum_{\substack{j \in E_i \\ j < M_0}} y_{i,j} g_j \\
& < g_{M_0} +  \sum_{i=1}^{h-1}  \sum_{ \substack{j \in W_i \\ j < M_0 } } (d_{j+1}-1)  g_j \\
& \leq a_0  +  \sum_{i=1}^{h-1}  \sum_{ \substack{j \in W_i \\ j < M_0 } } (d_{j+1}-1)  g_j.
\end{align*}
This is absurd, and so $b_0 \in A_{\mcg}(W_0)$.
It follows that the integer $n$ defined by~\eqref{MinBase:nn} has a unique representation
as the sum of $h$ elements of $A_{\mcg}(\mcw)$.  Therefore,
\beq             \label{MinBase:NotIn}
n \notin h \left(A_{\mcg}(\mcw) \setminus \{a\} \right).
\eeq

For all $i \in [1,h-1]$, there are infinitely many integers $M_i \in \mcm_i$
with $M_i \geq M_0 + t$,
and so infinitely many positive integers $n$ satisfying~\eqref{MinBase:NotIn}.
It follows that $A_{\mcg}(\mcw) \setminus \{a\}$ is not an asymptotic basis of order $h$
for all $a \in A_{\mcg}(\mcw)$.
Equivalently, $A_{\mcg}(\mcw)$ is a minimal asymptotic basis of order $h$.
This completes the proof.
\end{proof}

\bc
Let $\mcw = W_0 \cup W_1$ be a partition of $\N_0$  such that both $W_0$ and $W_1$ contain
infinitely many pairs of consecutive integers.
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence.
The set
\[
A_{\mcg}(\mcw) =  A_{\mcg}(W_1) \cup A_{\mcg}(W_2)
\]
is a minimal asymptotic basis of order $2$.
\ec

\begin{proof}
This is the case $h=2$ of Theorem~\ref{MinBasis:theorem:NewMinBasis}.
\end{proof}

\section{Open problems}
\begin{enumerate}
\item               \label{MinBasis:problem-a}
Let $h \geq 2$.  By Theorem~\ref{MinBasis:theorem:basis},
for every partition $\mcw = (W_i)_{i=0}^{h-1}$  of $\N_0$ and every \mcg-adic sequence, the set
\[
A = \{0\} \cup A_{\mcg}(\mcw)
\]
is a basis of order $h$ that is not a minimal asymptotic basis of order $h$.
\begin{enumerate}
\item
Determine the set of all integers $a\in A$ such that $A \setminus \{a\}$ is an asymptotic basis of order $h$.
\item
Determine the  set of all integers $a \in A$ such that $A\setminus \{a\}$ is a minimal asymptotic basis.
\end{enumerate}

\item                  \label{MinBasis:problem-b}
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence. Let $h \geq 2$.
\begin{enumerate}
\item
Construct partitions $\mcw = (W_i)_{i=0}^{h-1}$ of $\N_0$
such that the set $A_{\mcg}(\mcw)$ is a minimal asymptotic basis of order $h$.
\item
Construct partitions $\mcw = (W_i)_{i=0}^{h-1}$ of $\N_0$
such that the set $A_{\mcg}(\mcw)$ is not a minimal asymptotic basis of order $h$.
\end{enumerate}

\item                     \label{MinBasis:problem-c}
Let $(d_i)_{i=1}^{\infty}$ be a sequence of 2s and 3s.
Let  $\mcg = (g_i)_{i=0}^{\infty}$ be the \mcg-adic sequence defined by $g_0 =1$
and $g_i = \prod_{j=1}^i  d_i$ for $i \geq 1$.
Consider problem~\eqref{MinBasis:problem-b} with respect to this \mcg-adic sequence.
Of particular interest are the infinitely many \mcg-adic sequences $\mcg = (g_i)_{i=0}^{\infty}$
with quotients $\{ d_{2i-1}, d_{2i} \} = \{ 2,3 \}$ for all $i = 1,2,3,\ldots$.  In this case,
$g_{2i} = 6^i $ for all $i$.

\end{enumerate}

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config: arxiv


TITLE: Show that the Collatz Conjecture would be provable if we could solve the halting problem
QUESTION [3 upvotes]: This is problem 5.31 from Introduction to the Theory of Computation, 3rd Edition by Michael Sipser.

Let
  $$
f(x) =
\begin{cases}
3x+1, & \text{for odd $x$} \\
x/2,  & \text{for even $x$}
\end{cases}
$$
  for any natural number $x$. If you start with an integer $x$ and iterate $f$, you obtain a sequence, $x$, $f(x)$, $f(f(x))$, $\dots$ . Stop if you ever hit 1.
  For example, if $x = 17$, you get the sequence 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Extensive computer tests have shown that every starting point between 1 and a large positive integer gives a sequence that ends in 1. But the question of whether all positive starting points end up at 1 is unsolved; it is called the $3x + 1$ problem.
Suppose that $A_{TM}$ = $\{\langle M,w\rangle$∣ $M$ is a Turing Machine and $M$ accepts $w\}$ (Acceptance Turing machine) were decidable by a Turing Machine $H$. Use $H$ to describe a Turing Machine that is guaranteed to state the answer to the $3x + 1$ problem.

REPLY [5 votes]: I suppose the question is: if we can solve the halting problem, can we figure out the answer to the $3x+1$ (Collatz) problem?
Well, if we have a program $H$ that solves the halting problem (that is, it can decide whther some Turing machine $M$ with input $I$ halts or not), then do the following:
Create a Turing machine $F$ that takes a number $x$ and iterates through the $f(x)$ sequence, and stops once it reaches $1$. It is easy to show such a Turing machine $F$ exists .. here is its pseudocode:
F (input $i$)
$Begin$
$\quad$ While $i \not = 1$
$\quad \quad $ Case $i$ is even: $i = i/2$
$\quad \quad $ Case $i$ is odd: $i = 3*i+1$
$End$
(So this F routine will halt whenever the sequence ends with a 1 ... otherwise it will go on forever)
Now create a new Turing machine $C$ that starts with a given $i$, and that calls halting program $H$ on $F$ and $i$. If $H$ says that $F$ does not halt on $i$, then stop. If $H$ says that $F$ does halt $i$, then increase $i$ by $1$, and repeat the process. Again, it is easy to show that such a Turing machine $C$ exists, assuming $H$ exists .. here is its pseudocode:
C (input $i$)
$Begin$
$\quad$ While $H(F,i)$
$\quad \quad$ $i=i+1$
$End$
(so this routine $C$ will only stop if for some $i$, $H$ finds that $F$ does not halt on $i$. Effectively, $C$ looks for a counterexample to the Collatz conjecture. If there is one, then $C$ will eventually run into it. If there isn't, then $C$ will run forever)
Finally, call $H$ on $C$ and $1$. If $H$ says that $C$ with $1$ will not halt, then apparently the solution to the $3x+1$ is that the sequence will always end up with $1$ for any $x$ (that is, there is no counterexample to the Collatz conjecture ... so the Collatz Conjecture is true). If $H$ says that $C$ with $1$ does halt, then apparently the solution is that for some $x$ the sequence does not end up with $1$ (i.e. there is a counterexample to the Collatz conjecture). So, either way, you have solved the $3x+1$ Collatz problem.


TITLE: Identifying quotient rings
QUESTION [3 upvotes]: The question I have is from a past exam paper:
Identity if it is an integral domain, whether it admits a finite basis over a coefficient ring, or whether it is isomorphic to another ring.
The first one is
$$R = \mathbb{Z}[x]/(6x^2 -1, 2x-4)$$
My first angle of attack is to find if we can simplify the ideal $I=(6x^2-1,2x-4)$ in hopes of finding a monic irreducible polynomial by taking combinations of elements of the ideal to end up with $I = (f)$ to apply a substitution isomorphism, maybe this is too hopeful.
We start out by making a few computation on elements of the ring
\begin{align*}
6x^2-1 &= 0\\
2x*3x-1&=12x-1 = 0\\
\end{align*}
So we can use $12x-1=0$ and $2x-4$ to perhaps find if this ring is isomorphic to some modular ring.
$$6(2x-4) = 12x - 24 = 1-24 = 0 = 23 \in I$$
and
$$R = \mathbb{Z}[x]/(23,6x^2-1,2x-4) \cong \mathbb{Z}_{23}[x]/(6x^2-1,2x-4)$$
Next we can use this modulo $23$ in the following way to perform further computations:
$$4(6x^2-1) = 24x^2-4=x^2-4=0$$
We then can reduce the ideal to
$$I = (x^2-4,2x-4)$$
If $x^2 = 4$ and $2x=4$ we must have that $x = 2$ and $I = (x-2)$
We then write
$$R \cong \mathbb{Z}_{23}[x] /(x-2) \cong \mathbb{Z}_{23}[2] = \mathbb{Z}_{23}$$
Also, I know that this ring isn't an integral domain because we can take $23,1 \in R$ and get $23*1 = 0$, so therefore $23$ is a zero divisor.
Let me know if my working is correct, I feel a little bit iffy about reducing the ideal, thanks.
EDIT: A finite basis for $R$ would just be $\{a|a \in \mathbb{Z}\}$

REPLY [1 votes]: You're mixing up things and arrive to a wrong conclusion.
Your strategy is good, it just needs some adjustment in wording.
Let $\xi$ denote the image of $x$ in the quotient ring; then, by assumption, $6\xi^2-1=0$ and $2\xi-4=0$. Therefore
$$
1=3\xi\cdot 2\xi=3\xi\cdot 4=6\cdot 2\xi=24
$$
In particular, $24-1=0$ in the quotient ring, which means $23\in I$.
The next step is noting $4\cdot 6\xi^2=4$, so $\xi^2=4$. Together with $2\xi=4$, this implies $\xi^2=2\xi$. Also
$$
1=6\xi^2=6\cdot2\xi=12\xi
$$
and we conclude that $2$ is invertible in the quotient ring. From $2\xi=4$, we conclude $\xi=2$.
So, we have proved that $23\in I$ and $x-2\in I$. Now we can do the final step: proving $I=(23,x-2)$. Clearly, $2x-4\in(23,x-2)$. Also
$$
6x^2-1=(6x+12)(x-2)+23\in(23,x-2)
$$
Thus your ring is
$$
\mathbb{Z}[x]/(23,x-2)\cong(\mathbb{Z}/23\mathbb{Z})[x]/(x-2)
\cong\mathbb{Z}/23\mathbb{Z}
$$
is a field.


TITLE: Domain of a complex function
QUESTION [1 upvotes]: 1) Why do the domain of a complex function has to be a disk (circular neighborhood of Zo)?
$$
|z-z_0|<p
$$
2) Domain is an open connected set. An open set D is said to be connected if every pair of points $z_1$,$z_2$ in S can be joined by a polygonal path that lies entirely in S.
so why are the following sets are not domains?
a) $-1<Im( z )<= 1$
b) $ (Re(z))^2 >1$
Thank you

REPLY [3 votes]: The domain of a complex function does not have to be a disk. It can be any subset of $\mathbb C$.
As for the sets you describe, try identifying these sets in the complex plane and you will see why they do not qualify as domains. The first one is not open and the second one is not connected.