text:
\begin{document}

\title{Aronszajn Compacta
\footnote{
2000 Mathematics Subject Classification:
Primary  54D30, 03E35.
Key Words and Phrases:
Aronszajn tree, hereditarily separable, hereditarily Lindel\"of.
}}

\author{Joan E. Hart\footnote{University of Wisconsin, Oshkosh,
WI 54901, U.S.A.,
\ \ hartj@uwosh.edu}
\  and
Kenneth Kunen\footnote{University of Wisconsin,  Madison, WI  53706, U.S.A.,
\ \ kunen@math.wisc.edu}
\thanks{Both authors partially supported by NSF Grant
DMS-0456653.}
}

\maketitle

\begin{abstract}
We consider a class of compacta $X$ such that the maps from $X$
onto metric compacta define an
Aronszajn tree of closed subsets of $X$.
\end{abstract}

\section{Introduction}
\label{sec-intro}
All topologies discussed in this paper are assumed to be Hausdorff.
We begin by defining an \emph{Aronszajn compactum},
along with a natural tree structure,
by considering a space embedded into a cube.
An equivalent definition, in terms of elementary submodels,
is considered in Section \ref{sec-elem}.

\begin{notation}
\label{not-basic}
Given a product $\prod_{\xi < \lambda} K_\xi$:
If $\alpha \le \beta \le \lambda$, then
$\pi^\beta_\alpha$ denotes the natural projection from
$\prod_{\xi < \beta} K_\xi$ onto
$\prod_{\xi < \alpha} K_\xi$.
If we are studying a space $X \subseteq \prod_{\xi < \lambda} K_\xi$
then $X_\alpha$ denotes $\pi^\lambda_\alpha(X)$,
and $\sigma^\beta_\alpha$ denotes the restricted map
$\pi^\beta_\alpha \res X_\beta$; so
$\sigma^\beta_\alpha: X_\beta \onto X_\alpha$.
\end{notation}

\begin{definition}
\label{def-emb-ar}
An \emph{embedded Aronszajn compactum} is a closed subspace
$X \subseteq [0,1]^{\omega_1}$ with
$w(X) = \aleph_1$ and $\cchi(X) = \aleph_0$ such that for some club
$C \subseteq \omega_1$:
for each $\alpha \in C$
$\LL_\alpha := \{x \in X_\alpha : |(\sigma^{\omega_1}_\alpha)\iv\{x\}| > 1\}$
is countable.  For each such $X$, define
$T=T(X) := \bigcup\{\LL_\alpha : \alpha \in C\}$, and
let $\tleft$ denote the following order:
if $\alpha, \beta \in C$, $\alpha < \beta$, $x \in \LL_\alpha$
and $y \in \LL_\beta$, then $x \tleft y$ iff $x = \pi^\beta_\alpha(y)$.
\end{definition}

The $\sigma^{\omega_1}_\alpha$ for which
$|\LL_\alpha| \le \aleph_0$ are called
\textit{countable rank} maps in \cite{Fed,KT}.
Observe that $\langle T(X),\tleft \rangle $
is a tree.
Each level $\LL_\alpha$ is countable by definition, and is non-empty
because $w(X) = \aleph_1$; then $T$ is Aronszajn because $\cchi(X) = \aleph_0$.
Of course, a compactum of weight $\aleph_1$ may be embedded into
$[0,1]^{\omega_1}$ in many ways, but:

\begin{lemma}
\label{lemma-embed}
If $X,Y \subseteq [0,1]^{\omega_1}$, $X$ is
an embedded Aronszajn compactum, and $Y$ is homeomorphic to $X$,
then $Y$ is an embedded Aronszajn compactum.
\end{lemma}
\begin{proof}
Let $f: X \to Y$ be a homeomorphism.  Then use the fact that there
is a club $D \subseteq \omega_1$ on which $f$ commutes with projection;
that is, for $\alpha \in D$,
there is a homeomorphism $f_\alpha : X_\alpha \onto Y_\alpha$ such that
$\pi^{\omega_1}_\alpha \circ f  = f_\alpha \circ \pi^{\omega_1}_\alpha$.
\end{proof}

The proof of this lemma shows that the Aronszajn
trees derived from $X$ and from $Y$ are isomorphic on a club.

\begin{definition}
An \emph{Aronszajn compactum} is a
compact $X$ such that
$w(X) = \aleph_1$ and $\cchi(X) = \aleph_0$ and for some
\textup(equivalently, for all\,\textup) $Z \subseteq [0,1]^{\omega_1}$
homeomorphic to $X$, $Z$ is an embedded Aronszajn compactum.
\end{definition}

The next lemma is immediate from the definition.
Further closure properties of the class of Aronszajn compacta
are considered in Section \ref{sec-clos}.

\begin{lemma}
\label{lemma-subset}
A closed subset of an Aronszajn compactum is
either second countable or an Aronszajn compactum.
\end{lemma}

The Dedekind completion of an Aronszajn line is an Aronszajn compactum
(see Section \ref{sec-elem}),
and the associated tree is essentially
the same as the standard tree of closed intervals.
A special case of this is a compact Suslin line, which is
a well-known compact L-space; that is, it is
HL (hereditarily Lindel\"of) and not
HS (hereditarily separable).  The line derived from a special
Aronszajn tree is much different topologically,
since it is not even ccc.

In Section \ref{sec-constr} we shall prove:

\begin{theorem}
\label{thm-Aronsz-basic}
Assuming $\diamondsuit$, there is an Aronszajn compactum
which is both HS and HL.
\end{theorem}

Our construction is flexible enough
to build in additional properties for the space and its associated tree,
which may be either Suslin or special; see Theorem \ref{thm-Aronsz-refined}.
The form of the tree is (up to club-isomorphism)
a topological invariant of $X$, but seems to be unrelated
to more conventional topological properties of $X$;
for example, $X$ may be totally
disconnected, or it may be connected and locally connected,
with $\dim(X)$ finite or infinite.

\begin{question}
Is there, in ZFC, an HL Aronszajn compactum?
\end{question}

We would expect a ZFC example to be both HS and HL.
Note that an Aronszajn compactum is dissipated in the sense of \cite{KU2},
so it cannot be an L-space
if there are no Suslin lines by Corollary 5.3 of \cite{KU2}.

To refute the existence of an HL Aronszajn compactum,
one needs more than just an Aronszajn tree of closed sets,
since this much exists in the Cantor set:

\begin{proposition}
There is an Aronszajn tree $T$ whose nodes are closed subsets
of the Cantor set $2^\omega$.  The tree ordering is $\supset$,
with root $2^\omega$. Each level of $T$
consists of a pairwise disjoint family of sets.
\end{proposition}

The proof is like that of Theorem 4 of Galvin and Miller \cite{GM},
which is attributed there to Todor\v cevi\'c.

\section{Elementary Submodels}
\label{sec-elem}

We consider Aronszajn compacta from the point of view
of elementary submodels.
Assume that $X$ is compact, with $X$ (and its topology)
in some suitably large $H(\theta)$.
If $X$ is first countable, so that $|X| \le \cccc$ and its
topology is a set of size $\le 2^\cccc$, then $\theta$ can
be any regular cardinal larger than $2^\cccc$,
assuming that the set $X$ is chosen so that its transitive
closure has size $\le \cccc$.

If $X \in M \prec H(\theta)$,
then there is a natural quotient map $\pi = \pi_M: X \onto X/M$
obtained by identifying two points of $X$ iff they are not
separated by any function in $C(X,\RRR) \cap M$.
Furthermore, $X/M$ is second countable whenever $M$ is countable.

\begin{lemma}
\label{lemma-el}
Assume that $X$ is compact,
$w(X) = \aleph_1$, and $\cchi(X) = \aleph_0$.
Then the following are equivalent:
\begin{itemizz}
\item[1.] $X$ is an Aronszajn compactum.
\item[2.] Whenever $M$ is countable and
$X \in M \prec H(\theta)$,
there are only countably many $y \in X/M$
such that $\pi\iv\{y\}$ is not a singleton.
\item[3.] $(2)$ holds for all $M$ in some club of
countable elementary submodels of $H(\theta)$.
\end{itemizz}
\end{lemma}
\begin{proof}
For $(1)\to(2)$, note that $X \in M \prec H(\theta)$ implies
that $M$ contains some club satisfying Definition \ref{def-emb-ar}.
\end{proof}

For example, say that $X$ is a compact first countable LOTS.
Then the equivalence classes are all convex; and,
if $x < y$ then $\pi(x) = \pi(y)$ iff $[x,y] \cap M = \emptyset$.
Now consider Aronszajn lines:

\begin{definition}
\label{def-ar-line}
A \emph{compacted Aronszajn line} is a compact LOTS $X$
such that $w(X) = \aleph_1$ and $\cchi(X) = \aleph_0$
and the closure of every countable
set is second countable.
\end{definition}

By $\cchi(X) = \aleph_0$,
there are no increasing or decreasing $\omega_1$--sequences.
Note that our definition allows for the possibility
that $X$ contains uncountably many disjoint
intervals isomorphic to $[0,1]$.
The term ``\emph{compact} Aronszajn line'' is not
common in the literature.  An \emph{Aronszajn line} is usually
defined to be a LOTS of size $\aleph_1$
with no increasing or decreasing $\omega_1$--sequences
and no uncountable subsets of \emph{real type}
(that is, order-isomorphic to a subset of $\RRR$).
Such a LOTS cannot be compact; the Dedekind completions
of such LOTSes are the compact\emph{ed} Aronszajn lines of
Definition \ref{def-ar-line}.

\begin{lemma}
\label{lemma-ar-line}
A LOTS $X$ is an Aronszajn compactum iff $X$ is a compacted Aronszajn line.
\end{lemma}
\begin{proof}
For $\leftarrow$: suppose that
$X \in M \prec H(\theta)$ and $M$ is countable.
Then $X/M$ is a compact metric LOTS, and is hence order-embeddable
into $[0,1]$.  Suppose there were an uncountable $E \subseteq X/M$
such that $|\pi\iv\{y\}| \ge 2$ for all $y \in E$.
Say $\pi\iv\{y\} = [a_y, b_y] \subset X$ for $y \in E$,
where $a_y < b_y$.  If $D$ is a countable dense subset of $E$
then $\cl(\{a_y : y \in D\}) \subseteq X$ would not be second
countable, a contradiction.
\end{proof}

We use the standard definition of a \emph{Suslin line} as any
LOTS which is ccc and not separable; this is always an L-space.
Then a \emph{compact Suslin line} is just a Suslin line which
happens to be compact.  A compacted Aronszajn line may be a Suslin line,
but a compact Suslin line need not be a
compacted Aronszajn line.  For example, we may form $X$
from a connected compact Suslin line $Y$
by doubling uncountably many points lying in some Cantor subset
of $Y$.  More generally,

\begin{lemma}
\label{lemma-sus-ar}
Let $X$ be a compact Suslin line.  Then $X$ is
a compacted Aronszajn line iff
$D := \{x \in X : \exists y > x \,( [x,y] = \{x,y\} \}$
does not contain an uncountable subset of real type.
\end{lemma}
\begin{proof}
Note that $D$ is the set of all points
with a right nearest neighbor.
If $D$ contains an uncountable set $E$ real type,
let $B \subseteq E$ be countable and dense in $E$.
Then whenever $M$ is countable and
$X,B \in M \prec H(\theta)$,
there are uncountably many $y \in X/M$
such that $|\pi\iv\{y\}| \ge 2$,
so that $X$ is not an Aronszajn compactum.

Conversely, if $X$ is not an Aronszajn compactum,
consider any countable $M$ with
$X \in M \prec H(\theta)$ and
$A := \{y \in X/M : |\pi\iv\{y\}| \ge 2\}$ uncountable.
Let $A' := \{y \in X/M : |\pi\iv\{y\}| > 2\}$.
Since each $\pi\iv\{y\}$ is convex,
$A'$ is countable by the ccc, and the left points
of the $\pi\iv\{y\}$ for $y \in A \backslash A'$ yield an uncountable
subset of $D$ of real type.
\end{proof}

A zero dimensional compact Suslin line formed
in the usual way from a binary Suslin tree will
also be a compacted Aronszajn line.

\section{Normalizing Aronszajn Compacta}
\label{sec-norm}
The club $C$ and tree $T$ derived from an Aronszajn
compactum $X$ in Definition \ref{def-emb-ar} can depend on the
embedding of $X$ into $[0,1]^{\omega_1}$.
To standardize the tree, we  choose a nice embedding.
For $X \subseteq [0,1]^{\omega_1}$, $C$ cannot in general be $\omega_1$,
since $C = \omega_1$ implies that $\dim(X) \le 1$.
Replacing  $[0,1]$ by the Hilbert cube, however,
we can assume $C = \omega_1$, which simplifies our tree notation.
In particular, the levels will be indexed by $\omega_1$,
so that $\LL_\alpha$ will be level $\alpha$ of the tree
in the usual sense.

\begin{definition}
\label{def-cube}
$Q$ denotes the Hilbert cube, $[0,1]^\omega$. If
$X \subseteq Q^{\omega_1}$ is closed and $\alpha< \omega_1$, then
$\LL_\alpha = \LL_\alpha(X)=
\{x \in X_\alpha : |(\sigma^{\omega_1}_\alpha)\iv\{x\}| > 1\}$.
$\WW(X) = \{\alpha< \omega_1: |\LL_\alpha| \le \aleph_0\}$.
\end{definition}

So, $X$ is an Aronszajn compactum iff $\WW(X)$ contains a club;
$\WW(X)$ itself need not be closed, and $\WW(X)$ depends
on how $X$ is embedded into $ Q^{\omega_1}$.
Now, using the facts that $Q \cong Q^\omega$ and that
an Aronszajn tree can have only countably many finite levels:

\begin{lemma}
\label{lemma-nf}
Every Aronszajn compactum is homeomorphic to some $X \subseteq Q^{\omega_1}$
such that $\WW(X) = \omega_1$ and
$|\LL_\alpha| = \aleph_0$ for all $\alpha > 0$.
\end{lemma}

Of course, $\LL_0 = X_0 = \{\emptyset\} = Q^0$,
and $\emptyset$ is the root node of the tree.

\begin{definition}
If $X \subseteq Q^{\omega_1}$ is an Aronszajn compactum
and $\WW(X) = \omega_1$, let $\widehat \LL _\alpha =
\{x \in \LL_\alpha : w((\sigma^{\omega_1}_\alpha)\iv\{x\}) = \aleph_1\}$,
and let $\widehat T = \bigcup_\alpha \widehat \LL _\alpha$.
\end{definition}

Since $X$ is not second countable, each
$\widehat \LL_\alpha  \ne \emptyset$ and $\widehat T$ is
an Aronszajn subtree of $T$.
Repeating the above argument, we get

\begin{lemma}
\label{lemma-better-nf}
Every Aronszajn compactum is homeomorphic to some $X \subseteq Q^{\omega_1}$
such that $\WW(X) = \omega_1$, and
$|\widehat \LL_\alpha| = \aleph_0$ for all $\alpha > 0$,
and each $x \in \LL_ \alpha \backslash \widehat \LL_\alpha$ is a
leaf, and each $x \in \widehat \LL_\alpha$
has $\aleph_0$ immediate successors in $\widehat \LL_{\alpha +1}$.
\end{lemma}

This normalization can also be obtained with elementary submodels.
Start with a continuous chain of elementary submodels,
$M_\alpha \prec H(\theta)$, for $\alpha < \omega_1$,
with $X \in M_0$ and each $M_\alpha \in M_{\alpha+1}$.
Let $X_\alpha = X / M_\alpha$, let $\pi_\alpha: X \onto X_\alpha$
be the natural map, and let
$\LL_\alpha = \{y \in X_\alpha: |\pi_\alpha\iv\{y\}| > 1\}$.
We may view each $X_\alpha$ as embedded topologically
into $Q^\alpha$, in which case $\LL_\alpha$ has the same meaning
as before.
If $\pi_\alpha\iv\{y\}$ is second countable, then
(since $M_\alpha \in M_{\alpha+1}$), all the points in
$\pi_\alpha\iv\{y\}$ are separated by functions in
$C(X) \cap M_{\alpha+1}$, so
$y \in \LL_ \alpha \backslash \widehat \LL_\alpha$ is a leaf.

If $X$ is a compacted Aronszajn line, then
$X_{\alpha + 1}$ is formed by replacing
each $y \in \LL_\alpha$ by a compact interval $I_y$ of size at
least $2$.  If $y \in \LL_ \alpha \backslash \widehat \LL_\alpha$,
then $\pi_\alpha\iv\{y\}$ is second countable
and is isomorphic to $I_y$.
Note that the tree may have uncountably many leaves;
we do not obtain the conventional normalization of an Aronszajn tree,
where the tree is uncountable above every node.

Next, we consider the ideal of second countable subsets of $X$:

\begin{definition}
For any space $X$, $\II_X$ denotes the family of all $S \subseteq X$
such that $S$, with the subspace topology, is second countable.
\end{definition}

$\II_X$ need not be an ideal.  It is obviously closed under subsets,
but need not be closed under unions (consider $\omega \cup \{p\}
\subset \beta \omega $).

\begin{lemma}
\label{lemma-ideal}
Assume that $X \subseteq Q^{\omega_1}$ is an HL Aronszajn compactum,
as in Lemma \ref{lemma-nf}.  Then
$\II_X$  is a $\sigma$--ideal, and, for all $S \subseteq X$,
the following are equivalent:
\begin{itemizz}
\item[1.] $S \in \II_X$.
\item[2.] For some $\alpha < \omega_1$,
$\sigma^{\omega_1}_\alpha (S) \cap \LL_\alpha = \emptyset$.
\item[3.] There is a $G \supseteq S$ such that $G\in \II_X$
and $G$ is a $G_\delta$ subset of $X$.
\item[4.] There is an $f \in C(S,Q)$ such that $f$ is 1-1.
\item[5.] There is an $f \in C(S,Q)$ such that
$f\iv\{y\}$ is second countable for all $y \in Q$.
\end{itemizz}
\end{lemma}
\begin{proof}
It is easy to verify $(2) \to (3) \to (1) \to (4) \to (5)$.
In particular, for $(2) \to (3)$:  Fix $\alpha$ and let
$G = (\sigma^{\omega_1}_\alpha)\iv (X_\alpha \backslash \LL_\alpha)$.
Then $G$ is a $G_\delta$ set and
$\sigma^{\omega_1}_\alpha : G \onto X_\alpha \backslash \LL_\alpha$
is a 1-1 closed map, and hence a homeomorphism.

For $(1) \to (2)$:
Fix an open base for $S$ of the form
$\{V_n \cap S : n \in \omega\}$, where each $V_n$ is open in $X$.
$X$ is HL, so $V_n$ is an $F_\sigma$.
We can thus fix $\xi < \omega_1$ such that each
$V_n = (\sigma^{\omega_1}_\xi)\iv ( \sigma^{\omega_1}_\xi (V_n))$.
It follows that $\sigma^{\omega_1}_\xi$ is 1-1 on $S$.
We may then choose $\alpha$ with $\xi < \alpha < \omega_1$
such that $\sigma^{\omega_1}_\alpha(S) \cap \LL_\alpha = \emptyset$.

Now, $\II_X$  is a $\sigma$--ideal by  $(1) \leftrightarrow (2)$.

To prove $(5) \to (2)$:  Fix $f$ as in (5).
Let $\{U_n : n \in \omega\}$ be an open base for $Q$;
then $f\iv(U_n) = S \cap V_n$, where $V_n$ is open in $X$
and hence an $F_\sigma$ set.
We can thus fix $\alpha < \omega_1$ such that
$V_n = (\sigma^{\omega_1}_\alpha)\iv ( \sigma^{\omega_1}_\alpha (V_n))$.
It follows that $f$ is constant on
$S \cap (\sigma^{\omega_1}_\alpha)\iv \{z\}$ for all $z \in X_\alpha$.
Thus,
$S \cap (\sigma^{\omega_1}_\alpha)\iv \{z\}$ is second
countable for all $z \in X_\alpha$.
But then $S$ is contained in the union
of $\bigcup\{S\cap ( \sigma^{\omega_1}_\alpha )\iv\{y\} : y \in \LL_\alpha\}$
and $ ( \sigma^{\omega_1}_\alpha )\iv (X_\alpha \backslash \LL_\alpha) \cong
(X_\alpha \backslash \LL_\alpha) $, so $S \in \II_X$ because
$\II_X$  is a $\sigma$--ideal.
\end{proof}

This proof shows that every Aronszajn compactum
is an ascending union of $\omega_1$ Polish spaces:
namely, the
$(\sigma^{\omega_1}_\alpha)\iv  (X_\alpha \backslash \LL_\alpha)$.

We needed $X$ to be Aronszajn in Lemma \ref{lemma-ideal};
HS and HL are not enough to prove the equivalence of (1)(3)(4)(5).
If $S$ is the
Sorgenfrey line contained in the double arrow
space $X$, then (4)(5) are true but (1)(3) are false.
Similar remarks hold for similar spaces which are both HS and HL.
For example, assuming CH, Filippov \cite{Fil} constructed a
locally connected continuum which is HS and HL but not second
countable.  The space was obtained by replacing a Luzin set of
points in $[0,1]^2$ by circles.  If $S$ contains one point
from each of the circles, then S satisfies (4)(5) but
fails (1)(3).
In both examples, the space $X$ itself satisfies (5) but not (1)(3)(4).

More generally, any space $X$ that has an $f\in C(X,Q)$ as in (5)
cannot be an Aronszajn compactum.  Thus, a ZFC example of an
HL Aronszajn compactum would
settle in the negative the following well-known question of Fremlin
(\cite{Fr} 44Qc): is it consistent that for every HL compactum $X$,
there is an $f \in C(X,Q)$ such that
$|f\iv\{y\}| <  \aleph_0$  for all $y \in Q$?
In \cite{Gr}, Gruenhage gives some of the history
related to this question, and points out
some related results suggesting that the answer might be
``yes'' under PFA.

\section{Closure Properties of Aronszajn Compacta}
\label{sec-clos}
Closure under subspaces was already mentioned in
Lemma \ref{lemma-subset}.
For products, Lemma
\ref{lemma-el} implies:

\begin{lemma}
\label{lemma-Ar-prod}
Assume that $X$ is an Aronszajn compactum and $Y$ is an arbitrary space.
Then $X \times Y$ is an Aronszajn compactum iff $Y$ is compact and countable.
\end{lemma}

Regarding quotients, we first prove:

\begin{lemma}
\label{lemma-eq-pair}
Assume that $X,Y$ are compact, $\varphi: X \onto Y$, and
$X,Y,\varphi \in M \prec H(\theta)$.   Let $\sim$ denote the $M$ equivalence relation
on $X$ and on $Y$.  Then
\begin{itemizz}
\item[1.] If $x_0, x_1 \in X $ and $x_0 \sim x_1$, then
$\varphi(x_0) \sim \varphi(x_1)$;
so, the inverse image of an equivalence class of $Y$
is a union of equivalence classes of $X$.
\item[2.] If $y_0, y_1 \in Y $ and $x_0 \not\sim x_1$
for all $x_0 \in \varphi\iv\{y_0\}$ and
all $x_1 \in \varphi\iv\{y_1\}$, then
$y_0 \not\sim y_1$.
\end{itemizz}
\end{lemma}
\begin{proof}
For (1):  If $f \in C(Y) \cap M$ separates $\varphi(x_0)$
from $ \varphi(x_1)$ then
$f \circ \varphi \in C(X) \cap M$ separates $x_0$ from $x_1$.

For (2):  For each
$x_0 \in \varphi\iv\{y_0\}$ and $x_1 \in \varphi\iv\{y_1\}$,
there is an $f \in C(X, [0,1]) \cap M$ such that $f(x_0) \ne f(x_1)$.
By compactness of $\varphi\iv\{y_0\} \times \varphi\iv\{y_1\}$,
there are $f_0, \ldots, f_{n-1} \in C(X, [0,1]) \cap M$
for some $n\in \omega$ such that: for all
$x_0 \in \varphi\iv\{y_0\}$ and $x_1 \in \varphi\iv\{y_1\}$,
there is some $i < n$ such that
$f_i(x_0) \ne f_i(x_1)$.
These yield an $\vec f \in C(X, [0,1]^n) \cap M$ such that
$\vec f ( \varphi\iv\{y_0\} )  \cap  \vec f ( \varphi\iv\{y_1\} ) =
\emptyset  $.
Since $M$ contains a base for $[0,1]^n$,
there are open $U_0, U_1 \subseteq [0,1]^n$ with each $U_i \in M$
such that $\overline{U_0} \cap \overline{U_1} = \emptyset$
and each $\vec f ( \varphi\iv\{y_i\} )  \subseteq U_i$, so that
$\varphi\iv\{y_i\} \subseteq (\vec{f}\,)\iv (U_i)$.
Let $V_i = \{y \in Y : \varphi\iv\{y\} \subseteq (\vec f)\iv (U_i) \}$.
Then the $V_i$ are open in $Y$, each $V_i \in M$, each $y_i \in V_i$, and
$\overline{V_0} \cap \overline{V_1} = \emptyset$.
There is thus a $g \in C(Y) \cap M$
such that $g(\overline{V_0}) \cap g(\overline{V_1}) = \emptyset$,
so that $g(y_0) \ne g(y_1)$.  Thus, $y_0 \not\sim y_1$.
\end{proof}

\begin{theorem}
\label{thm-ar-quotient}
Assume that $X$ is an Aronszajn compactum,
$\varphi : X \onto Y$,
$w(Y) = \aleph_1$, and $\cchi(Y) = \aleph_0$.
Then $Y$ is an Aronszajn compactum.
\end{theorem}
\begin{proof}
It is sufficient to check that for a club of elementary submodels $M$,
all but countably many $M$--classes of $Y$ are singletons.
Fix $M$ as in Lemma \ref{lemma-eq-pair}; so
all but countably many $M$--classes of $X$ are singletons.
Then for all but countably many classes $K = [y]$ of $Y$:
all $M$--classes of $X$ inside of $\varphi\iv(K)$ are singletons,
so that, by the lemma, $K$ is a singleton.
\end{proof}

Note that we needed to assume that $\cchi(Y) = \aleph_0$.
Otherwise, when $X$ is not HL, we would get a trivial counterexample
of the form $X/K$, where $K$ is a closed set which is not a $G_\delta$.

Examining whether an Aronszajn compactum may be both HS and HL reduces
to considering zero dimensional spaces and connected spaces,
by the following lemma.

\begin{lemma}
Assume that $X$ is an HL Aronszajn compactum,
$\varphi : X \onto Y$.  Then either
$Y$ is an Aronszajn compactum or some $\varphi\iv\{y\}$ is an
Aronszajn compactum.
\end{lemma}
\begin{proof}
$Y$ will be an Aronszajn compactum unless it is second countable.
But if it is second countable, then some  $\varphi\iv\{y\}$
will be not second countable by Lemma \ref{lemma-ideal},
and then  $\varphi\iv\{y\}$ will be an Aronszajn compactum.
\end{proof}

\begin{corollary}
Suppose there is an Aronszajn compactum $X$ which is HS and HL.
Then there is an Aronszajn compactum $Z$ which is HS and HL
and which is either connected or zero dimensional.
\end{corollary}
\begin{proof}
Get $\varphi : X \onto Y$ by collapsing all connected components to points.
Then $Z$ is either $Y$ or some component.
\end{proof}

Note that the cone over $X$ is also connected, but is
not an Aronszajn compactum by Lemma \ref{lemma-Ar-prod}.

\section{Constructing Aronszajn Compacta}
\label{sec-constr}

We begin this section by constructing a space $X$  which proves
Theorem \ref{thm-Aronsz-basic}.
We construct $X=X_{\omega_1}$ as an inverse limit as
a closed subspace of $Q^{\omega_1}$.
To make $X$ both HS and HL,
we shall apply the following lemma:

\begin{lemma}
\label{lemma-hshl}
Assume that $X$ is compact and for all closed $F \subseteq X$,
there is a compact metric $Y$ and a map $g: X \onto Y$
such that $g \rest g\iv(g(F)) \;:\; g\iv(g(F)) \;\onto\; g(F) $
is irreducible.  Then $X$ is both HS and HL.
\end{lemma}
\begin{proof}
By irreducibility, $g\iv(g(F)) = F$,
so that $F$ is a $G_\delta$ and $F$ is separable.
Thus, $X$ is a compact HL space in which all closed
subsets are separable, so $X$ is HS.
\end{proof}

In applying the lemma to $X = X_{\omega_1}$,
$g$ will be some $\pi^{\omega_1}_\alpha\res X$.
We shall use $\diamondsuit$ to capture all
closed  $F \subseteq Q^{\omega_1}$ so that
all closed $F \subseteq X$ will be considered.
This method was also employed in \cite{HK2},
which constructed some compacta which were HS and HL
but not Aronszajn.

As in standard inverse limit constructions, we inductively construct
$X_\alpha \subseteq Q^\alpha$, for $\alpha \le \omega_1$.
To ensure that $X$ will be Aronszajn, at each stage $\alpha< \omega_1$,
we carefully select a countable set
$\EE_\alpha \subseteq X_\alpha$ of ``expandable points'',
and at each stage $\beta > \alpha$,
we construct $X_\beta\subseteq Q^\beta$ so that
$|(\sigma^{\beta}_\alpha)\iv\{x\}| = 1$
whenever $x \notin \EE_\alpha$.  Then the
$\LL_\alpha$ of Definition \ref{def-cube}
will be subsets of $\EE_\alpha$ and hence countable.

These preliminaries are included in
the following conditions:

\begin{conditions}
\label{cond-first-batch}
$X_\alpha$,  for $\alpha  \le \omega_1$, and
$\PP_\alpha, F_\alpha, \EE_\alpha, q_\alpha$,
for $0 < \alpha  < \omega_1$, satisfy:
\begin{enumctr}
\itemctr
 Each $X_\alpha$ is a closed subset of $Q^\alpha$.
\itemctr
 $\pi^\beta_\alpha(X_\beta) = X_\alpha$ whenever
$\alpha \le \beta \le \omega_1$.
\itemctr
 $\PP_\alpha$ is a countable family of closed subsets
of $X_\alpha$, and $F_\alpha \in \PP_\alpha$.
\itemctr
For all $P \in \PP_\alpha$: \\
a.
$\sigma^{\alpha+1}_\alpha \res ((\sigma^{\alpha+1}_\alpha) \iv (P)) \;:\;
(\sigma^{\alpha+1}_\alpha)\iv(P) \;\onto\; P$ is irreducible, and \\
b.
$(\sigma^{\beta}_\alpha)\iv(P) \in \PP_{\beta}$ whenever
$\alpha \le \beta < \omega_1$.
\itemctr
For all closed $F \subseteq X$, there is an
$\alpha$ with $0 < \alpha < \omega_1$ such that
$\sigma^{\omega_1}_\alpha(F) = F_\alpha$.
\itemctr
$\EE_\alpha$ is a countable dense subset of $X_\alpha$,
and $q_\alpha \in  \EE_\alpha$.
\itemctr
$\EE_\beta \subseteq (\sigma^{\beta}_\alpha)\iv(\EE_\alpha)$
whenever $0 < \alpha \le \beta < \omega_1$.
\itemctr
$|(\sigma^{\alpha + 1}_\alpha)\iv\{x\}| = 1$
whenever $0 < \alpha < \omega_1$ and
$x \in X_\alpha \backslash \{q_\alpha\}$.
\itemctr
$|(\sigma^{\alpha + 1}_\alpha)\iv\{q_\alpha\}| > 1$.
\end{enumctr}
\end{conditions}

We discuss below how to satisfy these conditions.
Conditions (1) and (2) simply determine our
$X = X_{\omega_1}  \subseteq Q^{\omega_1}$ with
each $X_\alpha=\pi^{\omega_1}_\alpha(X)$.
$\diamondsuit$ is used for (5).
Constructing an $X$ that satisfies Conditions (1 - 9)
is enough to prove
Theorem \ref{thm-Aronsz-basic}:

\begin{lemma}
Conditions $(1 - 9)$ imply that $X = X_{\omega_1}$ is an
Aronszajn compactum and is both HS and HL.
\end{lemma}
\begin{proof}
By (4) and induction on $\beta$,
$\sigma^{\beta}_\alpha \res ((\sigma^{\beta}_\alpha) \iv (P)) :
(\sigma^{\beta}_\alpha)\iv(P) \onto P$ is irreducible
whenever $\alpha \le \beta \le \omega_1$ and $P \in \PP_\alpha$.
Then $X$ is HS and HL by Lemma \ref{lemma-hshl} and (5)(3).

By (6)(7)(8) and induction,
$|(\sigma^{\beta}_\alpha)\iv\{x\}| = 1$
whenever $0 < \alpha \le \beta \le  \omega_1$
and $x \in X_\alpha \backslash \EE_\alpha$.
So, $\LL_\alpha :=
\{x \in X_\alpha : |(\sigma^{\omega_1}_\alpha)\iv\{x\}| > 1\}
\subseteq \EE_\alpha$, which is countable by (6).

Finally, $w(X) = \aleph_1$ by (9), and
$\cchi(X) = \aleph_0$ because $X$ is HL.
\end{proof}

To obtain Conditions $(1 - 9)$, we must add some further
conditions so that the natural construction avoids contradictions.
For example, satisfying Conditions (6) and (7) at stage $\beta$
requires $\bigcap_{\alpha < \beta}
(\sigma^{\beta}_\alpha)\iv(\EE_\alpha) \ne \emptyset$.
So we add Conditions (10 - 12) below making
the $\EE_\alpha$ into the levels of a tree;
the selection of the $\EE_\alpha$  will resemble the
standard inductive construction of an Aronszajn tree.

The sets $F_\alpha$ may be scattered or even singletons.
This cannot be avoided,
because we are using the $F_\alpha$ to ensure
that \emph{all} closed sets are $G_\delta$ sets, so that
$X$ is HL;  making just the perfect sets
$G_\delta$ could produce a Fedorchuk space (as in \cite{HK3}),
which is not even first countable.
If $x \in P \in \PP_\alpha$ and $x$
is isolated in $P$,  then the irreducibility condition
in (4) requires that $|(\sigma^{\alpha + 1}_\alpha)\iv\{x\}| = 1$,
but that contradicts (9) if $x = q_\alpha$.
Now, if every point of $\EE_\alpha$
is isolated in some $P \in \PP_\alpha$, then
we cannot choose $q_\alpha \in \EE_\alpha$, as required by (6).
We shall avoid these problems by requiring that if
$x \in \EE_\alpha$ and $P \in \PP_\alpha$, then
either $x \notin P$ or $x$ is in the perfect kernel of $P$.
This can be ensured by choosing $F_\alpha$ first
(as given by $\diamondsuit$), and then choosing $\EE_\alpha$;
for limit $\alpha$, our Aronszajn tree construction will give
us plenty of options for choosing the points of $\EE_\alpha$,
and we shall make $F_\alpha$ trivial for successor $\alpha$.
The additional conditions that handle this will employ
the notation in the following:

\begin{definition}
If $F$ is compact and not scattered, let $\ker(F)$ denote
the perfect kernel of $F$; otherwise, $\ker(F) = \emptyset$.
\end{definition}

To satisfy Condition (8), we construct
$X_{\alpha + 1}$ from $X_\alpha$
by choosing an appropriate
$h_\alpha \in C(X_\alpha \backslash \{q_\alpha\}, Q)$,
and letting $X_{\alpha+1}=\cl(h_\alpha)$.
Identifying
$Q^{\alpha+1}$ with $Q^\alpha \times Q$ and
$h_\alpha$ with its graph,
$h_\alpha(x)$ is the $y \in Q$ such that
$x \cat y \in X_{\alpha+1}$.
Note that $h_\alpha$ is indeed continuous because its graph
is closed.

Thus, to construct $X$ so that Conditions (1 - 9) are met,
we add the following:
\begin{conditions}
$h_\alpha$  and $r^n_\alpha$,
for $0 < \alpha  < \omega_1$ and $n < \omega$, satisfy:
\begin{enumctr}
\itemctr
$(\sigma^{\beta}_\alpha)(\EE_\beta) = \EE_\alpha$
whenever $0 < \alpha \le \beta < \omega_1$.
\itemctr
$|\EE_ {\alpha + 1}\cap(\sigma^{\alpha + 1}_\alpha)\iv\{q_\alpha\}| > 1$.
\itemctr
If $x \in \EE_\alpha$,
then $(\sigma^{\alpha + n}_\alpha)(q_{\alpha + n}) = x$ for
some $n \in \omega$.
\itemctr
$X_\alpha$ has no isolated points whenever $\alpha > 0$.
\itemctr
$F_\alpha = \emptyset$ whenever $\alpha$ is a successor ordinal.
\itemctr
$\PP_\beta = \{F_\beta\} \cup \{
(\sigma^{\beta}_\alpha)\iv(P): 0 < \alpha < \beta
\ \&\  P \in \PP_\alpha \}$.
\itemctr
$\EE_\alpha \cap (P \backslash \ker(P)) = \emptyset$
whenever $P \in \PP_\alpha$.
\itemctr
$r^n_\alpha \in X_\alpha \backslash \{q_\alpha \}$ and the sequence
$\langle r^n_\alpha : n \in \omega \rangle$ converges to $q_\alpha$.
\itemctr
$h_\alpha \in C(X_\alpha \backslash \{q_\alpha\}, Q)$, and
$ X_{\alpha+1} = \cl(h_\alpha)$.
\itemctr
If $q_\alpha \in P \in \PP_\alpha$,
then $ r^n_\alpha  \in \ker(P)$ for infinitely many $n$, and
every $y\in Q$ with $q_\alpha \cat y \in X_{\alpha + 1}$
is a limit point of the sequence
$\langle h_\alpha(r^n_\alpha ) :
n \in \omega \ \&\ r^n_\alpha  \in \ker(P) \rangle$.
\end{enumctr}
\end{conditions}

Observe that (10)(11)(12) will give us the following:

\begin{lemma}
\label{lem-EisL}
$\LL_\alpha = \EE_\alpha$ whenever $0 < \alpha < \omega_1$ .
\end{lemma}

In the tree $T(X)$, although only
the node $q_\alpha \in \LL_\alpha$  has more than one successor
in $\LL_{\alpha + 1}$, (12) ensures that
at limit levels $\gamma$, there are $2^{\aleph_0}$
choices for the elements of $\EE_\gamma$, so that we may
avoid the points in
$F_\gamma \backslash \ker(F_\gamma)$, as required by (16).

By (14)(15), $\emptyset \in \PP_\alpha$ for all $\alpha > 0$,
and non-empty sets are added into the $\PP_\alpha$ only
at limit $\alpha$.

The following proof gives the bare-bones construction; refinements
of it produce the spaces of Theorem \ref{thm-Aronsz-refined}.

\begin{proofof}{Theorem \ref{thm-Aronsz-basic}}
Before we start, use $\diamondsuit$ to choose a closed
$\widetilde F_\alpha \subseteq Q^\alpha$ for each $\alpha < \omega_1$,
so that
$\{\alpha < \omega_1 : \pi^{\omega_1}_\alpha(F)=\widetilde F_\alpha\}$
is stationary for all closed $F \subseteq Q^{\omega_1}$.
To begin the induction:
$X_0$ must be $\{\emptyset\} = Q^0$, and
$\PP_\alpha, F_\alpha, \ldots\ldots$ are only defined
for $\alpha > 0$.

Now, fix $\beta$ with $0 < \beta < \omega_1$, and assume
that all conditions have been met below $\beta$.
We define in order
$X_\beta$,
$F_\beta$,
$\PP_\beta$,
$\EE_\beta$,
$q_\beta$,
$r^n_\beta$,
$h_\beta$.

If $\beta$ is a limit, then $X_\beta$ is determined
by (1)(2) and the $X_\alpha$ for $\alpha < \beta$.
$X_1$ can be any perfect subset of $Q^1$.
If $\beta = \alpha+1 \ge 2$, then
$X_\beta = \cl(h_\alpha)$, as required by (18).
Now let $F_\beta = \widetilde F_\beta$ if
$\widetilde F_\beta \subseteq X_\beta$ and $\beta$ is a limit;
otherwise, let $F_\beta = \emptyset$.
$\PP_\beta$ is now determined by (15).

$\EE_1$ can be any countable dense subset of $X_1$.
If $\beta = \alpha+1 \ge 2$,
let $\EE_\beta =
(\sigma^{\beta}_\alpha) \iv (\EE_\alpha \backslash \{q_\alpha\} )
\cup \DD_\beta$, where $\DD_\beta$ is any subset of
$(\sigma^{\beta}_\alpha)\iv \{q_\alpha\}$ such that
$2 \le |\DD_\beta| \le \aleph_0$.
Observe that $\EE_\beta$ is dense in $X_\beta$
(without using $\DD_\beta$), so (6) is preserved,
and $\DD_\beta$ guarantees that (11) is preserved.
To verify (16) at $\beta$, note that by (15) at $\alpha$, every
non-empty set in $\PP_\beta$ is of the form
$\widehat P :=
(\sigma^{\beta}_\alpha) \iv (P)$ for some $P \in \PP_\alpha$.
So, if (16) fails at $\beta$, fix
$P \in \PP_\alpha$ and
$x \in \EE_\beta \cap (\widehat P \backslash \ker(\widehat P))$.
Then $x \in (\sigma^{\beta}_\alpha)\iv \{q_\alpha\}$,
so $q_\alpha \in P$, and hence $q_\alpha \in \ker(P)$;
but then by (19), $x$ is a limit of a sequence of elements
of $\ker(\widehat P)$, so that $x \in \ker(\widehat P)$.

For limit $\beta$, let
$\EE_\beta = \{x^* : x \in \bigcup_{\alpha < \beta} \EE_\alpha\}$,
where, $x^*$, for $x \in \EE_\alpha$,  is
some $y \in X_\beta$ such that $\pi^\beta_\alpha(y) = x$
and $\pi^\beta_\xi(y) \in \EE_\xi$ for all $\xi < \beta$.
Any such choice of the $x^*$ will satisfy (10).
But in fact, using (11)(12), for each such $x$ there
are $2^{\aleph_0}$ possible choices of $x^*$,
so we can satisfy (16) by avoiding the countable
sets $P \backslash \ker(P)$ for $P \in \PP_\beta$.

To facilitate (12), list each $\EE_\alpha$ as
$\{e_\alpha^j : j \in \omega\}$; let
$e_0^j = \emptyset \in X_0$.
Then, if $\beta$ is a successor ordinal of the form
$\gamma + 2 ^i 3^j$, where $\gamma$ is a limit or $0$,
choose $q_\beta \in \EE_\beta$ so that
$\sigma^\beta_{\gamma + i}(q_\beta) = e_{\gamma + i}^j$.
For other $\beta$, $q_\beta \in \EE_\beta$ can
be chosen arbitrarily.

Next, we may choose the $r^n_\beta$ to satisfy (19)
because if $q_\beta \in P \in \PP_\beta$,
then $q_\beta \in \ker(P)$ by (16), so that
$q_\beta$ is also a limit of points in $\ker(P)$.

Finally, we must choose
$h_\beta \in  C(X_\beta \backslash \{q_\beta\}, Q)$.
Conditions (18)(19) only require that
$h_\beta$ have a discontinuity at $q_\beta$ with the property that
every limit point of the function at $q_\beta$ is also
a limit of each of the sequences
$\langle h_\beta(r^n_\beta ) :
n \in \omega \ \&\ r^n_\beta  \in \ker(P) \rangle$.
Since $X_\beta$ is a compact metric space
with no isolated points, we may accomplish this by
making every point of $Q$ a limit point of each
$\langle h_\beta(r^n_\beta ) :
n \in \omega \ \&\ r^n_\beta  \in \ker(P) \rangle$.
\end{proofof}

If we choose each $h_\beta$ as above and also set $X_1 = Q$,
then our $X$ will be connected, and it is fairly easy to choose
the $h_\beta$ so that $X$ fails to be locally connected.
The next theorem shows how
to make $X$ connected and locally connected.
We construct $X$ so that each $X_\alpha$
is homeomorphic to the \textit{Menger sponge}, $\MS$,
and all the maps $\sigma^\beta_\alpha$ are monotone.
The Menger sponge \cite{Me} is a one dimensional
locally connected metric continuum; the properties of $\MS$ used
in inductive constructions such as these are summarized in \cite{HK3},
which contains further references to the literature.
A map is \emph{monotone} iff all point inverses are connected.
Monotonicity of the $\sigma^\beta_\alpha$ will imply that
$X$ is locally connected.

At successor stages,
to construct $X_{\alpha+1} \cong \MS$,
we assume that $X_\alpha \cong \MS$ and
apply the following special case of
Lemmas 2.7 and 2.8 of \cite{HK3}:

\begin{lemma}
\label{lem-MS}
Assume that $q \in X \cong \MS$ and that for each $j \in \omega$,
the sequence
$\langle r^n_j : n \in \omega \rangle$ converges to $q$,
with each $r^n_j \ne q$.  Let $\pi : X \times [0,1] \onto X$
be the natural projection.  Then there is a $Y \subseteq  X \times [0,1]$
such that:
\begin{itemizz}
\item[1.] $Y \cong \MS$ and $\pi(Y) = X$.
\item[2.] $|Y \cap \pi\iv\{x\}| = 1$ for all $x \ne q$.
\item[3.] $\pi\iv\{q\} = \{q\}\times [0,1]$.
\item[4.] Let $Y \cap \pi\iv\{r^n_j\} = \{(r^n_j, u^n_j)\}$.  Then,
for each $j$, every point in $[0,1]$ is a limit point
of $\langle u^n_j : n \in \omega \rangle$.
\end{itemizz}
\end{lemma}

Constructing $X$ as such an inverse limit of
Menger sponges will make $X$ one dimensional.
The results quoted
from \cite{HK3} about $\MS$ were patterned on an earlier
construction of van Mill \cite{vM}, which involved an inverse
limit of Hilbert cubes;
replacing $\MS$ by $Q$  here
would yield an infinite dimensional version
of this Aronszajn compactum.
The following summarizes several possibilities for
$X$ and its associated tree:
\begin{theorem}
\label{thm-Aronsz-refined}
Assume $\diamondsuit$.
For each of the following $2 \cdot 3 = 6$ possibilities,
there is an Aronszajn compactum $X$
with associated Aronszajn tree $T$ such that $X$ is HS and HL.
Possibilities for $T$:
\begin{itemizz}
\item[a.] $T$ is Suslin.
\item[b.] $T$ is special.
\end{itemizz}
Possibilities for $X$:
\begin{itemizz}
\item[$\alpha$.] $\dim(X) = 0$.
\item[$\beta$.] $\dim(X) = 1$ and $X$ is connected and locally connected.
\item[$\gamma$.] $\dim(X) = \infty$ and $X$ is connected and locally connected.
\end{itemizz}
\end{theorem}
\begin{proof}
We refine the proof of Theorem \ref{thm-Aronsz-basic},
To obtain $(a)$ or $(b)$, the refinement is in the choice
of the $\EE_\beta$ for limit $\beta$.
To obtain $(\alpha)$ or $(\beta)$ or $(\gamma)$,
the refinement is in the choice of $X_1$ and the functions $h_\alpha$.
Since these refinements are independent of each other,
the discussion of $(a)(b)$ is unrelated to the discussion of
$(\alpha)(\beta)(\gamma)$.

For $(a)$: We use $\diamondsuit$ to kill all
potential uncountable maximal antichains $A \subset T$.
Fix a sequence $\langle A_\alpha : \alpha < \omega_1\rangle$
such that each $A_\alpha$ is a countable subset of $Q^{< \alpha}$ and
such that for all $A \subseteq Q^{< \omega_1}$:
if each $A \cap Q^{< \alpha}$ is countable, then
$\{\alpha < \omega_1 : A\cap Q^{< \alpha} = A_\alpha\}$ is stationary.

Let $T_\beta= \bigcup\{\LL_\alpha: \alpha< \beta\}
= \bigcup\{\EE_\alpha: \alpha< \beta\}$ (see Lemma \ref{lem-EisL}),
and use $\tleft$ for the tree order.
For each limit $\beta < \omega_1$,
modify the construction of $\EE_\beta$
in the proof of Theorem \ref{thm-Aronsz-basic} as follows:
We still have
$\EE_\beta = \{x^* : x \in T_\beta\}$,
where, $x^*$, for $x \in T_\beta$, is chosen so that $x \tleft x^*$ and
$x^*$ defines a path through $T_\beta$.
But now, \emph{if} $A_\beta \subseteq T_\beta$ \emph{and}
$A_\beta$ is  a  maximal  antichain  in $T_\beta$,
\emph{then} make sure that each $x^*$ is above some element of $A_\beta$.
To do this, use maximality of $A_\beta$ first to choose
$x^\dag \in T_\beta$ so that $x \tleft x^\dag$ and $x^\dag$
is above some element of $A_\beta$, and then choose $x^*$ so that
 $x \tleft x^\dag \tleft x^*$.
There are still $2^{\aleph_0}$ possible choices for $x^*$,
so we can satisfy (16) by avoiding the countable
sets $P\! \setminus\!\ker(P)$ as before.
Now, the usual argument shows that $T$ is Suslin.

For $(b)$:  Let $\Lim$ denote the set of countable limit ordinals,
and let $T^\Lim = \bigcup\{\LL_\alpha: \alpha  \in \Lim\} =
\bigcup\{\EE_\alpha: \alpha  \in \Lim\}$.
To make $T$ special,
inductively define an order preserving map
$\varphi : T^\Lim \to \QQQ$.
To make the induction work, we also assume inductively:
\[
\forall \gamma,\beta \in \Lim \,  \forall x \in \LL_\gamma\,
\forall q \in \QQQ\;
[\gamma < \beta \; \&\;  q > \varphi(x)
\ \to \ \exists y \in \LL_\beta \, [x \tleft y \; \&\; \varphi(y) = q]]
\tag{$*$}
\]
To start the induction, $\varphi \res \LL_\omega : \LL_\omega \to \QQQ$
can be arbitrary.

For $\beta = \alpha + \omega$, where $\alpha$ is a limit ordinal:
First, determine the $x^*$ exactly as in the proof of
Theorem \ref{thm-Aronsz-basic}.  Then, note that for each $x \in \LL_\alpha$,
the set $S_x := \{y \in \EE_\beta : x \tleft y\}$ has size $\aleph_0$,
so we can let $\varphi \res S_x$ map $S_x$ \emph{onto}
$\QQQ \cap (\varphi(x), \infty)$.

For $\beta < \omega_1$ which is a limit of limit ordinals:
Let
\[
\EE_\beta = \{x^*_q : x \in T_\beta \;\&\;
q  \in \QQQ \cap (\varphi(x), \infty)\}
\ \ ,
\]
where each $x^*_q$  is chosen so that
$x \tleft x^*_q$ and
$x^*_q$ defines a path through $T_\beta$ and the $x^*_q$ are all different
as $q$ varies.
We let $\varphi(x_q^*) = q$, which will clearly preserve $(*)$,
but we must make sure that $\varphi$ remains order preserving.
For this, choose $x_q^*$ so that
$q > \sup\{\varphi(z) : z \in T^\Lim \; \&\ z \tleft x_q^*\}$.
Such a choice is possible using $(*)$ on $T_\beta$.
As before, there are $2^{\aleph_0}$ possible choices of $x_q^*$,
so we can still avoid the countable sets $P\!\setminus\!\ker(P)$.

For $(\alpha)$, just make sure that $X_\alpha$
is homeomorphic to the Cantor set $2^\omega$
whenever $0 < \alpha < \omega_1$.
In view of (13), this is equivalent to making $X_\alpha$
zero dimensional.  For $\alpha = 1$, we simply choose
$X_1$ so that $X_1 \cong 2^\omega$.
Then, for larger $\alpha$, just make sure that in (9),
we always have
$|(\sigma^{\alpha + 1}_\alpha)\iv\{q_\alpha\}| = 1$,
which will hold if in (18), we choose
$h_\alpha \in C(X_\alpha \backslash \{q_\alpha\}, 2)$
(identifying $2 = \{0,1\}$ as a subset of $Q$).
To make this choice, and satisfy (19):
First, let $A_j$, for $j \in \omega$, be disjoint
infinite subsets of $\omega$ such that for each $P \in \PP_\alpha$,
if $q_\alpha \in P$  then for some $j$,
$ r^n_\alpha  \in \ker(P)$ for all $n \in A_j$.
Next, let $X_\alpha = K_0 \supset K_1 \supset K_2 \supset \cdots$,
where each $K_i$ is clopen, $\bigcap_i K_i = \{q_\alpha\}$,
and, for each $j$, there are
infinitely many even $i$ and
infinitely many odd $i$ such that
$K_{i} \backslash K_{i + 1} \cap \{  r^n_\alpha : n \in A_j\} \ne \emptyset$.
Now, let $h_\alpha$ be
$0$ on $K_{i} \backslash K_{i + 1}$ when $i$ is even and
$1$ on $K_{i} \backslash K_{i + 1}$ when $i$ is odd.

For $(\beta)$,
construct $X$ so that each $X_\alpha$
is homeomorphic to the \textit{Menger sponge}, $\MS$,
and all the maps $\sigma^\beta_\alpha$ are monotone.
Then
$\dim(X) = 1$ will follow from the fact that $X$ is an inverse
limit of one dimensional spaces.

For monotonicity of the $\sigma^\beta_\alpha$,
it suffices to ensure that each
$\sigma^{\alpha + 1}_\alpha$ is monotone.  By Condition (8),
that will follow if we make
$(\sigma^{\alpha + 1}_\alpha)\iv\{q_\alpha\}$ connected;
in fact we shall make
$(\sigma^{\alpha + 1}_\alpha)\iv\{q_\alpha\}$ homeomorphic to $[0,1]$,
as in the proof of Theorem \ref{thm-Aronsz-basic}.
But we also need to verify inductively that
$X_\alpha \cong \MS$.  At limits, this follows from
Lemma 2.5 of \cite{HK3}.  At successor stages,
we assume that $X_\alpha \cong \MS$ and
identify $[0,1]$ as a subspace of $Q$, so that
$X_{\alpha+1}$ may be the $Y$ of Lemma \ref{lem-MS}.

$(\gamma)$ is proved analogously
to $(\beta)$.
Construct $X_\alpha \cong Q$ rather than $\MS$,
applying the results about $Q$ in \cite{vM}\S3.
As in \cite{vM}\S2, all the
$\sigma^\beta_\alpha$ are cell-like $Z^*$-maps.
\end{proof}

\section{Chains of Clopen Sets}

The double arrow space has an uncountable chain
(under $\subset$) of clopen sets of real type.
This cannot happen in an Aronszajn compactum:

\begin{lemma}
\label{lemma-no-real-type}
If $X$ is an Aronszajn compactum and $\EE$ is an
uncountable chain
of clopen subsets of $X$, then $\EE$ cannot be of real type.
\end{lemma}
\begin{proof}
Suppose that $\EE$ is such a chain.  Deleting some elements
of $\EE$, we may assume that $(\EE, \subset)$
is a dense total order.
Let $\DD$ be a countable dense subset of $\EE$.
Since $X$ is an Aronszajn compactum, there is a map
$\varphi : X \onto Z$, where $Z$ is a compact metric space,
$A = \varphi\iv(\varphi(A))$ for all $A \in \DD$,
and $\{y \in Z : |\varphi\iv\{y\}| > 1\}$ is countable.
Since $\DD$ is dense in $\EE$, the sets
$\varphi(B)$ for $B \in \EE$ are all different.
Each $\varphi(B)$ is closed, and only countably many
of the $\varphi(B)$ can be clopen.
Whenever $\varphi(B)$ is not clopen, choose
$y_B \in \varphi(B) \cap \varphi(X \backslash B)$.
Since $\DD$ is dense in $\EE$,
these $y_B$ are all different points, so
there are uncountably many such $y_B$.
But $\varphi\iv\{y_B\}$ meets both $B$ and $X \backslash B$,
so each $|\varphi\iv\{y_B\}| \ge 2$, a contradiction.
\end{proof}

Note that if this argument is applied with a chain of clopen sets in
the double arrow space, then
the  $|\varphi\iv\{y_B\}|$ will be exactly $2$.

\begin{lemma}
If $X$ is any separable space,
and $\EE$ is an uncountable chain
of clopen subsets of $X$, then $\EE$ must be of real type.
\end{lemma}
\begin{proof}
If $D \subseteq X$ is dense, then
$(\EE, \subset)$ is isomorphic to a chain in $(\PP(D), \subset)$.
\end{proof}

\begin{corollary}
If $X$ is a separable Aronszajn compactum and $\EE$ is a
chain of clopen subsets of $X$, then $\EE$ is countable.
\end{corollary}

Note that if $X$ is a zero dimensional compacted Aronszajn
line which is also Suslin (see Lemma \ref{lemma-sus-ar}),
then $X$ has
uncountable chain of clopen sets, but $X$ is not separable.

config: arxiv


text:
\begin{document}
\author{Serge E. \textsc{Salez}}
\title{The \Erdos\ conjecture\\
New modular equations \\
and checking up to $N=10^{17}$}
\date{}
\maketitle

\parindent=0pt

\begin{abstract}
In 1999 Allan Swett \cite{Swett} checked (in 150 hours) the  \Erdos\ conjecture up to $N=10^{14}$ with a sieve based on a single modular equation. After having proved the existence of a  "complete" set of seven modular equations (including three new ones), this paper offers an optimized sieve based on these equations. A program written in  C++ (and given elsewhere) allows then to make a checking whose running time, on a typical computer
\footnote{AMD TurionII Dual-Core Mobile M250 ($64$ bits, $16\,100$ MIPS).},
range from few minutes for $N=10^{14}$ to about 16 hours for $N=10^{17}$.
\end{abstract}

\section{Basic formulas}

A fraction is said to be \textit{$k$-Egyptian} if it is the sum of at most $k$ positive unit fractions (i.e with numerator equal to 1). The \Erdos\ conjecture claims that $4/n$ is a 3-Egyptian fraction for any $n>1$.

\subsection{Reduction} \label{reduction}

Through the identities
\[ \frac{1}{t}=\frac{1}{t+1}+\frac{1}{t(t+1)} \]

\[ \frac{2}{2t-1}=\frac{1}{t}+\frac{1}{t(2t-1)} \]
it is  equivalent (for $n>2$) to require having \textit{exactly} 3 \textit{different} unit fractions,  what we shall do thereafter.\\

On the other hand, the identities

\[ \frac{4}{3t-1}=\frac{1}{t}+\frac{1}{3t-1}+\frac{1}{t(3t-1)} \]

\[ \frac{4}{4t-1}=\frac{1}{t}+ \frac{1}{t(4t-1)} \]

\[ \frac{4}{8t-3}=\frac{1}{2t}+\frac{1}{t(8t-3)}+\frac{1}{2t(8t-3)}\]

show that the conjecture is verified  if $n=-1\mod 3$ or $n=-1 \mod 4$ or $n=-3 \mod 8$. Moreover, if $4/n$ is 3-Egyptian then $4/kn$ is too.\\

To conclude, it is then sufficient to prove that $4/p$ is 3-Egyptian, for any prime integer $p$ such that $p=1 \mod 24$.

\subsection{Rosati's formulas}

The following proposition is due (according to Mordell\footnote{
So, unlike most paper, we don't attribute to Mordell what Mordell himself attribute to others mathematicians. In his book \cite{Mordell}, often quoted, the four pages given to this conjecture doesn't introduce a personal work but report briefly some papers whose sources are scrupulously pointed out : hence, it is absolutely incorrect to speak of "Mordell's theorem" or of "Mordell's formulas". On a different scale, it should be better not to remake what was done with Pell's equation.})

to Rosati \cite{Rosati}. The proof needs only simple calculations and has been given many a time. This one is nevertheless original and standardize the notations.\\

We set $\mathcal{A}=\Z$. Let $\mathcal{A}_+$ be the set of strictly positive elements of $\Z$. In this context, we call \textit{prime element} an odd prime integer.

\begin{prop}  \label{Rosati}
Let $p$ a prime element. The fraction $4/p$ is 3-Egyptian if and only if there exists four elements of $\mathcal{A}_+$ denoted by $A$, $B$, $C$, $D$ such that
\begin{equation}  \label{Rosati_1}
                 4ABCD=A+B+pC  \et (ABD,p)=1
\end{equation}
or
\begin{equation}  \label{Rosati_2}
                 4ABCD=p(A+B)+C  \et (ABCD,p)=1
\end{equation}
\end{prop}

\begin{dem}
If we assume that $4/p$ is 3-Egyptian, then there exists 3 elements of $\mathcal{A}_+$ denoted by $X_1,X_2,X_3$ such that
\begin{equation}\label{Z}
   \dfrac{4}{p}=\dfrac{1}{X_1}+\dfrac{1}{X_2}+\dfrac{1}{X_3}
\end{equation}

The $X_i$ are not all divisible by $p$ for otherwise we would have
\[ 4=\dfrac{1}{X_1/p}+\dfrac{1}{X_2/p}+\dfrac{1}{X_3/p}\leqslant 3 \]

In view of (\ref{Z}) it follows that
\[ 4X_1X_2X_3=p(X_2X_3+X_3X_1+X_1X_2) \]
which shows that $p$ divides at least one of the $X_i$.
Hence we may set $x_i=X_i/p_i$ where
\[ p_1=p_2=1,\quad p_3=p \et (x_1x_2,p)=1\]
or
\[ p_1=p_2=p,\quad p_3=1 \et (x_3,p)=1\]
depending on $p$ divides exactly one or two $X_i$.\\

Therefore, since $p_2p_3=p$
  \[ 4p_1p\;x_1x_2x_3=p(p_2p_3x_2x_3+p_3p_1x_3x_1+p_1p_2x_1x_2) \]
and hence
  \[ 4x_1x_2x_3=p_3x_2x_3+p_3x_3x_1+p_2x_1x_2 \]
We set
 $D=(x_1,x_2,x_3)$ and $x'_i=x_i/D$. Then
  \[ 4Dx'_1x'_2x'_3=p_3x'_2x'_3+p_3x'_3x'_1+p_2x'_1x'_2 \]

At last we set
     \[ A=(x'_2,x'_3),\quad B=(x'_3,x'_1),\quad C=(x'_1,x'_2) \]

Since $(x'_1,x'_2,x'_3)=1$ it follows that $A,B,C$ are pairwise relatively prime. So, we may write
         \[x'_1=BCt_1,\quad x'_2=CAt_2,\quad x'_3=ABt_3\]
where $t_i\in\mathcal{A}_+$ are pairwise relatively prime. We note that
		\[(t_1,A)=(t_2,B)=(t_3,C)=1  \]
With these notations, we have
\[ 4D\,BC\,CA\,AB\,t_1t_2t_3=p_3CA\,AB\,t_2t_3 +  p_3AB\,BC\,t_3t_1 + p_2BC\,CA\,t_1t_2  \]
and hence
\[ 4ABCDt_1t_2t_3=p_3 At_2t_3 +  p_3 Bt_3t_1 + p_2 Ct_1t_2  \]

It follows that $t_1\mid p_3At_2t_3$  which reduce to $t_1\mid p_3$ and hence $t_1=1$ for $(x_1,p_3)=1$. Similar arguments lead to $t_2=1$ and $t_3=1$. Finally
\begin{equation} \label{Rosati_eq}
  4ABCD=p_3A+p_3B+p_2C
\end{equation}

Conversely, if we assume that $A,B,C,D$ verify (\ref{Rosati_eq}), we divide by $pABCD$ and then

\[ \frac{4}{p}=\frac{1}{p_1BCD}+\frac{1}{p_2ACD}+\frac{1}{p_3ABD} \]
which shows that $4/p$ is 3-Egyptian
\footnote{For all purpose, we may write $x=T/A$, $y=T/B$, $z=T/C$ where $T=ABCD$.}.
\end{dem}

We observe that if $p$ is not prime, (\ref{Rosati_eq}) is still \textit{sufficient} but no more \textit{necessary.}

\subsection{Notations}
Henceforth, we systematically make use of the notations of  Proposition~\ref{Rosati}.
We add also $E\in\mathcal A_+$ and $F\in\mathcal A_+$ as follow.\\

By (\ref{Rosati_eq}) and since $(C,p_3)=1$, we have $C\mid A+B$. If we write $E=(A+B)/C$ then $E\in\mathcal A_+$ and (\ref{Rosati_eq}) is equivalent to

\begin{equation} \label{Rosati_E}
 \begin{cases}
	4ABD=p_3 E + p_2\\
	A+B=CE
\end{cases}
\end{equation}

The relation (\ref{Rosati_eq}) may be rewritten $(4BCD-p_3)A=p_3 B+p_2 C $. We set $F=4BCD-p_3$  and then (\ref{Rosati_eq}) is equivalent to
\begin{equation} \label{Rosati_F}
 \begin{cases}
	FA=p_3 B+p_2 C2\\
	F+p_3=4BCD
\end{cases}
\end{equation}
The second equation of (\ref{Rosati_F}) shows that $F\in\mathcal A$, the first one that $F\in\mathcal A_+$.\\

Moreover, by (\ref{Rosati_E}) we have $ 4(CE-B)BD=p_3 E + p_2$ and then $ (4BCD-p_3)E=4B^2D+p_2 $. Whence
\begin{equation} \label{Rosati_FE}
 	FE=4B^2D+p_2
\end{equation}

\section{Generalization}

\subsection{Definitions}
Like for the integers, we say that a rational fraction is $k$-Egyptian if it is the sum of at most $k$ inverses of polynomials of $\Z[X]$.\\

We set $\mathcal{A}=\Z[X]$. Let $\mathcal{A}_+$ be the set of polynomials of $\Z[X]$ whose leading coefficient is strictly positive. In this context, we call \textit{prime element} an  irreducible polynomial of $\mathcal{A}_+$.\\

In the ring $\mathcal{A}$, the fundamental theorem of arithmetic is true and the GCD is unique if we request it has to be in $\mathcal{A}_+$. Hence, the Proposition~\ref{Rosati} holds also in this new context, without any change neither in the text nor in the proof. It is the same for $E$ and $F$ as well as the related equations.

\subsection{First application}

\begin{prop} \textnormal{(Schinzel's Theorem \label{Schinzel})}\\
Let $a>0$ and $b$ such as $(a,b)=1$. If $4/(at+b)$ is 3-Egyptian
\footnote{$at+b$ is supposed to be a polynomial (abuse of notation).}

then $b$ is a quadratic non residue modulo $a$.
\end{prop}

\begin{dem}
We write $p(t)=at+b$. There exists $\tau$ such as the polynomials $p,A,B,C,D,E$ take strictly positive values for $t>\tau$.\\

Depending on the case, the equation  $ 4(B-CE)BD=p_3 E + p_2$ may be written
  \[ 4(CE-B)BD=E+p  \ou 4(CE-B)BD=pE+1\]
So we have
  \[ p=(4BCD-1)E-4B^2D  \ou  pE^2=(4BCDE-1)E-4B^2DE\]
and then, if we write $D'=DE$
  \[ p=-4B^2D \mod 4BCD-1 \ou  pE^2=-4B^2D' \mod 4BCD'-1\]

If $b$ is a quadratic residue modulo $a$, there exists an integer $k>\tau$ such that $ak+b$ is a square.
If $t=k$, it follows from propriety of the Jacobi symbol
\footnote{The same notations  $p,A,B,C,D,E$ are still used for the values at $t=k$ of these polynomials. By the way, a similar calculation using the Kronecker symbol is made in the paper of Yamamoto\cite{Yamamoto}.}

  \[ \left( \dfrac{p}{4BCD-1} \right) =  \left( \dfrac{-4B^2D}{4BCD-1} \right) =-1\]
which contradicts the fact that $p$ is a square. Idem with $pE^2$.\\

More precisely, if we write $D=2^\alpha m$ where $m$ is odd, we obtain

 \[ \left( \dfrac{-4B^2D}{4BCD-1} \right)=-  \left( \dfrac{D}{4BCD-1} \right) =-\left( \dfrac{2}{4BCD-1}\right)^\alpha \left( \dfrac{m}{4BCD-1}\right) \]

If $\alpha >0$ then $4ABCD-1=7 \mod 8$ and this implies
 \[\left( \dfrac{2}{4BCD-1}\right)=1 \]
For the second factor, using the law of quadratic reciprocity, we have
 \[ \left( \dfrac{m}{4BCD-1} \right)=(-1)^{(m-1)/2}\left( \dfrac{4BCD-1}{m} \right)=(-1)^{(m-1)/2}\left( \dfrac{-1}{m} \right) \]
and then
 \[ \left( \dfrac{m}{4BCD-1} \right)= (-1)^{(m-1)/2}(-1)^{(m-1)/2}=1\]
\end{dem}

\subsection{Modular equations }

For greater convenience, we call \textit{modular equation} a modular equation (or a system of modular equations) \emph{with constant coefficients}.\\

Since $A$ and $B$ play symmetrical roles, we may suppose
\footnote{The arbitrary definition of $F$ ($A$ is factored out rather than $B$) was made in anticipation of this relation. Otherwise we could not have $\dg F=0$.}

that $\dg B\leqslant \dg A$, where $\dg$ is the degree of a polynomial.

\begin{lemme} \label{deg}
Let $p$ be a prime polynomial of degree 1.
\begin{enum}
\item[i)] If the relation (\ref{Rosati_1}) $ 4ABCD=A+B+pC$ holds, then

\begin{subequations}\label{deg1}
\begin{align}
\dg A=1 \qquad \dg B=0 \qquad \dg C=0 \qquad \dg D=0 \qquad \label{deg1a}\\
\dg A=0 \qquad \dg B=0 \qquad \dg C=0 \qquad \dg D=1 \qquad \label{deg1b}\\
\dg A=1 \qquad \dg B=0 \qquad \dg C=1 \qquad \dg D=0 \qquad \label{deg1c}
\end{align}
\end{subequations}

\item[ii)] If the relation (\ref{Rosati_2}) $4ABCD=p(A+B)+C$ holds, then

\begin{subequations}\label{deg2}
\begin{align}
\dg A=0 \qquad \dg B=0 \qquad \dg C=0 \qquad \dg D=1  \qquad \label{deg2a}\\

\dg A=1 \qquad \dg B=0 \qquad \dg C=0 \qquad \dg D=1  \qquad \label{deg2b}\\
\dg A=1 \qquad \dg B=0 \qquad \dg C=1 \qquad \dg D=0  \qquad \label{deg2c}\\
\dg A=2 \qquad \dg B=1 \qquad \dg C=0 \qquad \dg D=0  \qquad \label{deg2d}
\end{align}
\end{subequations}

\end{enum}
\end{lemme}

\begin{dem}
Since $\dg B\leqslant \dg A$ then $\dg (A+B)=\dg A$. Hence, by $C\,E=A+B$, we have $\dg C\leqslant \dg A$\\

i) By (\ref{Rosati_1}) it follows
\begin{equation}
     (4ABD-p)C=A+B  \label{P1}
\end{equation}
and
\begin{equation}
       (4BCD-1)A=B+Cp  \label{P2}
\end{equation}

By (\ref{P1}) we have $\dg (4ABD-p)\leqslant \dg A$ and hence  $\dg (4ABD-p)\leqslant \dg ABD$.\\

* Case $\dg (4ABD-p)=\dg ABD$ \\
By (\ref{P1}) we have  \[ \dg ABD+\dg C=\dg A \]
and then
    \[ \dg B+\dg C+\dg D=0  \label{P1.1} \]
This result implies, in view of (\ref{P2}), that
            \[\dg A=\dg p=1\]

* Case $\dg (4ABD-p)<\dg ABD$. In this case\\
  \[ \dg ABD=\dg p=1  \et   \dg (4ABD-p)=0 \]
We have, by the first equation
           \[ \dg B=0 \et \dg A+\dg D=1 \]
and by the second  together with (\ref{P1})
  \[ \dg C=\dg A \]

ii) By (\ref{Rosati_2}) it follows
\begin{equation}
     (4ABD-1)C=p(A+B)  \label{P3}
\end{equation}
and
\begin{equation}
       (4BCD-p)A=pB+C  \label{P4}
\end{equation}

By (\ref{P3}) we have
       \[ \dg ABD+\dg C=\dg A+\dg p \]
and then
       \[ \dg B+\dg C+\dg D=1 \]
Here $F=4BCD-p$. Then
   \[ \dg F \leqslant1 \et \dg (pB+C)=\dg pB\]
and together with (\ref{P4})
   \[  \dg F+\dg A=\dg B+1 \]

* Case $\dg F= 1$. In this case $\dg A=\dg B$. On the other hand, as $\dg F=1$, there exists $t_0\in\R$ such that $F(t_0)=0$. From $FE=4B^2D+1$, it follows $4B^2(t_0)D(t_0)+1=0$ and then $D(t_0)<0$. Therefore $\dg D=1$ and $\dg B=\dg C=0$.\\

* Case $\dg F= 0$. In this case $\dg A=\dg B+1 $.
\end{dem}

\begin{prop} \label{eqmod}
Let $p$ be a prime polynomial of degree 1. The fraction $4/p$ is 3-Egyptian if and only if one of the next 7 modular equations  holds.
\begin{subequations}\label{eqmod1}
\begin{align}
\hspace{1cm}& B+pC=0 \mod 4BCD-1   \label{eqmod1a} \\
\hspace{1cm}& p+E=0 \mod 4AB \et A+B=0 \mod E \qquad\label{eqmod1b} \\
\hspace{1cm}& p+E+4B^2D=0 \mod 4BDE   \label{eqmod1c}
\end{align}
\end{subequations}

\begin{subequations}\label{eqmod2}
\begin{align}
\hspace{1cm}&  pE+1=0 \mod 4AB \et A+B=0 \mod E  \label{eqmod2a} \\
\hspace{1cm}&  p+F=0 \mod 4BC \et pB+C=0 \mod F   \label{eqmod2b}  \\
\hspace{1cm}&  p+F=0 \mod 4BD \et 4B^2D+1=0 \mod F  \label{eqmod2c} \\
\hspace{1cm}&   p+F=0 \mod 4CD \et p^2+4C^2D=0 \mod F \label{eqmod2d}
\end{align}
\end{subequations}
where $(A,B)=(B,C)=(C,D)=(4ABD,E)=(4BCD,F)=1$.

\end{prop}

\begin{dem} $\quad$ The $[\:]$ refer to the equations of the Lemma.
\begin{itemize}
\item[(\ref{eqmod1})] Here "(\ref{Rosati_eq}) is equivalent to (\ref{Rosati_E})" is written
	 \[ 4ABCD=A+B+pC \iff (p+E=4ABD \et A+B=CE)\]

\begin{itemize}[leftmargin=1ex]
\item[{(\ref{eqmod1a})}] Case [\ref{deg1a}] : $B,C,D$ are constants. If we suppose that (\ref{Rosati_eq}) holds, then
  \[  B+pC=(4BCD-1)A=0 \mod (4BCD-1) \]
Conversely, we set
  \[ A=\dfrac{B+pC}{4BCD-1} \]

\item[(\ref{eqmod1b})] Case [\ref{deg1b}] : $A,B,E$ are constants. If we suppose that (\ref{Rosati_eq}) holds, then
  \[  p+E=4ABD=0 \mod 4AB \et A+B=CE=0 \mod E \]
Conversely, we set
  \[ D=\dfrac{p+E}{4AB} \et C=\dfrac{A+B}{E} \]

\item[(\ref{eqmod1c})] Case [\ref{deg1c}] : $B,D,E$ are constants.
If we suppose that (\ref{Rosati_eq}) holds, then
	\[ p+E=4(CE-B)BD \]
Hence
  \[ p+E+4B^2D=4BDEC=0 \mod 4BDE \]
Conversely, we set
  \[  A=\dfrac{p+E}{4BD} \et C=\dfrac{p+E+4B^2D}{4BDE} \quad
  \left( C E = A+B\right) \]
\end{itemize}

\item[(\ref{eqmod2})] Here "(\ref{Rosati_eq}) is equivalent to (\ref{Rosati_E})" is written
 \[ 4ABCD=p(A+B)+C \iff (4ABD=pE+1 \et A+B=CE) \]
and "(\ref{Rosati_eq}) is equivalent to (\ref{Rosati_F})" is written
\[ 4ABCD=p(A+B)+C \iff (p+F=4BCD \et pB+C=FA) \]
where $ F=4BCD-p \et FE=4B^2D+1 $

\begin{itemize}[leftmargin=1ex]
\item[(\ref{eqmod2a})] Case [\ref{deg2a}] : $A,B,E$ are constants. If we suppose that (\ref{Rosati_eq}) holds, then
  \[ pE+1=4ABD=0 \mod 4AB \et A+B=CE=0 \mod E \]
Conversely, we set
  \[ D=\dfrac{pE+1}{4AB} \et C=\dfrac{A+B}{E} \]

\item[\textit{In}]\textit{the next cases $\dg A=\dg B+1$ and then $\dg F=0$}.\\

\item[(\ref{eqmod2b})] Case [\ref{deg2b}] : $B,C,F$ are constants. If we suppose that (\ref{Rosati_eq}) holds, then
  \[ p+F=4BCD=0 \mod 4BC \et pB+C=FA=0 \mod F \]
Conversely, we set
  \[ D=\dfrac{p+F}{4BC} \et A=\dfrac{pB+C}{F} \]

\item[(\ref{eqmod2c})] Case [\ref{deg2c}] : $B,D,F$ are constants. If we suppose that (\ref{Rosati_eq}) holds, then
  \[ p+F=4BCD=0 \mod 4BD \et 4B^2D+1=EF=0 \mod F \]
Conversely, we set
  \[ C=\dfrac{p+F}{4BD} \quad E=\dfrac{4B^2D+1}{F} \et A=CE-B\]
We observe that $FA=pB+C$.

\item[(\ref{eqmod2d})] Case [\ref{deg2d}] : $C,D,F$ are constants. If we suppose that (\ref{Rosati_eq}) holds, then
    \[ p+F=4BCD=0 \mod 4CD \et pB+C=FA=0 \mod F \]
As
	\[ pB+C=p\dfrac{p+F}{4CD}+C=\dfrac{p^2+pF+4C^2D}{4CD} \]
it follows, since $(4CD,F)=1$,
  \[ p^2+4C^2D=0 \mod F \]
Conversely, we set
  \[ B=\dfrac{p+F}{4CD} \et A=\dfrac{pB+C}{F} \]
\end{itemize}

\end{itemize}
\end{dem}

We observe that, if $p$ is a prime polynomial of degree 1, the Lemme\ref{deg} shows that there are only 7 distinct cases, according to the degree of $A$, $B$, $C$, $D$ ($\dg B \leqslant\dg A $). By the Proposition~\ref{eqmod}, each case is connected to a modular equation. Hence, there exist only 7 distinct modular equations with \textit{constant coefficients}. So, we can build an algorithm giving the set (maybe empty) of \textit{all} the way to write $4/p$.

\subsection{Application to the integers}

The proof of the Proposition~\ref{eqmod} $\,$gives us formulas for $A,B,C,D$. These variables take strictly positive values when the given data are strictly positive and one of the equation (\ref{Rosati_1}) or (\ref{Rosati_2}) holds. Hence we have the following corollary.

\begin{corolaire}
Let $p$ be an odd prime integer. The fraction $4/p$ is 3-Egyptian if and only if one of the 7 modular equations of the Proposition~\ref{eqmod} holds.
\end{corolaire}

Thereafter, we call these equations \textit{reference equations} not only for the polynomials but for the integers too.

\subsubsection*{Comparison with previous results}

Four of these equations have been well known for a long time, but the others are new.

\begin{enum}
\item  Rosati \cite{Rosati} (1954) gives only one condition for (\ref{Rosati_1}) and one for (\ref{Rosati_2}). Although they are not written in a modular form, his equations (3) and (6)  are equivalent to (\ref{eqmod1a}) and (\ref{eqmod2a}).

\item Yamamoto \cite{Yamamoto} (1965) gives two conditions for (\ref{Rosati_1}) and two for (\ref{Rosati_2}). Written in a modular  form, his equations (3) to (6) are equivalent (not in the same order) to (\ref{eqmod1a}), (\ref{eqmod1b}), (\ref{eqmod2a}), (\ref{eqmod2b}).
\end{enum}

Polynomials explain why the Yamamoto equivalent equations give distinct results. Even better, they give us three new equations.

\subsubsection*{"Complete" set of modular equations}

Regarding prime polynomials of degree 1, the 7 reference equations  form a complete set
\footnote{Moreover, example~2 below shows that these equations are independent.},
that is, if a modular equation $n=b \mod a$ (where $(a,b)=1$) is not equivalent to one of the reference equations then $4/(at+b)$ cannot be an 3-Egyptian fraction. This feature does not hold for integers : it may exist a process using such an equation and leading to the conclusion that $4/n$ is a 3-Egyptian fraction. But, in this case, this process  has to be of a still unknown new type.

\subsection{Examples}

\textbf{Example 0.} Of course, we may find the identities of paragraph \ref{reduction}. Here, we don't look after \emph{all} the way to write $4/p$, just those given in the paragraph.

\begin{itemize}
\item $p=3t-1$ verifies (\ref{eqmod1a}) : $p+1=0 \mod 3$ \\
where $B=C=D=1$, and hence $A = (p+1)/3=t$.
\item $p=4t-1$ verifies (\ref{eqmod1b}) : $p+1 = 0 \mod 4$ \\
where $A=B=E=1$ and hence $C=2$ et $D=(p+1)/4=t$.
\item $p=8t-3$ verifies (\ref{eqmod1b}) : $p+3=0 \mod 8$ \\
where $A=1$, $B=2$, $E=3$ and hence $C=1$, $D=(p+3)/8=t$.
\end{itemize}

\textbf{Example 1.} $\quad p=24\cdot 5t-23 \qquad (p=1 \mod 24 \et p=2 \mod 5)$\\
We give all the way to write $4/p\,$ and the distinctive feature is that the 7 reference equations (shown in $[\;]$) are present. We don't know another analogous example where $p=1 \mod 24$.
\begin{enum}
\item[{[\ref{eqmod1a}]}] $\quad \dfrac{4}{p}=\dfrac{1}{p}\left(\dfrac{1}{4}+\dfrac{1}{4(16t-3)} \right)+\dfrac{1}{2(16t-3)}$\\

\item[{[\ref{eqmod1b}]}] $\quad \dfrac{4}{p}=\dfrac{1}{p}\left(\dfrac{1}{10(6t-1)}+\dfrac{1}{2(6t-1)} \right)+\dfrac{1}{5(6t-1)}$\\

\item[{[\ref{eqmod1c}]}] $\quad \dfrac{4}{p}=\dfrac{1}{p}\left(\dfrac{1}{10t}+\dfrac{1}{10t(6t-1)} \right)+\dfrac{1}{5(6t-1)}$\\

                         $\quad \dfrac{4}{p}=\dfrac{1}{p}\left(\dfrac{1}{2t}+\dfrac{1}{2t(15t-1)} \right)+\dfrac{1}{2(15t-1)}$\\

\item[{[\ref{eqmod2a}]}] $\quad \dfrac{4}{p}=\dfrac{1}{5(21t-4)}+\dfrac{1}{2(21t-4)} +\dfrac{1}{10(21t-4)\,p}$\\

\item[{[\ref{eqmod2b}]}] $\quad \dfrac{4}{p}=\dfrac{1}{5(6t-1)}+\dfrac{1}{2(6t-1)(100t-19)} +\dfrac{1}{10(6t-1)(100t-19)\,p}$\\

                     $\quad \dfrac{4}{p}=\dfrac{1}{5(6t-1)}+\dfrac{1}{10(6t-1)(20t-3)} +\dfrac{1}{2(6t-1)(20t-3)\,p}$\\

                     $\quad \dfrac{4}{p}=\dfrac{1}{2(15t-1))}+\dfrac{1}{(15t-1)(16t-3)} +\dfrac{1}{2(15t-1)(16t-3)\,p}$\\

\item[{[\ref{eqmod2c}]}] $\quad \dfrac{4}{p}=\dfrac{1}{5(6t-1)}+\dfrac{1}{10(6t-1)(21t-4)} +\dfrac{1}{10(21t-4)\,p}$\\

\item[{[\ref{eqmod2d}]}] $\quad \dfrac{4}{p}=\dfrac{1}{5(6t-1)}+\dfrac{1}{10(120t^2-43t+4)} +\dfrac{1}{10(6t-1)(120t^2-43t+4)\,p}$\\
\end{enum}

\textbf{Example 2.}$\quad$ In this example, each $p\,$ is of the form $p=24\cdot 583t+b$.  At the opposite of the example 1, the distinctive feature is that, for some $b$, there is only one way to write $4/p$. A value of $b$ is given for each reference equation.
\begin{enum}

\item[{[\ref{eqmod1a}]}] $\quad  p=24\cdot 583t-911 \quad  (p=1 \mod 24 \et p=255 \mod 583)$\\

      $\quad \dfrac{4}{p}=\dfrac{1}{p}\left( \dfrac{1}{2\cdot 73} +\dfrac{1}{6(16t-1)} \right)+\dfrac{1}{3\cdot 73(16t-1)}$\\

\item[{[\ref{eqmod1b}]}] $\quad p=24\cdot 583t-119 \quad  (p=1 \mod 24 \et p=464 \mod 583)$\\

      $\quad \dfrac{4}{p}=\dfrac{1}{p}\left( \dfrac{1}{66t} +\dfrac{1}{53t} \right)+\dfrac{1}{66\cdot 53t} $\\

\item[{[\ref{eqmod1c}]}] $\quad p=24\cdot 583t-1127 \quad  (p=1 \mod 24 \et p=39 \mod 583)$\\

     $\quad \dfrac{4}{p}= \dfrac{1}{p}\left(\dfrac{1}{22t} +\dfrac{1}{ 2t(159t-11)}\right) +\dfrac{1}{ 22(159t-11)} $\\

\item[{[\ref{eqmod2a}]}] $\quad p=24\cdot 583t-1799 \quad  (p=1 \mod 24 \et p=533 \mod 583)$\\

     $\quad \dfrac{4}{p}= \dfrac{1}{50\cdot 1749(70t-9)} +\dfrac{1}{50(70t-9)} +\dfrac{1}{1749(70t-9)\,p}$\\

\item[{[\ref{eqmod2b}]}] $\quad p=24\cdot 583t-11159 \quad  (p=1 \mod 24 \et p=501 \mod 583)$
\begin{multline*}
     \dfrac{4}{p}= \dfrac{1}{ 22(159t-125)}
     +\dfrac{1}{ 8(242t-193)(159t-125)}\\
     +\dfrac{1}{88(242t-193)(159t-125)\,p}
\end{multline*}

\item[{[\ref{eqmod2c}]}] $\quad p=24\cdot 583t-503 \quad  (p=1 \mod 24 \et p=80 \mod 583)$\\

     $\quad \dfrac{4}{p}= \dfrac{1}{6\cdot 583t} +\dfrac{1}{ 6\cdot 53t(306t-11)} +\dfrac{1}{583(306t-11)\,p} $\\

\item[{[\ref{eqmod2d}]}] $\quad p=24\cdot 583t-6407 \quad  (p=1 \mod 24 \et p=6 \mod 583)$
\begin{multline*}
     \dfrac{4}{p}= \dfrac{1}{22(159t-71)}
     +\dfrac{1}{22(13992t^2-12655t+2861)}\\
       +\dfrac{1}{11(159t-71)(13992t^2-12655t+2861)\,p}
\end{multline*}

\end{enum}

\section{Modular sieve}

The algorithms setting, for a given integer $n>2$, at least one way (and even more) to write $4/n$ are interesting. However, regarding the checking of the conjecture, an efficient algorithm needs an another point of view
\footnote{This point of view has been used since Rosati's paper (or maybe before).}.\\

We denote by $\N_0$ the set of the integers $n\in\N$ verifying the condition $n=1 \mod 24$. The process described below takes account specifically of the fact that the checked integers are in $\N_0$. On the other hand, we let down the condition that $n$ is prime, which needs too much running time. Regarding the polynomial $at+b$, the correlated conditions are $at+b =1 \mod 24$ (which is equivalent to $a=0 \mod 24$ and $b=1 \mod24$) and the cancellation of the condition $(a,b)=1$.

\subsection{Modular filters}
\textit{Definition : } A \textit{sieve} is a sorted set of filters.\\

\textit{Definition : } A \textit{filter}
\footnote{We use the terminology given by Swett. If an integer $n\in\N_0$ is such that $n\% m\in F$ then $n$ verifies the conjecture and $n$ is "trapped" by the filter. Otherwise $n$ "pass through".}
\textit{modulo} $m$ is a set $F$ such that for any $n\in\N_0$
\[ n\%m\in F \Rightarrow 4/n \;\textnormal{is 3-Egyptian}\]
where  $n\%m$ is the residue of $n$ modulo $m$ (notation borrowed from C language).\\

For $a>0$, we denote by $\Omega_a$ the set of $b\in\Z$ such that
$4/(at+b)$ is a 3-Egyptian fraction. If $m$ is odd, we set
$ S_m=\left(\Omega_{[m,24]}\cap\N_0\right)\%m $
where $[u,v]=\ppcm(u,v)$. It follows some obvious proprieties.

\begin{enumerate}[label =\textit{\roman*)}]
\item  if $q\mid a$ then $\Omega_q\subset\Omega_a$.
\item if $b_1=b_2 \mod a$ then
$b_1\in\Omega_a \Rightarrow b_2\in\Omega_a$.
\item if $n\in \Omega_a$ ($n>0$) then $4/n$ is a 3-Egyptian fraction.
\item if $n\in\N_0$ then $\left( n\in\Omega_{[m,24]} \iff n\%m\in S_m \right)$, which shows that $S_m$ is a filter modulo $m$.
\item \label{reduit} if $n\in\N_0$ and if $q \mid m$ then $n\%q \in S_q \Rightarrow n\%m \in S_m$.
\end{enumerate}

\textit{Définition :} We say that $n\in\N_0$ is \emph{certified} if there exists $m$ such that $n\%m\in S_m$. We also say that $n$ is certified by $m$ or that $m$ is a \emph{modular certificate} of $n$ (vocabulary borrowed from the complexity theory).\\

\textit{The first results with prime integers} :
\begin{multicols}{2}
\begin{enum}
\item[] $S_5=\{0,2,3\}$
\item[] $S_7=\{0,3,5,6\}$
\item[] $S_{11}=\{0,7,8,10\}$
\item[] $S_{13}=\{0,5,6,8,11\}$
\item[] $S_{17}=\{0,10,11,12,14\}$
\item[] $S_{19}=\{0,8,12,14,15,18\}$
\item[] $S_{23}=\{0,7,10,11,15,17,19,20,21,22\}$
\item[] $S_{29}=\{0,14,18,19,21,26,27\}$
\item[] $S_{31}=\{0,15,22,23,24,27,29,30\}$
\item[] $S_{37}=\{0,5,15,18,22,23,29,32,35\}$
\end{enum}
\end{multicols}

\textit{Some results with odd composite integers} :
\begin{enum}
\item[]$S_{15}=\{7, 10, 13\}$
\item[]$S_{35}=\{0,2,3,5,6,7,8,10,12,13,14,15,17,18,19,20,21,22,23,24 $\\ \hphantom{m} \hfill $,25, 26,27,28,30,31,32,33,34\}$
\item[]$S_{55}=\{0,2,3,5,7,8,10,11,12,13,15,17,18,19,20,21,22,23 $\\ \hphantom{m}\hfil $,24,25,27,28,29,30,32,33,35,37,38,39,40,41,42,43$\\ \hphantom{m}\hfill $,44,45,47,48,50,51,52,53,54\} $

\end{enum}

\subsection{Shortened filters}
If $m$ is composite, some integers $n\in\N_0$ are certified both par $m$ and by one of its divisors (cf. the propriety~\ref{reduit} above). The next definition allows us to point out what is particular to $m$.\\

\textit{Definition} : The \textit{shortened filter} $S^*_m$ is the set of all $x\in S_m$ such that
\begin{center}
	$x\%q\notin S_q$ for any $q\mid m$, $q\neq m$
\end{center}

We observe that if $m$ is prime then $S^*_m=S_m$.\\

\textit{The first (no empty) results}

\begin{multicols}{2}
\begin{enum}
\item[] $S^*_{55}=\{24,39\}$
\item[] $S^*_{65}=\{54,59\}$
\item[] $S^*_{77}=\{46,72\}$
\item[] $S^*_{85}=\{54,74\}$
\item[] $S^*_{95}=\{29,59,79,89\}$
\item[] $S^*_{99}=\{61,79,94\}$
\item[] $S^*_{117}=\{85,106\}$
\item[] $S^*_{119}=\{23,39,57,58,71,88,107,109\}$
\end{enum}
\end{multicols}

\section{Checking of the conjecture}

\subsection{Choice of the progressions}

The checked integers $n$ are in an arithmetic progression, namely they are of the form $n=24k+1$. We call \textit{gap} of the progression the difference between two consecutive terms. Here the gap is $G_0=24$ but if we use some filters $S_m$ we may obtain other progressions whose gap is bigger.\\

With $S_5=\{0,2,3\}$ we check only $n$ such that
\[ n\%24 = 1 \et n\%5\in \{1,4\} \]
and hence, by the Chinese remainder theorem
\[ n\%120\in \{1,49\} \]
The new gap is $G_1=120$, and there are 2 residues : then the mean gap is $g_1=60$. In comparison to 24, we check $2.5$ times fewer integers ($60/24=2.5$).\\

Next, with  $S_7=\{0,3,5,6\}$ we check only $n$ such that
\[ n\%120 \in \{1,49\} \et n\%7 \in \{1,2,4\} \]
and hence
\[ n\%840 \in R_2 \]
where $R_2=\{1,121,169,289,361,529\}$ is the set of the residues
\footnote{It was the choice made by Swett.}.
The new gap is $G_2=840$ and the mean gap is $g_2=140$. In comparison to 24, we check nearly 6 times fewer integers ($140/24=35/6$).\\

We may keep on and use others $S_m$. The checked integers are then of the form
\[ n\%G_i\in R_i \]
where the first values of $G_i=G_{i-1}m_i$ and $\#R_i$ (the number of elements of $R_i$) are set out in the following table.

\[\begin{array}{|c|c|c|c|c|}
\Hline i & m_i &       G_i     & \#R_i &   g_i \\
\Hline 1 & 5 &           120 &         2 &   60 \\
\Hline 2 & 7 &           840 &         6 &   140 \\
\Hline 3 & 11 &       9\,240 &        34 &   272 \\
\Hline 4 & 13 &      120\,120 &       192 &   626 \\
\Hline 5 & 17 &   2\,042\,040 &    1\,507 &  1\,355 \\
\Hline 6 & 19 & 38\,798\,760 &   13\,380 &  2\,900 \\
\Hline 7 & 23 & 892\,371\,480 &   147\,348 &  6\,056 \\
\hline
\end{array}\]\\

Three comments about this table.
\begin{itemize}
\item The first concerns the reduction of $R_i$ (done in the table). If $n=r \mod G_i$ then for any $q$ divisor of $G_i$ we have
$n\%q = r\%q$. Hence, we may remove the residues $r\in R_i$ verifying $r\%q \in S_q$. This reduction is essential, otherwise it's just a useless complicated process.

\item The second concerns the last column : the mean gap $g_i=G_i/\#R_i$ is a good speed indicator. By example, as $6\,056/140\approx 43$, then using $G_7$ rather than $G_2$ leads to check about $43$ times fewer integers and the running time is shortened accordingly.

\item The last concerns the choice of the $m_i$. The usual order is misleading~: each other set of seven integers seems to give a worse $g_7$. Next, with height integers we expect to add $31$ (rather than $29$). However, these two propositions have to be confirmed.
\end{itemize}

\subsection{Optimized sieve}
We denote by $\N_i$ the set of all the integers $n\in\N$ verifying $n\%G_i\in R_i$. As the conjecture is verified for any integer $n\notin N_i$, we have just to check the prime integers of $N_i$.\\

Let $N=10^{17}$ and $M$ the set of all the odd integers $m<5\,000$. We claim that each $n\in\N_7$ has a modular certificate in $M$ if $n<N$ and if $n$ is not a square. It is equivalent to say that $ \N_7 \setminus \bigcup_{m\in M}\Omega_{[m,24]}$ has not any element $n<N$, except squares.\\

We could use this $M$ to prove that the conjecture is verified up to $N$. However, if we want a running time as fewer as possible, we have to optimize the sieve. For this purpose, we remove the useless elements and sort $M$ in order to have at first the most efficient filters
\footnote{The approach mostly hinge on experiments and make use of the shortened filters.}.
By example for  $N=10^{17}$, we give below the set $M=MOD$ which is used in our C++ program.\\

$ MOD =\{$ 3, 5, 7, 11, 13, 17, 19, 23, 4495, 2491, 2627, 4661, 4223, 1505, 4355, 3355, 4509, 4775, 2629, 4565, 4599, 4585, 3955, 3535, 3857, 3115, 3419, 3949, 3395, 3353, 1391, 1199, 3775, 4325, 4031, 2799, 1639, 4475, 2159, 4795, 2961, 1727, 4075, 1791, 4743, 2849, 3595, 1115, 3445, 3263, 2155, 2065, 2515, 2681, 4195, 3223, 2519, 4103, 3731, 4345, 3743, 2439, 1055, 2951, 1799, 4193, 1991, 3047, 2933, 3951, 4147, 1631, 2219, 4615, 3913, 3679, 1535, 2959, 1655, 4123, 1439, 3839, 1319, 3695, 4255, 3895, 1351, 2495, 1835, 2855, 2335, 4529, 1917, 1079, 1559, 1735, 1679, 2165, 4367, 4555, 2359, 2723, 3065, 3899, 3295, 3035, 4927, 3359, 4437, 3635, 4315, 2735, 3241, 4319, 4105, 4069, 1039, 4059, 1247, 3095, 4571, 3665, 1007, 1583, 4895, 1847, 2435, 1765, 2807, 3647, 1343, 2651, 3965, 1511, 2655, 4403, 1151, 887, 2935, 3545, 2879, 1967, 2815, 2399, 4419, 1159, 4487, 3119, 1223, 2039, 4745, 2305, 1103, 4077, 3215, 3715, 2279, 4915, 4873, 1031, 1475, 3865, 2483, 1399, 1823, 3173, 3305, 2241, 3985, 3563, 1349, 1259, 3959, 4415, 3455, 2615, 1487, 3599, 3935, 1759, 3505, 1871, 4879, 4535, 3199, 2045, 1367, 1493, 1919, 3787, 2111, 1975, 2053, 4739, 1231, 4151, 1837, 1213, 3655, 2183, 4135, 4939, 1019, 3023, 3995, 1855, 4265, 4079, 3983, 2575, 1063, 2351, 4985, 2687, 3167, 2447, 2725, 4631, 4595, 4115, 4175, 4055, 4679, 1013, 2239, 4385, 1091, 3429, 1909, 1719, 2365, 3415, 3079, 4955, 1147, 1133, 3191, 3475, 2759, 4405, 2207, 4765, 3431, 1139, 4471, 2727, 4145, 3247, 1279, 1751, 3755, 1087, 4835, 1733, 4645, 1979, 4711, 1177, 1073, 3055, 3239, 2999, 2087, 4855, 4039, 1703, 3527, 4295, 4799, 4207, 4505, 1187, 1109, 1567, 1379, 2119, 2911, 2591, 2015, 3785, 1651, 3155, 1819, 4751, 3719, 4735, 2345, 2831, 2099, 4995, 1427, 2059, 1333, 1069, 1663, 2719, 2063, 4285, 2231, 1093, 1607, 1423, 1411, 1027, 3805, 1769, 1121, 1903, 4063, 4759, 1363, 1973, 4715, 2663, 3863, 1433, 2479, 4703, 3299, 1451, 2339, 1613, 1471, 1619, 3671, 2287, 2367, 3845, 3537, 1591, 3733, 4463, 1271, 1931, 4619, 2903, 2135, 4921, 4685, 4705, 1003, 1429, 1193, 4067, 3275, 4311, 1327, 3015, 1499, 2413, 1237, 1181, 4045, 4081, 3605, 3779, 3103, 2837, 1579, 3439, 1033, 3799, 2333, 1829, 1241, 4393, 2357, 4159, 2699, 3791, 2453, 3625, 2579, 4945, 4127, 1649, 4741, 4871, 1667, 2177, 3835, 1043, 3407, 4919, 4885, 2267, 2693, 2507, 4967, 2327, 4639, 1691, 1549, 2583, 1123, 1717, 1999, 1807, 1933, 4553, 1049, 3479, 1553, 1853, 2543, 4343, 1501, 2743, 3699, 1787, 3989, 1129, 1525, 4445, 1675, 1993, 1301, 2273, 1217, 1843, 4003, 2411, 3245, 3401, 1117, 1789, 3379, 3901, 1831, 1957, 4085, 1507, 1987, 3373, 3893, 1621, 1943, 3937, 1291, 1543, 1571, 2143, 2533, 2767, 3253, 4883, 2551, 2833, 1229, 1877, 1949, 2009, 4391, 1643, 2251, 2729, 3915, 1907, 2243, 2603, 2669, 2897, 3043, 3313, 3739, 1171, 1361, 1817, 1879, 2659, 3623, 4283, 4859, 1537, 2003, 2161, 2389, 2869, 4439, 1099, 1415, 2269, 2293, 2943, 3233, 3967, 4181, 4261, 4559, 4699, 1447, 1895, 1921, 2195, 2939, 3293, 3565, 3607, 3749, 4247, 4591, 4829, 1157, 1417, 1951, 1997, 2179, 2225, 2619, 2785, 3041, 3717, 4583, 4783, 4887, 1283, 1517, 1721, 1747, 1961, 2033, 2117, 2129, 2741, 2803, 2893, 3161, 3589, 3613, 1211, 1273, 1459, 1483, 1811, 1867, 1889, 1971, 2043, 2069, 2149, 2213, 2423, 2709, 2779, 3013, 3149, 3551, 4013, 4097, 4363, 4399, 4589, 4681, 1021, 1097, 1145, 1197, 1297, 1373, 1397, 1555, 1609, 1723, 1773, 1777, 1783, 1801, 2123, 2191, 2259, 2291, 2371, 2407, 2443, 2671, 2845, 3389, 3493, 3725, 4021, 4171, 4351, 4999 $\}$

\subsection{Results}
The checked integers are of the form $n=r+k\times G_7$ where $r\in R_7$ and $0 \leqslant k < K$. With $N=10^{17}$, we take $K=112\,066\,560$. Therefore we check $16\,512\,783\,482\,880$ integers including $51\,732\,427$ squares.\\

For each $m\in MOD$, the number of integers certified by $m$ is given at the same rank in the table below. We may observe that the sum of these numbers added with the number of squares is equal to the number of checked integers.\\

$ \{$ 0, 0, 0, 0, 0, 0, 0, 9223757362766, 3739609092281, 1565954748220, 739166512371, 397180210351, 249398230928, 169050837573, 104088377604, 69101085771, 54368854713, 42523071218, 33206924179, 23406992663, 18890746142, 15211918708, 11968966501, 9473482721, 7560449664, 6273004978, 5196086887, 4344239727, 3485872879, 2944121141, 2498890993, 2067185415, 1765012627, 1499112458, 1259328652, 1044404123, 874512654, 723079141, 617340453, 515245196, 452563855, 390773540, 343076561, 300065591, 260653549, 229207022, 198772233, 174906642, 153551008, 135203129, 118673167, 99017032, 88208940, 78571579, 69928806, 62430095, 55603526, 48999877, 42755472, 38618483, 34775913, 31335757, 27560576, 24702471, 22410294, 19685100, 17852098, 16081935, 14466854, 13141729, 11726132, 10640116, 9491430, 8477371, 7732328, 6982821, 6272905, 5702788, 5268793, 4722801, 4390120, 4019516, 3650026, 3398755, 3140726, 2945648, 2736821, 2552135, 2394011, 2241501, 2100950, 1967613, 1834764, 1721795, 1606462, 1497392, 1396075, 1313066, 1211933, 1135277, 1058550, 992002, 932632, 867721, 807226, 759519, 707804, 665956, 625412, 586324, 552400, 520156, 484882, 459951, 434799, 408981, 385709, 365271, 343865, 322175, 305617, 290089, 275265, 260444, 247858, 235278, 223480, 212702, 201616, 190609, 179783, 171406, 163172, 153416, 146567, 138229, 131604, 125502, 118787, 113170, 106706, 101331, 96415, 91243, 87184, 83366, 79429, 75744, 72027, 68424, 65511, 62568, 59796, 57362, 54579, 52058, 49809, 47397, 45319, 43371, 41545, 39868, 38029, 36324, 34724, 33179, 31866, 30561, 29163, 27958, 26789, 25721, 24727, 23684, 22631, 21616, 20727, 19908, 19064, 18235, 17562, 16604, 15933, 15087, 14379, 13832, 13303, 12763, 12305, 11871, 11311, 10807, 10411, 9972, 9650, 9215, 8834, 8518, 8200, 7898, 7612, 7318, 7021, 6760, 6493, 6232, 5968, 5727, 5528, 5308, 5119, 4884, 4693, 4538, 4369, 4224, 4063, 3940, 3786, 3653, 3530, 3414, 3246, 3117, 3013, 2912, 2787, 2685, 2559, 2455, 2384, 2269, 2185, 2103, 2048, 1985, 1908, 1831, 1777, 1725, 1671, 1621, 1563, 1517, 1465, 1423, 1375, 1335, 1285, 1245, 1196, 1156, 1122, 1076, 1048, 1012, 977, 943, 920, 892, 860, 834, 803, 783, 761, 743, 718, 701, 679, 654, 637, 602, 586, 571, 559, 539, 521, 510, 488, 473, 461, 445, 435, 426, 414, 402, 383, 374, 362, 352, 346, 335, 324, 317, 313, 303, 295, 284, 277, 270, 261, 252, 245, 242, 232, 226, 223, 217, 213, 205, 200, 197, 191, 187, 180, 177, 174, 169, 166, 162, 157, 153, 151, 147, 142, 138, 135, 133, 131, 129, 127, 124, 119, 115, 114, 111, 110, 105, 102, 100, 98, 95, 94, 92, 90, 86, 85, 83, 82, 79, 78, 78, 76, 75, 72, 70, 69, 67, 65, 64, 61, 60, 58, 57, 57, 55, 55, 54, 52, 51, 49, 48, 47, 46, 44, 44, 43, 42, 42, 41, 40, 39, 39, 38, 37, 37, 35, 35, 34, 34, 33, 32, 32, 30, 30, 30, 29, 29, 29, 27, 27, 26, 25, 25, 25, 24, 24, 24, 23, 23, 22, 22, 22, 21, 21, 20, 20, 19, 19, 19, 18, 18, 18, 17, 17, 17, 17, 16, 16, 16, 15, 15, 15, 15, 14, 14, 14, 13, 13, 13, 13, 13, 13, 13, 13, 12, 12, 11, 11, 11, 11, 11, 10, 10, 10, 10, 9, 9, 9, 9, 9, 9, 9, 9, 8, 8, 8, 8, 8, 8, 8, 8, 7, 7, 7, 7, 7, 7, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 $\}$


TITLE: How to show a set is compact in a function space?
QUESTION [4 upvotes]: I have a question asking if $\{f_n\}$ is a compact in $C_b([0,\infty))$ (bounded continuous) with $||\cdot||_{L^\infty}$.
The sequence is
$$f_n (t) = \sin\sqrt{t+(2n\pi)^2},$$
I have showed that

$f_n \rightarrow 0$ pointwise.
The convergence is uniform when looking at $t$ in a bounded interval,
and it is not uniform when $t\in[0,\infty)$.
$\{f_n\}$ is equi-continuous.

Questions:

Now assume that the space is $C_b([0,N])$, can I use Arzelà–Ascoli theorem to show $\{f_n\}$ is sequential compact? Which implies compactness since $C_b([0,N])$ is a metric space?
How would I approach this kind of problem, which definition of compactness should I use when working on function spaces? Complete and totally bounded; every bounded sequence has a convergent subsequence ?

Thanks a lot!

REPLY [1 votes]: You proved that $\sup_{x\in [0,N]}|f_n(x)|\to 0$, hence $f_n\to 0$ in the space $C[0,1]$. If $(x_n)_{n\geqslant 1}$ is a convergent sequence in a metric space, then the set $\{x_n,n\geqslant 1\}$ has a compact closure.
Since the potential limit point can only be the null function and it does not belong to $\{f_n,n\geqslant 1\}=:S$, the set $S$ cannot be compact. But it may have a priori a compact closure. We can investigate whether a converging subsequence exists. Since such a sequence should converge uniformly to $0$, and no subsequence is uniformly convergent (take a suitable $t$), we conclude that the closure of $S$ is not compact.


TITLE: What is the $ds^2$ notation in relativistic physics?
QUESTION [3 upvotes]: Could someone please explain me intuitively how $ds^2$ represents distance in relativistic physics?

REPLY [6 votes]: If you wanted to find the length of a curve in 3d you could break it into pieces and replace each piece with a straight line segment and add up the lengths of each piece. This would be an under estimate, but it would get closer and closer as you broke it into smaller and smaller pieces.
To find the length of each piece you could find the $x,$ $y,$ and $z$ coordinates of each end of the line segment and then compute the differences and call them $\Delta x=x_2-x_1,$ $\Delta y=y_2-y_1,$ and $\Delta z=z_2-z_1$. Then you could compute $\Delta s^2=\Delta x^2+\Delta y^2+\Delta z^2$. Then the length is $\sqrt{\Delta s^2}$.
So what's the difference between $\mathrm
ds^2$ and $\Delta s^2$? The latter depends on two points whereas the former holds at just one point. For each line segment the latter is correct. But if you start breaking it into small segments you could start using $\mathrm ds^2$ instead and you'll get the wrong answer for thebline segment, but they will approach the right answer as the segments get small. It's a calculus thing. But it's really hard to see the difference if you use the $x,$ $y,$ $z$ coordinate system. So you could use altitude $r$ (measured in meters) and angle from the north pole $\theta$ and angle from the prime meridian $\phi$ (both angles measured in radians, a unit of angle designed to make calculus formulas easier).
Then $$\mathrm ds^2=\mathrm dr^2+r^2\left(\mathrm d\theta^2+(\sin^2\theta)\mathrm d\phi^2\right).$$
To evaluate it you need one $r,$ and $\theta$ to evaluate $r^2$ and $\sin^2\theta$ and then you can compute $\Delta r$ to use for $\mathrm dr$ and compute $\Delta \theta$ for $\mathrm d\theta$ and compute $\Delta \phi$ for $\mathrm d\phi$. This doesn't give you the actual length of that line segment. But the limit of the sum of the lengths will be the right answer.
Now we can talk about relativity. You still want to be able to compute lengths. So you still compute a $\mathrm ds^2$ for each small segment and now you either have to take $\sqrt{-\mathrm ds^2}$ or $\sqrt{\mathrm ds^2}$ to get the length of a little piece because some $\mathrm ds^2$ are positive and some $\mathrm ds^2$ are negative. If you know your curve will have a positive or negative $\mathrm ds^2$ then you'll know which to do.
If you don't know. It might be trickier. But usually you do know. Just like you usually know if you have a ruler or a clock that you are measuring with.
So the $\mathrm d$ part tells you to only use it for small segments and only expect the right answer in the limit and smaller and smaller segments. The square in $\mathrm ds^2$ tells you that you need to take a square root to get a length of the little segment. So the notation is supposed to remind you what to do and how to use it.

REPLY [2 votes]: ds is an infinitesimal distance along the geodesic.  In General Relativity all particles and photons travel along a geodesic, calculated by solving a tensor differential equation, and applying initial conditions.


TITLE: Time required to achieve change in velocity when acceleration varies with velocity.
QUESTION [1 upvotes]: This question is similar, but I am a bit out of practice on mathematics and struggle to apply the steps to my problem.
How can the $\triangle$$t$ be calculated for a given $\triangle$$v$ when acceleration changes as a function of $v$?
Given:
$a(v) =- \frac{8.69+0.0252v^{2}}{200}$
My current goal is to find the time required to change from $v=70$ to $v=60$. (velocity: $\frac{km}{hr}$, acceleration: $\frac{km}{hr/s}$)

My approach:
I have tried to use the linked question to rewrite $a(v)$ as $v(t)$.  Then I would set the equation equal to 70 and then 60 respectively and solve for $\triangle$$t$.
I do not understand where $U$ comes from in the linked problem.

Semi-related question, are the steps above necessary for expressing $a(v)$ in terms $a(t)$?

REPLY [1 votes]: $$
\frac{dv}{dt}=a+b v^2\implies \frac{dv}{a+bv^2}=dt\implies
\Delta t=\frac1{\sqrt{ab}}\left[\arctan\sqrt{\frac ab}v\right]_{v_1}^{v_2}.
$$


TITLE: Is the matrix $[g_{\mu\nu}]$ of metric tensor a linear operator?
QUESTION [1 upvotes]: Starting in an arbitrary coordinate system with basis vectors $\textbf{e}_\mu$ and metric components $g_{\mu\nu}$, we can always diagonalize the matrix $[g_{\mu\nu}]$ so that the components of the metric are those of the canonical form $\eta_{\mu\nu}=\text{diag}(-1,-1,...,+1,+1,..,0,0,...,0)$.
This is similar to a coordinate transformation where the new basis vectors $\textbf{e}'_\mu$ satisfy $\textbf{g}(\textbf{e}'_\mu,\textbf{e}'_\nu)=\eta_{\mu\nu}$  ($\textbf{g}$ is the metric tensor).
In linear algebra, I learned that diagonalization of matrices to find orthogonal basis vectors is only valid for square matrices that are represent linear operators. However, the metric tensor is not a linear operator as it acts on two arguments instead of one. Why can we still use the diagonalization process to find orthogonal basis vectors?

REPLY [3 votes]: As you may know, the metric tensor is a bilinear 2-form. It accepts two vectors from vector space $V$ and gives back a real number in $\mathbb R$. It is linear in both arguments, hence 'bilinear'. The metric tensor is interpreted as a linear operator in the sense that it maps one of its arguments (either one; doesn't matter because it's symmetric) to a dual vector in $V^*$. This dual vector is interpreted as a functional on $V$ (the traditional definition of the dual space of $V$), which acts on the second vector to give a scalar value. So $g$ is a linear map from $V$ to $V^*$. When you write $g$ as a matrix and operate on a column vector $v$, transpose the resulting vector to make it a row vector and you have the dual vector $v^*$.
From a general point of view, the metric tensor is a rank 2 tensor, specifically a rank $(0,2)$ tensor. In general, a rank $(n,m)$ tensor is a multilinear functional which acts on an ordered collection of vectors in $V$ and dual vectors in the dual space $V^*$. For a vector space $V$ over a field $\mathbb F$ (usually $\mathbb R$ or $\mathbb C$), a tensor $T$ is a multilinear map of the form
$$ T : V^m \times V^{*n} \rightarrow \mathbb F .$$
Rank $(0,2)$ tensors over the real numbers, like $g_{\mu \nu}$,
$$ g : V \times V \rightarrow \mathbb R$$
are particularly interesting as they often appear in mathematics and physics. This is because they define inner products. The inner product between two vectors $\begin{pmatrix}a_1\\a_2\end{pmatrix}$ and $\begin{pmatrix}b_1\\b_2\end{pmatrix}$ in an inner product space $V$ is
$$\begin{pmatrix}a_1\\a_2\end{pmatrix} \cdot \begin{pmatrix}b_1\\b_2\end{pmatrix} = \begin{pmatrix}a_1&a_2\end{pmatrix} \begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix} \begin{pmatrix}b_1\\b_2\end{pmatrix}$$
where $\mathbf A$ forms a symmetric positive-definite matrix (symmetric with positive real eigenvalues). By convention we normally write vectors in $V$ in an orthonormal basis, which is a basis that diagonalises $\mathbf A$ to the identity matrix, and so we usually omit $\mathbf A$ entirely when taking inner products because of this orthonormal choice of basis:
$$\begin{pmatrix}a_1\\a_2\end{pmatrix} \cdot \begin{pmatrix}b_1\\b_2\end{pmatrix} = \begin{pmatrix}a_1&a_2\end{pmatrix} \begin{pmatrix}b_1\\b_2\end{pmatrix}$$
when the vectors are written in an orthonormal basis.
These inner products $\mathbf A$ are basically the same thing as metric tensors $g$. Two terms for one concept. Of course in pseudoriemannian geometry, $\mathbf A$/$g$ need not be positive-definite. It is clear how $\mathbf A$ should be interpreted as a linear operator though, right? It maps the vector $\mathbf b$ to its dual vector $\mathbf b^*$ like so:
$$\mathbf b^* (\mathbf a) = \mathbf a \cdot \mathbf b \tag{definition of dual vector space $V^*$}$$
$$\begin{align}\mathbf a \cdot \mathbf b &= \begin{pmatrix}a_1&a_2\end{pmatrix} \begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix} \begin{pmatrix}b_1\\b_2\end{pmatrix} \\ &= \left[ \begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix} \begin{pmatrix}b_1\\b_2\end{pmatrix} \right]^{\mathrm T} \begin{pmatrix}a_1\\a_2\end{pmatrix} \\ &\Rightarrow \quad \mathbf b^* = \left[ \begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix} \begin{pmatrix}b_1\\b_2\end{pmatrix} \right]^{\mathrm T} \end{align}.$$
Having to take the transpose makes this a little confusing, but it should be clear that $\mathbf A$ defines a dual vector $\mathbf b^*$ for each vector $\mathbf b$.
The concept of a metric tensor is basically the same thing, but with a different notation. I could not say earlier that $\mathbf A$ maps a vector to its dual, but that it defines such a map. This is because I had to use the transpose operation. Matrices are a notation designed to express vectors $V$ and linear operators $M : V \rightarrow V$, and the notation is not flexible enough to express a linear map $V \rightarrow V^*$. The notation used for expressing metric tensors (upper/lower index notation; tensor notation; not sure if it has a better name) is more flexible. The metric is denoted $g$, and by writing it with two lower indices as $g_{\mu \nu}$ we are designating it as a rank $(0,2)$ tensor that maps $V \times V \rightarrow \mathbb R$. By giving $g_{\mu \nu}$ just one argument and leaving the other empty, we are left with a map $V \rightarrow \mathbb R$, which is the same thing as a dual vector in $V^*$. We write vectors by their components, $x^\mu$, and then $g$ defines a linear map $g : V \rightarrow V^*$ like so:
$$g : \mathbf x \mapsto \mathbf x^*, \quad x_\mu = \sum_{\nu} g_{\mu \nu} x^{\nu}.$$
The notation $x^\mu$ expresses the components of the vector $\mathbf x$ in the chosen basis of $V$, and $x_\mu$ expresses the components of the dual vector $\mathbf x^*$ in the dual basis of $V$, i.e. the corresponding basis in $V^*$. The notation is frequently heavily abused for brevity, so you may see expressions like
$$g : x^\mu \rightarrow g_{\mu \nu} x^\nu \tag{implied summation over $\nu$}$$
to mean the same thing as I said above.
You may notice the similarity with matrix multiplication:
$$(b^*)_\mu = \sum_{\nu} A_{\mu \nu} b_\nu.$$
When $g$ is expressed as a matrix as in your question, it simply maps the components $x^\mu$ to the components of its dual vector, $x_\mu$. It very much is a linear map, $g : V \rightarrow V^*$ and all the associated tools of linear analysis may be applied.