TITLE: Tension in a rotating object
QUESTION [3 upvotes]: I'm somewhat stuck on this very simple question, but I can't convince myself of a satisfactory answer. I tried to look through the previous questions asked to avoid asking a duplicate I hope.
Any rotating object... say a rod in this case, sitting stationary in free space, when pushed at one end will cause a rotation about the center of mass. This means the pushed end and other end will move in opposite directions of each other.
The only way that the rod can transmit force down its length is through the interaction of each particle pushing/pulling on the next (tension perpendicular to the length) as it travels down the rod. Once you reach the center of mass, there is no induced rotation. How do the particles past the center of mass receive a push or pull at all if there is a point in the center which has no movement?
Additionally, how does this original force in the positive direction turn into movement in the negative direction past the center of mass? Essentially both of these questions can be answered by an explanation of how the particles inside an object behave when a torque is applied... which I cant find an answer to.
Edit: So I found this other post which asks a very similar question: Why does a rigid body rotate and not simply translate when pushed with an instantaneous force?
I feel that the corded particles explanation answer there is of the form I am imagining for this, but it seems that it doesn't explain the other side rotating backwards.
The particles in the rod near the push start to move forward, which "pull" on the neighboring particles via tension. Then at some point, this tension turns into compression, effectively pushing the particles past the center of mass downward.
Does anyone have a satisfactory internal particle explanation involving tension/compression caused by an initial force at one end which results in the final behavior of rotation of a rod? My understanding of rotational motion feels incomplete to me if I can't explain intuitively why something rotates when pushed off-center besides "if one side goes up, the other must go down".

REPLY [0 votes]: Newton’s laws; applied to translation and rotation:
Consider a system of relatively small particles. They may be atoms. They may, or may not, be bound into a rigid structure.  The particles may be subject to forces from other particles within the system or to external forces coming from outside of the system. To simplify the notation, assume that all, r’s, v’s, and a’s in this discussion represent vectors.  Small letters refer to individual particles, and capital letters refer to the total mass and the center of mass.  The, r’s, are measured from the origin of an arbitrarily chosen inertial coordinate system (it is often convenient to do calculations in terms of the x,y, and z components of vectors).    Vectors with primes are measured in a similar system that is centered on (and moving with) the center of mass.  An, x, indicates a vector product and the symbol, Σ, indicates a vector sum which includes all of the particles in the system.
The vector, R, which locates the center of mass is defined by the expression MR = Σmr.  The derivative with respect to time give a more intuitive expression: MV = Σmv.  The total momentum of the system is given by the vector sum of the momenta of all of the particles and can be expressed as the total mass times the velocity of the center of mass. Another derivative gives: MA = Σd (mv)/dt.  Since the internal forces occur as equal and opposite pairs, the changes in momentum associated with these forces are also equal and opposite, and drop out of the summation. Only the component of the changes in momentum caused by the external forces remain so: MA = Σm dv/dt = Σma = Σf. (The acceleration of the center of mass is determined by the vector sum of the external forces.)
The angular momentum of the system (relative to the origin of the inertial system) is L =  Σ r x mv.  In the center of mass coordinates it is: Σ [(R + r’) x m(V + v’)] = Σ (R x mV) + Σ (R x mv’) + Σ (mr’ x  V) + Σ (r’ x  mv’) =  (R x MV) + Σ(r’ x mv’)  In the second (of four terms) the summation determines the velocity of the center of mass in the center of mass system (which is zero).  In the third term, the summation determines the position of the center of mass in that system (also zero). Conclusion:  The angular momentum of the system can be considered as a combination of two parts.  One part is associated with the motion of the center of mass; the other part is associated with the rotation of the system about the center of mass.
Again, taking a derivative:  dL/dt = (dR/dt x MV) + (R x M dV/dt) + Σ (dr’/dt x mv’) + Σd(r’x mv’)/dt) but dR/dt is V and dr’/dt  is v’ so the vector products in the first and third term are each zero. And again, since the internal forces occur as equal and opposite pairs, any changes in angular momentum caused by the internal forces cancel out. Then dL/ dt  = R x MA + Σ r’ x  f.  Thus the change in angular momentum associated with the motion of the center of mass depends on its acceleration which is caused by the vector sum of the external forces; and the change in the angular momentum about the center of mass is caused by the vector sum of the torques about the center of mass caused by the external forces.  (i.e.  The external torques acting about the center of mass cause an angular acceleration around the center of mass.)


TITLE: Infinite solutions of system
QUESTION [0 upvotes]: For which values of λ does the system:
8x + 7y + 5z = 0
4x + 5y + 6z = 0
7x + 8y + λz = 0

has infinite values? Which is the way for solving these type of exercises?

REPLY [1 votes]: A linear system may behave in any one of three possible ways:

1.The system has infinitely many solutions.
2.The system has a single unique solution.
3.The system has no solution.
        Any system of equations is inconsistent if the rank of the augmented matrix is greater than the rank of the coefficient matrix. If, on the other hand, the ranks of these two matrices are equal, the system must have at least one solution. The solution is unique if and only if the rank equals the number of variables.
       Otherwise there are an infinit of solutions.
And for homogenous systems of equation has at least one solution, known as the zero solution (or trivial solution), which is obtained by assigning the value of zero to each of the variables. If the system has a non-singular matrix (det(A) ≠ 0) then it is also the only solution. If the system has a singular matrix then there is a solution set with an infinite number of solutions. This solution set has the following additional properties.
        That is if deteterminant of matrix is zero then system has infinite solution. So only take determinant equal to zero and find value of λ.


TITLE: simulating a fair random process with an unfair one.
QUESTION [4 upvotes]: Let's say I have a stochastic process that outputs $1$ or $0$ with probability $p$ or $1-p$ respectively, $p\neq 1/2$. Let's assume this is a repeatable iid process. So I can generate $X_1,X_2\dots$ which are each $1$ or $0$ as above. Can I create a proposition (a logical construction out of the $X_i$s) with probability a half. That is, can I simulate a fair process with this unfair one.
If $p=1$ or $p=0$ then obviously no. But (without loss of generality let's assume $1>p>1/2>1-p>0$. Is there always some such proposition? I can get arbitrarily close, sure. Think about the set of binary strings of length $n$ (you can think of my stochastic process as generating these sequences with certain probabilities). For large enough $n$, even the most likely of them is less than a half, and by picking the right ones to go in my "proposition" I can get as close to $1/2$ as I like, as long as I can increase $n$ as much as I want.
For some values of $p$ you can get to exactly $1/2$. For example, take $p(2-p)=1/2$ and solve for $p$. For this value, $X_1 \lor X_2$ has probability exactly $1/2$. Is it always possible to find a proposition that has exactly probability a half? Or is there a characterisation of the values of $p$ that you can do this for?
(Wasn't entirely sure what tags were appropriate)

REPLY [1 votes]: Yes, and the idea of doing things with a target distribution given only access to a different distribution is related to concepts like rejection sampling and importance sampling.


TITLE: Calculation of area using definite integral
QUESTION [0 upvotes]: Find area bounded by function $\left\lfloor x \right \rfloor dx$ within limits $-3 $ to $4$,
where $\left\lfloor x \right \rfloor $ is the floor function.
I have tried finding it but the answer is $0$.

REPLY [0 votes]: From the picture:
The area on $[0,4] = 6$ units. The "signed" area on $[-3,0] = -6$ units, while "unsigned" area is just $6$ units.
The area on $[-3,4]$ is thus either $12$ or $0$, depending on if you mean unsigned area or signed area respectively.


TITLE: Is the disk quasiconformally isomorphic to the plane?
QUESTION [4 upvotes]: (This question might turn out to be too elementary for this site,

if so I'm sorry, but I can't find the answer anywhere.)

Does there exist a function $\; f : \{z\in \mathbb{C} : |z| < 1\} \to \mathbb{C} \;$

such that $f\hspace{.01 in}$ is a quasiconformal bijection and $f^{-1}$ is quasiconformal?

REPLY [4 votes]: If you mean whether there is a quasiconformal mapping from the unit disk $D$ onto the whole complex plane $\mathbb{C}$, then the answer is NO. As you have noticed, the inverse mapping of a quasiconformal mapping is always quasiconformal. One can easily prove that $f(\mathbb{R}^n)=\mathbb{R}^n$ when $f$ is an entire quasiconformal mapping. (Here one can use the fact that quasiconformal mappings are locally quasisymmetric and so it must be global quasisymmetric (since the domain is $\mathbb{R}^n$). For quasisymmetric mapping, it is easy to see that it must send unbounded set to unbounded set.)


TITLE: Value of an indefinitely long expression : Quadratic Equation Approach
QUESTION [1 upvotes]: I came across this problem and have been able to get the correct answer for this although I tried many times on my own.
$1st$ Approach : $8+\frac{8}{8+\frac{1}{8+y}} = y$ and when I solve this I get to the quadratic equation i.e. $584 + 7y = 8y^2$ which does not lead to the correct solution as the roots for this quadratic equation are $y = \frac{7}{16} \pm \frac{\sqrt(18737)}{16}$ .
$2nd$ Approach : $8+\frac{8}{8+\frac{1}{y}} = y$ and when I solve this I get to the quadratic equation i.e. $8+ 71y = 8y^2$ which also does not lead to the correct solution as the roots for this quadratic equation are $y = \frac{71}{16} \pm \frac{\sqrt(5297)}{16}$ .
Please help me on this ! I am stuck with this problem for a while.
Thanks in advance !
PS : This question is a MCQ from where I picked up this question and the options avaliable are

$6\frac{2}{3}$
$4+\sqrt17$
$8+\sqrt8$
9

I am expecting an analytical and step-by-step solution as I was trying to do.

REPLY [1 votes]: I am assuming that there is a typo in the question as the pattern is not obvious.
Let $$8 + \frac{\color{red}{1}}{8+\frac{1}{8+\ldots}}=y$$
$$8 + \frac1y = y$$
$$y^2-8y-1=0$$
$$y = \frac{8 \pm \sqrt{64+4}}{2}$$
Since $y > 0$,   $y = \frac{8 + \sqrt{68}}{2}=4 + \sqrt{17}$

I did consider other possibilities but I did not hit one of the options:
$$8 \left( 1+ \frac1{8 + \frac1{8 + \dots}}\right) = y$$
Now, first, let's study what do we get from $\frac1{8 + \frac1{8 + \dots}}=z$
$$\frac1{8+z}=z$$
$$z^2+8z-1=0$$
$$z=\frac{-8 +\sqrt{64+4}}{2}=-4 + \sqrt{17}$$
$$8 (-3+\sqrt{17})=y$$


TITLE: Why is $2^{\aleph_\omega} = \left( \sum_{\mu < \aleph_\omega} 2 ^{\mu} \right)^{\aleph_0}$?
QUESTION [3 upvotes]: Why is $2^{\aleph_\omega} = \left( \sum_{\mu < \aleph_\omega} 2 ^{\mu} \right)^{\aleph_0}$?

The above equality is a part of a proof that I am trying to understand.
I know that $$\aleph_\omega = \bigcup_{n < \omega} \aleph_n$$
I tried writing $$2^{\aleph_\omega} = 2^{\bigcup_{n < \omega} \aleph_n} = \prod_{n < \omega} 2^{\aleph_n}$$
but I don't know if I can write it like this or if this is helpful.
Please be relatively elementary in the explanation.

REPLY [2 votes]: There is in general a stronger equality which states that $2^\kappa=(2^{<\kappa})^{cf(\kappa)}$ where you have $2^{<\kappa}=\lim_{\nu<\kappa}2^\nu=\sum_{\nu<\kappa}2^\nu$
Proof:
Let $\{\kappa_{\alpha}\}_{\alpha < cf(\kappa)}$ be a set of cardinals where $\kappa_{\alpha}<\kappa$ and such that:
\begin{equation*} \kappa=\sum_{\alpha < cf(\kappa)} \kappa_{\alpha} \end{equation*}
Then:
\begin{equation*} 2^{\kappa}=2^{\sum_{\alpha < cf(\kappa)} \kappa_{\alpha}}=\prod_{\alpha<cf(\kappa)}2^{\kappa_{\alpha}}\leq(2^{<\kappa})^{cf(\kappa)}\leq 2^{\kappa \cdot cf(\kappa)}=2^{\kappa} \end{equation*}
In your case we have $cf(\aleph_{\omega})=\aleph_0$ which you just plug in the general case.

REPLY [2 votes]: First let's argue $2^{\aleph_\omega}\geq \left(\sum_{\mu<\aleph_\omega}2^\mu\right)^{\aleph_0}$:
Clearly $2^{\aleph_\omega}\geq 2^\mu$  for any $\mu<\aleph_\omega$, so also $2^{\aleph_\omega}\geq \sum_{\mu<\aleph_\omega}2^\mu$. Then also $
\left(2^{\aleph_\omega}\right)^{\aleph_0}\geq \left(\sum_{\mu<\aleph_\omega}2^\mu\right)^{\aleph_0}$. But we have:
$$
\left(2^{\aleph_\omega}\right)^{\aleph_0}=
2^{\aleph_\omega\cdot \aleph_0}=2^{\aleph_\omega}
$$

Next let's argue $2^{\aleph_\omega}\leq \left(\sum_{\mu<\aleph_\omega}2^\mu\right)^{\aleph_0}$:
Suppose I have a subset $X\subseteq \omega_\omega$, then I can see $X$ as a union of subsets of $\omega_n$ by considering $X_n=\{x\in X\mid x\in \omega_n\}$: we have $X=\bigcup_{n\in\omega} X_n$. Furthermore, if $X,X'$ are distinct subsets, then $X_n\neq X'_n$ for some $n\in\omega$.
So, in this way, we can send each subset $X\subseteq \omega_\omega$ injectively to an $\omega$-sequence $\langle X_n\mid n\in\omega\rangle$ where each $X_n\subseteq \omega_n$. Hence, this is an injective function from $\mathcal P(\omega_\omega)\to\left(\bigcup_{n\in\omega} \mathcal P(\omega_n)\right)^\omega$.
We're now done, since $|\mathcal P(\omega_\omega)|=2^{\aleph_\omega}$ and
$$\left|\big(\bigcup_{n\in\omega} \mathcal P(\omega_n)\big)^\omega\right|=\left(\sum_{n<\omega}2^{\aleph_n}\right)^{\aleph_0}=\left(\sum_{\mu<\aleph_\omega}2^\mu\right)^{\aleph_0}.$$


text:
\begin{document}

\title[]{Group actions on treelike compact spaces}

\author[]{Eli Glasner}
\address{Department of Mathematics,
	Tel-Aviv University, Ramat Aviv, Israel}
\email{glasner@math.tau.ac.il}
\urladdr{http://www.math.tau.ac.il/$^\sim$glasner}

\author[]{Michael Megrelishvili}
\address{Department of Mathematics,
	Bar-Ilan University, 52900 Ramat-Gan, Israel}
\email{megereli@math.biu.ac.il}
\urladdr{http://www.math.biu.ac.il/$^\sim$megereli}

\date{May 27, 2019}

\thanks{This research was supported by a grant of the Israel Science Foundation (ISF 668/13)}

\keywords{amenable group, dendrite, dendron, ends, fragmentability, median pretree, null system, proximal action, Rosenthal Banach space, shadow topology, treelike structures, tame dynamical system}

\subjclass[2010]{Primary 37Bxx; Secondary 54H20; 54H15; 22A25}

\begin{abstract}
	We show that group actions on many treelike compact spaces are not too complicated dynamically.
We first observe that an old argument of Seidler \cite{Seid-90}, implies that every action of a topological group $G$
on a regular continuum is null and therefore also tame. As every local dendron is regular, one concludes that every action of $G$ on a local dendron is null. We then use a more direct method to show that every continuous group action of $G$ on a dendron is Rosenthal representable, hence also tame. Similar results are obtained for median pretrees.
As a related result we show that Helly's selection principle can be extended to bounded monotone sequences
defined on median pretrees (e.g., dendrons or linearly ordered sets).
 Finally, we point out some applications of these results to continuous group  actions on dendrites.
\end{abstract}

\maketitle

\setcounter{tocdepth}{1}
 \tableofcontents

\section{Introduction and preliminaries}

One of the

motivations for writing the present work was to show
that group actions on dendrons are Rosenthal representable, hence tame (see Definition \ref{d:repr} below).
Representations on Banach spaces with ``good" geometry lead to a
natural hierarchy in the world of continuous actions $G \curvearrowright X$
of topological groups $G$ on topological spaces $X$.
In particular, representations on Banach spaces without a copy of $l_1$ (we call them {\it Rosenthal} Banach spaces) play a very important role in this hierarchy.

According to the Rosenthal $l_1$-dichotomy \cite{Ros0},
and the corresponding Dynamical Bourgin-Fremlin-Talagrand dichotomy \cite{GM-HNS,GM-survey},
there is a sharp dichotomy for

metrizable

dynamical systems;
either their enveloping semigroup is of cardinality smaller or equal to that of the continuum, or it
is very large and contains a copy of the Stone-\v{C}ech compactification of $\N$ (the set of natural numbers).

When $X$ is compact

metrizable,

in the first case, such a dynamical system $(G,X)$ is called {\it tame}.
By A. K\"{o}hler's \cite{Koh} definition, tameness
 means that for every continuous real valued function
$f: X \to \R$ the family of functions $fG:=\{fg\}_{g \in G}$ is ``combinatorially small";
namely, $fG$ does not contain an {\it independent sequence}.

Recall the classical definition from \cite{Ros0}: a sequence $f_n$ of real valued functions on a set $X$
is said to be \emph{independent} if
there exist real numbers $a < b$ such that
$$
\bigcap_{n \in P} f_n^{-1}(-\infty,a) \cap  \bigcap_{n \in M} f_n^{-1}(b,\infty) \neq \emptyset
$$
for all finite disjoint subsets $P, M$ of $\N$.

As in \cite{GM-tame,GM-tLN} we say that a bounded family of real valued functions is a
{\it tame family} if it does not contain an independent sequence.
 For

 compact
 metrizable systems $X$ the two concepts discussed above coincide:
 $(G,X)$ is tame iff it is Rosenthal representable.
 The case of metrizable $X$ admits
 also interesting enveloping semigroup characterization: $(G,X)$ is tame

 iff every element

  $p \in E(G,X)$ is a Baire class 1 function $p: X \to X$ (see  \cite{GMU,GM-survey}).

 Recall that the {\it enveloping semigroup} $E(G,X)$ of a compact $G$-system $X$ is the pointwise closure of the set of all $g$-translations $X \to X$ ($g \in G$) in $X^X$.

We next recall the definition of a Banach representation of a dynamical system.
For a Banach space $V$ denote by $Is(V)$ the topological group (equipped with the strong operator topology) of all linear isometries $V \to V$.  The group $Is(V)$ acts on the dual space $V^*$ and its weak-star compact subsets.

 \begin{defin} \label{d:repr} (See \cite{Me-nz,GM-HNS,GM-survey})

	A \emph{representation} of an action  $G \curvearrowright X$ on a Banach space $V$ is a pair $(h,\a)$, where
	$h: G \to Is(V)$ is a strongly continuous co-homomorphism

	and $\a: X \to V^*$ is a weak-star continuous bounded map

	into the dual of $V$ such that $(h,\a)$ respect the  original action and the induced dual action. That is,
	$$
	\lan v,\a(gx) \ran = \lan h(g)(v),\a(x) \ran   \ \ \ \text{for all} \  v \in V, g \in G, x \in X.
	$$
\end{defin}

We say that $(G,X)$ is {\it Rosenthal representable}, or WRN (Weakly Radon-Nikodym),  if there exist: a Rosenthal Banach space $V$ and a representation $(h,\a)$ as above such that $\a$ is a topological embedding.
 For compact $G$-systems $X$ this concept was defined in \cite{GM-rose}.

  The class of all Rosenthal representable dynamical systems and its subclass of tame systems are quite large.
  When considering the trivial action of $G$ on a space $X$, this definition reduces to the purely topological notion
  of a WRN space.

  For motivation, properties and examples see, for example,
  \cite{GM-rose,GM-survey,GM-tLN} and the references thereof.
A relevant recent result is that, for
 every circularly  (in particular, linearly) ordered compact space $K$, the action $H_+(K) \curvearrowright K$, of the topological group $H_+(K)$ of all order-preserving homeomorphisms of $K$ on $K$,
 is Rosenthal representable, \cite{GM-c}.

\br

 Recall that a \textit{continuum} is a compact Hausdorff connected space.
 A continuum $D$ is said to be a \textit{dendron} \cite{Mill-Wattel} if every pair of distinct points
 $u,v$ can be separated in $D$ by a third point $w$.

A metrizable dendron is called a \textit{dendrite}.
The class of dendrons is an important class of 1-dimensional

treelike
compact spaces.

A  compact space is said to be a \textit{local dendron} if each of its points admits
a closed neighborhood which is a dendron. Similarly, a compact space is called a
\textit{local dendrite}
if each of its points admits a closed neighborhood which is a dendrite. Clearly, the circle is a local dendrite.

In a dendron $X$, for every pair of points $u,v$ in $X$  one defines a ``generalized arc"
	$$
	[u,v]=\{x \in X: \ x \ \text{separates $u$ from $v$}\} \cup \{u,v\}.
	$$
In dendrites the

generalized arc $[u,v]$ is a {\em real arc},

i.e. a topological copy of an interval in the real line.
This is not necessarily the case for dendrons. Indeed, note that any connected linearly ordered compact space,
in its interval topology,  is an example of a dendron.

	A topological space $X$ is called {\em regular} if every point has a local base for its topology,
	each member of which has finite boundary.
	For compact Hausdorff spaces this is equivalent to saying that each open cover admits a finer (finite) open cover each member of which has finite  boundary (see e.g. \cite{Kur}).
	Some works refer to such compact spaces as being  {\em rim-finite}  (see e.g. \cite{Ward}).
Every (local) dendron is regular \cite[Theorem 21]{Ward}.
For more information on dendrons and dendrites
see, for example, \cite{Charatoniks, Mill-Wattel,DM-16}.

 \sk

Our goal in the first part of this paper is to prove the following two results

(see Theorems \ref{t:nullness} and \ref{t:Dendron} for the proofs).

 \begin{thm} \label{t:nullnessINTRO}
 	Any action of a group $G$ on a regular

 	continuum is null, hence also tame.
 \end{thm}

\begin{thm} \label{thm-tame}
Let $D$ be a dendron. For every topological group $G$ and continuous action
$G \curvearrowright D$, the dynamical $G$-system $D$ is Rosenthal representable, hence also tame.
\end{thm}

In the proof of Theorem \ref{t:nullness} we use a method of Seidler \cite{Seid-90}.
Note that by a result of Kerr and Li \cite{KL} every null dynamical $G$-system is tame.
This explains the tameness conclusion in Theorem \ref{t:nullnessINTRO}.

\sk

In the proof of Theorem \ref{t:Dendron}
we use the following useful characterization.

\begin{thm} \label{t:criter1} \cite[Theorem 6.10]{GM-rose}
A dynamical $G$-system $X$ is Rosenthal representable if and only if there exists a
$G$-invariant point-separating bounded family $F$ of continuous functions $X \to \R$ which is
tame as a family of functions.

\end{thm}

We consider monotone (not necessarily, continuous) functions (Definition \ref{d:monot2}) on dendrons $D$.
In Theorem \ref{t:MonotFragm} we show that any such function is fragmented (Baire class 1, on dendrites).
In order to apply Theorem \ref{t:criter1} for dendrons $D$,
in the role of the family $F$ we consider the set of all continuous monotone functions $D \to [0,1]$.

Later in Theorem \ref{t:pretree} we generalize Theorem \ref{thm-tame} to \textit{compact median pretrees}
with monotone group actions. This approach

 provides
the following corollary (see Corollary \ref{c:Ends}).

\begin{cor}
Let $X$ be a $\Z$-tree. Denote by $Ends(X)$ the set of all its ends. Then for every monotone group action $G \curvearrowright X$ by homeomorphisms, the induced action of $G$ on the compact space
$\widehat{X}:=X \cup Ends(X)$
is Rosenthal representable.
\end{cor}

As a related result we show, in Theorem \ref{GenHellyThm}, that
Helly's selection principle can be extended to bounded monotone
sequences of real valued functions defined on median pretrees (e.g., dendrons or linearly ordered sets).

Also in Theorem \ref{t:FragmPretr} we show that
every monotone real valued function $f: X \to \R$ on a compact Hausdorff pretree is fragmented.

\sk

In the second part of the paper,
as applications of Theorems \ref{t:nullnessINTRO} and \ref{thm-tame},
and building on ideas and results from \cite{DM-16} and \cite{MN-16},
we easily recover some old results, and prove some new ones,
concerning dynamical systems defined on (local) dendrons.

In Section \ref{cons} we show that
when a group $G$ acts on a dendrite $X$ with no finite orbits and
$M \subset X$ is the unique minimal set, then the action on $X$ is strongly proximal
and the action on $M$ is extremely proximal.
In Section \ref{discrete-spectrum} we show that for an amenable group $G$
every infinite minimal set in a dendrite system

$(G,X)$
is almost automorphic.
This result was strengthened recently by Shi and Ye \cite{SY-18}, who have shown
that it is actually equicontinuous. In the final section we comment on the special case
where the acting group is the group of integers.

\sk

\nt \textbf{Acknowledment.}
Thanks are due to Nicolas Monod for enlightening conversations concerning tameness of
actions on dendrites. We also thank him and the organizers of the conference
entitled ``Structure and dynamics of Polish groups", held at the CIB in Lausanne, March 2018,
for the invitation to participate in this conference. The successful conference contributed a great deal
to the progress of this work.
Thanks are due also to Jan van Mill for his helpful advise.

\br

\section{Actions of groups on a regular continuum are null}

Following Goodman \cite{Go-74},
for a sequence $S = \{s_1, s_2, \dots\} \subset G$ we define the topological sequence entropy of
a dynamical system $(G, X)$,
with respect to $S$ and a finite open cover $\bA$ of $X$, by
$$
h_{top}(X,\bA ; S) = \lim_{n \to \infty} n^{-1} \log \left( N \left (\bigvee_{i=1}^{n} s_i(\bA)\right)\right),
$$
where $N(\cdot)$ denotes the minimal cardinality of a subcover.
We say that $(G, X)$ is {\em null} if $h_{top}(X,\bA ; S) =0$ for all open covers
$\bA$  of $X$ and all sequences $S$ in $G$.

\sk

The proofs in this section are taken, almost verbatim, from  Seidler's paper
\cite{Seid-90}. For a subset  $A \subset X$ we denote its boundary by $\partial(A)$.

\begin{lem}\label{l-1}
Let $\bA$ be an open cover of a compact continuum $X$ and let $P$ be the set of
boundary points of elements of $\bA$. If $\bA$ is a minimal cover then the number of
elements of $\bA$ is at most the number of elements in $P$.
\end{lem}

\begin{proof}

 Assume that $\bA$ does not have a proper subcover.
Then $\bA$ is a finite collection because $X$ is compact. Let $a$ be the number of elements of $\bA$ so
that $\bA = \{A_i: 1 \leq i \leq a \}$.
Let $A_j \in \bA$, and let $V = \bigcup_{i\not= j} A_i$.
Note that $V$ is open as a union of open sets.
Because $\bA$ doesn't have a proper subcover we see that
$X \setminus V \not= \emptyset$, so that  $\partial(V) \not= \emptyset$,
since $X$ is connected.
Further $\partial(V) \subset P$ because $V$ is a finite union of elements of $\bA$.
As $V$ is open, $\partial(V) \cap V = \emptyset$, so that $\partial(V) \subset A_j$.
Thus, for each element $A_j$ of $\bA$,
there must exist a nonempty subset of $P$ contained in $A_j$
but disjoint from every other element of $\bA$.
This implies that the number of elements of $\bA$ be at most the number of elements in $P$.

\end{proof}

\sk

\begin{lem}\label{l-2}
Let $G$ be a countable infinite group acting on a regular compact continuum $X$.
Let $S = \{s_0, s_1, s_2, \dots\}$ be a sequence of elements of $G$.
Let $\bA$ be a minimal open cover of $X$ of at least two elements such that every
element of $\bA$ has finite boundary. Let $L_{\bA}$ be the total number of boundary points
of elements of $\bA$.
For each positive integer $n$,
Let $\bM_n$ be a subcover of
minimum cardinality of $\bigvee_{i=0}^{n-1} s_i(\bA)$
and let $P_n$ be the collection of boundary points
of elements of $\tilde{\bA} = \bigcup_{i=0}^{n-1} s_i(\bA)$. Then
\begin{enumerate}
\item
 For each positive integer $n$ every boundary point of an element of $\bM_n$ is in $P_n$.
\item
$N(\bigvee_{i=0}^{n-1} s_i(\bA)) \le n L_{\bA}$.
\end{enumerate}
\end{lem}

\sk

\begin{proof}
(1) \
Let $n$ be a positive integer and let $x$
be a boundary point of $M \in \bM_n$.
By the definition of $\bM_n$,
for each positive integer $0 \le i <  n$, there exists $V_i \in \tilde{\bA}$
such that $M = \bigcap_{i=0}^{n-1} V_i$.
Let $x$ be a boundary point of $M$
and let $B$ be an open set containing $x$;
clearly $B \cap M \not=\emptyset$ so that $B \cap V_i \not=\emptyset$ for each $i$.
For each $V_i$, this requires that either $x \in \partial(V_i)$ or $x \in V_i$. Suppose that every
$V_i$ contains $x$. Then $x \in M$ by definition. But this contradicts $x \in \partial(M)$,
because $M$ is open. Thus $x$ is a boundary point of at least one $V_i$ and thus $x \in P_n$.

\sk

 (2)\
Let $n$ be a positive integer. Because each $s_j$ is a homeomorphism the number of boundary
points of elements of $s_j(\bA)$  is $L_{\bA}$ for every integer $j$.
This requires that the number of elements of $P_n$ be at most $nL_{\bA}$.
As every boundary point of an element of $\bM_n$ is in $P_n$ and as $\bM_n$ doesn't
have a proper subcover, Lemma \ref{l-1} implies that there exist at most $nL_{\bA}$ elements in $\bM_n$.
The desired result then follows from the definition of $\bM_n$.

\end{proof}

\sk

\begin{thm} \label{t:nullness}
Every action

of a group $G$ on a regular

continuum is null, hence a fortiori tame.
\end{thm}

\begin{proof}
Let $X$ be a regular compact space on which $G$ acts.
Let $\bA$ be a minimal open cover of
$X$ containing at least two elements such that every element of $\bA$ has finite boundary.
Let $L_{\bA}$ be the total number of boundary points of elements of $\bA$.
Given $S = \{s_1, s_2, \dots\} \subset G$,
we have then, from Part (2) of Lemma \ref{l-2}, that
\begin{align*}
h_{top}(X,\bA ; S) & = \lim_{n \to \infty} n^{-1} \log \left( N \left (\bigvee_{i=1}^{n} s_i(\bA)\right)\right)\\
&  \le \lim_{n \to \infty} n^{-1} \log(n L_{\bA}) = 0.
\end{align*}

Thus $h_{top}(X; S)= 0$ and this shows that the system $(G, X)$ is null.
By a theorem of Kerr and Li \cite{KL} it is also tame.
\end{proof}

\begin{remark}
In \cite[Theorem 12.2]{KL} the authors demonstrate with a simple proof,
that every action of a convergence group $G$ on a compact space $X$
(in particular, any hyperbolic group acting on its Gromov boundary) is null.
\end{remark}

\br

\section{Dendrons, monotone functions and group actions}

\subsection{Standard betweenness relation and dendrons}

All the topological spaces in this work are assumed to be Hausdorff.
	Let $X$ be a connected
	topological space and $u,v \in X$.
	As usual, we say that a point $w$ \textit{separates} $u$ and $v$ in $X$
	if there exist in $X$ open disjoint neighborhoods $U,V$ of $u$ and $v$
	respectively such that $X \setminus\{w\}=U \cup V$.

	For every $u,v$ in
	$X$ define the ``generalized arc"
	$$
	[u,v]=\{x \in X: \ x \ \text{separates $u$ from $v$}\} \cup \{u,v\}.
	$$

	By definition $[u,v]=[v,u]$.

\begin{defin} \label{d:BrelationD}
	Let $X$ be a topological space and $u,w,v \in X$.
	We say that $w$ is \textit{between} $u$ and $v$ in $X$ if $w \in [u,v]$.
	That is, $w$ separates $u$ and $v$ or $w \in \{u,v\}$.
	This defines a natural betweenness ternary relation on $X$.
	Denote by $R_B$ this ternary relation. Sometimes we write

$\langle u,w,v \rangle$ instead of
	$(u,w,v) \in R_B$.
\end{defin}

\begin{lem} \label{l:D-prop}
	Let $D$ be a dendron.

	\ben
	\item \cite{Mill-Wattel} \  Then $D$ is locally connected and the intersection
	of arbitrary family of subcontinua of $D$ is either empty or is a continuum.
	\item \cite[Cor. 2.15.1]{Mill-Wattel} \  $[u,v]$ is the smallest subcontinuum of
	$D$ containing $u$ and $v$. That is,
	$$
	[u,v]=\cap \{C \subseteq D: \ u,v \in C \ \text{and} \ C \ \text{is a subcontinuum}\}
	$$
	\item \cite{Bank13} \

	$[a,b] \subseteq [a,c] \cup [c,b]$ for every $a,b,c \in D$.
	\een
\end{lem}

Every dendron with its standard betweenness relation is a \textit{pretree} (Section \ref{s:pretrees}). This provides another explanation of \ref{l:D-prop}.3.
The following proposition can be derived from a result of Bankston \cite[Theorem 3.1]{Bank15},
which, in fact, shows that this assertion holds for every locally connected continuum.
The direct proof given below, for dendrons, was explained to us by Nicolas Monod.

\begin{prop} \label{p:Monod}
	Let $D$ be a dendron. Then the
	betweenness relation  $R_B$ on $D$ is closed.
\end{prop}
\begin{proof}
	Suppose that in $D$ we have converging nets,  $\lim u_i =u, \lim_i w_i =w, \lim v_i =v$
	such that $\langle u_i,w_i,v_i \rangle$ for every $i \in I$.
	We have to show that $\langle u,w,v \rangle$.
	Assuming
	the contrary, we have $w \notin [u,v]$. Since $[u,v]$ is compact there exist
        neighborhoods $U$, $W$, $V$ of $u,w,v$ respectively such that $W \cap (U \cup [u,v] \cup V) =\emptyset$.
	Since $D$ is locally connected, we can assume, in addition, that $U$ and $V$ are connected and closed.
	There exists $i_0$ such that $w_{i_0} \in W, u_{i_0} \in U,v_{i_0} \in V$.

	By Lemma \ref{l:D-prop}.2 we have
	$[u_{i_0},u] \subset U, [v,v_{i_0}] \subset V$.
	Then $[u_{i_0},v_{i_0}] \subset  [u_{i_0},u] \cup [u,v] \cup [v,v_{i_0}]$ by Lemma \ref{l:D-prop}.3. By our choice $W$ does not meet $[u_{i_0},u] \cup [u,v] \cup [v,v_{i_0}]$,  hence, $w_{i_0} \notin [u_{i_0},v_{i_0}]$.
	This contradiction completes the proof.
\end{proof}

Note that in the \textit{comb space} (which is not locally connected) the relation $R_B$ is not closed (Bankston  \cite[Exercises 3.4(ii)]{Bank15}).

\sk \sk
\subsection{Monotone functions}

 	\begin{defin} \label{d:monot2}
 	Let us say that a (\textbf{not necessarily continuous}) map $f: X \to Y$ between two connected topological spaces is:

 	\ben
 	\item  \textit{B-monotone}

 	if it respects the betweenness relations $R_B$ (from Definition \ref{d:BrelationD}) of $X$ and $Y$.
	Meaning that $\langle u,w,v \rangle$ implies $\langle f(u),f(w),f(v) \rangle$.

	It is equivalent to the requirement that $f$ be \textit{interval preserving}: $f[u,v] \subseteq [f(u),f(v)]$.

 	 Notation: $f \in M_B(X,Y)$. For $Y=\R$ we write $ M_B(X)$.
 	 	\item
 	 \textit{C-monotone}

	 (or, simply, \textit{monotone})
	 if the preimage $f^{-1}(A)$ of every connected subset $A \subset Y$ is connected.
 	 Notation: $f \in M_C(X,Y)$. For $Y=\R$ we write $M_C(X)$.
 	\een
 \end{defin}

For continuous maps on continua definition (2) is well known.

See, Kuratowski \cite[Section 46]{Kur}.

Not every B-monotone continuous function is C-monotone.
For a concrete example consider the distance (continuous)
 function
	$$f: [0,1]^2 \to \R, x \mapsto d(x,K), \ K:=([0,\frac{1}{3}] \cup [\frac{2}{3},1]) \times [0,1].$$

The fiber $f^{-1}(0)=K$ is not connected. So $f$ is not monotone.
On the other hand, $f$ is B-monotone because $[0,1]^2$ has no separating points.

\sk

\begin{lem} \label{l:propM} \ Let $X$ be a connected space.
	\ben
	\item Composition of B-monotone (C-monotome) functions is B-monotone (resp., C-monotone).

	\item Let $G \curvearrowright X$ be an action of a group $G$ on $X$ by homeomorphisms.
	For every $g \in G$ and every $f \in$ $M_B(X)$ ($f \in$ $M_C(X)$) we have $fg \in$ $M_B(X)$ (resp., $fg \in$ $M_C(X)$),

	where $(fg)(x):=f(gx)$.

	\item The set
	$M_B(X,D)$ is a pointwise closed (hence, compact) subset of $D^X$ for every dendron $D$.
	\item The set
	$M_B(X,[c,d])$ is a pointwise closed (hence, compact) subset of $[c,d]^X$.

	\een
\end{lem}
\begin{proof}
	(1) and (2) are straightforward.

	(3)
	Let $f: X \to D$ be the pointwise limit of the net $f_i: X \to D$, $i \in I$ where each

	$f_i \in M_B(X, D)$.
	We have to show that $f$ is also B-monotone. That is, \nl $\langle u, w, v \rangle$ in $X$ implies that
	$\langle f(u), f(w), f(v) \rangle$ in $D$.
	Since every $f_i$ is B-monotone we have $\langle f_i(u),f_i(w),f_i(v) \rangle$.
	As we already mentioned (Proposition \ref{p:Monod}) the betweenness relation $R_B$ is closed in $D^3$. So,	since $f$ is the pointwise limit of $f_i$ we get $\langle f(u), f(w), f(v) \rangle$.

	(4) is a particular case of (3).
\end{proof}

\begin{lem} \label{l:monotonicities}
	For every dendrons $X, Y$ we have $M_C(X,Y) = M_B(X,Y)$ and $M_C(X) = M_B(X)$.
\end{lem}
\begin{proof} (1)
	$M_C(X,Y) \subseteq M_B(X,Y)$.

	Assuming the contrary let $f: X \to Y$ be C-monotone but not B-monotone. Then there exist $u,w,v \in X$ such that $\langle u, w, v \rangle$ but
	$\neg  \langle f(u), f(w), f(v) \rangle$.
	 This means that $w$ separates the points $u,v$ but $f(w)$ does not separate the pair $f(u), f(v)$ in $Y$ and $f(w) \notin \{f(u),f(v)\}$. So,
	$$
	f(w) \notin C:=[f(u),f(v)].
	$$
	Since $Y$ is a dendron the generalized arc $C=[f(u),f(v)]$ is connected (Lemma \ref{l:D-prop}).
	By the C-monotonicity of the function $f$ the preimage $f^{-1}(C)$ is connected in $X$.
	As

	$w$ separates $u,v$ in $X$ we have

	$$
	X \setminus \{w\} = U \cup V,
	$$
	where $U,V$ are

	disjoint open neighborhoods of $u$ and $v$ respectively. Then
	$u \in f^{-1} (C) \cap U$ and $v \in f^{-1} (C) \cap V$  are disjoint,
	nonempty (and open in $f^{-1}(C)$). The union of these subsets is
	$f^{-1}(C)$ (because $w \notin f^{-1}(C)$). So,
	we get that $f^{-1}(C)$ is not connected, a contradiction to the fact that $f$ is C-monotone.

	\sk (2)
	 $M_C(X,Y) \supseteq M_B(X,Y)$.

	Assuming the contrary let $f \in M_B(X,Y)$ such that $f \notin M_C(X,Y)$.
	Then there exists a connected subset $C \subset Y$ such that $f^{-1}(C)$ is not connected in $X$.
	So there exist (distinct) $u,v \in f^{-1}(C)$ such that the ``generalized arc"
	$[u,v]$ (which is connected

	by Lemma \ref{l:D-prop}
	because $X$ is a dendron) is not contained in $f^{-1}(C)$.
	Therefore, there exists $w \in [u,v]$ such that $w \notin f^{-1}(C)$.
	Then $\langle u,w,v \rangle$ but it is not true that
	$\langle f(u),f(w),f(v) \rangle$ because
	$f(u), f(v) \in C   \subset Y \setminus \{f(w)\}$, and $C$ is connected.

	This proves $M_C(X,Y) = M_B(X,Y)$.
	In order to conclude that $M_C(X) = M_B(X)$ use the linear order equivalence
	$\R \to (0,1) \subset Y:=[0,1], \ x \mapsto \frac{x}{1+|x|}$.
\end{proof}

\begin{remark}
Lemma \ref{l:monotonicities} suggests dropping the subscripts ``C" and ``B"
and writing simply $M(D_1,D_2)$. We write $CM(D)$ for the

set of continuous monotone real valued functions on

$D$.
\end{remark}

Recall that the set $\F(X)$ of
fragmented real valued functions on $X$ is a vector space over $\R$.

For the definition and properties of fragmented functions we refer to
\cite{Me-nz,GM-HNS,GM-rose}. In the present paper we only use fragmentability in the case of real valued
functions $f: X \to \R$ defined on compact $X$. In this case the
 fragmentability of $f$ is equivalent to the PCP-property (see \cite{GM-HNS}).

Meaning that for every closed nonempty subset $Y \subseteq X$ the restriction $f|_Y: Y \to \R$ has a point of continuity.
 For Polish $X$, $\F(X)$ coincides with
the set $\mathcal{B}_1(X)$ of Baire  class 1 functions.

A function $f: X \to Y$ is said to be \emph{Baire class 1 function} if the inverse image $f^{-1}(O)$ of every open set
$O \subset Y$ is $F_\sigma$ in $X$.

In \cite{GM-tame, Me-Helly} we proved that every linear order preserving function on a compact linearly ordered topological space is fragmented.
The following theorem is a result in the same spirit.

\begin{defin} \label{d:stable}
Let $R$ be an abstract ternary relation on a Hausdorff topological space $(X,\tau)$.
\begin{enumerate}
\item
We say that $R$ is \textit{$\tau$-stable} if for every pair of distinct points $u,v \in X$,
there are,	a point $w \in X$ with $w \in [u,v] \setminus \{u,v\}$,
	(where $[u,v]:=\{x \in X: \langle u,w,v \rangle\} \cup \{u,v\}$),
	and neighborhoods $U,V$ of $u,v$ respectively,
	such that $w \in [x,y]$ for every $x \in U, y \in V$.
\item
We say that $R$ is

 \textit{weakly $\tau$-stable}
  if for every infinite subset
$K \subset X$ there exist: a pair of distinct points $u,v \in K$,
a point $w \in X$ with $w \in [u,v] \setminus \{u,v\}$,
and neighborhoods $U,V$ of $u,v$ respectively,
such that $w \in [x,y]$ for every $x \in U \cap K, y \in V \cap K$.
\end{enumerate}
\end{defin}

Let $R$ be a ternary relation on $X$ and let
$f: X \to \R$ respect $R$ and the standard betweenness relation of the reals.
We will say that $f$ is an  R-\textit{monotone} function.

\begin{thm} \label{t:GenFragm}
	Let $(X,\tau)$ be a compact space and $R$ a ternary relation on $X$ which is

	weakly $\tau$-stable.
	Then every $R$-monotone function $f: X \to \R$ is fragmented.
\end{thm}
\begin{proof}

	If $f$ is not fragmented then, by \cite[Lemma 3.7]{Dulst} there exists a closed nonempty subspace

	$L \subset X$ and real numbers $\alpha < \beta$

	such that the subsets
	$f^{-1}(-\infty,\alpha) \cap L$ and $f^{-1} (\beta,\infty) \cap L$

	are dense in $L$.
	That is,
	\begin{equation} \label{e:dense}
	cl(f^{-1}(-\infty,\alpha) \cap L)=	cl(f^{-1} (\beta, \infty) \cap L)=L.
	\end{equation}
	$L$ is necessarily infinite

	(because $f^{-1}(-\infty,\alpha) \cap L$ and $f^{-1} (\beta, \infty) \cap L$ are disjoint).

	Since $X$ is

	weakly $\tau$-stable
	we may choose distinct points
	$u,v \in L$, $w \in X$ with $w \in [u,v] \setminus \{u,v\}$, and $\tau$-neighborhoods
	$U$ and $V$ of $u$ and $v$ respectively such that $\lan x,w,y \ran$
	for every $x \in U \cap L, y \in V \cap L$.

	By Equation \ref{e:dense} there exist $u',u'' \in U \cap L$ and $v', v'' \in V \cap L$ such that
	\begin{equation} \label{eq:green-red}
	f(u') <\a, f(v')< \a, \ \ \ \beta < f(u''), \beta < f(v'')
	\end{equation}
	It is enough to show that $f$ is not $R$-monotone. There are two cases

	to check:

	\sk

	Case 1. $f(w) \leq \beta$.
	\sk
	Then $\langle u'',w, v'' \rangle$ but $f(w)$ is not between  $f(u''), f(v'')$ in $\R$.

	\sk
	Case 2. $\beta < f(w)$.
	\sk
	Then $\langle u',w,v' \rangle $ but $f(w)$ is not between   $f(u'), f(v')$ in $\R$.

\end{proof}

\begin{thm} \label{t:MonotFragm} Let $D$ be a dendron. Every monotone function $f: D \to \R$ is fragmented
(Baire class 1, when $D$ is a dendrite).
It follows that $M(D) \subset \F(D)$.
\end{thm}
\begin{proof}
For a dendron $D$ the standard betweenness relation (Definition \ref{d:BrelationD}) is stable.
Indeed, by the definition of dendrons for every distinct $u \neq v$ we have a separation by a point $w$. So, $D\setminus \{w\}=U \cup V$, where $U$ and $V$ are open disjoint neighborhoods of $u$ and $v$. Hence, $w$ separates any pair $x \in U, y \in V$.

	\end{proof}

For dendrites we have
$M(D) \subset \mathcal{B}_1(D)=\F(D)$.

\begin{cor} \label{c:tameFamily}
	For every dendron $D$ the family $F:=CM(D,[0,1])$ is tame.

\end{cor}
\begin{proof}
By Lemma \ref{l:propM}.4, every function $\varphi: D \to [0,1]$ from the pointwise closure of
	$F$ in $[c,d]^D$ is a (not necessarily continuous)
	B-monotone function.
	By Lemma \ref{l:monotonicities},
	$\varphi \in M(D,[0,1])$. By Theorem \ref{t:MonotFragm} we know that $M(D,[0,1]) \subset \F(D)$

and we
conclude that $cl_p(F) \subset \F(D)$. This means that $F$ is a Rosenthal family, in terms of \cite{GM-rose}.

This is the same as saying
that $F$ does not contain an independent sequence (for a detailed proof see for example \cite[Theorem 2.12]{GM-tame}).
\end{proof}

In \cite[Cor. 2.15]{Mill-Wattel} van Mill and Wattel proved the following remarkable result.
We thank Jan van Mill for providing us
the short proof below.
	For dendrites Theorem \ref{r:Mill} is well known, \cite[Theorem 1.2]{Charatoniks}.

\begin{thm} \label{r:Mill}
For every dendron $D$ and every subcontinuum $C$ there exists a naturally
defined continuous retraction $r_C: D \to C$.
Moreover, this retraction is always monotone.
\end{thm}

\begin{proof}
The map $r_C: D \to C$ is defined by
$$
r_C(x)= \cap \{[x,c] \cap C: \ c \in C\}.
$$

	 We discuss only the monotonicity of $r_C$. Other details see in \cite[Cor. 2.15]{Mill-Wattel}.
Let $x \in D$ and $y=r_C(x)$.
If $p \in [x,y]$ then $[p,y]$ is contained in $[x,y]$. Since $y\in C$,
the formula for $r_C(p)$ gives us that $r_C(p) \in [p,y] \cap C = {y}$, hence $r_C(p) =y$.

Thus all the points in the fiber of the point $y\in C$ can be connected to $y$ by a continuum in the fiber.

So, $r_C^{-1}(y)$ is connected for every $y \in C$. Since $r_C: D \to C$ is a continuous closed map then
by Section 46, subsection I, Theorem 9 in \cite{Kur} the map $r_C$ is C-monotone (Definition \ref{d:monot2}).
The subcontinuum $C$ is also a dendron. So, $r_C$ is also $B$-monotone (Lemma \ref{l:monotonicities}).
\end{proof}

This result leads to the following

\begin{lem} \label{l:retracts}
	On every dendron $D$ and

every pair $u,v$ of distinct points in $D$ there exists a continuous

monotone
function $f: D \to [0,1]$ such that $f(u)=0, f(v)=1$.

\end{lem}
\begin{proof} By Theorem \ref{r:Mill}
	for every pair of distinct points $u,v \in D$
	we have a monotone continuous retraction $r_{[u,v]}: D \to [u,v]$.
	Now recall that the ``generalized arc" $[u,v]$ is a linearly ordered compact
	connected space, \cite{Mill-Wattel}.

	 By results of Nachbin \cite[p. 48 and 113]{Nach} we have an
	 order-preserving (hence, monotone in the sense of Definition \ref{d:monot2}.1) continuous map

$h : [u, v] \to [0,1]$.

	The  composition $f =  h \circ r_{[u,v]}$
	 is the required continuous monotone map $D \to [0,1]$ which separates $u$ and $v$.
\end{proof}

\begin{thm} \label{t:Dendron}
	Let $D$ be a dendron. For every topological group $G$ and continuous action
	$G \curvearrowright D$, the dynamical $G$-system $D$ is Rosenthal representable (hence, also tame).
	It follows that the topological group $H(D)$ is Rosenthal representable.
\end{thm}
\begin{proof}
	 By Lemma \ref{l:propM}.2 we have $fg \in CM(D,[0,1])$
	for every $g \in G$ and every $f \in CM(D,[0,1])$. So, if $F:=CM(D,[0,1])$ then $FG=F$ is a $G$-invariant bounded family of continuous functions.
	By Corollary \ref{c:tameFamily}, $F$ is a tame family.
	 By Lemma \ref{l:retracts}, $CM(D,[0,1])$ separates the points of $D$.
	So one may apply Theorem \ref{t:criter1} and we obtain that the dynamical $G$-system $D$ is Rosenthal representable.

	Finally note that it is straightforward to see that Rosenthal representability of any compact dynamical $H(K)$-system $K$ implies that the topological group $H(K)$ is Rosenthal representable (for details see for example \cite[Lemma 3.5]{GM-tame}).

\end{proof}

\begin{thm} \label{t:MonotFr}
	Every monotone map $f: D_1 \to D_2$ between dendrons is fragmented (Baire class 1 map, if $D_1,D_2$ are dendrites).
\end{thm}

\begin{proof}
	By Lemma \ref{l:retracts} there exists a family of continuous monotone maps $\{q_i: D_2 \to [0,1]: i \in I\}$ which separates the points of $D_2$. Then any composition  $q_i \circ f: D_1 \to [0,1]$ is monotone, hence fragmented by Theorem \ref{t:MonotFragm}. Now, using \cite[lemma 2.3.3]{GM-rose} we obtain that the original map $f: D_1 \to D_2$ is also fragmented.
	\end{proof}

The tameness of any continuous group actions on dendrons
$G \curvearrowright D$ can be derived

from Theorem \ref{t:MonotFr} using the following

\begin{cor} \label{c:fragm}
Let $G$ act on a dendron $D$ then every element of the enveloping semigroup $E(G, D)$ is
fragmented, hence the system $(G, D)$ is tame.
\end{cor}
\begin{proof}

Using Lemma \ref{l:propM}.3 we obtain that
every element $p \in E(G, D)$ is a monotone map $p: D \to D$.
By Theorem \ref{t:MonotFr}, $p$ is a fragmented map and it follows that

the $G$-system $D$ is tame by the enveloping semigroup characterization of

tameness, \cite{GM-rose}.
\end{proof}

Theorem \ref{t:Dendron} implies also the following purely topological nontrivial fact.

\begin{cor} \label{c:DisWRN}
	Every dendron $D$, as a topological space, is Rosenthal representable. That is, $D$ is WRN.
\end{cor}

As a related result note that by \cite[Theorem 6.6]{Mill-Wattel}
a Hausdorff space can be embedded in a dendron if and only if it possesses a \textit{cross-free}
(see \cite{Mill-Wattel} for definitions)
closed subbase.

\begin{remark} \label{r:semigr}

Theorem \ref{t:Dendron} remains true also for continuous \textit{monotone} monoid actions

$S$ on $D$

such
that for all $s \in S$ the corresponding $s$-translation $D \to D$ is monotone.
Clearly continuous group actions on dendrons are always monotone.

\end{remark}

\sk

\section{Monotone actions on median pretrees}
\label{s:pretrees}

In this section we consider actions on a pretree;
a useful treelike structure that naturally generalizes several important structures

 including linear orders and the betweenness relation on dendrons.

By a \textit{pretree} (see for example \cite{B,Mal-14}) we mean a pair $(X,R)$, where
$X$ is a set and $R$ is a ternary relation on $X$
(we write $\lan a,b,c \ran$ to denote $(a,b,c) \in R$) satisfying the following three axioms:
\begin{itemize}
	\item [(B1)] $\lan a,b,c \ran \Rightarrow \lan c,b,a \ran$.
	\item [(B2)] $\lan a,b,c \ran \wedge \lan a,c,b \ran \Leftrightarrow b=c$.
	\item [(B3)] $\lan a,b,c \ran \Rightarrow \lan a,b,d \ran  \vee \lan d,b,c \ran$.
\end{itemize}
In \cite{AN} such a ternary relation is called a \textit{B-relation}.

It is convenient to use also an interval approach. For every $u,v \in X$ define
$$
[u,v]:=\{x \in X: \lan u,x,v \ran\}
$$

In the list of properties below the first four conditions (A0),(A1),(A2),(A3),
as a system of axioms, is equivalent to the above definition via (B1), (B2), (B3)
(see \cite{Mal-14}).

\begin{lem} \label{l:PropPretree} \
	In every pretree $(X,R)$ we have
	\begin{itemize}
	\item [(A0)] $[a,b] \supseteq \{a,b\}$.

	\item [(A1)] $[a,b]=[b,a]$.
		\item [(A2)] If $c \in [a,b]$ and $b \in [a,c]$ then $b=c$.

		\item [(A3)]  $[a,b] \subseteq [a,c] \cup [c,b]$ for every $a,b,c \in X$.

	\item [(A4)] $[a,b] = [a,c] \cup [c,b]$ for every $a,b \in X, c \in [a,b]$.
	\item [(A5)]  If $b \in [a,c]$ and $c \in [a,d]$ then $c \in [b,d]$.

	\end{itemize}
\end{lem}

Following \cite{Mal-14} we define the so-called \textit{shadow topology} $\tau_s$ on $(X,R)$.

Given an ordered pair $(u,v) \in X^2, u \neq v$,
let
$$
S^v_u:=\{x \in X: u \in [x,v]\}
$$
be the \textit{shadow} in $X$ defined by the ordered pair $(u,v)$.

Pictorially, the shadow $S^v_u$ is

cast
by a point $u$ when the light source is located at the point $v$.
The family $\mathcal{S} = \{S^v_u: u,v \in X, u \neq v\}$ is a subbase for the closed sets

of the topology $\tau_s$.

 In the case of a linearly ordered set we get the interval topology.
In general, for an abstract pretree the shadow topology is

often
(but not always) Hausdorff.
Furthermore, by \cite[Theorem 7.3]{Mal-14} a pretree equipped with its shadow topology is Hausdorff
if and only if, as a topological space, it can be embedded into a dendron.

So, by Corollary \ref{c:DisWRN} we can deduce the following:
\begin{cor}
Every Hausdorff pretree (e.g., linearly ordered topological space) is a WRN topological space.
\end{cor}

\sk

For every triple $a,b,c$ in a pretree $X$ the {\it median} $m(a,b,c)$ is the intersection
$$
m(a,b,c):=[a,b] \cap [a,c] \cap [b,c].
$$
When it is nonempty the median is a singleton (see for example, \cite[p. 14]{B}).
A pretree $(X,R)$ for which this intersection is always nonempty is called a \textit{median pretree}.

A {\it Median algebra} (see for example \cite[p. 14]{B} or, \cite{Vel}) is a pair $(X,m)$,
where the function $m: X^3 \to X$ satisfies the following three axioms:
\begin{itemize}
	\item [(M1)]  $m(x,x,y)=x$.
	\item [(M2)]  $m(x,y,z)=m(x,y,z)=m(y,z,x)$.
	\item [(M3)]  $m(m(x,y,z),u,v)=m(x,m(y,u,v),m(z,u,v)).$
\end{itemize}

\begin{remark} \label{r:med} \
	\ben
	\item Every median pretree is a \textit{median algebra}.

	\item A map $f: X_1 \to X_2$ between two median algebras is monotone (i.e., interval preserving)

	if and only if $f$ is median-preserving (\cite[page 120]{Vel}) if and only if $f$ is convex (\cite[page 123]{Vel}). Convexity of $f$ means that the preimage of a convex subset is convex.

	\item Every median pretree is Hausdorff
	in its shadow topology (\cite[Theorem 7.3]{Mal-14}).
	\een
\end{remark}

A \textit{compact (median) pretree} is a (median) pretree $(X,R)$ for which the shadow topology  $\tau_s$ is compact.

\begin{examples} \label{ex:median} \
	\ben
	\item
	Every dendron $D$ is a compact median pretree with respect to the standard betweenness relation $R_B$.
		Its shadow topology is just the given compact Hausdorff topology on $D$ (see \cite{Mal-14,Mill-Wattel}).

	\item Every linearly ordered set is a median pretree. Its shadow topology is just the interval topology of the order.

	\item Let $X$ be a \textit{$\Z$-tree} (a median pretree with finite intervals $[u,v]$).
	Denote by $Ends(X)$  the set of all its ends.
	According to \cite[Section 12]{Mal-14} the set $X \cup Ends(X)$ carries a natural
	$\tau_s$-\textit{compact} median pretree structure.
	\een
\end{examples}

\begin{prop} \label{p:mRetr}
	Let $(X,R)$ be a median pretree. Then the retraction map
	$$
	\phi_{u,v}: X \to [u,v], \ x \mapsto m(u,x,v)
	$$
	is monotone and continuous in the shadow topology for every $u, v \in X$.
\end{prop}
\begin{proof}
Note that always $c \in [a,b]$ if and only if $med(a,c,b)=c$. So, $\phi_{u,v}(x)=x$ for every $x \in [u,v]$.
This means that $\phi_{u,v}$ is a retraction.

	A well known property of median algebras, namely \cite[Equation 8.7]{Shol}, directly implies that
	$$
	m(m(u,x_1,v), m(u,x_2,v), m(u,x_3,v)) = m (u,m(x_1,x_2,x_3),v).
	$$
	This means that $m(\phi(x_1), \phi(x_2), \phi(x_3))= \phi(m(x_1,x_2,x_3))$.
	So every $\phi_{u,v}$ is a median preserving map (hence, monotone, as a map between pretrees).

	Now we check the continuity of $\phi_{u,v}$.
	If $u=v$ then $\phi_{u,v}$ is constant. So, we can suppose that $u \neq v$.
	Every interval $[a,b]$ is a convex subset of $X$. Hence its interval topology and its topological subspace topology of $(X,\tau_s)$ are the same
	(see \cite[Prop. 6.5]{Mal-14}).
	 It is enough to show that the preimage of a closed subbase

	 element in the space $[u,v]$ is closed in the shadow topology. We prove that in fact
$$	\phi_{u,v}^{-1}[u,w] = S^v_w \ \ \ \forall \ w \in [u,v) \ \ \text{and} \ \  \phi_{u,v}^{-1}[w,v] = S^u_w \ \ \ \forall \ w \in (u,v].$$
First we show that $\phi_{u,v}^{-1}[u,w] \subseteq S^v_w$.

Let $x \in \phi_{u,v}^{-1}[u,w]$. This means that $m:=m(u,x,v) \in [u,w]$. Since
$w \in [u,v]$, by (A5) we have $w \in [m,v]$. So, $[m,w] \cup [w,v]=[m,v]$ by (A4). Again, by (A4) we obtain $[x,m] \cup [m,v]=[x,v]$. So, $w \in [x,v]$. Hence, $x \in S^v_w$.

Now we show $\phi_{u,v}^{-1}[u,w] \supseteq S^v_w$.
Let $x \in S^v_w$. This means that $w \in [x,v]$. We have to show that $m \in [u,w]$ (where $m:=m(u,x,v)$).

 Assuming the contrary, suppose $m \notin [u,w]$.
 Clearly, $m \in [u,v]$. By (A3), $[u,v] \subseteq [u,w] \cup [w,v]$. So, we have $m \in (w,v]$. Since $w \in [v,x]$ and $m \in [v,w]$, by (A5) we get $w \in [m,x] = [x,m]$. Similarly, since $m \in [v,w]$ and $ w \in [v,u]$, by (A5) we get $w \in [m,u]$.

 Now taking into account that $m \neq w$

by (A2) we have $m \notin [x,w]$ and also $m \notin [w,u]$. So,  $m \notin [x,w] \cup [w,u]$.
Then (A3) guarantees that $m \notin [x,u]$. This contradicts the fact that $m$, being the median of the triple $x,u,v$, belongs to $[x,u]$.

\sk
Similarly we can check also that $\phi_{w,v}^{-1}[u,w] = S^u_w$.

	\end{proof}

\begin{thm} \label{p:ConvFun}

	Let $X$ be a median pretree. Then every pair of monotone (equivalently, convex)
	real valued functions $f_i: X \to \R$, $i \in \{0,1\}$ is not independent.
\end{thm}
\begin{proof}  Assuming that $\{f_1, f_2\}$ is an independent pair, there exist real numbers $a < b$
	such that
	$$
	A_1 \cap A_2 \neq \emptyset, \ A_1 \cap B_2 \neq \emptyset, \ B_1 \cap A_2 \neq \emptyset, \ B_1 \cap B_2 \neq \emptyset, \
	$$
	where $A_i=f_i^{-1}(-\infty,a), \ B_i=f_i^{-1}(b, \infty)$, $i \in \{0,1\}$.

	Choose four points $$a \in A_1 \cap B_2, \ b \in A_1 \cap A_2, \ c \in A_2 \cap B_1, \ d \in B_1 \cap B_2.$$
	By the monotonicity of the functions $f_1, f_2$ the preimages of convex subsets are convex. So,
	$A_i, B_i$ are convex. Then $$[a,b] \subset A_1, [b,c] \subset A_2, [c,d] \subset B_1, [d,a] \subset B_2.$$

	Consider the median $m:=med(c,a,d)=[a,c] \cap [c,d] \cap [d,a]$. Then
	$$m \in [a,c] \cap B_1 \cap B_2.$$
	Using (A3) we get
	$$
	[a,c] \subset [a,b] \cup [b,c] \subset A_1 \cup A_2.
	$$
	Therefore, $$m \in (A_1 \cup A_2) \cap (B_1 \cap B_2).$$
	Clearly, $A_1 \cap B_1 = A_2 \cap B_2 =\emptyset$. So, $(A_1 \cup A_2) \cap (B_1 \cap B_2)= \emptyset$, a contradiction.

	\end{proof}

\sk

For a compact median pretree $X$ we denote by $H_+(X)$ the topological group of
$R$-monotone (equivalently, median-preserving) homeomorphisms.
We treat $H_+(X)$ as a topological subgroup of the full homeomorphism group $H(X)$.

The following result generalizes Theorem \ref{t:Dendron}.
In the case of a dendron $D$ we have $H_+(D)=H(D)$.

\begin{thm} \label{t:pretree}
	For every compact median
	pretree $X$ and its automorphism group $G=H_+(X)$ the action
	of the topological group $G$ on $X$ is Rosenthal representable.
	It follows that the topological group $H_+(X)$ is Rosenthal representable.
\end{thm}
\begin{proof}

	Recall again that a median pretree is Hausdorff

	in its shadow topology (Remark \ref{r:med}.3).

	By Proposition \ref{p:mRetr} the retraction map
	$$
	\phi_{u,v}: X \to [u,v], \ x \mapsto m(u,x,v)
	$$
	is monotone and continuous in the shadow topology for every $u, v \in X$.

	Every $[u,v]$ is a linearly ordered set with respect to the order: $x \leq y$
	whenever $\lan x,y,v \ran$ (see for example, \cite[p.14]{B}).
	Moreover $[u,v]=\phi(X)$ is
	compact in the subspace topology which coincides with the interval topology of the linear order.
	So, as in Lemma \ref{l:retracts}, using Nachbin's result we see that the set $CM(X)$ of continuous monotone functions separates the points. Also, $CM(X)$ is $G$-invariant because the action $G \curvearrowright X$ is monotone. Theorem \ref{p:ConvFun} guarantees that  $CM(X)$ is a tame family.
	A median pretree is always Hausdorff
	in its shadow topology (Remark \ref{r:med}.3).

	The rest is similar to the proof of Theorem \ref{t:Dendron}.
\end{proof}

 By Example \ref{ex:median}.4, Theorem \ref{t:pretree}
 applies when $X$ is a $\Z$-tree and we get the following:

\begin{cor} \label{c:Ends}
	Let $X$ be a $\Z$-tree. Denote by $Ends(X)$ the set of all its ends.
	Then for every monotone group action
	$G \curvearrowright X$ with continuous transformations
	the induced action of $G$ on the compact space $\widehat{X}:=X \cup Ends(X)$
	is Rosenthal representable.
\end{cor}

	Such compact spaces $\widehat{X}$ as in Corollary \ref{c:Ends} are  often zero-dimensional.
So, at least, formally this case cannot be deduced from the dendron's case.

\br

\subsection{Generalized Helly's selection principle}

\begin{thm} \label{GenHellyThm}
Let $X$ be a median pretree and $\{f_n: X \to \R\}_{n \in \N}$ be a bounded sequence of convex
(equivalently, monotone) real valued maps. Then there exists a pointwise converging subsequence.
\end{thm}
\begin{proof}
Combine Theorem \ref{p:ConvFun}

with the following form of Rosenthal's theorem:
Let $f_n: S \to \R$ be a bounded sequence of functions on a set $S$.
Then it has a subsequence which is pointwise converging or

has a subsequence which is independent;
see Theorem 1 in Rosenthal's classical paper \cite{Ros0}.
In Rosenthal's original formulation he states a weaker statement ``$l_1$-subsequence" (instead of ``independent").
However, as Rosenthal's proof shows he proves in fact a little bit more
(see text above Lemma 5 on page 2413),

namely
that there exists an
independent subsequence.

\end{proof}

\begin{cor} \label{c:Helly}
Let $X$ be either a dendron or a linearly ordered set.
Then the pointwise compact set of all monotone maps $M(X,[c,d])$ into the real interval $[c,d]$ is sequentially compact.
\end{cor}

\begin{remark}
	For linearly ordered sets Theorem \ref{GenHellyThm} can be extended to
	sequences of real valued functions with bounded total variation \cite{Me-Helly}.
	This suggests the idea that one should search for a right analog for bounded variation functions, defined on dendrons, or more generally on (median) pretrees.
\end{remark}

  \br

\subsection{Fragmentability of monotone functions on compact Hausdorff pretrees}

\begin{lem} \label{l:stabPretr}
	If the shadow topology $\tau_s$ on a pretree $(X,R)$ is Hausdorff (e.g., median pretree). Then the pretree relation $R$ is

	weakly
	$\tau_s$-stable (Definition \ref{d:stable}.2).
\end{lem}
\begin{proof}
	We have to show that for every infinite subset $K \subset X$ there exist:
	a pair $u,v \in K$, a point $w \in X$ with $w \in [u,v] \setminus \{u,v\}$,
	neighborhoods $U,V$ of $u,v$ respectively,
	such that $w \in [x,y]$ for every $x \in U \cap K, y \in V \cap K$.
	In fact, we show this for every $x \in U, y \in V$.

	First of all note that there exist distinct $u,v \in K$, $w \in X$
	such that $w \in [u,v] \setminus \{u,v\}$. If not then
	$K$ is a \textit{star subset} in terms of \cite{Mal-14}. Since $\tau_s$ is
	Hausdorff there is no \textit{infinite} star subset in $X$ by \cite[Theorem 7.3]{Mal-14}.

	Consider the standard shadow topology neighborhoods
	$$u \in U:=X \setminus S^u_w \  \text{and} \ v \in V:=X \setminus S^v_w.$$
	We are going to show that
	$w \in [x,y]$ for every $x \in U, y \in V$.

	By (B3) and $\lan u,w,v \ran$ we have $\lan u,w,x \ran  \vee \lan x,w,v \ran$.
	By our choice of $x \in U=X \setminus S^u_w$ it is impossible that $w \in [x,u]$.
	So we necessarily have the second condition $\lan x,w,v \ran$. Now apply
	again (B3) but now for the triple $\lan x,w,v \ran$ and a point $y \in V$.
	Then by the definition of $V:=X \setminus S^v_w$ it is impossible that $\lan y,w,v \ran$.
	Therefore, we necessarily have $\lan x,w,y \ran$.

\end{proof}

Lemma \ref{l:stabPretr} and Theorem \ref{t:GenFragm} imply

\begin{thm} \label{t:FragmPretr}
	Let $(X,\tau_s)$ be a compact Hausdorff pretree (e.g., median pretree).
	Then every monotone real valued function $f: X \to \R$ is fragmented.
\end{thm}

\begin{remark}
	By a result of Malyutin \cite[Theorem 7.3]{Mal-14}
	every $\tau_s$-Hausdorff pretree $X$ can be topologically embedded into a dendron $D$.
	This leads to a natural question if there exists an \textit{equivariant version} of this embedding.
	Consider the automorphism group $G:=\Aut(X,R)$
	(of all monotone homeomorphisms) as a topological subgroup of $H(X, \tau_s)$.
	Is it true that there exists an equivariant embedding of $(G,X)$ into $(H(D),D)$
	with some dendron $D$?

	It seems that this is true under an additional assumption that the shadow subbase $\mathcal{S}$ of $\tau_s$ is \textit{connected} in the sense of \cite{Mill-Wattel}. Then it is possible to use results from \cite{Mill-Wattel} and \textit{superextensions} (as in \cite[Theorem 7.3]{Mal-14}).
\end{remark}

\br

\section{Some consequences of the tameness of actions on dendrites}\label{cons}

In the following sections we would like to apply Theorem \ref{thm-tame} in order
to strengthen several known results on actions of
groups on dendrites, mainly from \cite{DM-16}, \cite{MN-16} and \cite{Na-12}.
For the following definitions see e.g. \cite{Gl-76}.

\begin{defin}
Let $(G, X)$ be a dynamical system.
\begin{enumerate}
\item
A pair of points $x, y \in X$ is said to be {\em proximal}
if there is a net $g_i \in G$ and a point $z \in X$ with
$\lim g_i x = \lim g_iy =z$.
The system $(G, X)$ is {\em proximal} if every pair of points in $X$ is proximal.
\item
The system $(G, X)$ is {\em strongly proximal}
if for every probability measure $\mu$ on $X$
there is a net $g_i \in G$ and a point $z \in X$ with
$\lim g_i = \del_z$.
\item
An infinite minimal dynamical system $(G, X)$ is said to be {\em extremely proximal} if for
every nonempty closed subset $A \subsetneq X$ and any nonempty open subset $U \subset X$
there is an element $g \in G$ with $gA \subset U$.
\end{enumerate}
\end{defin}

An action of a group $G$ on a dendron $D$ is called {\em dendro-minimal}
if every $G$-invariant subdendron $C \subset D$ is either all of $D$ or the emptyset.
By Zorn's lemma every group action on a dendron admits a (nonempty) dendro-minimal subdendron.
An arc $C$ in a dendrite $D$ is a {\em free arc} if it contains no end points:
$C \cap End(D) = \emptyset$.

Except for the extreme proximality claim the following results were proven (independently) in
\cite{S-12}, \cite{Mal-13}, \cite{DM-16} and \cite{SY-17}.

\begin{thm}\label{EP}
Suppose $G$ acts on a dendrite $X$ with no finite orbits:
\begin{enumerate}
\item
There is a unique infinite minimal set $M \subset X$.
\item
If the action is dendro-minimal and $X$ has no free arc then $M=X$.

\item
The action on $M$ is extremely proximal.
\item
The action on $X$ is proximal.
\end{enumerate}
\end{thm}

\begin{proof}
(1) \
Claim (1) is proved in \cite[Corollary 4.3]{MN-16}.
In fact, it suffices to assume that there is no fixed point (see remark 4.2 in \cite{DM-16}).

For (2) see Remark 4.7 in \cite{DM-16}.

(3) \
By Lemma 4.4 of \cite{DM-16} the set of end points of $X$ is contained (and is dense) in $M$.
If $x$ is an endpoint then it has a basis for its neighborhoods which consists
of connected open sets $U$ with $|\partial (U)| = 1$ (see Lemma 2.3 in \cite{DM-16}).
Let $K \subsetneq M$ be a closed subset.
Thus there is a $x \in End(X) \cap K^c$
and a connected open neighborhood $x \in U$ with $|\partial (U)| = 1$.
As $X$ is a dendrite we have that $\overline{U^c}$ is also a dendrite
and, by Lemma 4.3 of \cite{DM-16} there is $g \in G$ with
$gK \subset g U^c \subset U$.

(4) \
Given $x, y \in X$ there is a sequence $g_j \in G$ such that
the limits $g_j x = x'$ and $\lim g_j y = y'$ exist and are elements of $M$.
Since $x'$ and $y'$ are proximal, it follows that $x$ and $y$ are proximal as well.
\end{proof}

As corollaries we obtain the following:

\begin{cor}
Suppose $G$ acts on a dendrite $X$ with no finite orbits and let
$M \subset X$ be the unique minimal set, then:
\begin{enumerate}
\item
The action of $G$ on $M$ is strongly proximal.
\item
$G$ contains a free group on two generators. In particular $G$ is not amenable.
\item
(\cite[Corollary 8.3]{DM-16})
The amenable radical of $G$ acts trivially on $M$.
\end{enumerate}
\end{cor}

\begin{proof}
It is shown in \cite{Gl-74} that extreme proximality implies strong proximality
and that a group admitting a nontrivial extremely proximal action has a free
subgroup on two generators (see also \cite{Gl-76}).
To prove part (3) consider the action of the amenable radical, say, $R \nor G$ on $M$.
This admits an invariant probability measure, say $\mu$.
By strong proximality there is a sequence $g_i \in G$ and a point $y \in M$
such that $\lim g_i \mu = \del_y$. As $R$ is a normal subgroup it follows that
each translate $g_i\mu$ as well as the limit $\del_y$ are all $R$-invariant measures.
Next we deduce similarly that every point mass $gy, \ g \in G$, is an $R$ fixed point,
and finally conclude that $R$ acts trivially on $M$.
\end{proof}

\sk

\begin{prop}\label{prop}
Let $(G, X)$ be a metric tame dynamical system.
\begin{enumerate}
\item
For any element $p \in E(G, X)$ the set $pX$ is an analytic,
hence universally measurable, subset of $X$.
\item
For every $p \in E(G,X)$ and any probability measure $\mu$ on $X$ we have
$\mu(pX)=1$.
\end{enumerate}
\end{prop}

\begin{proof}
(1)
The system $(G, X)$ being tame,

by the enveloping semigroup characterization of metric tame systems \cite{GMU},
 we have that every element $p \in E(G,X)$ is

Baire class 1, hence

 Borel measurable.
It follows that the set $pX$ is an analytic, hence a universally measurable, subset of $X$.

(2)
As $(G,X)$ is tame there is a {\bf sequence}

$g_n \in G$ with
$\lim g_n \to p$ in $E(G,X)$.
Given any function $f \in C(X)$ we have
$\lim f (g_n x) = f(px)$ for every $x \in X$.
We also have that the function $f \circ p$ is Baire class 1,
hence Borel measurable.
By Lebesgue's bounded convergence theorem it follows that
$$
\int f\, d\mu  = \lim \int (f \circ g_n)\, d\mu =
 \int \lim (f \circ g_n)\, d\mu
  = \int (f \circ p)\, d\mu = 1.
 $$
Taking $f = 1_X$ we conclude that $\mu(pX)=1$.
\end{proof}

\sk

The next theorem strengthens Theorem 10.1 of \cite{DM-16}.

\begin{thm}\label{SP}
Suppose $G$ acts on a dendrite $X$ with no finite orbits,
then the system $(G, X)$ is strongly proximal.
\end{thm}

\begin{proof}
It follows from Theorem \ref{EP} that $E(G, X)$ has a unique minimal ideal $I \subset E(G, X)$
and that $uX$ is a singleton for every minimal idempotent $u \in I$.
And conversely, for every $x \in M$ there is a minimal idempotent $u \in I$
with $uX=\{x\}$.
Applying part (2) of Proposition \ref{prop}

and Theorem \ref{thm-tame}
 to $u$ and denoting $uX =\{x\}$, we have $\delta_x(uX) =1$,
hence $u_*\mu = \delta_x$.
\end{proof}

\begin{examples} \
	\begin{enumerate}
		\item For each one of Wa\.{z}ewski's universal dendrites $X= D_n$, $n=3,4,\dots,\infty$ (\cite{Wa-23}) the dynamical system $(H(X), X)$, where $H(X)$ is the group of homeomorphisms of $X$, is minimal, tame and extremely, hence also strongly, proximal.

		Recently Kwiatkowska \cite{Kw} gave a description of the universal minimal $G$-system $M(G)$ for the topological group $G=H(X)$.
		It would be interesting to check if $M(G)$ as a dynamical $G$-system is tame.
		This space $M(G)$ is not a dendrite (being zero-dimensional) but perhaps
		it admits some suitable betweenness relation.

			\item Let $T$ be the $\R$-tree built on the Cayley graph of the free group on two generators $F_2$
			with $S = \{a, b, a^{-1}, b^{-1}\}$ as a set of generators.
			Let $X = T \cup Y$ be the natural compactification of $T$ obtained by adding the
			boundary $Y$ comprising the infinite reduced words on the generators $\{a, b, a^{-1}, b^{-1}\}$.
			Then $X$ is a dendrite

			 and the corresponding dynamical system $(F_2, X)$  is tame, with $Y$ as its unique minimal subset,
			and the system $(F_2, Y)$ is extremely proximal.

				\item Let $T$ be the $\R$-tree built on the increasing array of finite groups $\Z/ 2^k\Z$
				and let $Y = \lim_{\leftarrow}\Z/ 2^k\Z$ be the inverse limit of this array
				which can be identified with the dyadic adding machine.
				Let $X = T \cup Y$ be the corresponding compactification of $T$.
				The dynamical system $(\Z, X)$ is a $\Z$-action on a

				 dendrite,
				 hence tame, with the (equicontinuous) adding machine $Y$
				as its unique minimal subset.
	\end{enumerate}
\end{examples}

\br

\section{Tame actions of Amenable groups on dendrites}\label{discrete-spectrum}

Our starting point is the structure theorem for minimal metric tame dynamical systems
of amenable groups \cite{Gl-18} (see also \cite{H}, \cite{KL} and \cite{Gl-str}).

\begin{thm}\label{amen}
Let $\Ga$ be any group and $(G, X)$ a metric tame minimal system that
admits an invariant probability measure $\mu$.
Then
$X$ is almost automorphic,
i.e. it has the structure $X \overset{\iota}{\to} Z$, where $Z$ is equicontinuous and $\iota$
is an almost one-to-one extension. Moreover, $\mu$ is unique and the map $\iota$ is a measure theoretical isomorphism $\iota : (X, \mu, \Ga) \to (Z, \la, \Ga)$, where  $\la$ is the Haar measure on the
homogeneous space $Z$.

Thus, when $\Ga$ is amenable, since every $G$-system admits an invariant probability measure,
the claim above holds for any minimal metric tame $G$-system.
\end{thm}

\vspace{.5cm}

\begin{thm}
Let $(G, X)$ be a tame dynamical system.
Let $E = E(G, X)$ denote its enveloping semigroup
and $I \subset E$ be any minimal left ideal in $E$.
Let $\mathcal{M}$ be the collection of minimal subsets of $X$ and set
$$
\tilde{M} = \bigcup \{M \subset X :
M \in \mathcal{M}\}.
$$
Let $P_G(X)$ be the space of $G$-invariant probability measures on $X$.
Then
\begin{enumerate}
\item
For any element $p \in E$ the set $pX$ is an analytic,
hence universally measurable, subset of $X$.
\item
$\mu(pX)=1$ for every $\mu \in P_G(X)$ and $p \in E$.
\item
For any element $p \in I$ the set $pX$ is contained in $\tilde{M}$.
\item
If $\mu \in P_G(X)$ is ergodic then there exists $M \in \mathcal{M}$
with $\mu(M)=1$. Thus the minimal subsystem $(M,\mu,G)$
is uniquely ergodic and satisfies the conclusions of Theorem \ref{amen}.
\item
For any $\mu \in P_G(X)$
the measure dynamical system $(X,\mu,G)$ has discrete spectrum;
i.e. $L_2(\mu)$ is spanned by the collection of matrix
coefficients of finite dimensional unitary representations of $G$.
\end{enumerate}
\end{thm}

\begin{proof}

For claims (1) and (2) repeat the proofs of the corresponding claims in Proposition \ref{prop}.

Claim (3) is clear.

(4)
Suppose $\mu \in P_G(X)$ is ergodic. Let $p$ be any
element of a minimal ideal in $E(G, X)$.
By (2) and (3) $\mu(pX)=1$ and we can find a $\mu$-generic point
$x \in pX$. Let $M =\overline{Gx} \in \mathcal{M}$
\footnote{
For an amenable group $G$ this has the usual meaning. When $G$ is
not amenable we can still consider
a probability measure $m$ on the discrete group $G$
such that $\mu(\{g\}) > 0$ for every $g \in G$, and then find a generic point
for the Markov operator on $L_1(\mu)$ defined by $K f = m * f$.
}.
Then clearly
$\mu(M) =1$.
As the subsystem $(G, M)$ is minimal and tame, it follows
from Theorem \ref{amen}
that it is uniquely ergodic and satisfies the conclusion of
that theorem.

(5)
Let $\mu \in P_G(X)$ be any invariant measure.
As in the proof of part (2) we see that the map
$V_p : f \mapsto f \circ p$ is a linear operator on $L_2(\mu)$
of norm $\le 1$.
Since $\ch \circ p = \ch$ it follows that $\|V_p\| =1$.
Moreover, the map $V : p \mapsto V_p$ is
a continuous semigroup homomorphism from $E$ into the
semigroup of linear contractions of $L_2(\mu)$ equipped with its strong operator topology.
Let $u$ be an idempotent in $I$. Then by (2) $\mu(uX)=1$
and for every $x \in uX$ and every $p \in E$ we have $px = (pu)x$.
It follows that the image of $E$ under $V$ coincides with the image
of $I$, $\{V_p : p \in E\} = \{V_p : p \in I\}$.
Moreover, for every $p \in I$ we have $\mu(pX \cap uX) =1$,
hence $px = (up)x$ $\mu$-a.e. Thus $V_p = V_{up}$ and we
conclude that $\{V_p : p \in E\} = \{V_{up} : p \in I\}$.
Now $uI$ is a {\bf group} and it follows that this image,
which is also the closure of the Koopman group $\{V_g : g \in G\}$,
is a compact group of unitary operators.
The Peter-Weyl theorem completes the proof.
\end{proof}

Now using results of this section and Theorem \ref{thm-tame} we get

\begin{cor}
Suppose an amenable group $G$ acts on a dendrite $X$, then
\begin{enumerate}
\item
Each minimal subset  $M \subset X$ is as described in Theorem \ref{amen}
(but see Theorem \ref{SY} below).
\item
Thus an ergodic invariant probability measure on $X$ is either
 a uniform distribution on a finite set, or it is the
uniqeuly ergodic measure on a minimal infinite almost automorphic $M \subset X$.
\end{enumerate}
\end{cor}

\sk

\begin{remarks} \label{r:Mobius} \
	\ben
	\item

 In \cite{HWY} the authors show that every tame cascade satisfies the ``M\"{o}bius disjointness conjecture".
 Their proof is based on the fact that tame cascades have discrete spectrum.
 	\item

In \cite{AAM} it was shown that every monotone cascade on a local dendrite satisfies the

``M\"{o}bius disjointness conjecture".
\item
Theorem \ref{t:nullness} shows that (2) can be derived from (1), at least for every
invertible cascade on a local dendrite.
\een
\end{remarks}

In the first version of our work, posted 26 June,  2018, on the Arxiv, we
posed the following question:

\begin{question}
Is there an amenable group $G$, an action of $G$ on a dendrite $X$,
and a minimal subset $Y \subset X$, such that the system $(G, Y)$ is almost
automorphic but not equicontinuous?
\end{question}

On 4 July, 2018, Shi and Ye provided a negative answer \cite{SY-18}:

\begin{thm}\label{SY}
Let $G$ be an amenable group acting on a dendrite $X$.
Suppose $K$ is a minimal set in $X$.
Then

$(G,K)$ is equicontinuous, and $K$ is either finite or homeomorphic to the Cantor set.
\end{thm}

\br

\section{Cascades on dendrites}\label{casc}

Let $(T, X)$ be

a cascade; i.e. a $\Z$-dynamical system where $T : X \to X$
is the homeomorphism of $X$ which corresponds to $1 \in \mathbb{Z}$.

We recall the following results of Naghmouchi \cite{Na-12};
in all of them $f: X \to X$ is a monotone dendrite map.
The sets $P(f)$, $R(f)$ and $UR(f)$ denote, respectively, the
set of periodic points, recurrent points and uniformly recurrent points of $f$.
The set $\La(f)$ is the union of the $\om$-limit sets.

\begin{thm}\label{omega}
For any $x \in X$, we have:
 \begin{enumerate}
 \item
 $\om_f(x)$ is a minimal set.
 \item
$\om_f(x) \subset \overline{P(f)}$.
 \end{enumerate}
\end{thm}

\begin{thm}\label{cantor}

For any $x \in X$, $\om_f(x)$ is either a finite set or a minimal Cantor set.
In particular, f is not transitive.
\end{thm}

\begin{thm}\label{UR}

The following equalities hold
$$
UR(f) = R(f) = \La(f) = \overline{P(f)}.
$$
\end{thm}

\begin{thm}\label{distal}

The restriction map
$f \rest R(f)$ is a distal homeomorphism.
\end{thm}

Of course every self homeomorphism $T : X \to X$ of a dendrite $X$ is monotone.

We can now augment some of Naghamouchi's results in \cite{Na-12} as follows
(see also \cite{MN-16} and \cite{MN-18}):

\begin{thm}
Let $T  : X \to X$ be a self homeomorphism of a dendrite $X$
and consider the $\Z$-system $(T, X)$.
\begin{enumerate}
\item
If $M \subset X$ is a minimal subset then it is either finite or an adding machine.
\item
The union of minimal sets
$$
\tilde{M} = \bigcup \{M \subset X : M \ {\text {is minimal}}\}
$$
is closed.
\item
Every point in $X \setminus \tilde{M}$ is asymptotic to $\tilde{M}$;
i.e. $\om_T(x) \subset \tilde{M}$.
\end{enumerate}
\end{thm}

\begin{proof}
(1)\
By Theorem \ref{thm-tame}, the system $(T, X)$ is tame, and so is $M$.
By Theorem \ref{amen} $M$ is almost automorphic. By Theorem \ref{distal} it is distal,
and by Theorem \ref{cantor} it is a Cantor set.
These facts put together imply that $M$ is  a minimal  equicontinuous cascade on a Cantor set;
i.e. an adding machine.

(2)\  Follows from Theorem \ref{UR}.

(3)\  Follows from Theorem \ref{omega}.

\end{proof}

\vskip 0.7cm

\bibliographystyle{amsplain}

config: arxiv


TITLE: Why all such polynomials have $-1$ as a root?
QUESTION [0 upvotes]: Why all polynomials of this form have $-1$ as a root?
$ x^5+x^4+x^3+x^2+x+1 $    and similar polynomials like
   $ x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1$

REPLY [2 votes]: Those polynomial functions can be compressed using geometric sums,
$$
x^{2n-1}+x^{2n-2}+...+x+1=\frac{x^{2n}-1}{x-1}
$$
The numerator allows another factorization
$$
x^{2n}-1=(x^2-1)(x^{2(n-1)}+x^{2(n-2)}+...+x^2+1)
$$
so that the original polynomial is equal to
$$
(x+1)(x^{2(n-1)}+x^{2(n-2)}+...+x^2+1)
$$
This could of course also be seen directly by grouping the even number of terms in the original polynomial.


TITLE: algorithm to find the root of a real-valued function $f$
QUESTION [1 upvotes]: I see in a book the following algorithm to find the root of a real-valued function $f$
$$ \theta_{n+1} = \theta_{n} + \epsilon f(\theta_n); \epsilon >0 $$
 with the condition that the initial point will be sufficiently close the solution. What is the proof ?

REPLY [3 votes]: This will tend to converge on roots where $f'$ is negative. $\Delta \theta$ in the vicinity will point toward the zero on either side. This has loosely similar behaviour to gradient descent algorithms, but those target minima rather than zeroes.
The algorithm will diverge from roots where $f'$ is positive. $\Delta \theta$ will point away from such a root.


TITLE: From $X=\bigcup_{j=1}^n O_{y_j}$ to $\bigcap_{j=1}^n O_{y_j} \supset X$?
QUESTION [0 upvotes]: Why $\bigcap_{j=1}^n O_{y_j} \supset X$ in the proof ?
My understanding is $X=\bigcup_{j=1}^n O_{y_j}$ since $\{Oy_j:j\in\{1,...,n\}\}$ is a finite subcover of $X$.

Every Hausdorff space is normal.
$\textbf{Main proof:}$  (Made by D.ex-Machina user)
Let $T$ be a compact Hausdorff topological space and let $X$ and $Y$ be disjoint closed sets in $T$. By the lemma, for any $y\in Y$ there exists a neighborhood $U_y\ni y$ and an open set $O_y$ containing $X$ such that $U_y\cap O_y =\varnothing$.
Since $Y$ is a closed subset of a compact set it is itself compact, and therefore the cover $\left\lbrace U_y\right\rbrace _{y\in Y}$ of $Y$ has a finite subcover $U_{y_1},U_{y_2},\cdots , U_{y_n}$.
  The open sets
\begin{align*}
O^1=\bigcap_{j=1}^n O_{y_j} \supset X,  \qquad O^2=\bigcup_{j=1}^n U_{y_j} \supset Y,
\end{align*}
  are then disjoint open sets containing $X$ and $Y$, proving that every compact Hausdorff space is normal.

REPLY [1 votes]: For each $y\in Y$, the open set $O_{y}$ is such that $X\subseteq O_{y}$. In particular, $X\subseteq O_{y_{j}}$ for all $j=1,...,n$, so $X\subseteq\displaystyle\bigcap_{j=1}^{n}O_{y_{j}}$.
Note that $y_{j}$ is just a specific $y$ taken in $Y$, so $O_{y_{j}}$ is a specific $O_{y}$.


TITLE: How are particles in a collision chosen?
QUESTION [2 upvotes]: In synchrotron particle colliders, how are the particles which are collided chosen?  For the most part, collisions of different types of particles don't do anything like you might expect in a video game; there is no secret recipe list of cool things, each which require different types of particles.  So, what makes certain particles more favorable than others?

REPLY [2 votes]: In rough lines, when one is planning a particle collider experiment, one has a theory which will be tested by the experiments in the collider.
There are two streams :discovery machines, as was the Tevatron and now the LHC, which by discovering new predicted by the theory particles validate the theories, and accuracy machines , as was LEP, the electron positron collider, and the future ILC.
Hadronic collisions have much higher cross sections and as the other answer explains can go, due to their mass, to much higher energies, opening channels not available to the lower energy (per force) electron positron colliders.But the theory is very messy because of QCD and no accurate theory predictions can be checked or fitted. The electron positron collisions have much simpler feynman diagram contributions in the calculations, and thus a theory can be fitted with small errors, as was the standard model fitted at LEP.
I think it was Feynman( or maybe somebody else, I like it as an analogy) who said :"if you want to study the inside of a clock you do not bang two clocks against each other and count the fallen gears, you use a screw driver " .
So it depends on what the people proposing the machines want to do. Go for accuracy, or for spectacular new particles, which will herald supersymetry, the proposed extension of the standard model.


TITLE: Sub-algebra generating a $\sigma$-algebra is dense in this latter
QUESTION [0 upvotes]: Let $(\Omega,\mathscr{F},\mu)$ be a measured set with $\mu$ a probability measure.
Using $1$ as the indicator function we define a distance
$$ d(A,B) = \int_\Omega{|1_A - 1_B | d\mu} ~~ , ~~ A,B\in\mathscr{F}.$$
I wanted to know if we have a $\sigma$-algebra $\mathscr{A}$ which generates $\mathscr{F}$ then $\mathscr{A}$ is dense in $\mathscr{F}$ according to $d$.
I thought it would be useful to show that the set $\{ F \in \mathscr{F} | \forall \varepsilon > 0 \exists A \in \mathscr{A}, d(A,B) < \varepsilon \}$ is equal to $\mathscr{F}$.
One way is obvious.
The other one I don't see what to do. Any hint ?

REPLY [0 votes]: $d$ is not a metric. Indeed, take $\Omega =\{1,2,3,4\}$ and $\mu\{1\}=\mu\{2\}=\frac{1}{2}$. Then $d(\{3\},\{4\})=0$ but $\{3\}\neq \{4\}$.


TITLE: Probability density function of $\log(1-|X|/2)$ where $X$ is a uniformly distributed variable
QUESTION [0 upvotes]: I'm a bit confused on how to solve this. This is my work so far:

But the last step I have done is apparently wrong. Any help would be appreciated!

REPLY [1 votes]: $\mathsf P(\lvert X\rvert > n)=\mathsf P(X>n \;\color{blue}\bigcup\; X<-n)$ for any $n>0$
That is: OR not AND.


TITLE: Finding global minimizer and maximizer
QUESTION [0 upvotes]: Let $f(x,y)=x^4-8x^2+y^4-18y^2$
Find the set of global minimizers of f?
Does f have a global maximizer?Justify?
I first calculated the gradient of f and then let partial derivative of x and y to be equal to 0.
Thereby the critical points I found are (0,0)(0,3)(0,-3)(6,0)(6,3)(6,-3).
I think a global maximizer doesn't exist as when limit of function goes to infinity from both x and y ,function goes to infinity.
But how to find if global minimizers are there?

REPLY [0 votes]: The global minimizer has to be one of these
(0,0)(0,3)(0,-3)(6,0)(6,3)(6,-3). Check each to see which one makes $f$ the least value.
Because of symmetry, you can save some work.
But....
Are you sure of your answers?
$$
4 x^3 - 16x = 0 \Rightarrow x=0, \pm 2 \\
4 y^3 - 36 y=0 \Rightarrow y=0, \pm 3
$$
So you have 9 critica points: (0,0), (0,3), (0,-3), (2,0), (2,3), (2,-3), (-2,0), (-2,3), (-2,-3)


TITLE: Characterization of a collection of sets containing identity in a group so that it generates a topology?
QUESTION [0 upvotes]: Start with a group $G$ without topology, we want to generate a topology on the group by a collection of sets containing identity.
To be more specific, let $\mathcal{S}_e$ be a collection of sets containing identity $e$. Let $\mathcal{S}_x:= \{L_x(U) : x\in G, U \in \mathcal{S}_e\}$, where $L_x$ is the left multiplication by $x$. What I want to ask is under what conditions (if such conditions exist at all) of $\mathcal{S}_e$, $\mathcal{S}= \bigcup_{x\in G} \mathcal{S}_x $ is a base so that it generates a topology for $G$.
The same question can be asked in the manner of "subbase" if we do finite intersection first. Define $\mathcal{B}_e:= \{U : U \text{ is a finite intersection of sets in } S_e\}$. $\mathcal{B}_x:= \{L_x(U) : x\in G, U \in \mathcal{B}_e\}$. Under what conditions (if such conditions exist at all) of $\mathcal{S}_e$, $\mathcal{B} = \bigcup_{x\in G} \mathcal{B}_x$ is a base so that it generates a topology for $G$.
Note here if $\mathcal{S}$ or $\mathcal{B}$ is a base then the topology it generates is automatically translation-invariant. To simplify the question, I do not require $G$ is a topological group under this topology so we do not need to worry about taking inverse is continuous or not.
I think A random collection doesn't work, in either case. A counterexample in my mind is $G = Z_5$, if we define $\mathcal{S}_0 = \{4,0,1\}.$ $\mathcal{B}=\mathcal{S}$ = {{4,0,1},{0,1,2},{1,2,3},{2,3,4},{3,4,0}}. It is obviously not a base for any topology.

REPLY [1 votes]: A classic basic theorem that characterises local bases of $e$: a non-empty collection of $\mathcal{V}$ of subsets of $G$ is a local base at $e$ iff

$\forall U \in \mathcal{V}: e \in U$ (obviously)
$\forall U,V \in \mathcal{V}: \exists W \in \mathcal{V}: W \subseteq U \cap V$ (filterbase property)
$\forall U \in \mathcal{V}: \exists V \in \mathcal{V}: VV^{-1} \subseteq U$
$\forall U \in \mathcal{V} \forall x \in U: \exists V \in \mathcal{V}: xV \subseteq U$
$\forall U \in \mathcal{V}: \forall a \in G: \exists W \in \mathcal{V}: aWa^{-1} \subseteq U$.

And to characterise Hausdorff topological groups replace 1 by 1':
1'. $\bigcap \mathcal{V} = \{e\}$.
Under these circumstances $\{xV, Vx: x \in G, V \in \mathcal{V}\}$ forms a base for the topology on $G$ that makes it into a (Hausdorff) topological group.
I quoted this version from Grupos Topológicos, by Tkachenko et al. (in Spanish, bought it in Mexico), but I believe Bourbaki has a version of this too, probably many other books as well.


TITLE: What is the order of the constant $K$ in the multidimensional Dvoretzky-Kiefer-Wolfowitz inequality($Ke^{-c z}$)?
QUESTION [6 upvotes]: Let $F_n$ be the empirical distribution obtained from an i.i.d. sample
of the distribution $F:R ^d \to [0, 1]$.
Kiefer (1961) shows that the convergence of the empirical distribution is like
$$
P\left( \left\lVert F_n - F\right\rVert_\infty > z \right) \le K \exp \left( - \left(2 - \epsilon\right) n z^2 \right)
$$
where the constant $K$ depends on the dimension $d$ and the threshold $\epsilon >0$.
What is the order of the constant $K$?
Is it like $2^d$ or is it much smaller?
In the prood for the case $d=1$, I see that $K$ is approximatively the tail of some distribution, so it is small.
For higher dimensions, the proof proceed by induction and I don't see how the value of $K$ evolves.
Definition of $F_n$:
For a vector $x$ in $R^d$, DKW defines the empirical distribution as the number of sample point $X_i$ satisfying $X_i < x$, coordinate-wise; divided by the number of sampled points $n$.
Reference:
DKW wiki
DKW d=1: Dvoretzky, Kiefer, Wolfowitz 1958
DKW d>1: Kiefer, Wolfowitz, 1958
DKW d>1, sharp exponent: Kiefer, 1961

REPLY [1 votes]: I wrote the paper and the answer is 2d, at least asymptotically. In my paper, I actually prove that Kiefer’s lower bound argument is wrong. There are a handful of papers that also reach the incorrect conclusion. Largely based on Kiefer’s faulty 1958 counter example.


TITLE: Prove or disprove the existence of inverse element in $H=\{x=2k+1\}$ of the group $(\Bbb{Q}\setminus\{0\},\cdot)$
QUESTION [2 upvotes]: Let $(\Bbb{Q}\setminus\{0\},\cdot)$ be a group. Is $H=\{x=2k+1,\;k\in\Bbb{Z}\}$ a subgroup of $(\Bbb{Q}\setminus\{0\},\cdot)$?

I need help with the existence of inverse element.
First of all I think $x$ in $H$ is $x\color{red}{\in\Bbb{Q}\setminus\{0\}}$ i.e. $x=2k+1\in\Bbb{Q}\setminus\{0\}$. Not only $x=2k+1\color{blue}{\in\Bbb{Z}}$. Why? Because of the following:
I need to prove that for all $x\color{red}{\in\Bbb{Q}\setminus\{0\}}$ there exists a $x^{-1}\color{red}{\in\Bbb{Q}\setminus\{0\}}$ such that $$x\cdot x^{-1}=x^{-1}\cdot x=e,$$ where $e=1\in\Bbb{Q}\setminus\{0\}$. (The definition of existence of inverse element.)
So let $x=2k+1\color{red}{\in\Bbb{Q}\setminus\{0\}}$. Pick $x^{-1}=\frac{1}{2k+1}\color{red}{\in\Bbb{Q}\setminus\{0\}}$. The denominator is not $0$ since $k=-1/2\notin\Bbb{Z}$. Then for all $k\in\Bbb{Z}$ we have $$(2k+1)\cdot\frac{1}{2k+1}=\frac{1}{2k+1}\cdot(2k+1)=1=e,$$ so $H$ has an inverse $x^{-1}=1/(2k+1)$.
So my questions are:

Is the $\color{red}{\text{red}}$ part correct? Or should we use the $\color{blue}{\text{blue}}$ part?
Is my inverse correct?

REPLY [2 votes]: $H$ is a set of integers. Use your knowledge of what inverses of integers look like in $\mathbb{Q}$. They are numbers of the from $\frac{1}{z}$ for z $\in \mathbb{Z}$.
Now $\frac{1}{z}$ in general will not be an integer unless $z = 1$ or $z = -1$. Hence if you find any element in $H$ that is not $1$ or $-1$, you know that there is no inverse of that element in $H$, as $H$ only contains integers.
Hence $H$ is not closed under inverses, hence $H$ is not a subgroup.


TITLE: Number of integer triangles.
QUESTION [1 upvotes]: Number of integer isosceles or equilateral triangle none of whose
  sides exceed $2c$ is?

I substituted $c =3$ checked, and got $27= 3 \times (3^2)$ i.e. $3c^2$ triplets. It gives the right answer($3c^2$) but how do I write a proper formal proof of this?
I tried it this way: We have three vacant places for three integers. First place can be filled in $2c$ ways, 2nd one in $2c$ ways, 3rd one in $2c$ way but it involves over-counting, is it possible to eliminate the extras?

REPLY [1 votes]: Let $b$ be the length of the base and $a$ be the length of the legs of such a triangle. Given $c\geq1$ we want
$$1\leq b\leq 2c, \quad {b\over2}<a\leq 2c\ .\tag{1}$$
If $b=2k-1$ with $1\leq k\leq c$ is odd the condition $(1)$ enforces $k\leq a\leq 2c$ and allows of $2c-(k-1)$ different integer values for $a$.
If $b=2k$ with $1\leq k\leq c$ is even then $(1)$ enforces $k+1\leq a\leq 2c$ and allows of $2c-k$ different integer values for $a$.
The total number $N$ of isosceles triangles having sides $\leq2c$ therefore is given by
$$N=\sum_{k=1}^c\bigl(2c-(k-1)\bigr)+\sum_{k=1}^c(2c-k)=4c^2-\sum_{k=1}^c(2k-1)=3c^2\ .$$


TITLE: In a group $G$, if every elements are of finite order then $G$ is at most countable.
QUESTION [2 upvotes]: Let us consider a group $G$.
If $G$ is a finite group then every element of $G$ has finite order. But the converse need not be true, i.e., every elements in a group has finite order doesn't imply the group is finite.
Examples: The factor group $\mathbb{Q}/\mathbb{Z}$, the Prüfer $p$-group etc. (all these groups are countable).

Conjecture $1$: In a group $G$, if every element has finite order
then $G$ is at most countable.
Conjecture $2$: Every uncountable groups contain an infinite cyclic
group.

Let $S$ is a countable subset of $G$ (possible because of Axiom of Choice). Since $G$ is not countably generated i.e $G\neq \langle S\rangle$, $\exists g_1\in G$ such that $g_1\notin \langle S\rangle$.
Again $\langle S, g_1\rangle$  is countable implies $G\neq \langle S, g_1\rangle $.Hence $\exists g_2\in G$ such that $g_2\notin \langle S, g_1\rangle $
$\vdots$
Hence $\exists (g_n) \in G$ such that $g_{n+1}\notin \langle S,  g_1,\ldots,g_n\rangle $
Let $H=\{g_n\}_{n\in\Bbb{N}}$
Claim: $\langle H\rangle\cong \mathbb{Z}$
I have no way to proceed further. Is it possible to define an isomorphism between $H$ and $\mathbb{Z}$ ?
Conjecture $1$ and $2$ are equivalent. I want to prove the second one using the fact that "an uncountable group can't be countably generated".
One final thought : Does $ G=\Pi_{i\in \Bbb{N}} \mathbb{Z}_2$ provide a counter example?

REPLY [0 votes]: As we noted,   $$G=\prod_{i\in \Bbb N}\Bbb Z_2$$ would indeed suffice for a counterexample.   If you're familiar with the binary representation of real numbers; or by Cantor,  $$\mid G\mid=2^{\aleph_0}\gt\aleph_0 $$ it's an uncountable group. (@egreg's answer covers this I believe)