text:
\section{Introduction}

In this paper we will study the F\o lner function for Thompson's group $F$.
Recall that a finite subset
$A$ of a finitely generated group $G$ is \emph{$\varepsilon$-F\o lner} (with
respect to a finite generating set $\Gamma \subseteq G$) if
\[
\sum_{\gamma \in \Gamma} |(A \cdot \gamma) \ \triangle A| < \varepsilon |A|
\]
where $\triangle$ denotes symmetric difference.
The \emph{F\o lner function} of $G$ (with respect to $\Gamma$) is defined by
\[
\Fol_{G,\Gamma}(n) = \min \{|A| : A \subseteq G \textrm{ is } \frac{1}{n}\textrm{-F\o lner w.r.t. } \Gamma\}
\]
with $\Fol_{G,\Gamma} (n) = \infty$ if there is no $1/n$-F\o lner set with respect to $\Gamma$.
Notice that $G$ is amenable if and only if its F\o lner function (with respect to any finite generating set
$\Gamma$) is finite valued.

Thompson's group $F$ has many equivalent definitions;
we will use the formulation in terms of tree diagrams defined below.
The standard presentation of $F$ is infinite, with generators $x_i$ $(i \in \Nbb)$
satisfying $x_i^{-1} x_n x_i = x_{n+1}$ for all $i < n$.
It is well known, however, that $F$ admits the finite presentation
\[
\Seq{A,B \ |\ [AB^{-1},A^{-1}BA] = [AB^{-1},A^{-2}BA^2]= \mathrm{id}}
\]
(see \cite{CFP}).
Geoghegan conjectured that $F$ is not amenable \cite{OP_inf-dim_top}
and at present this problem remains open.
The goal of this paper is to establish the following lower bound on the
F\o lner function for $F$.
\begin{thm} \label{tower_growth}
For every finite symmetric generating set $\Gamma \subseteq F$
there is a constant $C>1$ such that if $A \subseteq F$
is a $C^{-n}$-F\o lner set with
respect to $\Gamma$, then $A$ contains at least $\exp_{n} (0)$ elements.
In particular $\Fol_{F,\Gamma}$ is not eventually dominated $\exp_p(n)$ for any
finite $p$.
\end{thm}
Here $\exp_p(n)$ is the $p$-fold composition of the exponential function
defined by $\exp_{0} (n) = n$ and $\exp_{p+1}(n) = 2^{\exp_p (n)}$.
If it turns out that $F$ is amenable, then Theorem \ref{tower_growth}
would be a step toward answering (negatively)
the following question of Gromov \cite[p. 578]{entropy_isoperimetry}.

\begin{question}
Is there a primitive recursive function which eventually dominates every
F\o lner function of an amenable finitely presented group?
\end{question}
While it is known that the F\o lner functions of amenable finitely generated groups can grow arbitrarily
fast \cite{pw_auto_groups}, this is not the case for finitely presented groups
(since there are only countably many such groups).
See \cite{iso_profiles_fg}
for what can be accomplished via wreath products of $\Zbb$.

This note is organized as follows.
In Section \ref{background},
I will review some of the basic definitions associated with $F$ and
fix some notational conventions.
In Section \ref{marginal_section}, I will introduce the notion of a \emph{marginal set}
and prove some basic lemmas about them.
These are sets which must have small intersections with F\o lner sets.
They play a central role in the proof of the main result of the paper.
Section \ref{act_on_Tcal} recasts the amenability of $F$ in terms of its partial right
action on the finite rooted ordered binary trees $\Tcal$.
Section \ref{partial_section}
defines an operation on elements of $\Tcal$ which exponentially decreases their size
and commutes with the partial right action of $F$.
It is shown that the trees which are trivialized by this operation are
marginal and it is this that allows the proof of Theorem \ref{tower_growth}.

\section{Notation and background}

\label{background}

I will use \cite{CFP} as a general reference for Thompson's group $F$, although the reader is warned
that the notation in the present paper will differ somewhat from that of \cite{CFP}.
Let $\Tcal$ denote the collection of all \emph{finite rooted binary trees}.
For concreteness, we will view elements $T$ of $\Tcal$ as finite sets of
binary sequences which have the following property:
\begin{itemize}

\item For infinite binary sequence $x$, there is a unique element of $T$ which is an initial
part of $x$.

\end{itemize}
Thus an element $T$ of $\Tcal$ is a record of
the addresses of the leaves of the tree which it represents.
The \emph{trivial tree} is the set which consists only of the sequence of length $0$.

The collection of finite binary sequences is equipped with the operation of
concatenation (denoted $u \tie v$), the partial order $\subseteq$ of extension (defined
by $u \subseteq u \tie v$), and the lexicographic order (denoted $u <_\lex v$).
Note that \emph{$u$ extends $v$} includes the possibility that $u = v$.
$\Seq{}$ will be used to denote the sequence of length $0$.
$\Tcal$ can also be characterized as being those finite sets $T$ of binary sequences such that:
\begin{itemize}

\item
no element of $T$ is an initial part of another element of $T$ and

\item
if $u$ is a binary sequence, then $u \tie \seq{0}$ has an extension in $T$ if and only
if $u \tie \seq{1}$ has an extension in $T$.

\end{itemize}
If $U$ and $V$ are in $\Tcal$, then we will say that $U$ is \emph{contained in} $V$
if every element of $U$ has a extension in $V$.
Even though this is a somewhat unorthodox usage of \emph{containment}, it should cause no confusion
as $U \subseteq V$ can occur for $U, V \in \Tcal$ only when $U = V$.
If $T$ is in $\Tcal$ and $u$ a finite binary sequence,
define
\[
T/u = \{s : u \tie s \in T\}.
\]
If this set is non-empty, then it is again an a member of $\Tcal$
(in which case $T/u$ is the tree of descendants of $u$).
Elements of $\Tcal$ come with a canonical ordering provided by $<_\lex$
and phrases such
as \emph{the $i{}^{\textrm{th}}$ element} and \emph{the minimum element}
will always refer to the $<_\lex$-order in this context.
In this paper $\Nbb$ contains $0$ and in particular counting will always start at 0.
The letters $i$, $j$, $k$, and $l$ will always be used to denote natural numbers.

A \emph{tree diagram} is a pair $(L,R)$ of elements of $\Tcal$ such that
$|L| = |R|$.
We view a tree diagram as describing a map of sequences defined by
$$s_i \tie x \mapsto t_i \tie x$$
where $s_i$ and $t_i$ are the $i{}^{\textrm{th}}$ elements of $L$ and $R$ respectively and
$x$ is any binary sequence.
This map is defined not only on all infinite length binary sequences but also on all but finitely many
finite binary sequences.
The value of the associated map $f$ at a sequence $t$ will be denoted $t \cdot f$ when it
is defined.
If $t$ is a finite sequence, then we will say that $f$
\emph{acts properly} on $t$ if $t \cdot f$ is defined and
the final digit of $t \cdot f$ agrees with that of $t$.
Notice that $f$ acts properly on $t$ unless $s\cdot f$ is undefined for every
proper initial part $s$ of $t$.
In particular, if $f$ acts properly on $t$, it acts properly on any extension of $t$.

If two tree diagrams define the same map on infinite sequences, then they are said
to be equivalent.
Every tree diagram is equivalent to
a unique minimal tree diagram;
such a tree diagram is said to be \emph{reduced}.
Furthermore a tree diagram $(S,T)$ is reduced if and only if whenever
$i < |S| - 1$ it is not the case that both $s_i$ and $t_i$ end with $\seq{0}$ and
$s_{i+1}$ and $t_{i+1}$ end with $\seq{1}$ (where $s_i$ and $t_i$ are the $i{}^{\textrm{th}}$ elements
of $S$ and $T$ respectively).
See \cite{CFP} for details.

Thompson's group $F$ is the collection of reduced tree diagrams with
the operation defined by $f \cdot g = g \circ f$ (i.e. ``$f$ followed by $g$'').
$F$ is generated by $\{x_0,x_1\}$ where $x_0$ and $x_1$ are specified by:
\[
x_0 =
\begin{cases}
\seq{00} \mapsto \seq{0} \\
\seq{01} \mapsto \seq{10} \\
\seq{1} \mapsto \seq{11} \\
\end{cases}
x_1 =
\begin{cases}
\seq{0} \mapsto \seq{0} \\
\seq{100} \mapsto \seq{10} \\
\seq{101} \mapsto \seq{110} \\
\seq{11} \mapsto \seq{111} \\
\end{cases}
\]
In our discussion of $F$
``generator'' will mean an element of the set $\Gamma = \{x_0,x_1,x_0^{-1},x_1^{-1}\}$
(this is really only relevant in Lemma \ref{doubling}).
This includes the usage of \emph{F\o lner} in Sections \ref{act_on_Tcal} and \ref{partial_section}.
If $x_n$ is defined by $x_n = x_0^{-n} x_1 x_0^n$ for $n > 1$, then we obtain
the generators which yield the infinite presentation of $F$ mentioned in the introduction.

Notice that the existence of constant $C$ satisfying the conclusion of Theorem \ref{tower_growth}
for $\Gamma$ implies the main theorem for all finite generating sets.
This is because if $\Gamma'$ is any other finite generating set, there is a constant $K >0$ such
that any set which is $\varepsilon$-F\o lner with respect to $\Gamma'$ is $K\varepsilon$-F\o lner
with respect to $\Gamma$.

I will also identify elements of $F$ with the corresponding maps on sequences.
If $T$ is in $\Tcal$ and $f$ is in $F$ with $f$ defined on all of $T$,
then $T \cdot f$ is the pointwise image of $T$ under $f$.
It is easily checked that this results in an element of $\Tcal$ and hence
this defines a partial right action of $F$ on $\Tcal$.
If $T$ is in $\Tcal$ and $f$ is in $F$, then $f$ acts properly on $T$ if
it acts properly on the elements of $T$.
Observe that if $(S,T)$ is a reduced tree diagram which represents $g$ and $f$ acts properly
on $T$, then $(S,T \cdot f)$ is reduced and represents $f \circ g$.
If $f$ is in $F$, I will write $(L_f,R_f)$ to denote the reduced tree diagram
for $f$.

\section{Marginal sets}

\label{marginal_section}
In this section we will introduce the notion of a \emph{marginal set} and
collect some basic lemmas which we will use in Sections \ref{act_on_Tcal} and \ref{partial_section}.
Throughout this section ``right'' in an implicit adjective whenever
applicable, although all statements have their corresponding ``left'' analogs.
Fix, for the duration of the section,
a group $G$ with a finite symmetric generating set $\Gamma$.

\begin{defn}
A \emph{partial action} of $G$ on a set $S$ is a partial function
$\cdot: S \times G \to S$ such that:
\begin{itemize}

\item $x \cdot e = x$ for all $x \in S$;

\item $x \cdot g = y$ if and only if $x = y \cdot g^{-1}$ for all $g \in G$ and $x,y \in S$;

\item
$x \cdot (gh) = (x \cdot g) \cdot h$ for all $g,h\in G$ and all $x \in S$ for which
all computations involving $\cdot$ are defined.

\end{itemize}
If $E \subseteq S$ and $g$ is in $G$, I will write $E \cdot g$ to denote $\{x \cdot g : x \in E\}$.
\end{defn}

\begin{remark}
Since the exponentiation is defined in $G$, it is entirely possible in
general for $x \cdot g^i$ to be undefined while $x \cdot g^j$ is defined for
some $i < j$.
\end{remark}

\begin{defn}
A \emph{weighted $\varepsilon$-F\o lner set with respect to $\Gamma$} is a function $\mu$ from a
finite subset of $S$ into $(0,\infty)$
which satisfies
\[
\sum_{\gamma \in \Gamma} \sum_{s \in S} |\mu(s \cdot \gamma) - \mu (s)| < \varepsilon \sum_{s \in S} \mu(s)
\]
where we adopt with the conventions that
$\mu (s) = 0$ if $s$ is not in the domain of $\mu$ and
$\mu(s \cdot g) = 0$ if $s \cdot g$ is undefined.
The set $\{s \in S : \mu(s) > 0\}$ will be referred to as the \emph{support}
of $\mu$.
The function $\mu$ induces a finitely supported measure on $S$, also denoted
$\mu$, defined by $\mu(A) = \sum_{s \in A} \mu(s)$.
In the remainder of the paper, the generating set
will always be clear from the context
and we will often suppress mention of it.
\end{defn}

Notice that an $\varepsilon$-F\o lner set is a set $A \subseteq S$ such that
the characteristic function $1_A$ is a weighted $\varepsilon$-F\o lner set.
$\mu \restriction A$ will be used to denote $\mu \cdot 1_A$.
We will need the following property of weighted F\o lner sets, which is
what justifies their added generality in the present paper.

\begin{lem} \label{Folner_hom_prop}
Suppose $G$ acts partially on sets $S$ and $T$ and
that $\mu$ is a weighted $\varepsilon$-F\o lner set with respect to the action on $S$.
If $h:S \to T$ satisfies that $h(s \cdot \gamma) = h(s) \cdot \gamma$
(with both quantities defined) whenever
$\mu(s) + \mu(s \cdot \gamma) > 0$ and $\gamma \in \Gamma$,
then $\nu(t) = \sum_{h(s) = t} \mu(s)$ ($s$ ranges over $S$) defines a weighted $\varepsilon$-F\o lner
set (with respect to the action on $T$).
\end{lem}

\begin{proof}
Let $G$, $S$, $T$, $\mu$, and $\nu$ be as in the statement of the lemma.
We need to verify that
\[
\sum_{\gamma \in \Gamma} \sum_{t \in T} |\nu(t \cdot \gamma) - \nu (t)| < \varepsilon \nu(T)
\]
Let $\gamma \in \Gamma$ and $t \in T$ be fixed for the moment.
First observe that if $|\nu(t \cdot \gamma) - \nu (t)| > 0$,
then $t \cdot \gamma$ is defined.
This is because otherwise it must be the case that $\nu(t) > 0$ and
hence there must be an $s$ such that
$h(s) = t$ and $\mu(s) > 0$.
In particular, this implies $t \cdot \gamma$ is defined.

Next we have that
\[
\sum_{h(u) = t \cdot \gamma} \mu(u) = \sum_{h(u) \cdot \gamma^{-1} = t} \mu(u) =
\sum_{h(u \cdot \gamma^{-1}) = t} \mu (u)
= \sum_{h(s) = t} \mu(s \cdot \gamma)
\]
The first equality is justified by the properties of a partial action;
the second equality is justified by our assumption that $\mu(u) > 0$ implies
$h(u \cdot \gamma^{-1}) = h(u) \cdot \gamma^{-1}$ with both quantities defined;
the third equality is justified by the properties of a partial action and our assumption
that $\mu(u) > 0$ implies $u \cdot \gamma^{-1}$ is defined.
Now it follows that
\[
\sum_{\gamma \in \Gamma} \sum_{t \in T} |\nu(t \cdot \gamma) - \nu (t)| =
\sum_{\gamma \in \Gamma} \sum_{t \in T} |\sum_{h(s) = t} \mu(s \cdot \gamma) - \mu(s)|
\]
\[
\leq \sum_{\gamma \in \Gamma} \sum_{s \in S} |\mu(s \cdot \gamma) - \mu(s)| < \varepsilon \mu(S) = \varepsilon \nu(T).
\]
\end{proof}

Fix a partial action of $G$ on a set $S$ for the duration of this section.
If $g$ is in $G$, let $d_g$ be the minimum length of a word in $\Gamma$
which evaluates to $g$.
We will need the following lemma.
\begin{lem} \label{dg_Folner}
If $\varepsilon > 0$, $\mu$ is a weighted $\varepsilon$-F\o lner set and
$g$ is in $G$, then
\[
\sum_{s \in S} |\mu(s \cdot g) - \mu(s)| < \varepsilon d_g \mu(S).
\]
\end{lem}

\begin{proof}
Let $\gamma_i$ $(i < d_g)$ be elements of $\Gamma$
such that $g = \prod_{i < d_g} \gamma_i$.
Let $g_j = \prod_{i < j} \gamma_i$.
\[
\sum_{s \in S} |\mu(s \cdot g) - \mu(s)| \leq \sum_{s \in S} \sum_{i < d_g} |\mu(s \cdot g_{i+1}) - \mu(s \cdot g_i)|.
\]
Notice that it may be that $s \cdot g_{i+1}$ is defined even though $s \cdot g_i$ is not.
If this is the case, however, then
\[
|\mu(s \cdot g_{i+1}) - \mu(s \cdot g_i)| = |\mu((s \cdot g_{i+1}) \cdot \gamma_i) - \mu(s \cdot g_{i+1})|.
\]
It follows that
\[
\sum_{i < d_g} \sum_{s \in S} |\mu(s \cdot g_{i+1}) - \mu(s \cdot g_i)| \leq
\sum_{i < d_g} \sum_{\sigma = \pm 1} \sum_{s \in S} |\mu(s \cdot \gamma_i^\sigma) - \mu(s)|
\]
which is less than $\varepsilon d_g \mu (S)$.
\end{proof}

We will be interested in the following notion which ensures that a set
has small intersection with any F\o lner set.
The definition is motivated by the following simple observation.
If $\mu$ is a finitely additive invariant probability measure on a group $G$
and $E \subseteq G$ satisfies that, for some $g \in G$,
$\{E \cdot g^i : i \in \Nbb\}$ is a pairwise disjoint family, then
$\mu(E) = 0$.

\begin{defn}
If $g \in G$, $I \subseteq S$, and $E \subseteq S$, then
\emph{$g$ marginalizes $E$ off $I$} if for every $x \in E$
if $x \cdot g^k \in E$ and $k > 0$, then there is an $i < k$ such that
$x \cdot g^i$ is in $I$ or is undefined.
If $I$ is the empty-set, then I will write $g$ \emph{marginalizes} $E$.
\end{defn}

\begin{defn}
The $k$-\emph{marginal sets} for the partial action of $G$ on $S$ are
defined recursively as follows.
The empty-set is $0$-marginal.
If there is a decomposition
$E = \bigcup_{i < l} E_i \subseteq S$ and for each $i < l$, there is
a $g_i \in G$ and a $k$-marginal set $I_i$ such that
$g_i$ marginalizes $E_i$ off $I_i$, then $E$ is $(k+1)$-marginal.
$E \subseteq S$ is \emph{marginal}
if it is $k$-marginal for some $k < \infty$.
\end{defn}

We will need the following lemma which shows that marginal sets have small
intersections with F\o lner sets.

\begin{lem} \label{marginal}
Suppose $\varepsilon > 0$ and
$\mu$ is a weighted $\varepsilon$-F\o lner set with support $A$.
If $E \subseteq S$ and $g$ marginalizes $E$ off $S \setminus A$,
then $\mu(E) < d_g \varepsilon \mu(S)$.
\end{lem}

\begin{proof}
For each $x \in E \cap A$, let
$\Phi(x)$ denote the set of all $x \cdot g^i$ such that
\begin{itemize}

\item
$x \cdot g^j$ is defined and in the support of $\mu$
for all $j \leq i$ and

\item
$\mu(x \cdot g^{i+1}) < \mu (x \cdot g^j)$ for all $j \leq i$.

\end{itemize}
Observe that each $\Phi(x)$ is finite and non empty.
Because $E$ is marginalized off $A$ by $g$,
$\{\Phi(x) : x \in E\}$ is a pairwise disjoint family.
Observe that for a fixed $x \in E$
\[
\mu(x) \leq
\sum_{y \in \Phi(x)} \mu(y) - \mu(y \cdot g) =
\sum_{y \in \Phi(x)} |\mu(y \cdot g) - \mu (y)|.
\]
Summing over $x \in E$ and applying Lemma \ref{dg_Folner}
now gives the desired estimate.
\end{proof}

\begin{lem}
If $E \subseteq S$ is marginal, then there is a constant $C$ such that
if $\varepsilon > 0$ and $\mu$ is a weighted $\varepsilon$-F\o lner set, then
$\mu(E) < C \varepsilon \mu(S)$.
\end{lem}

\begin{proof}
I will prove by induction on $k$ that if $E$ is $k$-marginal, then the
conclusion of the lemma holds.
If $k = 0$, then $E$ is in fact empty and $\mu(E) = 0$.
Next suppose that $E$ is $(k+1)$-marginal.
Let $I_i$ $(i < l)$, $g_i$ $(i < l)$, and $E_i$ $(i < l)$ be such that:
\begin{itemize}

\item for each $i < l$, $g_i$ is in $G$ and marginalizes $E_i$ off $I_i$;

\item for each $i < l$, $I_i$ is $k$-marginal;

\item $E = \bigcup_{i < l} E_i$.

\end{itemize}
By our inductive assumption, there are $C_i$ $(i < l)$ such that
if $\mu$ is a weighted $\varepsilon$-F\o lner set, then
$\mu(E_i) < C_i \varepsilon \mu(S)$.
Set $C = \sum_{i < l} C_i + d_{g_i}$.
Now if $\mu$ is a weighted $\varepsilon$-F\o lner set,
\[
\mu(E) \leq \mu(\bigcup_{i < l} I_i) + \mu(\bigcup_{i < l} E_i \setminus \bigcup_{i < l} I_i)
\leq \sum_{i < l} C_i \varepsilon \mu(S) + d_{g_i} \varepsilon \mu(S) = C \varepsilon \mu(S).
\]
\end{proof}

\begin{lem} \label{Folner_refine}
If $\mu$ is a weighted $\varepsilon$-F\o lner set and $\nu$ is a function from a finite subset
of $S$ into $[0,\infty)$ with $\nu \leq \mu$ pointwise and $\nu(S) \geq (1-\delta) \mu(S)$,
then $\nu$ is a weighted $(\varepsilon + |\Gamma| \cdot \delta)/(1-\delta)$-F\o lner set.
\end{lem}

\begin{proof}
Let $\mu$ and $\nu$ be given as in the statement of the lemma.
Fix an $s \in S$ and $\gamma \in \Gamma$.
If $\nu(s \cdot \gamma) \geq \nu (s)$, then
\[
|\nu(s \cdot \gamma) - \nu(s)| \leq \mu(s \cdot \gamma) - \mu(s) + \delta \mu(s).
\]
If $\nu(s \cdot \gamma) \leq \nu(s)$, then
\[
|\nu(s \cdot \gamma) - \nu(s)| \leq \mu(s) - \mu(s \cdot \gamma) + \delta \mu(s \cdot \gamma).
\]
In either case, we have
$|\nu(s \cdot \gamma) - \nu(s)| \leq |\mu(s \cdot \gamma) - \mu(s)| + \delta (\mu(s) + \mu(s \cdot \gamma))$.
Summing over $s \in S$ and $\gamma \in \Gamma$ and combining this with our
hypotheses we obtain:
\[
\sum_{\gamma \in \Gamma} \sum_{s \in S} |\nu(s \cdot \gamma) - \nu(s)| < \varepsilon \mu(S) + 2 |\Gamma| \delta \mu(S)
\leq \frac{ \varepsilon + 2|\Gamma| \delta}{1 - \delta} \nu(S).
\]
\end{proof}

\begin{lem} \label{marginal->Folner}
If $E \subseteq S$ is marginal, then there is a constant $C$ such that
if $\mu$ is a weighted $\varepsilon$-F\o lner set,
then $\mu \restriction (S \setminus E)$ is a weighted $C \varepsilon$-F\o lner set.
\end{lem}

\begin{proof}
Let $E$ be marginal and let $C_0$ be such that if $\varepsilon > 0$ and $\mu$ is a weighted
$\varepsilon$-F\o lner set, then
$\mu(E) < C_0 \varepsilon \mu(S)$.
Set $C = 2(1 + 2 C_0 |\Gamma|)$ and observe that if
$C \varepsilon \leq 1$, then
\[
\frac{\varepsilon + 2 |\Gamma| C_0 \varepsilon}{1-C_0 \varepsilon} \leq
2(1 + 2 |\Gamma| C_0) \varepsilon.
\]
Hence by Lemma \ref{Folner_refine}, if
$\mu$ is $\varepsilon$-F\o lner, $\mu \restriction (S \setminus E)$ is
$C \varepsilon$-F\o lner.
\end{proof}

\begin{defn}
A subset $A \subseteq G$ is \emph{$\Gamma$-connected}
if whenever $x$ and $y$ are in $A$,
there are $\gamma_i$ $(i < l)$ in $\Gamma$ such that, setting
$x_0 = x$ and $x_{i+1} = x_i \cdot \gamma_i$, then $x_i$ is defined
for each $i \leq l$ and $y = x_l$.
A maximal connected subset of a given $B \subseteq G$
is said to be a \emph{$\Gamma$-connected component} of $B$.
\end{defn}

\begin{lem} \label{Folner_connected}
If $\varepsilon > 0$ and $\mu$ is a weighted $\varepsilon$-F\o lner set, then the support
of $\mu$ contains a $\Gamma$-connected component $A$
such that $\mu \restriction A$ is $\varepsilon$-F\o lner.
\end{lem}

\begin{proof}
Let $A_i$ $(i < l)$
enumerate the connected components of the support of $\mu$.
Then since
\[
 \sum_{i < l} \sum_{\gamma \in \Gamma} \sum_{s \in A_i} |\mu(s \cdot \gamma) - \mu(s)| <
\varepsilon  \sum_{i < l} \mu (A_i),
\]
there must exist an $i < l$ such that
\[
\sum_{\gamma \in \Gamma} \sum_{s \in A_i} |\mu(s \cdot \gamma) - \mu(s)| <
\varepsilon  \mu (A_i).
\]
Since $A_i$ is connected,
$\mu (s \cdot \gamma) = \mu \restriction A_i (s \cdot \gamma)$ whenever
$s$ is in $A_i$ and therefore $A_i$ is $\varepsilon$-F\o lner.
\end{proof}

\section{F\o lner sets of trees}

\label{act_on_Tcal}

Rather than studying F\o lner sets in $F$ directly, it will
be easier to deal with weighted F\o lner sets in the partial right action of
$F$ on $\Tcal$ which I will
refer to as \emph{weighted F\o lner sets of trees}.
These are essentially weighted right F\o lner sets consisting of positive elements of $F$ (positive
with respect to the infinite presentation mentioned above).
The reformulation of the amenability problem for $F$ in terms of the existence of F\o lner
sets of positive elements is well known \cite{growth_amen_semi} and part of a more general phenomenon
(see \cite[1.28]{amenability}),
but we will need the more precise analytic consequences of the following lemmas.

\begin{lem} \label{not_in_marginal}
For every finite binary sequence $u$, the set $E_u$ of all $f \in F$ such that
$u$ is not extended by an element of $R_f$ is marginal with respect to the right action
of $F$ on itself.
\end{lem}

\begin{proof}
Let $u$ be fixed and
let $g$ be the element of $F$ which is the identity on
sequences which do not extend $u$ and which satisfies
\[
(u \tie v) \cdot g = u \tie (v \cdot x_0)
\]
for all sequences $v$.

Now let $f$ be an element of $E_u$ and let $t$ be the initial part of $u$ which is
in $R_f$ (such a $t$ exists by our assumption that $u$ does not have an extension
in $R_f$).
Let $s = t \cdot f^{-1}$.
Observe that $f \cdot g^n$ can be represented by the following tree diagram:
\[
(L_f \cup \{s \tie v : t \tie v \in L_{g^n}\},R_f \cup R_{g^n}).
\]
By applying the characterization of reduced tree diagrams mentioned in Section \ref{background},
it follows that this tree diagram is reduced since
$(L_f,R_f)$ and $(L_{g^n},R_{g^n})$ are reduced and since
the sets
\[
\{s \tie v : t \tie v \in L_{g^n}\} \textrm{ and }
\{t \tie v : t \tie v \in R_{g^n}\}
\]
begin (end) with sequences ending in a $\seq{0}$ (respectively $\seq{1}$).
Consequently $u$ is extended by an element of $R_{f \cdot g^n}$ for all $n > 0$ and
hence $g$ marginalizes $E_u$.
\end{proof}

\begin{lem} \label{Folner->trees}
There is a constant $C$ such that if $A \subseteq F$ is a (right)
$\varepsilon$-F\o lner set, then there is a weighted $C \varepsilon$-F\o lner set
of trees supported on a subset of $\{R_f : f \in A\}$.
\end{lem}

\begin{proof}
Let $U$ consist of all binary sequences of length 4, noting that
if $T$ contains $U$, then every element of $\Gamma$ acts properly on $T$.
By Lemma \ref{not_in_marginal},
the set of $f \in F$ such that $U$ is not contained in $R_f$
is marginal.
By Lemma \ref{marginal->Folner},
there is a constant $C > 1$ such that if $A$ is $\varepsilon$-F\o lner and
\[
A_0 = \{f \in A : U \textrm{ is contained in } R_f\},
\]
then $A_0$ is $C \varepsilon$-F\o lner.
Observe that if $\gamma$ is a generator and $f$ is in $A_0$,
then $L_{f \cdot \gamma} = L_f$ and $R_{(f \cdot \gamma)} = (R_f) \cdot \gamma$.
We are now finished by Lemma \ref{Folner_hom_prop} applied to
$h(f) = R_f$ and $\mu  = 1_{A_0}$.
\end{proof}

\section{An operation on elements of $\Tcal$}

\label{partial_section}

In this section I will define an operation $\partial$
on elements of $\Tcal$ which reduces
their size logarithmically.

If $T$ is in $\Tcal$, then the \emph{end points} of $T$ are the maximum and
minimum elements of $T$.
All other elements of $T$ are said to be \emph{interior}.

\begin{defn}
Suppose that $T$ is in $\Tcal$.
$\partial T$ is the maximum $U \in \Tcal$ contained in $T$
which satisfies the following \emph{defining conditions}:
\begin{enumerate}

\item \label{has_int} $U$ contains extensions of both $\seq{01}$ and $\seq{10}$;

\item \label{monotone_cond}
one of the following holds:
\begin{itemize}

\item
if $u <_\lex v$ are interior elements of $U$,
then $2 |T/u| \leq |T/v|$;

\item
if $u <_\lex v$ are interior elements of $U$,
then $2 |T/v| \leq |T/u|$;

\end{itemize}

\item \label{int_cond}
the minimum (maximum) interior elements of $U$ terminate
with
a $\seq{1}$ (respectively $\seq{0}$).

\end{enumerate}
If no such $U$ exists, then $\partial T$ is defined to be the trivial tree.
\end{defn}

The following lemma justifies the use of \emph{maximum} in the definition of
$\partial T$.

\begin{lem}
If there is a $U$ satisfying the defining conditions for $\partial T$,
then there is a maximum such $U$.
\end{lem}

\begin{remark}
\emph{Condition \ref{int_cond}} is necessary to ensure the uniqueness
of maximal elements
of $\Tcal$ which satisfying the defining conditions for $\partial T$.
\end{remark}

\begin{proof}
Suppose for contradiction that $U$ and $V$ are distinct
maximal elements of $\Tcal$, each contained in $T$,
which satisfy the defining conditions of $\partial T$.
First I claim that
\emph{condition \ref{monotone_cond}} is satisfied in the same way for $U$ and $V$.
For this, it is sufficient to show that if $W$ is in $\Tcal$, contains extensions
of both $\seq{01}$ and $\seq{10}$,
and satisfies the first (second) option of \emph{condition \ref{monotone_cond}}
for $T$, then
$|T/\seq{01}| < |T/\seq{10}|$ (respectively $|T/\seq{10}| < |T/\seq{01}|$).
Suppose that $W$ satisfies the first option (the other case is symmetric).
Let $w$ be the greatest element of $W$ extending $\seq{01}$ and $w'$ be the least element
of $W$ extending $\seq{10}$.
Observe that \emph{condition \ref{monotone_cond}} implies that if $u_i$ $(i \leq l)$ are
the elements of $W$ which extend $\seq{01}$, then
\[
\sum_{i < l} |T/u_i| < |T/u_l| = |T/w|
\]
and hence
$|T/\seq{01}| < 2|T/w|$.
It follows that
\[
|T/\seq{01}| < 2|T/w| \leq |T/w'| \leq |T/\seq{10}|.
\]

By replacing $U$ and $V$ with their mirror images if necessary,
we will assume that the quantity $|T/u|$ is increasing as
$u$ increases in $U$ (or equivalently as $u$ increases in $V$).
Since $U \triangle V$ is non-empty, there is a minimum element
of $U$ which either properly extends an element of $V$ or is properly extended
by an element of $V$.
By exchanging the roles of $U$ with $V$ if necessary, we may assume the former occurs.

First suppose that the minimum elements of $U$ and $V$ are the same.
Let $u$ be the greatest element of $U$ such that it and all of its
$<_\lex$-predecessors extend an element of $V$ (this
extension may not be proper in the case of the predecessors of $u$).
Let $W$ be obtained by adding to $V$ those elements of $U$ which are extended
by $u$ or one of its predecessors.
If $w$ is an element of $W$ which occurs at or before $u$, then $|T/w| \leq |T/v|$,
where $v$ is the element of $V$ which $w$ extends.
Observe that if $x$ is in $W \setminus V$ and $y$ is in $W \setminus u$, then $x \leq_\lex u <_\lex v$
which in turn implies $|T/x| < |T/y|$.
Consequently, since both $U$ and $V$ satisfy \emph{condition \ref{monotone_cond}}, it follows that
$W$ does as well.
Next notice that $U$ and $W$ have the same minimum interior element.
Also, since $V$ is not contained in $U$, $u$ is not the maximum
interior element of $W$.
Therefore $W$ satisfies \emph{condition \ref{int_cond}}.
But now $W$ satisfies the defining conditions for $\partial T$,
contradicting the maximality of $V$.

Now suppose that the minimum elements of $U$ and $V$ differ.
By our choice of $U$, $V$, and $u$, we have that $u$ is the minimum element of $U$.
Moreover, $u$ extends the minimum element of $V$.
Let $v$ be such that $v \tie \seq{0}$ is the minimum element of $V$, noting
that $v \tie \seq{1}$ is the minimum interior element of $V$.
Since $v \tie \seq{0}$ is extended by an element of $U$, it must be that $v \tie \seq{1}$ is also
extended by an element of $U$.
Let $W$ be the union of $V$ together with all elements of $U$ which extend
$v \tie \seq{0}$.
Since $v \tie \seq{0}$ is not in $U$, it must be that both $v \tie \seq{00}$ and
$v \tie \seq{01}$ have extensions in $U$ and in particular,
the minimum interior element of $W$ is the same as the one of $U$.
Since $V$ is not contained in $U$, $v \tie \seq{1}$ is not the maximum element of
$V$ or of $W$.
Since every element of $W$ not extending $v \tie \seq{0}$ is in $V$, it follows
that the maximum interior element of $W$ is the same as the one in $V$ and
and therefore $W$ satisfies \emph{condition \ref{int_cond}}.
Also, for the same reasons
as above, $W$ satisfies \emph{condition \ref{monotone_cond}}.
Again, $W$ satisfies the defining conditions for $\partial T$,
contradicting the maximality of $V$.
\end{proof}

\begin{lem} \label{partial_growth}
If $\partial T$ has $n$ elements, then $T$ has more than
$2^{n-2}$ elements.
\end{lem}

\begin{proof}
There are $n-2$ interior elements of $\partial T$ and by our assumption, the total
number of elements of $T$ which extend an interior element of $\partial T$ is at least
$\sum_{i=1}^{n-2} 2^{i-1} = 2^{n-2} - 1$.
Since $\partial T$ is contained in $T$, there are at least two elements of $T$ remaining
to be counted, putting the total greater than $2^{n-2}$.
\end{proof}

\begin{lem} \label{partial_compatible}
If $g$ is in $F$, $T$ is in $\Tcal$,
and $g$ acts properly on $\partial T$,
then $\partial (T \cdot g) = (\partial T) \cdot g$.
\end{lem}

\begin{proof}
First I will verify that if $g$ and $T$ are as in the statement of the lemma,
then $(\partial T) \cdot g$ is contained in $\partial (T \cdot g)$.
Since the action of $g$ on the elements of $\partial T$
does not change their final digit,
$(\partial T) \cdot g$ satisfies \emph{condition \ref{int_cond}}.
Since the action of $g$ preserves lexicographic order and extension of sequences,
$(\partial T) \cdot g$ satisfies \emph{condition \ref{monotone_cond}}.
Finally, the hypothesis of the lemma implies that $\partial T \cdot g$
satisfies \emph{condition \ref{has_int}}.
It follows that $(\partial T) \cdot g$ is contained in $\partial (T \cdot g)$.

Next observe that if $g$ acts properly on $T$, then $g^{-1}$ acts
properly on $T \cdot g$.
It follows that
$\partial (T \cdot g) \cdot g^{-1}$ is contained in
$\partial (T \cdot g \cdot g^{-1}) = \partial T$.
Since $g$ and $g^{-1}$ are injections, it follows that $\partial (T \cdot g)$ and
$(\partial T) \cdot g$ have the same elements and hence are equal.
\end{proof}

\begin{defn}
Let $\Ecal$ be the set of all $T \in \Tcal$ such that neither of
the following inequalities hold:
\[
(+) \qquad |T/\seq{001}| < |T/\seq{01}| < |T/\seq{10}|
\]
\[
(-) \qquad |T/\seq{001}| > |T/\seq{01}| > |T/\seq{10}|
\]
\end{defn}

\begin{lem} \label{E_marginal}
$\Ecal$ is marginal.
\end{lem}

\begin{proof}

Define the following elements of $F$:
\[
a =
\begin{cases}
\seq{000} \mapsto \seq{000} \\
\seq{0010} \mapsto \seq{001} \\
\seq{0011} \mapsto \seq{0100} \\
\seq{01} \mapsto \seq{0101} \\
\seq{100} \mapsto \seq{011} \\
\seq{101} \mapsto \seq{10} \\
\seq{11} \mapsto \seq{11} \\
\end{cases}
b =
\begin{cases}
\seq{000} \mapsto \seq{000} \\
\seq{0010} \mapsto \seq{001} \\
\seq{0011} \mapsto \seq{01} \\
\seq{01} \mapsto \seq{100} \\
\seq{10} \mapsto \seq{101} \\
\seq{11} \mapsto \seq{11} \\
\end{cases}
\]
(i.e. $a = x_0^2 x_1 x_4 x_2^{-2} x_0^{-2}$ and $b = x_0^2 x_1 x_3^{-1} x_0^{-2}$).
Define
\[
\Ecal_a = \{T \in \Tcal : \max (|T/\seq{001}|,|T/\seq{10}|) = |T/\seq{01}| \}
\]
\[
\Ecal_b = \{T \in \Tcal : \max (|T/\seq{001}|,|T/\seq{10}|) < |T/\seq{01}| \}
\]
Observe that for all $T \in \Tcal$ such that $T \cdot a$ is defined, we have:
\[
|(T\cdot a)/\seq{001}| = |T/\seq{0010}| < |T/\seq{001}|
\]
\[
|(T\cdot a)/\seq{01}| = |T/\seq{0011}| + |T/\seq{01}| + |T/\seq{100}| > |T/\seq{01}|
\]
\[
|(T\cdot a)/\seq{10}| = |T/\seq{101}| < |T/\seq{10}|
\]
Therefore if $k > 0$ and $T$ is in $\Ecal_a \cdot a^k$, then
\[
\max( |T/{\seq{001}}|, |T/{\seq{10}}| ) < |T/{\seq{01}}|
\]
and, in particular, $T$ is not in $\Ecal_a$.
This shows that $a$ marginalizes $\Ecal_a$.

Observe that for all $T \in \Tcal$ such that $T \cdot b$ is defined, we have:
\[
|(T\cdot b)/\seq{001}| = |T/\seq{0010}| < |T/\seq{001}|
\]
\[
|(T\cdot b)/\seq{01}| = |T/\seq{0011}| < |T/\seq{001}|
\]
\[
|(T\cdot b)/\seq{10}| = |T/\seq{01}| + |T/\seq{10}| > |T/\seq{10}|
\]
Therefore if $k > 0$ and $T$ is in $\Ecal_b \cdot b^k$, then
\[
\max( |T/{\seq{001}}|, |T/{\seq{01}}| ) < |T/{\seq{10}}|
\]
and, in particular, $T$ is not in $\Ecal_b$.
This shows that $b$ marginalizes $\Ecal_b$.

Now notice that $T$ is in $\Ecal_a \cup \Ecal_b$ if and only if
\[
|T/\seq{01}| = \max (|T/\seq{001}|,|T/\seq{01}|,|T/\seq{10}|)
\]
Clearly the elements of $\Ecal \setminus (\Ecal_a \cup \Ecal_b)$ lie in one
of the following sets:
\[
\Ecal_1 = \{T \in \Tcal : |T/{\seq{01}}| = |T/{\seq{10}}| < |T/{\seq{001}}|\}
\]
\[
\Ecal_2 = \{T \in \Tcal : |T/{\seq{01}}| = |T/{\seq{001}}| < |T/{\seq{10}}|\}
\]
\[
\Ecal_3 = \{T \in \Tcal : |T/{\seq{01}}| < \min(|T/{\seq{10}}| , |T/{\seq{001}}|)\}
\]
It is therefore sufficient to show that each of these sets is marginalized by
$x_0$ off $\Ecal_a \cup \Ecal_b$.
As the argument is similar for each of these sets, I will focus on $\Ecal_1$.
Let $T$ be in $\Ecal_1$ and suppose that $k > 0$ is such that for all $i < k$, $T \cdot x_0^i$
is defined and is not in $\Ecal_a \cup \Ecal_b$.
Consider the sequence $n_i$ for $0 \leq i \leq k+1$
defined by
\[
n_i = |T/{\seq{0}^{i+1} \seq{1}}| = |(T \cdot x_0^i)/\seq{01}|
\]
Since $T$ is in $\Ecal_1$, $n_0 < n_1$.
Also, if $0 < i < k$ and $n_{i-1} < n_i$, then
$n_i < n_{i+1}$ since
\[
n_{i-1} = |(T \cdot x_0^i)/\seq{10}|
\]
\[
n_i = |(T \cdot x_0^i)/\seq{01}|
\]
\[
n_{i+1} = |(T \cdot x_0^i)/\seq{001}|
\]
and $T \cdot x_0^i$ is not in $\Ecal_a \cup \Ecal_b$.
It follows inductively that $n_{i+1} > n_i$ for all $i < k$.
Consequently, $T \cdot x_0^k$ is not in $\Ecal_1$.
This shows that $x_0$ marginalizes $\Ecal_1$ off $\Ecal_a \cup \Ecal_b$.
\end{proof}

\begin{defn}
Let $\Ecal^*$ be the set of all $T \in \Tcal$ such that neither of
the following inequalities hold:
\[
(2\times) \qquad 2 |T/\seq{001}| \leq |T/\seq{01}| \leq \frac{1}{2} |T/\seq{10}|
\]
\[
(\frac{1}{2} \times) \qquad \frac{1}{2} |T/\seq{001}| \geq |T/\seq{01}| \geq 2 |T/\seq{10}|
\]
\end{defn}

Observe that if $T$ is not in $\Ecal^*$, then $\partial T$ is not the trivial
tree since $\{\seq{00},\seq{01},\seq{10},\seq{11}\}$ is then an element of $\Tcal$
which satisfies the defining conditions for $\partial T$.

\begin{lem} \label{doubling_inv}
If $T$ satisfies $(+)$ and $\gamma$ is a generator,
then either $T \cdot \gamma$ is undefined,
$T \cdot \gamma$ is in $\Ecal$, or else $T \cdot \gamma$
satisfies $(+)$ (and similarly for $(-)$).
In particular, if $\Acal \subseteq \Tcal \setminus \Ecal^*$ is connected,
then either all elements of $\Acal$ satisfy $(2 \times)$ or all elements
of $\Acal$ satisfy $(\frac{1}{2} \times)$.
\end{lem}

\begin{proof}
This follows from the following equalities which hold whenever
the relevant action is defined:
\[
T/\seq{001} = (T \cdot x_0)/\seq{01}
\]
\[
T/\seq{01} = (T \cdot x_0)/\seq{10}
\]
\[
T/\seq{001} = (T \cdot x_1^{\pm})/\seq{001}
\]
\[
T/\seq{01} = (T \cdot x_1^{\pm})/\seq{01}.
\]
\end{proof}

\begin{lem} \label{doubling}
$\Ecal^*$ is marginal.
\end{lem}

\begin{remark}
Lemmas \ref{doubling_inv} and \ref{doubling} have non trivial qualitative
consequences for $F$-invariant probability measures on $\Tcal$.
If $\mu$ is an $F$-invariant probability measure on $\Tcal$, then $\mu(\Ecal^*) = 0$.
Furthermore, $\mu(\Tcal^+ \triangle (\Tcal^+ \cdot \Gamma)) = 0$ and therefore if $\mu$ is
additionally ergodic, then it must be that $\mu$ assigns measure 1 either to the set of
elements of $\Tcal$ which satisfy $(2 \times)$ or else to the set of those which
satisfy $(\frac{1}{2} \times)$.
\end{remark}

\begin{proof}
Define the following elements of $F$:
\[
c =
\begin{cases}
\seq{000} \mapsto \seq{00} \\
\seq{001} \mapsto \seq{01} \\
\seq{01} \mapsto \seq{100} \\
\seq{10} \mapsto \seq{101} \\
\seq{11} \mapsto \seq{11} \\
\end{cases}
d =
\begin{cases}
\seq{000} \mapsto \seq{00} \\
\seq{001} \mapsto \seq{010} \\
\seq{01} \mapsto \seq{011} \\
\seq{10} \mapsto \seq{10} \\
\seq{11} \mapsto \seq{11} \\
\end{cases}
\]
(i.e. $c = x_0^2 x_2^{-1} x_0^{-1}$ and $d = x_0^2 x_1^{-1} x_0^{-1}$).
Let $\Tcal^\pm$ denote those elements of $\Tcal$ which satisfy $(\pm)$.
Define
\[
\Xcal = \{T \in \Tcal^+ : 2 |T/\seq{01}| \leq |T/\seq{10}|\}
\]
\[
\Ecal_4 = \{T \in \Tcal^+ : 2 |T/\seq{01}| > |T/\seq{10}| \}
\]
\[
\Ecal_5 = \{T \in \Tcal^+ : 2|T/\seq{001}| > |T/\seq{01}| \}.
\]
I first claim that $\Tcal^+ \cdot c \subseteq \Xcal \cup \Ecal$.
To see this, suppose that $T$ is in $\Tcal^+$.
Then
\[
2|(T\cdot c)/\seq{01}| = 2 |T/\seq{001}| <
|T/\seq{01}|+|T/\seq{10}| = |(T \cdot c)/\seq{10}|.
\]
and hence if $T\cdot c$ is in $\Tcal^+$, it is in $\Xcal$.
Since
\[
|(T \cdot c)/\seq{01}| < |T/\seq{001}| < |T/\seq{10}| < |(T \cdot c)/\seq{10}|,
\]
$T \cdot c$ is not in $\Tcal^-$ and hence is either in $\Tcal^+$ or in $\Ecal$.

Thus we have that if $T$ is in $\Ecal_4$ and $k > 0$,
then either $T \cdot c^k$ is in $\Xcal$ or there is an $i < k$
such that either $T \cdot c^i$ is undefined or in $\Ecal$.
It follows that $c$ marginalizes $\Ecal_4$ off $\Ecal$ and hence that
$\Ecal_4$ is marginal.
Also, by Lemma \ref{doubling_inv},
$\Ecal_5 \cdot x_0 \subseteq \Ecal_4 \cup \Ecal$ and therefore
$\Ecal_5$ is marginal.

Next define
\[
\Ycal = \{T \in \Tcal^- : 2 |T/\seq{01}| \geq |T/\seq{10}|\}
\]
\[
\Ecal_6 = \{T \in \Tcal^- : |T/\seq{01}| < 2|T/\seq{10}| \}
\]
\[
\Ecal_7 = \{T \in \Tcal^- : |T/\seq{001}| < 2|T/\seq{01}| \}.
\]
I will now show that $\Tcal^- \cdot d \subseteq \Ycal \cup \Ecal$.
If $T$ satisfies $(-)$,
then we have that
\[
2|(T\cdot d)/\seq{10}| = 2 |T/\seq{10}| <
|T/\seq{001}|+|T/\seq{01}| = |(T \cdot d)/\seq{01}|.
\]
Arguing as above, $d$ marginalizes $\Ecal_6$ off $\Ecal$ and
$\Ecal_7 \cdot x_0 \subseteq \Ecal_6 \cup \Ecal$.
Consequently both $\Ecal_6$ and $\Ecal_7$ are marginal.
Since $\Ecal^* = \Ecal \cup \bigcup_{i=4}^7 \Ecal_i$, $\Ecal^*$ is marginal and
the proof is complete.
\end{proof}

\begin{lem} \label{proper_act}
The set of all $T \in \Tcal$ such that there is a $\gamma \in \Gamma$ which
does not act properly on $\partial T$ is marginal.
\end{lem}

\begin{proof}
By Lemma \ref{doubling}, $\Ecal^*$ is marginal.
Moreover, if $k$ is any natural number
$\bigcup_{i = 0}^k \Ecal^* \cdot x_0^i$ is marginal.
Hence by Lemma \ref{doubling_inv}, the set of all $T$ such that
neither of the following conditions holds
\[
(\forall i < k) \quad 2|T/\seq{1}^i\seq{0}| < |T/\seq{1}^{i+1}\seq{0}|
\]
\[
(\forall i < k) \quad 2|T/\seq{1}^{i+1}\seq{0}|  < |T/\seq{1}^i\seq{0}|
\]
is marginal.
Now let $U$ consist of all binary sequences of length 4,
noting that $U$ satisfies \emph{conditions \ref{has_int} and \ref{int_cond}} and
if $T$ contains $U$ then every element of $\Gamma$ acts properly on $U$.
Set $L = \{\seq{1}^k\} \cup \{\seq{1}^i\seq{0} : i < 15\}$ and let
$g$ denote the element of $F$ defined by the tree diagram $(L,U)$.
If $T$ is not in $\bigcup_{i = 0}^{15} \Ecal^* \cdot x_0^i$, then $T \cdot g$
has the property that $U$ satisfies the defining conditions for $\partial(T \cdot g)$ and
hence $\partial (T \cdot g)$ is non-empty.
Therefore the set of $T$ such that $U$ does not satisfy the defining conditions for
$\partial T$ is marginalized by $g$ off $\bigcup_{i = 0}^{15} \Ecal^* \cdot x_0^i$.
\end{proof}

\begin{lem} \label{partial_marginalize}
There is a constant $C$ such that if $\mu$ is a weighted $\varepsilon$-F\o lner
set of trees, then there is a weighted $C \varepsilon$-F\o lner set of trees
which is supported on a subset of
\[
\{\partial T : (\mu(T) > 0) \land (\partial T \textrm{ is non trivial})\}.
\]
\end{lem}

\begin{proof}
By Lemmas \ref{marginal->Folner} and \ref{proper_act},
there is a $C$ such that if $\mu$ is a weighted
$\varepsilon$-F\o lner set of trees and
\[
\Acal = \{T \in \Tcal : (\mu(T) > 0) \land \forall \gamma \in \Gamma (\gamma \textrm{ acts properly on } \partial T)\},
\]
then $\mu \restriction \Acal$ is $C \varepsilon$-F\o lner.
Now let such a $\mu$ be given and define $\Acal$ as above.
By Lemma \ref{partial_compatible},
\[
(\partial T) \cdot \gamma = \partial (T \cdot \gamma)
\]
whenever $\gamma$ is a generator and $T$ is in $\Acal$.
Applying Lemma \ref{Folner_hom_prop} to
$\mu \restriction \Acal$ and $h = \partial$ gives the desired conclusion.
\end{proof}

Now we are ready to complete the proof of Theorem \ref{tower_growth}.
This is proved by exhibiting
\begin{description}

\item[$(*)$] There exists a constant $C > 1$ such that
if $A \subseteq F$ is a $C^{-n}$-F\o lner set,
then $A$ contains an element with a tree diagram whose pairs contain
at least $\exp_{n} (0)$ elements.
\end{description}
To see why this is sufficient, first recall that there is a linear
relationship between the distance of an element $f$ of $F$
to the identity and the size of the trees occurring in a
tree diagram for $f$ \cite{metric_Thompson}.
By Lemma \ref{Folner_connected},
every $\varepsilon$-F\o lner set contains a $\Gamma$-connected component
which is an $\varepsilon$-F\o lner set.
Furthermore, a (right) $\varepsilon$-F\o lner set for $F$ can be assumed to contain the identity
by replacing it by an appropriate left translate.
Therefore $(*)$
implies that, for some $C > 1$, both the diameter
and the cardinality of
a $C^{-n}$-F\o lner set are at least $\exp_n(0)$.

Now I will prove $(*)$.
By Lemmas \ref{Folner->trees} and \ref{partial_marginalize} there is a constant $C$ such that:
\begin{itemize}

\item if $A \subseteq F$ is an $\varepsilon$-F\o lner set,
then there is a weighted $C \varepsilon$-F\o lner set of trees
$\mu$ with support contained in $\{R_f:f \in A\}$;

\item if $\mu$ is a weighted $\varepsilon$-F\o lner set of trees,
then there is a weighted $C \varepsilon$-F\o lner set of trees
$\nu$ which is supported on
\[
\{\partial T : (\mu(T) > 0) \land (\partial T \textrm{ is non trivial})\}
\]

\end{itemize}
Notice that if $A \subseteq F$ is $C^{-(n+1)}$-F\o lner, then
there is an $A' \subseteq A$ such that $\{R_f : f \in A'\}$ is $C^{-n}$-F\o lner
and hence
there is a $f \in A$ such that $\partial^{n} R_f$ is non trivial.
Let $k_i = |\partial^{n-i} R_f|$ and observe that by Lemma \ref{partial_growth},
$k_0 = 4$ and $k_{i+1} > 2^{k_i - 2}$.
It follows by induction that $\exp_i (0) + 2 < k_i$ and in particular that
that $R_f$ contains at least $\exp_{n}(0)$ elements.
Finally, by adjusting the constant $C$ if necessary, it may be arranged that if
$A$ is $C^{-n}$-F\o lner, then $|A| \geq \exp_n(0)$.
This finishes the proof of Theorem \ref{tower_growth}.

\def\leavevmode\lower.6ex\hbox to 0pt{\hskip-.23ex \accent"16\hss}D{\leavevmode\lower.6ex\hbox to 0pt{\hskip-.23ex \accent"16\hss}D}

redpajama_set_name: RedPajamaArXiv


Having a small yard or outside dwelling area does not imply that you may't have a fantastic backyard. My choice is to burn, not solely as a result of it cost nothing to burn, which is cheaper than hiring a skip but in addition the ensuing potash is good for the garden, especially in the event you grow your personal greens; potatoes in particular love potash so even in burning the wooden would not go to waste but helps in growing greens in your again backyard for use within the kitchen.
Try our gallery and if there may be anything you want give us a call and we are going to help you decide how you could possibly incoporate it into your backyard. Leaving the garden the place my youngsters performed from their earliest years, we rode over the Williams-burg Bridge, chatting about cycling and the tradition of the city, zooming south past the Japanese Parkway.
They can be discovered for round £30 each, and three must be lots for a median sized backyard. That isn't to say you cannot grown things in a modern backyard; attempt experimenting with newly developed plant hybrids or flowers in extravagant hues, or trying out a modern gardening technique comparable to hydroponic gardening.
Her backyard appears like a wonderful oasis within the midst of a big city, and it is stuffed with inspiration and the names of the various native vegetation they've included of their garden. Get these 'pollinator gardens' rising and send me a photograph on our Fb web page (-Florida-Gardener-197232317050577/) and I will gladly share them by the summer to assist inspire different Florida gardeners.
A sloped garden may at first seem as a problem however offers great opportunities to incoporate raised beds and terraces which in themselves provide great interest, it doesn't matter what time of year. If you are considering creating your own pallet backyard, listed below are some useful pallet garden ideas that ought to get you going.


text:
The Huntington Thesis:The Dogma Of The Far-Right And The Ideology Of White Supremacist Terror
in World — by T T Sreekumar —	March 18, 2019
The shocking context of the white supremacist hateful terror attack that killed 50 Muslims and injured as many as 34 others offering prayers at the Al Noor mosque, in Christchurch, New Zealand,cannot be seen in isolation from the ideological strands that get woven into the anti-migrant, anti-Muslim narrative that was primarily triggered by the publication of Samuel Huntington's book, "Clash of Civilizations". New Zealand Prime Minister Jacinda Arden and few others have confirmed that they received a copy of the suspected Christchurch shooter's "manifesto" minutes before the attack started. The 87-page document, was also posted on social media before the shooting. It is a document that is more or less bursting with anti-immigrant, anti-Muslim harangue that reminds one of Anders Breivik, a far-right terrorist who committed the 2011 Norway attacks instigated by similar anti-immigrant white supremacist dogmas that instigated Brenton Harrison Tarrant, the committer of the Al Noor Mosque attack. The writer in fact refers to Breivik in the manifesto. CNN and some international media are trying to us the fact that the terrorist told his online viewers to subscribe to the YouTube channel of PewDiePie, and the fact the document tries to tie up several internet in-jokes, memes and trolls into its narrative as an indicator that the Internet should also blamed for the crime. There were quotes appearing form experts in the news media who were inclined to evaluate the role of popular culture and video gaming in instigating people to carry out murderous attacks on others.This is probably done with the motivation to help divert attention from the larger political and ideological backdrop of white supremacist terror globally unleashed by the far- right. If one carefully reads the manifestos of both Breivik and Tarrant and the various similar documents that have been circulated globally by white supremacist groups and terrorists it can be very clearly seen that there are strong traces of compelling complicity of the Huntington thesis in the perpetuation and replications of the ideological perspective that constitute the core of the detestable world-view shared by the far-right, which includes, apparently, the American President Donald Trump himself.
In his controversial thesis about the Clash of Civilizations, Samuel Huntington's hypothesis was that the fundamental source of conflict in the new world will not be ideological or economical, but cultural. He asserts that during the cold-war period, the clash was between the Western countries and wasfought over ideological categories like democracy and communism.Cold War was an aberration according to him and it would be followed by a shift in the nature of world conflicts that would be fought at the fault-lines between the major civilizations of today. Out of the nine contemporary civilizations that he identifies, three monopolizes his attention; West, Islam and Sinic. That is, Huntington proposes that the major clashes of the world henceforth would occur due to the threat for the West from Islam and China. Though the theory obfuscated the realities of unequal power, the deepest divisions between rich and poor countries, though it was carelessly constructed, ahistorical and contradictory, Huntington's thesis dominated public discourse since it was first launched in 1993 in Foreign Affairs, and later published as a book and was subsequently adopted as a primary framework for foreign policy formulations in US. Apparently this popularity was more due to its appeal to the Western, particularly, American imagination that required to reinitiate a new enemy to feed its fear psychosis.
For his thesis (that Edward Said calls 'belligerent'), Huntington largely depended on the 1990 article by veteran Orientalist, Bernard Lewis, whose stance is manifest in his very title 'The Roots of Muslim Rage'. Without delving into the deeper implications of 'civilizational identities' Huntington provides three yardsticks for the future conflicts: between 'core states', between 'fault-lines' of the civilizations and fault-lines between groups within the states.
The significance of Huntington's thesis perhaps lies primarily in revisiting the multiple ways in which it was countered and critiqued. Apart from arguments against the proposal of monolithic identities for civilizations, one particular stream of criticism against Huntington's theory that deserves mention is the theory of 'clash of ignorance'(Karim and Eid,2012, 'Clash of Ignorance' Global Media Journal — Canadian Edition Volume 5, Issue 1, pp. 7-27).This proposes the antithetic position that clashes arise primarily due to the fault-lines in the knowledge production itself, due to failures of intercultural communication and understanding. This mode of critique not only provides a unique epistemological potential for multidisciplinary research in intercultural interactions, but proves to be extremely handy in confronting the many historical and cultural misrepresentation of facts within cultural communities. In fact, as Karim and Eid argue, 'religious illiteracy' and cultural ignorance about the 'Other' lies at the root of all 'clashes of ignorance'.
To take an example of the Western and Muslim societies, a vast store of knowledges about many productive interactions between the two cultures over the course of centuries have been brutally and stealthily put aside. The common Abrahamic root of Judaism, Christianity and Islam and contemporary Western Philosophy is one such. West and Islam are conceptualized in most dominant discourses as inimical geopolitical entities locked inside their own boundaries. But we know that Muslim presence in Europe is 1300 years old (which is not a very small period to speak about considering how young humanity itself is) And as Edward Said asked, in his legendary book 'Orientalism', how can we speak of Western Civilization except as the ideological fiction implying the superiority of a handful of ideas and values that itself has no meaning outside the history of conquest, immigration and travel? Similarly, misleading is the idea of Islam as a singular civilization of geo-political bloc. Islam has appeared as a timeless phenomenon, unchanging and static in both Muslim and Non-Muslim discourses. However, religion like any other socio cultural entity can never be static and timeless. Karim and Eid and many other scholars point to the confluence of faiths, cultures and ethnicities comprising of Ummayad, Abassid, Safavid, Fatimid, Mughal and Ottoman territories that makes the idea of a monolithic Muslim civilization highly redundant.
But far from all these complexities, Huntington's theory proposed a renovation of the cold war paradigm, i.e., West vs the rest. And ever since September 11, every carefully devised pathological slaughter has been held up as justification for Huntington's theory.The normally sober British weekly 'The Economist', for instance, in its issue of September 22-28, also reached out for generalization, praising Huntington extravagantly for his "cruel and sweeping, but nonetheless acute" observations about Islam, as pointed out by Edward Said. Even as more and more of America's newspapers and magazines added to the same vocabulary of apocalypse through the alien, more and more Churchillian rhetoric of defense and nationalism was revoked to its aid. When Bush administration took up the democracy promotion in Arab countries as a kind of crusade, he even declared that there is no clash of civilizations when it comes to the 'common rights and needs of the people'(Ronald Inglehart and Pippa Norris, 2003, 'The True Clash of Civilizations', Foreign Policy, No. 135. pp. 62-70). The very assertion, its need and urgency itself bears testimony to how much Huntington's theory had become part of the entire warfare that was designed for the 'protection' of Arab world. The theory became the essence and soul of the othering process of which American military was the corporal end. Echoing Huntington, Polly Toynbee wrote in the Guardian that a 'globalized force of extremists from many countries' are bound together in their hatred of Western values. A central argument in the 'protection' rhetoric was that 'democracy' (understood as a western cultural product) is unattainable for the Arab countries. This basically springs from Huntington's theory that Islam lacks the core political values that gave birth to representative democracy in the western civilization. The statistical data that very few Muslim countries have electoral democracy as a form of governance do not reveal anything about the underlying democratic believes of the Muslim population spread across these countries. Citing a World value survey conducted in the turn of the millennium, a paper by Inglehart and Norris asserts that throughout the world, the value of democracy is preferred at this point of time. The paper however further tries to prove that it is over the values of 'eros' rather than 'demos' that the polaritybetween civilizations lies. This theory of 'Sexual clash of civilizations' is equally problematic as it merely imposes the ethical content and value system of one people over the other. It becomes all the more problematic when a certain form of 'progressiveness' is understood in terms of some sexual values and projected as a derivative of the Western civilization
While 'Clash of Civilization' in itself was a theory that was propounded to justify the violent military pursuits of America over the Arab nations— the gigantic and stealthy step to capture the monopoly over oil— most of the critiques that came up against the theory tried to address its faults and thereby helped to legitimize the discourse about a 'clash'. It is during the colonial era that Europeans appropriated the term 'civilized' as a derivative of 'civilization' and applied it to themselves. The concept that powerfully 'civilized' nations had the right to acquire the guardianship of the lesser privileged continents like Asia and Africa formed a basis of most colonial discourses. On the theoretical realm this formed the foundation to assume certain 'standards of civilization' and practically it gave the colonial invaders the right to 'legally' subject the natives to discriminatory treatment. A Nietzschean genealogy of this morale of 'protection' and 'correction' would easily reveal how ubiquitous and permeated this savior syndrome is in the Foreign policies of European countries and America. Even the idea of 'modernity', as the theoretical epitome of progress and standard that came up from the cradle of Western Enlightenment, is replete with this urge to be the 'light' for the 'dark' regions of the world.
The othering of Islam against a civilization of 'Christendom' looks all the more molded up because till recently Jews had been excluded from this same civilization, were regarded as the polluters of civilization, the same civilization that is now being re-read as Judeo-Christian(Mohammed Ayoob, 2012, 'Was Huntington Right? Revisiting the Clash of Civilizations', Insight Turkey Vol. 14 / No. 4 / pp. 1-11). The main reasons for this shift could be either the Christian Europe's guilt over Holocaust or the political clout of the American Jewish community, or perhaps both, as manifested in the imposed tragedy of Palestine. But with 1990s political Islam began to increasingly replace the 'mortal enemy' (Hitler's description for the Jew) and the 'dog to be spurned and spat upon' (Merchant of Venice) as the main challenger to the Western consciousness. "One of the greatest achievements of the Israel lobby and of Jewish Americans, has been to transform the idea of America based on the Anglo-Saxon variety of Christian civilization into one based on a Judeo-Christian civilization," observes Mohammed Ayoob. He observes that after the cold war Israel is no longer treated as the surrogate of America in Arab provinces, but in a reverse order US has become a willing tool to Israel as evident from the foreign policies of both States. Ayoob endorses the 'cultural kinship' theory of Huntington by observing that nothing but the cultural kinship between US and Israel as Judeo-Christian representatives can explain the alacrity with which Obama submitted to Netanyahu's will. A re-reading of the theory thus discusses how Palestinian crisis stemmed from a civilizational prejudice against a people who were treated as 'non-people' in order to give home to a people with no land. Ayoob also cites the same assumed 'civilizational enmity' as one of the strong reasons for Europe's denial of Turkey and its acceptance of several former Soviet bloc nations of dubious records into EU. Turkey's attempts to improve the relations with its neighbors and to gather support for Gaza was even termed as 'neo-Ottomanism'. In short even the argument of those who strongly criticize Huntington's ideology inadvertently supported his basic premise that a clash over cultural kinship would be prominent in the post-cold war era.
The continuing white supremacist terror is a practical application of the Huntington thesis. Huntington's theory is understood in a white supremacist framework as implying an impending danger to the western civilization caused by Islam, just as the cold war era problematized the freedom of the west as being ensured only by keeping the spread of communism around the globe. The west now not only have to defend a civilization, but had to resist an impending onslaught from an enemy civilization, identified as Islam in the current global context, that is gaining strength day by day.The irony is that many contemporary global discourses, upsetting feuds, devastating wars, appeasing dialogues, diplomatic initiatives and patchwork policies are built up over this theory about a significant civilizational other. 2001 was proclaimed by UN General Assembly as the 'Year of Dialogue among Civilizations' which was followed by an attempt for an initiative for an 'alliance of civilizations'. Mostly when universal civilizational values like democracy, human rights and rule of law are brought in, 'clash of civilization' become accepted like a self-fulfilling prophecy. The complicity of the Huntington thesis in the programmatic perpetuation of white supremacist terror cannot be undermined just by linking the narrative of the so-called manifestos of the terrorists with Internet discourses and their influences. Its ideological leanings are clearly rooted in the 1990s' reconfiguring of the unipolar world that emerged after the collapse of the Soviet Union as an imaginary bipolar one with Islam and the west as its two hostile poles.
Dr T T Sreekumar, Professor , The English and Foreign Languages University, Hyderabad
Tags: Christchurch Terror Attack

redpajama_set_name: RedPajamaCommonCrawl


text:
Home > News: Press > Patient Engagement, Practice Tools > Froedtert & the Medical College of Wisconsin health network improves experience and care with PatientWisdom
Froedtert & the Medical College of Wisconsin health network improves experience and care with PatientWisdom
By PR Newswire Published at: November 15, 2018
MILWAUKEE, Wis. (November 15, 2018) /PRNewswire/ —Approximately 10,000 people cared for by the Froedtert & MCW health network have used PatientWisdom, a mobile-responsive digital site which helps patients share 'stories' about what matters to them, to improve health and care by listening, individually and at scale.
People are using this secure web-based tool to let providers know what matters to them ahead of clinical encounters to improve patient experience and the delivery of care. A one-screen summary with what matters to each individual, such as health priorities, goals, barriers to staying healthy, pressures, and worries, is delivered via the electronic health record to every member of the care team ahead of the patient visit.
From the perspectives people share, care teams can more fully and quickly understand how to deliver care that is individualized for the whole person. Froedtert & MCW is one of the first health systems in the world to offer a digital health tool like PatientWisdom.
"Electronic health records can help us record and manage clinical data, but they rarely give the full story of who a person is," says Bradley Crotty MD, MPH, an internist with Froedtert & MCW health network; Assistant Professor, Medicine, MCW. "By inviting our patients to share who they are, and what's important to them, it helps doctors and care team members better connect at an individual level."
Crotty, one of 300 clinicians incorporating PatientWisdom into care, also leads research on the program's impact. "Our early data show that PatientWisdom is helping to make appointments go smoother, and it helps build and strengthen relationships between patients and their care team," Crotty adds.
"Our team developed PatientWisdom to help leading organizations like Froedtert & MCW create more vibrant relationships with patients. We know that listening is the real key to improving health and care," said Gregory Makoul, CEO and Founder of PatientWisdom. Makoul and the PatientWisdom team are collaborating closely with Froedtert & MCW health network to improve the tool and reach more patients.
PatientWisdom is one of many services Froedtert & MCW health network has introduced with the goal of improving access, quality, and the experience of health care. The health network's innovation accelerator, Inception Health, partnered with PatientWisdom for its clinical evidence and personal, intuitive user experience. PatientWisdom was recently named "Most Patient-Centered" and an "Audience Favorite" at the Health Tech StandOut! Competition during the 2018 Connected Health Conference.
Froedtert & the Medical College of Wisconsin
The Froedtert & the Medical College of Wisconsin regional health network is a partnership between Froedtert Health and the Medical College of Wisconsin supporting a shared mission of patient care, innovation, medical research and education. Our health network operates eastern Wisconsin's only academic medical center and adult Level I Trauma Center at Froedtert Hospital, Milwaukee, an internationally recognized training and research center engaged in thousands of clinical trials and studies. The Froedtert & MCW health network, which includes five hospitals, one of which is a nationally ranked academic medical center, more than 1,600 physicians and nearly 40 health centers and clinics, draws patients from throughout the Midwest and the nation. In our most recent fiscal year, outpatient visits exceeded 1.1 million, inpatient admissions to our hospitals were 49,250 and visits to our network physicians totaled 932,000. For more information, visit froedtert.com.
About PatientWisdom, Inc.
PatientWisdom, Inc. has developed a digital health platform that improves health and care by listening to patients, providers, and community members – individually and at scale. The company is based in New Haven, CT and offers three powerful solutions: PatientWisdom makes it easy to understand what matters to patients ahead of clinical encounters to improve patient experience and the delivery of care; ProviderWisdom obtains real-time, real-world insights from clinicians and staff to enhance engagement, reduce burnout and streamline care; CommunityWisdom produces a deep view of the community to give health organizations clear lines-of-sight for developing or partnering with programs people will use. In 2017, the company was named Venture of the Year at the CT Entrepreneur Awards and Froedtert & the Medical College of Wisconsin earned the Vizient Clinical Innovation Award for its approach to implementing the PatientWisdom digital platform. In 2018, The Tech Tribune named PatientWisdom, Inc. one of the Best Tech Startups.


Great to have old friends help out. Our new KIOSK bags will be hot off the presses from The Power Plant, a home grown roof top printing enterprise! Not sure what's going on with the preserved vegetable, but… Exciting stuff!


Your hair is always your greatest accessory, but never more so on your Big Day. Therefore, before choosing between the traditional updo or the blowout, think about which style will complement your dress and wedding style to maximum effect. There are numerous beautiful options to try but read our handy guide below for a head start in finding the perfect one for you.
If you have chosen a modern, minimalist dress you have complete creative freedom when it comes to your hair. Former first daughter, Barbara Bush recently wed in a custom Vera Wang slip gown of ivory silk crepe and wore her hair down in a very natural style. Alternatively, when Carolyn Bessette married John F. Kennedy Jr., she paired her ultra simple slip with a low, messy chignon. A sleek bob or a polished ballerina bun with on-trend earrings would also create a very contemporary tone.
Whether your wedding is fully vintage themed or even if your dress just hints to a bygone era with capped sleeves or a sweetheart neckline, reflect this vibe with a birdcage veil or tumbling Hollywood curls. For her wedding, Gwen Stefani channeled her signature retro look with a high coiffure (think Grease) and trademark red lipstick. Pin curl updos are newly popular and offer an alternative to the classic updo; they also work for all kinds and lengths of hair.
When supermodel Kate Moss married, she wore a custom Dior bias-cut 1920s dress featuring a medley of Jazz Age details including a lace overlay, gold embroidery, sheer accents and metallic detail. Kate wore her shoulder length hair in soft waves but a finger wave bob would have also completed her stunning dress. Real or faux (long hair pinned into the style) is easy to wear and looks picture perfect accessorized with diamond brooches and sparkly hair slides.
A bride wanting to create a boho-chic look might opt for a free-flowing gown with pretty embroidery, a cheeky 1960s inspired shift or even a Bianca Jagger pantsuit. So when it comes to hair, loose curls, accessorized with fresh flowers or a floral crown, French braids twisted into a messy updo or a wispy loose bun all work to match the relaxed tone. A high bouffant bun will reflect a 1960s feel and long, center-parted locks recall 1970s glamour.
Layers of diaphanous chiffon, delicate swirls of tulle and pretty lace overlays are all hallmarks of a romantic wedding dress. Soft, feminine hair makes the perfect partner. Tumbling curls work well with a lower neckline or a strapless dress. A higher neckline or halter neck (think Meghan Markle's dress for her evening reception) is complemented by a soft, messy updo or chignon. For the bride looking to add whimsy, a swinging ponytail with wisps of hair framing the face can complete this dreamy ensemble.
The new classic has become an off-the-shoulder figure-hugging silhouette with a ball gown or fishtail skirt. Bare shoulders leave plenty of opportunity to wear hair long and down, curly or straight, accessorized with a tiara or individual fresh flowers. A timeless effect can also be achieved with a beautiful messy chignon dressed on the nape or to the side of the head; delicate slides or pins add a personal touch. A half updo, think Kate Middleton's wedding day style, is relaxed and sophisticated at the same time and is easy to take down for the evening festivities.


OPINION — Check your morning news and you are likely to read distressing stories about the threat of a bad flu season, the consequences of natural disasters like wildfires in California, unacceptably high maternal and infant death rates, or the opioid epidemic.
All these emerging challenges occur on top of our nation's chronic public health issues, like heart disease, cancer and HIV/AIDS, which continue to take a toll on the length and quality of life for people in the United States. This also takes a toll on the health and vitality of our communities and comes at great cost to our federal and state health care budgets.
As we face these challenges, we turn to local public health agencies to protect us: by addressing policy solutions and building the multisector collaborations it takes to lead a community response.
A public health system with foundational capabilities is one with the ability to track the health of a community through data, case findings, and lab tests; the capacity to respond to emergencies of all kinds; the experience to translate public health science into appropriate policy and regulation; the adeptness to communicate effectively with the public with timely, science-based information; and the power to harness and align community resources and actors to advance the health of the entire community.
The American people should rightly expect that where you live should not determine your level of public health protection — but current data tell us it does. Only 51 percent of the U.S. population is served by a comprehensive public health system.
Because daily public health demands and emergent challenges like epidemics and natural disasters do not recognize state or city boundaries, America should be protected by a strong public health infrastructure. Health protection should not depend on local decisions or stop abruptly at political borders. Instead, it should be more reliable, like the interstate highways. Foundational public health capabilities should be part of our nation's infrastructure.
Organized around this common-sense notion that all people in America deserve public health protection, the Public Health Leadership Forum, a project of RESOLVE, with funding from the Robert Wood Johnson Foundation, brought together a panel of experts to inform recommendations on how we might commit to building our nation's public health infrastructure.
Using a national study conducted by the University of Kentucky, the panel found that supporting foundational public health capabilities across the country would cost an estimated $9.5 billion a year.
That sounds like a lot of money, but it comes to only $32 per person per year. Compare that to the estimated $11,193 per person we spend on health care, and it is a great return on investment.
Indeed, total spending for foundational public health capabilities is less than one-fourteenth of the increase in per capita health care spending in the last year.
Based on current investments by all levels of government, the University of Kentucky estimated there is a $4.5 billion gap between what is now in place and what is needed to achieve a truly nationwide public health infrastructure.
To close this gap, we recommend creating a Public Health Infrastructure Fund. This recommendation is part of a new white paper we are releasing this week during an event at the Bipartisan Policy Center. The paper calls for investing $4.5 billion annually in establishing and sustaining foundational public health capabilities.
All people in America should be served by a public health agency that ensures equitable access to and protection by foundational public health capabilities.
Financing of foundational capabilities is a governmental responsibility and should be assured through sustainable, dedicated revenue streams. Local, state, and federal governments should share in financing those capabilities as all levels of government ultimately benefit from them.
Foundational public health capabilities should be assessed and provided in every community based on national standards — and maintaining those standards should be a condition of continued funding.
We recommend that this Public Health Infrastructure Fund should be a mandatory funding stream, awarded through the states to state and local health departments on a per capita basis. The fund would have a 10 percent state matching requirement, with a set aside for technical assistance, research, and demonstration project.
As a new Congress is seated next month, there is no doubt that rebuilding our nation's overall infrastructure as well as taking on the new (and old) threats to our nation's health and prosperity will be high on its agenda. We hope Congress will also see investment in a Public Health Infrastructure Fund as a critical step to assuring all people in America have the public health protection they should expect and deserve.
Karen DeSalvo, MD, is professor of medicine and population health, Dell Medical School, University of Texas, Austin, and former HHS acting assistant secretary for health.
Jeffrey Levi, PhD, is professor of health policy and management at the George Washington University's Milken Institute School of Public Health.
Both formerly served on Bipartisan Policy Center task forces.


Johannes has more than 15 years of experience in the IT industry. In his role, Johannes manages a team tasked with helping organizations create analytics solutions that help them transform the way they do business for the better while empowering individual knowledge workers with better decision making capabilities.
He first joined Qlik in 2007 in Sweden and held multiple roles in the organisation from operations planning to consulting services. Johannes migrated to Singapore in 2009 when Qlik opened their Singapore office and has remained in the region since.
Currently the Regional Presales Manager for Korea and the Asean region, Johannes develops unique solutions with his broad industry and technical knowledge. Key verticals he focuses on include Banking and Insurance along with data intensive industries like telco and retail.
Johannes graduated from Lund University, Sweden. Outside of work, Johannes tries to lead an active lifestyle running marathons and balancing this activity with the responsibilities of raising a family.


Whether you're looking to learn to play guitar or serious about getting one for recording and giving concerts, the guitar you choose can have a big impact. Getting a guitar that is capable of projecting your musical vision and sound is key, but professional grade guitars tend to be expensive and are therefore out of reach for many people. In cases like these, financing can allow you to start learning on an high quality instrument you can grow with or give you the opportunity to start recording professional quality audio and give memorable concerts now. But, according to cfed, 'more than half of Americans are either credit invisible, or have poor credit,' which means financing may be a real challenge for many people. In this article, I've put together a list of some of the absolute best financing options for people of all credit types with a list of pros and cons for each that I think will be helpful.
Zzounds is my personal top choice for financing musical instruments interest-free on a payment plan. In fact, I wrote an in-depth comparison earlier on why I pick them over other music stores. I've personally shopped with them and had nothing but good experiences. They've also always held true to their promises, including their best price guaranty. With Pay as You Play* you can currently finance from $59 up to $1000 in guitar goodness with no credit check. With a credit check, zZounds will allow you to spread out a purchase over 12 months with zero interest. Another plus is that zZounds has a huge selection of both electric and acoustic guitars to choose from. All you have to do is add whatever eligible items you want to your cart, then select Pay as You Play to finance it on monthly payments at checkout. They also offer 6, 8, and 12 payment terms, but for these you'll need to be approved through a credit check. I love that zZounds spreads out financing over 4 payments rather than 3, like many of the other music stores, because it makes financing a guitar more affordable to consumers.
Another great place to get musical instruments without a credit check is Sweetwater. While their best financing option is currently the Sweetwater Card, which does require a credit check, they also offer a 3 Easy Payments option for financing most of the items they sell, including their guitars. Sweetwater offers a super high threshold on their 3 Easy Payments option - it's currently up to $4,000, but you'll need to be able to afford paying as much as $1,333.33 per month if you want to push this to its maximum limit. This is also another place I've personally shopped before. In fact, I purchased several guitars from them. While not every experience has been perfect, they have gone out of their way to make things right when problems arose.
American Musical Supply is another music supplier that offers interest-free monthly payments through their EzPay plan, which currently doesn't require a credit check if the purchase is between $49-$1500. While, like Sweetwater, we do wish they'd offer longer payment terms without a credit check, this is still definitely one of the best options for people looking to finance a guitar. I've personally shopped with them and my experience has always been a positive one. It's cool that AMS offers up to $1500 in credit check free financing - $500 more than zZounds does currently, but keep in mind that if you choose to finance more, you'll have to be able to make higher payments over a 3-payment term to swing it. AMS also offers 5, 8, and 12 payment options, but they do require a credit check and approval.
Eastwood Guitars now offers interest-free credit-check-free financing on guitars as well through their Play Now, Pay Later option. While we weren't able to locate the specifics regarding their minimum and maximum qualifying order amounts, it does appear that they currently offer payments spread out over a period as long as 6 months, which is pretty exciting because it makes getting more costly gear much more attainable for consumers. While I haven't personally shopped with them and I wasn't able to locate BizRate survey information on them, they do appear to have lots of good customer feedback on their website.
Lane Music also offers financing with no credit needed through their Progressive Leasing option. According to the information currently provided on their site, Progressive Leasing allows you to pay off music equipment over a 90 day term -or- make low payments over the course of a year on purchases up to $2,500. It's worth noting that you do not own the musical instruments you buy from Lane Music until they are entirely paid off. While they claim to have 'no up-front-fees,' it's unclear from their current info page whether or not fees are charged at some point or interest is applied. Given the generous 1-year term for payment, it may be something worth writing or calling them about to check into.
Alto Music is yet another online musical instrument store that allows you to finance finance a guitar on a payment plan. However, if you're looking for payment options without a credit check, Alto Doesn't offer a true pay as you play option. Instead, they offer terms up to 6 months on layaway, allowing you to pay for the instrument and receive it after it's paid in full. AltoMusic also offers Progressive Leasing, which they currently claim does not require a high credit score. They also let you make payments via Affirm or the AltoMusic card, both of which require a credit check.
While you'll only have access to Kiesel brand guitars, Kiesel does currently offer a 90 day payment option with no credit needed through their Progressive Leasing option. You can quickly apply online for approval for this in-house financing option and Kiesel does appear to make some pretty badass guitars so you likely won't be disappointed.
While Reverb doesn't currently offer any payment options without a credit check, they do offer financing on lots of new and used instruments, including electric and acoustic guitars, through Affirm. According to NerdWallet, Affirm currently has no minimum credit score or minimum credit history requirement so this may be a good guitar financing option for people with low credit scores, but we can't say for sure.
While it's probably going to be harder to qualify for than options I've already listed, Sam Ash is an amazing musical instrument store with years of experience and tons and tons of guitars to choose from. I've had plenty of positive past experiences with them and still have fond memories of visiting their store at King of Prussia Mall. Sam Ash does currently offer some pretty exceptional financing options through the Sam Ash Card, but according to one of the forums I referenced, it does appear to be a harder card to qualify for if you have poor to bad credit. If you're feeling luck or adventurous, it may still be worth trying to apply for.
Guitar Center is another musical instrument store I've had lots of positive experiences with. In fact, I bought my first guitar at one of their stores back when I was a teen. They also offer some pretty awesome financing options with their Gear Card, which does require a credit check. According to WalletHub, people with fair credit have a pretty good chance of getting approved for financing at Guitar Center so this may be a good alternative to the Sam Ash card.
While most people don't think of Amazon as a place to buy musical instruments, let alone a guitar, the fact is, they have looooooads of options to choose from and some of the best prices out there. Amazon also currently offers special zero-interest financing in 6 and 12 month terms through the Amazon Store Card. According to WalletHub, you'll need good, but not necessarily excellent credit to qualify for financing with them.
I haven't personally shopped at Musician's Friend, but people I know have raved about them for years. While they don't offer payment options without a credit check for your next guitar purchase, they definitely do currently offer some pretty awesome options for financing, including 6, 12 , 24, and 48 month payment terms with no-interest through the Musician's Friend Platinum Card. According to the forums I referenced, including HarmonyCentral, it appears that it's not a very difficult to card to qualify for. This may be a good option for people who want longer payment terms with no interest, but have poor to bad credit.
Zager makes beautiful custom guitars that can be purchased directly from their website. While they don't offer their own financing, they do allow you to pay with PayPal Financing. However, according to Pocket Sense, the average score for people who qualify is about 720, which is rather high if you have poor to bad credit.
Like Amazon, Walmart is another place that might not immediately come to mind when you think of financing a guitar. However, you'll find they have tons of great guitars to choose from on their website. Another plus is that, according to WalletHub, you may only need to have fair credit to qualify for the Walmart Store Card, making this a potentially awesome option for people with less than ideal credit scores.
Kraft Music is another well established musical instrument supplier that offers a good selection of guitars to choose from at excellent prices. They currently offer a number of interest-free financing options including terms of 12, 18, and 36 months. They do, however, require a credit check. While I wasn't able to locate any information on what kind of credit scores usually qualify, it might be worth reaching out to them by phone or email to see what sort of credit scores currently have a good chance of being approved.
The Kraft guarantee helps ensure that you get the best price and can refund you up to $50 if you find a better price somewhere else and let them know about it within 30 days.
Fingerhut is well known for offering a great store card that helps with rebuilding bad credit, but what many people don't know is that they also have a decent selection of guitars. While there is a credit check required, WalletHub currently reports that people with bad credit can get approved. If you do a quick price comparison, you may find that Fingerhut tends to have slightly higher prices on their items than other retailers, but they're a great store with a long standing history of helping people rebuild or establish a good credit standing, so they're totally worth considering as an option for financing your guitar.
They're well established. According to the Better Business Bureau, Finger-hut is a brand of BlueStem Brands Inc, which has been in business for 15 years.
HSN is another store that's fairly well known for their accessible payment plans. They offer 4 payments on most of the items they sell through FlexPay. While a quick comparison may show that guitar prices at HSN tend to be a bit higher than other options, their payment plan is quick and easy to qualify for and they offer a limited selection of Guitars. If you're a beginner looking for something to start out on, this may be a good option for you to consider.
They're well established. According to the Better Business Bureau, they've been in business for 32 years.
Quest Musique offers a great selection of guitars and other musical instruments. For financing, they currently have three different options, Pay in Full, a layaway plan with an 8 week payment term that doesn't require a credit check, and financing through Desjardins Accord D Financing Program. It's worth noting that the last option, according to the site, may require a well established credit history for approval.
LeaseVille offers a wide array of products, including loads of guitars you can choose from. They allow you to lease your guitar and pay it off while you play it. According to their site, not having credit isn't a problem, so if you're having difficulty qualifying for other no-credit-check/bad credit options, this may be a good one to consider.
Gettington is another online store that offers a small but good selection of musical instruments, including acoustic and electric guitars. Wallethub currently reports that people with fair credit can qualify for financing with them, making them a great option for people who are building their credit scores or trying to establish credit.
Flexshopper is another online merchant that allows you to rent music equipment until you own it. The process looks quick and easy and they have a good selection of both acoustic and electric guitars to pick from. FlexPay also offers options like early purchase, which helps bring down the overall cost of a purchase. If other options haven't worked for you, this may be a good one to consider.
Emporium is another retailer that offers a buy-now-pay-later option along with a surprisingly good selection of guitars, despite not being a music store. They currently offer up to a $5000 credit limit and state that less than perfect credit is accepted.
*This article contains affiliate links, click here to learn more.
Hi guys, my credit score is going up! And I was able to finally purchase the guitar of my dreams. I could not have done it without this site. You helped me a lot with credit. I thank my friend for referring me here.
This was really helpful for finding the right place to finance a guitar. Thanks!


text:
North Korea Threatens 'Stronger' Response as U.S. Seeks Sanctions Over Missile Assessments
SEOUL — North Korea warned on Friday that it will take "stronger and sure response" if america helped impose extra sanctions on the North in response to its latest collection of missile checks.
The assertion by the North's International Ministry got here after a proposal by america that the U.N. Safety Council place recent sanctions on North Korea following its six ballistic and different missile checks since September 2021.
The tit for tat between North Korea and america raised tensions at a delicate time within the area, as China equipped for internet hosting the Winter Olympics in Beijing subsequent month and South Korea for its presidential election on March 9.
Individually on Wednesday, the Biden administration blacklisted 5 North Korean officers lively in Russia and China who Washington stated had been accountable for procuring items for North Korea's weapons of mass destruction and ballistic missile-related packages.
North Korea resumed testing missiles in September after a six-month hiatus. It has since performed at the least six missile checks, which concerned a long-range strategic cruise missile, ballistic missiles rolled out of mountain tunnels and a mini submarine-launched ballistic missile. In two checks this month, it launched what it referred to as hypersonic ballistic missiles with removable gliding warheads that made them tougher to intercept as a result of they might change course throughout flight.
All of the checks violated U.N. Safety Council resolutions that banned North Korea from growing or testing ballistic missile applied sciences or applied sciences used to make and ship nuclear weapons. However the North's International Ministry insisted on Friday that it was exercising "its proper to self-defense" and that the missile checks had been "a part of its efforts for modernizing its nationwide protection functionality."
"The U.S. is deliberately escalating the state of affairs even with the activation of unbiased sanctions, not content material with referring the D.P.R.Okay.'s simply exercise to the U.N. Safety Council," the ministry stated in a press release, utilizing the acronym for the North's official title, the Democratic Folks's Republic of Korea. It added, "If the U.S. adopts such a confrontational stance, the D.P.R.Okay. will likely be compelled to take stronger and sure response to it."
The assertion didn't elaborate on North Korea's potential future actions. However the nation has resumed missile checks since conferences between its chief, Kim Jong-un, and Donald J. Trump, then president, ended with out an settlement on the right way to roll again the North's nuclear weapons program or when to carry sanctions.
These checks indicated that the North was growing extra refined methods of delivering nuclear and different warheads to South Korea, Japan and American bases there on its shorter-range missiles, in response to protection analysts. Among the missiles it has examined since 2019 have used stable gasoline and have made midair maneuvers, making them tougher to intercept, the analysts stated.
North Korea has not resumed testing any long-range missiles of the type that would straight threaten the continental United States because it performed three intercontinental ballistic missile checks in 2017. However because the Kim-Trump diplomacy collapsed, North Korea has warned that it not felt certain by its self-imposed moratorium on nuclear and long-range missile checks. It has since unveiled its largest-ever, still-untested ICBM throughout a army parade and exhibition.
Throughout a Staff' Celebration assembly in January final yr, Mr. Kim vowed to make extra refined short-range nuclear missiles, hypersonic missiles, giant ICBMs and submarine-launched long-range missiles, in addition to to put army spy satellites into orbit.
On Friday, North Korea reiterated that its missile checks "didn't goal any particular nation or pressure and it didn't do any hurt to the safety of neighboring nations." However within the take a look at on Tuesday, the North's hypersonic missile traversed the nation from west to east after which veered to the northeast, flying over the waters between the Russian Far East and Japan towards the Pacific, in response to its trajectory graphic in one of many photographs launched in North Korean state media.
The missile hit a goal 621 miles away, the North stated. And because the missile hurtled out of North Korea at as much as 10 occasions the pace of sound, aviation regulators briefly halted flights out of some airports on the U.S. West Coast as a precaution.
It was the primary missile take a look at that Mr. Kim had attended since March 2020, in response to stories in North Korean media.
The take a look at prompted South Korea to reassure its residents this week that its army can detect and intercept the North's new missiles.
Washington has repeatedly urged North Korea to return to talks, however the nation has stated it will not till it was satisfied that america would take away its "hostile" coverage, together with sanctions.
China, which might veto Washington's try and impose extra sanctions on the Safety Council, referred to as for dialogue.
"Willful sanctions don't assist resolve the Korean Peninsula subject, however solely worsen the confrontational temper," Wang Wenbin, a spokesman of the Chinese language International Ministry, stated throughout a information briefing on Wednesday.
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text:
<?php

namespace Mremi\Dolist\Authentication;

/**
 * Authentication token context class
 *
 * @author Rémi Marseille <marseille.remi@gmail.com>
 */
class AuthenticationTokenContext
{
    /**
     * @var integer
     */
    private $accountId;

    /**
     * @var string
     */
    private $key;

    /**
     * Constructor
     *
     * @param integer $accountId An account identifier
     * @param string  $key       An authentication key
     */
    public function __construct($accountId, $key)
    {
        $this->accountId = $accountId;
        $this->key       = $key;
    }

    /**
     * Gets the account identifier
     *
     * @return integer
     */
    public function getAccountId()
    {
        return $this->accountId;
    }

    /**
     * Gets the authentication key
     *
     * @return string
     */
    public function getKey()
    {
        return $this->key;
    }

    /**
     * Gets an array representation
     *
     * @return array
     */
    public function toArray()
    {
        return array(
            'AccountID' => $this->accountId,
            'Key'       => $this->key,
        );
    }
}

redpajama_set_name: RedPajamaGithub