text:
\section{Introduction}

The theory of dynamical systems deal with dynamical properties of group actions. In more restricted case, a system is called discrete or continuous depending on whether the group under consideration is $\mathbb{Z}$ or $\mathbb{R}$. Although the shadowing property was originated from the work \cite{A} of Anosov in differentiable dynamics, it is one of the extensively studied properties for homeomorphisms or continuous maps in topological dynamics. The underlying idea is to trace each approximate orbit by an actual orbit so that the behaviour of the actual orbit reflects into the approximate one. This property plays significant role in the numerical investigation of various modern dynamical systems occurred in different areas of Applied Mathematics, Physics, Engineering and so on. Though the shadowing is very strong property, mathematicians have found several important dynamical systems satisfying the same. For instance, the tent map on the unit interval, the period doubling map of the circle and the shift map of the shift space on two symbols satisfy this property. Another important dynamical property in topological dynamics is widely known as expansivity. Although the expansive property of symbolic flows was recognized earlier, the first formal introduction of this property for arbitrary homeomorphisms was done in \cite{U1} by Utz. The book \cite{A1} can impart a great deal of knowledge to the reader on expansivity and the shadowing property for homeomorphisms and continuous maps.
\medskip

In 1970, P. Walters proved \cite{U2} that Anosov diffeomorphisms are topologically stable, a property in the presence of which continuous perturbations does not affect the nature of the system. In 1971, Z. Nitecki proved [\cite{N}, p. 108] that Anosov diffeomorphisms are expansive. In 1975, Bowen proved [\cite{B}, p. 74] that Anosov diffeomorphisms also satisfy the shadowing property. These facts naturally guided Walters to prove \cite{U3} that expansive homeomorphisms with the shadowing property on compact metric spaces are topologically stable. These two results regarding topological stability are popularly known as Walters stability theorems. In \cite{LM}, authors introduced shadowing and topological stability for Borel measures and proved that an expansive measure \cite{M} with shadowing is topologically stable. In \cite{A3}, authors replaced the usual $C^{0}$-distance with the $C^{0}$-Gromov-Hausdorff distance to define the $GH$-stability for homeomorphisms on a compact metric space in the class of all homeomorphisms on possibly different compact metric space. They proved that expansive homeomorphisms with the shadowing property on compact metric space are $GH$-stable. Smale proved \cite{S1} that the non-wandering set of Anosov diffeomorphisms can be decomposed into union of finitely many disjoint closed invariant sets on each of which the diffeomorphism is transitive. Aoki extended this result \cite{A21} to homeomorphisms on compact metric spaces satisfying expansivity and the shadowing property. These two results are popularly known as Smale's spectral decomposition theorems.
\medskip

Since the dynamics of a homeomorphism on a compact metric space can be seen as the dynamics of the corresponding induced $\mathbb{Z}$-action, it is natural to try to understand the dynamics of finitely generated group actions. One of the most interesting study on expansive actions was published by Hurder in \cite{H1}. Although $\mathbb{S}^{1}$ does not admit expansive homeomorphism, he was able to provide example of expansive action on $\mathbb{S}^1$. Recently, authors of \cite{BDS} studied properties of expansive group actions. In \cite{P1}, authors introduced the shadowing property for $\mathbb{Z}^{p}$-action and obtained sufficient conditions for a linear $\mathbb{Z}^{p}$-action on $\mathbb{C}^{m}$ to have this property. In \cite{O2}, authors extended the shadowing property to finitely generated group actions. In \cite{C3}, authors proved that finitely generated group actions with expansivity and the shadowing property are topologically stable. In \cite{DKY}, authors proved a measurable version of this result for finitely generated group actions. Since spectral decomposition \cite{A1} is a fundamental property of a homeomorphism with expansivity and the shadowing property, it must be interesting to investigate the validity of such a decomposition for finitely generated group actions. In this regard, the study of chain recurrent set is very important.
\medskip

In this paper, we continue exploring the dynamics of finitely generated group actions on compact metric spaces satisfying expansivity, the shadowing property and sequential shadowing property. In section 2, we recall preliminaries on finitely generated group actions required for subsequent sections. In section 3, we study the chain recurrent set for actions satisfying the shadowing property and characterize expansivity in terms of existence of generators. We discuss the projecting and lifting problems for actions satisfying expansivity as well as the shadowing property. In section 4, we introduce the property of strong $GH$-stability and $GH$-stability for finitely generated group actions and prove that expansive actions with the shadowing property are strongly $GH$-stable and hence, $GH$-stable. In the final section, we introduce the property of sequential shadowing for finitely generated group actions and prove that the phase space of any action with the sequential shadowing property can be decomposed into finitely many closed invariant sets on each of which the action is transitive.

\section{Preliminaries}
Throughout the paper, $G$ denotes a finitely generated group with finite symmetric generating set $S=\lbrace s_{i}:1\leq i\leq n\rbrace$, $O(G)$ denotes the cardinality of $G$ and $X$, $Y$, $Z$ denote metric spaces. For $k\in\mathbb{N}$, set $G_{k}=\lbrace g\in G:L(g)\leq k\rbrace$, where $L(g)$ denotes the length of $g$ with respect to the generating set $S$. We assume that the length of the identity element $e$ is $0$. Note that, if $G$ is a finite group, then there exists the least $k\in \mathbb{N}$ such that $G_{k+i}= G$, for all $i\geq 1$ and if $G$ is a countably infinite group then $G_{k}\setminus G_{k-1}\neq\phi$, for all $k\in \mathbb{N}$.
\medskip

A map $\Phi:G\times X\rightarrow X$ is said to be a continuous action of $G$ on $X$ if the following conditions hold.
\begin{enumerate}
\item [(i)] $\Phi_g=\Phi(g,.)$ is a homeomorphism, for every $g\in G$.
\item [(ii)] $\Phi_{e}(x)=x$, for all $x\in X$ , where $e$ is the identity element of the group $G$.
\item [(iii)] $\Phi_{g_1g_2}(x)=\Phi_{g_1}(\Phi_{g_2}(x))$, for all $x\in X$ and for all $g_1,g_2\in G$.
\end{enumerate}
\medskip

An action $\Phi$ is said to be uniformly continuous if for each $g\in G$, $\Phi_g$ is uniform equivalence (i.e. both $\Phi_g$ and $\Phi_{g^{-1}}$ are uniformly continuous). We denote the set of all uniformly continuous actions of $G$ on $X$ by $Act(G,X)$. If the phase space is compact, then continuous actions and uniformly continuous actions are the same. An action $\Phi$ is said to be commutative if $\Phi_{s}\circ \Phi_{s'}= \Phi_{s'}\circ \Phi_{s}$ for all $s, s'\in S$. A subset $W\subset X$ is said to be $\Phi$-invariant if $\Phi_g(W)\subset W$, for all $g\in G$.
\medskip

An action $\Phi\in Act(G,X)$ is said to be expansive, if there exists a constant $\mathfrak{e}>0$ such that for any pair of distinct points $x,y\in X$, there exists $g\in G$ such that $d(\Phi_g(x),\Phi_g(y))>\mathfrak{e}$. Such a constant $\mathfrak{e}$ is said to be an expansive constant for $\Phi$.
\medskip

Let $\Phi\in Act(G,X)$ and let $\delta>0$. A map $f:G\rightarrow X$ is said to be a $\delta$-pseudo orbit for $\Phi$ (with respect to the generating set $S$) if $d(\Phi_s(f(g)),f(sg))<\delta$, for all $s\in S$, $g\in G$. On the other hand, a map $f:G\rightarrow X$ is said to be $\delta$-traced by some point $x\in X$ if $d(\Phi_g(x),f(g))<\delta$, for all $g\in G$. An action $\Phi$ is said to have shadowing property (with respect to the generating set $S$) if for every $\epsilon>0$ there exists $\delta>0$ such that every $\delta$-pseudo orbit is $\epsilon$-traced by some point in $X$ \cite{O2}.
\medskip

Two actions $\Phi\in Act(G,X)$ and $\Psi\in Act(G,X)$ are said to be uniformly conjugate if there exists a uniform equivalence $h:X\rightarrow Y$ such that $h\circ \Phi_g=\Psi_g\circ h$, for all $g\in G$. The map $h$ is called a uniform conjugacy between $\Phi$ and $\Psi$.
\medskip

The properties of finitely generated group actions which are invariant under uniform conjugacy are called uniform dynamical property. It is known that the shadowing property and expansivity are uniform dynamical property.
\medskip

A point $x\in X$ is said to be a non-wandering point for $\Phi\in Act(G, X)$ if for every non-empty open neighbourhood $U$ of $x$, there exists a non-identity element $g\in G$ such that $\Phi_{g}(U)\cap U\neq \phi$. The set of all non-wandering points is denoted by $\Omega(\Phi)$. An action $\Phi\in Act(G, X)$ is said to be transitive if for every pair of non-empty open sets $U$ and $V$ of $X$, there exists a non-identity element $g\in G$ such that $\Phi_{g}(U)\cap V\neq \phi$.
\medskip

A continuous onto map $\pi : Y\rightarrow X$ is said to be locally isometric covering map if for each $x\in X$, there exists a neighbourhood $U(x)$ of $x$ such that $\pi^{-1}(U(x)) = \cup_{\alpha}U_{\alpha}$, where $\lbrace U_{\alpha}\rbrace$ is a pairwise disjoint family of open sets and $\pi|_{U_{\alpha}} : U_{\alpha}\rightarrow U(x)$ is an isometry for each $\alpha$.
\medskip

An isometry between $(X,d_X)$ and $(Y,d_Y)$ is an onto map $i : X\rightarrow Y$ satisfying $d_{Y}(i(x_{1}), i(x_{2}))$ $= d_{X}(x_{1}, x_{2})$, for all $x_{1}, x_{2}\in X$. We say that $X$ and $Y$ are isometric if there exists an isometry between them.
\medskip

For the metric space $(X,d_X)$, define a bounded metric $d^{X}(x, y)=$ min$\lbrace d_{X}(x, y), 1\rbrace$. Note that, if $\Phi\in Act(G, (X, d_{X}))$ then $\Phi\in Act(G, (X, d^{X}))$. For $A, B\subset X$, we define $d^{X}(A, B) = $inf$\lbrace d^{X}(a, b) \mid a\in A, b\in B\rbrace$. If $A = \lbrace a\rbrace$ then we write $'a'$ only. The Hausdorff distance between two subsets $A,B\subset X$ is given by $d_{H}^{X}(A, B) =$ max $\lbrace$ sup$_{a\in A}d^{X}(a, B)$, sup$_{b\in B}d^{X}(A, b)\rbrace$.
\medskip

For $\delta > 0$, a non-necessarily continuous map $i: X\rightarrow Y$ between $X$ and $Y$ is said to be a $\delta$-isometry if max$\lbrace d^{Y}_{H}(i(X), Y)$, sup$_{x_{1},x_{2}\in X}|d^{Y}(i(x_{1}), i(x_{2})) - d^{X}(x_{1}, x_{2})|\rbrace < \delta$.

\section{Expansivity and the Shadowing Property}

In this section, the purpose is to prove the validity of some interesting discrete dynamical results on expansivity and the shadowing property for finitely generated group actions.
\medskip

Let $x,y\in X$ and $\delta>0$. We say that a map $f:G\rightarrow X$ is a $\delta$-pseudo orbit of $\Phi\in Act(G,X)$ from $x$ to $y$ if there exists distinct $h,k\in G$ such that $f(h)=x$ and $f(k)=y$. Then, $x$ is said to be $\delta$-related to $y$ (written as $x R^S_{\delta} y$) if there exists a $\delta$-pseudo orbit of $\Phi$ from $x$ to $y$. If $x R^S_{\delta} y$ for each $\delta>0$, then we say that $x$ is related to $y$ (written $x R^S y$). The set $CR^S(\Phi)=\lbrace x\in X\mid x R^S x\rbrace$ is called the chain recurrent set of $\Phi$. An action $\Phi\in Act(G,X)$ is said to be chain transitive with respect to $S$ if $x R^S y$ for all $x,y\in X$.
\medskip

Because of the following result we denote the chain recurrent set by $CR(\Phi)$.

\begin{theorem}
Let $\Phi\in Act(G, X)$. Then, chain transitivity of $\Phi$ and the chain recurrent set of $\Phi$ are independent of the generating set.
\label{4.7}
\end{theorem}
\begin{proof}
Let $S$ and $S'$ be two generating sets of $G$. Since any two word norms are bi-Lipschitz equivalent, we can choose a constant $C \geq 1$ such that $\frac{|g|_{S'}}{C}\leq |g|_{S} \leq C|g|_{S'}$ for all $g\in G$. Let $\epsilon > 0$ and $0 < \delta < \frac{\epsilon}{C}$ be such that  $d(\Phi_{g}(a), \Phi_{g}(b)) < \frac{\epsilon}{C}$, whenever $d(a,b) < \delta$ for all $a,b\in X$ and $g\in G$ satisfying $|g|_{S'}\leq C$. Let $f$ be a $\delta$-pseudo orbit of $\Phi$ from $x$ to $y$ with respect to $S'$ i.e. $d(\Phi_{s'}(f(g)), f(s'g)) < \delta$ for all $s'\in S$, $g\in G$.
\medskip

Then, for any $s'_{1},...,s'_{C}\in S'$, $g\in G$
\begin{align*}
d(\Phi_{s'_{C}...s'_{1}}(f(g)), f(s'_{C}...s'_{1}g)) &\leq d(\Phi_{s'_{C}...s'_{2}}(\Phi_{s'_{1}}f(g)), \Phi_{s'_{C}...s'_{2}}(f(s'_{1}g)))\\
&+ d(\Phi_{s'_{C}...s'_{2}}(f(s'_{1}g)), \Phi_{s'_{C}...s'_{3}}(f(s'_{2}s'_{1}g)))\\
&+...+ d(\Phi_{s'_{C}}(\Phi_{s'_{C-1}}f(s'_{C-2}...s'_{1}g)), \Phi_{s'_{C}}f(s'_{C-1}...s'_{1}g))\\
&+ d(\Phi_{s'_{C}}f(s'_{C-1}...s'_{1}g), f(s'_{C}...s'_{1}g)) \\
&<  \frac{\epsilon}{C} + \frac{\epsilon}{C} + ...\delta < \epsilon
\end{align*}

Thus, $d(\Phi_{h}(f(g), f(hg)) < \epsilon$ for all $g\in G$, $h\in G$ satisfying $|h|_{S'} \leq C$. Since any $s\in S$ satisfies $|s|_{S'}\leq C$, $f$ is an $\epsilon$-pseudo orbit of $\Phi$ from $x$ to $y$ with respect to $S$. So for any pair $x,y\in X$, $x R^{S'} y$ implies $x R^S y$. Therefore, if $\Phi$ is chain transitive with respect to $S'$ then it is chain transitive with respect to $S$ and we conclude that chain transitivity of $\Phi$ is independent of the generating set.
\medskip

Further, the argument implies that $CR^{S'}(\Phi) \subset CR^{S}(\Phi)$ and in a similar manner, the converse also holds. Therefore, we conclude that the chain recurrent set is independent of the generating set.
\end{proof}

\begin{theorem}
Let $\Phi\in Act(G, X)$ and $\Psi\in Act(G, Y)$ be conjugate. Then, $\Phi$ is chain transitive if and only if $\Psi$ is chain transitive.
\label{4.8}
\end{theorem}
\begin{proof}
Let $h:X\rightarrow Y$ be a conjugacy between $\Phi$ and $\Psi$. Suppose that $\Phi$ is chain transitive. Let $\epsilon>0$ and $y_{1}, y_{2}\in Y$. Then there exists $x_{1}, x_{2}\in X$ such that $h(x_{1}) = y_{1}$, $h(x_{2}) = y_{2}$. Corresponds to $\epsilon$, choose $0 < \delta < \epsilon $ by the uniform continuity of $h$. By chain transitivity of $\Phi$, there exists a $\delta$-pseudo orbit $f:G\rightarrow X$ for $\Phi$ from $x_{1}$ to $x_{2}$ satisfying $f(g_{1}) = x_{1}$ and $f(g_{2}) = x_{2}$ for some $g_1,g_2\in G$. Then, the map $hf:G\rightarrow Y$ satisfies $p(\Psi_{s}(hf(g)), hf(sg)) = p(h\Phi_{s}(f(g)), h(f(sg))) < \epsilon$ for all $s\in S$, $g\in G$. Thus, $hf$ is $\epsilon$-pseudo orbit for $\Psi$ from $y_{1}$ to $y_{2}$. Therefore, we conclude that $\Psi$ is chain transitive. One can prove the converse similarly.
\end{proof}

\begin{theorem}
If $\Phi\in Act(G,X)$, then the following statements are true
\begin{enumerate}
\item[(i)] $CR(\Phi)$ is closed.
\item[(ii)] If $\Phi$ is commutative, then $CR(\Phi)$ is $\Phi$-invariant.
\end{enumerate}
\label{4.9}
\end{theorem}
\begin{proof}
(i) Let $\epsilon>0$ and $0<\delta<\frac{\epsilon}{3}$ be such that $d(a,b)<\delta$ implies $d(\Phi_{s}(a),\Phi_{s}(b))<\frac{\epsilon}{3}$, for all $s\in S$. Let $\lbrace x_{n}\rbrace$ be a sequence in $CR(\Phi)$ such that $x_{n}$ converges to $x$.
Choose $k>0$ such that $d(x,x_{n})<\delta$ for all $n\geq k$. Since $x_{k}\in CR(\Phi)$, there exists $\delta$-pseudo orbit $f_{1}:G\rightarrow X$ from $x_{k}$ to $x_{k}$ satisfying $f_{1}(h_{1})=x_{k}=f_{1}(h_{2})$ for distinct $h_{1},h_{2}\in G$. Define $f_2:G\rightarrow X$ by $f_{2}(g)=f_{1}(g)$ if $g\in G\setminus \lbrace h_{1},h_{2}\rbrace$ and $f_{2}(g)=x$ if otherwise. We claim that $f_{2}$ is an $\epsilon$-pseudo orbit from $x$ to $x$.
\begin{enumerate}
\item[Case 1] Let $sg,g\in G\setminus \lbrace h_{1},h_{2}\rbrace$. Then $d(\Phi_{s}(f_{2}(g)),f_2(sg))=d(\Phi_{s}(f_{1}(g)),f_1(sg))< \delta <\epsilon$ for all $s\in S$.

\item[Case 2] Let $sg\in G\setminus \lbrace h_{1},h_{2}\rbrace$, $g\in \lbrace h_{1},h_{2}\rbrace$.
We have $d(\Phi_{s}(f_{1}(g),f_1(sg))<\delta$ for all $s\in S$
and $d(\Phi_{s}(f_{1}(g)),\Phi_{s}(x))=d(\Phi_{s}(x_{k}),\Phi_{s}(x))<\frac{\epsilon}{3}$ by the uniform continuity of $\Phi_s$.
Thus, we have $d(f_{2}(sg), \Phi_{s}(f_{2}(g)))= d(f_{1}(sg),\Phi_{s}(x))\leq d(f_{1}(sg),\Phi_{s}(f_{1}(g))+$ $d(\Phi_{s}(f_{1}(g)), \Phi_{s}(x))<\epsilon$.

\item[Case 3] Let $sg\in \lbrace h_{1},h_{2}\rbrace$, $g\in G\setminus \lbrace h_{1},h_{2}\rbrace$.
Then, $d(\Phi_{s}(f_{2}(g)),f_2(sg))=d(\Phi_{s}(f_{1}(g),x)\leq d(x,x_{k})+d(x_{k},\Phi_{s}(f_{1}(g)))=d(x,x_{k})+d(f_{1}(sg),\Phi_{s}(f_{1}(g)))<\epsilon$.

\item[Case 4] Let $sg,g\in \lbrace h_{1},h_{2}\rbrace$. Then, $d(f_{2}(sg),\Phi_{s}(f_{2}(g)))=d(x,\Phi_{s}(x))\leq d(x,x_{k})+d(x_{k}, \Phi_{s}(f_{1}(g)))+d(\Phi_{s}(f_{1}(g)),\Phi_{s}(x)) =
d(x,x_{k})+d(f_{1}(sg),\Phi_{s}(f_{1}(g))) + d(\Phi_{s}(x_k),
\\
\Phi_{s}(x))<\epsilon$. This completes a proof.
\end{enumerate}
\medskip

(ii) Let us fix $p\in S$ and $y\in \Phi_{p}(CR(\Phi))$, then there exists $x\in CR(\Phi)$ such that $y=\Phi_p(x)$. Let $\epsilon>0$ be given and $\delta>0$ such that $d(a,b)<\delta$ implies $d(\Phi_{s}(a),\Phi_{s}(b))<\epsilon$, for all $s\in S$. Since $x\in CR(\Phi)$, there exists a $\delta$-pseudo orbit $f:G\rightarrow X$ from $x$ to $x$. Note that $\Phi_{p}f:G\rightarrow X$ is an $\epsilon$-pseudo orbit from $\Phi_{p}(x)$ to $\Phi_{p}(x)$ and hence $\Phi_p(CR(\Phi))\subset CR(\Phi)$. Since the generating set is symmetric, $CR(\Phi)\subset \Phi_{p}(CR(\Phi))$ for all $p\in S$ and hence $\Phi\in Act(G, CR(\Phi))$.
\end{proof}

\begin{theorem}
Let $X$ be a compact metric space. If $\Phi\in Act(G,X)$ is a commutative expansive action and $\Phi|_{CR(\Phi)}$ has the shadowing property, then $CR(\Phi)$ is isolated.
\label{4.10}
\end{theorem}
\begin{proof}
Let $\mathfrak{c}$ be an expansive constant for $\Phi$. For $0 <\beta < \frac{\mathfrak{c}}{2}$, choose $\alpha > 0$ by the shadowing property of $\Phi|_{CR(\Phi)}$. Further, choose $0 < \gamma < min\lbrace \frac{\alpha}{2}, \frac{\mathfrak{c}}{2} \rbrace$ such that $d(x,y) < \gamma$ implies that $d(\Phi_{s}(x), \Phi_{s}(y)) < \frac{\alpha}{2}$, for all $s\in S$. Set $U = \lbrace y\in X\mid d(y, CR(\Phi)) \leq \gamma \rbrace$ and $\cap_{g\in G}\Phi_{g}(U) = U_{\Phi}$. Clearly, $U$ is compact being a closed subset of a compact metric space. It is also easy to see that $CR(\Phi)\subset U_{\Phi}$. For the reverse containment, choose $y\in U_{\Phi}$. Since $\Phi_{g}(y)\in U$ for all $g\in G$, there exists $x_{g}\in CR(\Phi)$ such that $ d(\Phi_{g}(y) , x_{g}) \leq \gamma$, for all $g\in G$. Thus $d(\Phi_{s}(x_{g}), x_{sg}) \leq d(\Phi_{s}(x_{g}), \Phi_{sg}(y)) + d(\Phi_{sg}(y), x_{sg}) < \frac{\alpha}{2} + \gamma < \alpha$, for all $g\in G$ and all $s\in S$. Thus, $f : G\rightarrow X$ defined by $f(g) = x_{g}$ forms a $\alpha$-pseudo orbit for $\Phi|_{CR(\Phi)}$. Let $x\in CR(\Phi)$ be a $\beta$-tracing point for $f$ but this implies that $d(\Phi_{g}(x), \Phi_{g}(y))\leq d(\Phi_{g}(x), x_{g}) + d(x_{g}, \Phi_{g}(y)) < \beta + \gamma < \mathfrak{c}$. By the expansivity of $\Phi$, we get that $x = y$ and therefore $U_{\Phi} \subset CR(\Phi)$. Hence, $CR(\Phi)$ is isolated.
\end{proof}

We now extend the concept of generators \cite{K1} for homeomorphisms to group actions and characterize expansivity in terms of existence of generators.
\begin{Definition}
A finite open cover $\Lambda= \lbrace A_{1}, A_{2}, . . ., A_{n} \rbrace$ of $X$ is said to be
\\
(i) generator for $\Phi\in Act(G,X)$ if for any family $\lbrace U_{g} \rbrace _{g \in G}$ of members of $\Lambda$, $\cap_{g \in  G} \Phi_{g^{-1}}(\oversl{U_{g}})$ contains at most one point.
\\
(ii) weak generator for $\Phi$ if for any family $\lbrace U_{g} \rbrace _{g \in G}$ of members of $\Lambda$, $\cap_{g \in G} \Phi_{g^{-1}}(U_{g})$ contains at most one point.
\label{3.3}
\end{Definition}

\begin{theorem}
Let $X$ be a compact metric space and $\Phi \in Act(G,X)$. Then, the following statements are equivalent
\begin{enumerate}
\item[(i)] $\Phi$ is expansive.
\item[(ii)] $\Phi$ admits a generator.
\item[(iii)] $\Phi$ admits a weak generator.
\end{enumerate}
\label{3.4}
\end{theorem}
\begin{proof}
$(i)\Rightarrow(ii)$
Suppose that $\Phi$ is expansive with expansive constant $\mathfrak{c}>0$. Let $\beta$ be a finite open cover of $X$ consisting of open balls of radius $\mathfrak{c} / 2$ and $\lbrace B_{g}\rbrace$ be an arbitrary family of members from $\beta$. If $x ,y \in \cap_{g \in G}(\Phi_{g^{-1}}(\oversl{B_{g}}))$, then $\Phi_{g}(x) , \Phi_{g}(y) \in \oversl{B_{g}}$, for all $g \in G$ or $d(\Phi_{g}(x) , \Phi_{g}(y)) \leq \mathfrak{c}$, for all $g \in G$. By expansivity of $\Phi$, we must have $x=y$. Hence, $\beta$ is a generator for $\Phi$.
\medskip

$(ii)\Rightarrow(iii)$ Follows from the definition.
\medskip

$(iii)\Rightarrow(i)$
Let $\alpha$ be a weak generator for $\Phi$ and let $\mathfrak{c}$ be the Lebesgue number for $\alpha$.
Our claim is that $\Phi$ is expansive with expansive constant $\mathfrak{c}$. Let $x,y\in X$ such that $d(\Phi_{g}(x) , \Phi_{g}(y)) \leq \mathfrak{c}$ for all $g\in G$. Since $\mathfrak{c}$ is the Lebesgue number for $\alpha$, there exists $A_{g'} \in \alpha$ such that $\Phi_{g}(x) ,\Phi_{g}(y)\in A_{g^{'}}$ for each $g \in G$. Consider $\lbrace A_{g} \rbrace_{g\in G}$ such that $A_{g}= A_{g'}$ for all $g \in G$. Therefore, $x , y \in \cap_{g \in G}(\Phi_{g^{-1}}(A_{g}))$ which is a contradiction to our assumption. Hence $x = y$ and thus $\Phi$ is expansive with expansive constant $\mathfrak{c}$.
\end{proof}

The following example justifies the fact that Theorem \ref{3.4} does not hold for non-compact spaces.
\begin{Example} Let $X=\mathbb{Z}$ be the set of integers equipped with the discrete metric and $G=(\mathbb{Z}^{2}, +)$ with symmetric generating set $\lbrace e_{1}, e_{2}, e_{3}, e_{4}\rbrace$, where $e_{1}= (1, 0)$, $e_{2}= (0, 1)$, $e_3= (-1, 0)$ and $e_4= (0, -1)$. Let $\Phi$ be the action generated by $\Phi_{e_{1}}(x)= 1-x = \Phi_{e_{2}}(x)$ for all $x\in X$. One can observe that the action is expansive. Further, the action does not admit a generator. If possible, suppose $\alpha = \lbrace V_{1}, V_{2}, . . . V_{n} \rbrace$ is a generator. Note that $\oversl{V_{j}} = V_{j}$ for all $1\leq j\leq n$ and without loss of generality, we can assume that $V_{1}$ is infinite. Choose $U_{g} = V_{1}$ to show that, if $x\in V_{1}$ then $(1-x)\notin V_{1}$. Similarly, we can show that for every $2\leq j\leq n$ and every pair of distinct points $x,y\in V_{1}$, either $(1-x)\notin V_{j}$ or $(1-y)\notin V_{j}$, which is not possible. Thus $\Phi$ does not admit a generator.
\label{3.6}
\end{Example}

\begin{Definition}
Let $\Phi\in Act(G,X)$ be commutative and $m$ be a positive integer. The $m$-times self composition of $\Phi$ is the action $\Phi^{m} :G\times X\rightarrow X$ given by $\Phi^{m}_{g}(x)=\Phi_{g}\Phi_{g}...\Phi_{g}(x)$. The inverse of $\Phi$ is the action $\Phi^{-1}:G\times X\rightarrow X$ given by $(\Phi^{-1})_{g}(x)=\Phi_{g^{-1}}(x)$.
\label{3.7}
\end{Definition}

\begin{theorem}
Let $\Phi\in Act(G,X)$. Then,
\begin{enumerate}
\item [(i)] $\Phi$ is expansive if and only if $\Phi^{-1}$ is expansive.
\item [(ii)] If $\Phi^{m}$ is expansive for some $m\in\mathbb{Z}\setminus\lbrace 0\rbrace$, then $\Phi$ is expansive.
\item [(iii)] If $\Phi$ is commutative expansive action, then $\Phi^m$ is expansive for all $m\in \mathbb{Z}\setminus\lbrace 0\rbrace$.
\end{enumerate}
\label{3.9}
\end{theorem}
\begin{proof} (i) and (ii) are easy to verify using the definition and the fact that $\Phi^m_g(x)=\Phi_{g^m}(x)$ for all $g\in G$ and all $x\in X$.
\medskip

(iii) Fix $m$, $i\in \mathbb{N} $ and $h\in G_{i}$. Since $\Phi_{h^{-1}}$ is uniformly continuous, there exists $\gamma^{h}_{i} >0$  such that $d(x,y)\geq \mathfrak{c}$ implies that $d(\Phi_{h}(x),\Phi_{h}(y)) = d(\Phi^{-1}_{h^{-1}}(x),\Phi^{-1}_{h^{-1}}(y)) > \gamma^{h}_{i}$. Since the cardinality of $G_{i}$ is finite, choose $\gamma_{i}=min_{h\in G_{i}}\lbrace \gamma^{h}_{i}\rbrace$ and set $\gamma= min_{i\in \lbrace 1, 2,..., nm\rbrace}\lbrace \gamma_{i}\rbrace$.
Thus $d(x,y)\geq \mathfrak{c}$ implies that $d(\Phi_{h}(x),\Phi_{h}(y))>\gamma$, for all $h\in G_{nm}$.
Let $x\neq y$ in $X$ and $g\in G$ such that $d(\Phi_{g}(x),\Phi_{g}(y)) > \mathfrak{c} $.
Since $\Phi$ is commutative, we can write $\Phi_{g}= \Phi_{s^{ k_{1}m }_{1}}...\Phi_{s^{ k_{n}m }_{n}}\Phi_{s^{r_{1}}_{1}}...\Phi_{s^{r_{n}}_{n}}$, for some $k_{1},..., k_{n}\in \mathbb{N} \cup \lbrace 0 \rbrace$ and $r_{1},..., r_{n} \in \lbrace 0,...,m-1\rbrace$. As $s^{m-r_{1}}_{1}...s^{m-r_{n}}_{n}\in G_{nm}$, we have
$d(\Phi_{s^{m-r_{1}}_{1}...s^{m-r_{n}}_{n}}\Phi_{g}(x),\Phi_{s^{m-r_{1}}_{1}...s^{m-r_{n}}_{n}}\Phi_{g}(y))>\gamma$ or $d(\Phi_{s^{(k_{1}+1)m}_{1}}$...$\Phi_{s^{(k_{n}+1)m}_{n}}(x),\Phi_{s^{(k_{1}+1)m}_{1}}$...
$\Phi_{s^{(k_{n}+1)m}_{n}}(y)) > \gamma$ or $d(\Phi^{m}_{s^{k_{1}+1}_{1}...s^{k_{n}+1}_{n}}(x),\Phi^{m}_{{s^{k_{1}+1}_{1}}...s^{k_{n}+1}_{n}}(y))
\\
>\gamma$. Hence, $\Phi^{m}$ is expansive with expansive constant $\gamma$.   Now, use part (i) to complete the proof.
\end{proof}
Recall that a point is said to be periodic under an action if the orbit of that point is finite. Since the set of all periodic points of any expansive homeomorphism on a compact metric space is at most countable, it is obvious to ask whether the set of all periodic points of any expansive action is countable or not. We give an affirmative answer for commutative action of infinite group.

\begin{theorem}
The set of all periodic points of any commutative expansive action on a compact metric space is at most countable.
\label{3.11}
\end{theorem}
\begin{proof}
For $\Phi\in Act(G,X)$, set $F(\Phi) = \lbrace x\in X : \Phi_{g}(x) = x$ for all $g\in G\rbrace$. Then, for $m\in \lbrace j > 1 : O(G)$ does not divide $j\rbrace$, set $P(\Phi^m)=\lbrace F(\Phi^{m})\setminus \lbrace F(\Phi^{k}):1\leq k \leq (m-1)$ and $O(G)$ does not divide $k \rbrace$. Since $F(\Phi^{m})$ is a finite set for all $m > 0$ and $P(\Phi)=P(\Phi^m)\cup F(\Phi) \subset \cup_{m\in (\mathbb{N}\setminus \lbrace m : O(G)| m\rbrace} F(\Phi^{m})$, $P(\Phi)$ is atmost countable. To complete the proof, it is enough to show that $x$ is a periodic point if and only if $x\in P(\Phi)$. To see, assume that $x\in P(\Phi^{m})$ for some $m > 0$. Then, $\Phi_{g^{m}}(x) = x$ for all $g\in G$ and hence $\Phi_{s^{m}}(x) = x$ for all $s\in S$. Denote the set $\mathbb{O}(x)=\lbrace \Phi_{s^{j_{1}}_{1}} \Phi_{s^{j_{2}}_{2}}$ ... $\Phi_{s^{j_{n}}_{n}}(x) : j_{i} \in \lbrace 0, 1, m-1 \rbrace$ for $1\leq i\leq n\rbrace$. For $g\in G$, $\Phi_{g}(x) = \Phi_{s^{k_{1}}_{1}} \Phi_{s^{k_{2}}_{2}}$ ... $\Phi_{s^{k_{n}}_{n}}(x) = \Phi_{s^{r_{1}}_{1}} \Phi_{s^{r_{2}}_{2}} ... \Phi_{s^{r_{n}}_{n}}(x)$, where $k_{i} = t_{i}m + r_{i}$ for some $t_{i}\geq 0$, $r_{i}\in \lbrace 0, 1, ... ,m-1 \rbrace$ and $1\leq i\leq n$. Thus, $\Phi_{g}(x)\in \mathbb{O}(x)$ which is a finite set. Hence the orbit of $x$, $\mathcal{O}(x)$ is finite and coincides with $\mathbb{O}(x)$.
\medskip

Conversely, assume that $x\in X$ is a periodic point, thus the orbit of $x$ is a finite set. Since the order of the group is infinite, the orbit of $x$ is finite and $\Phi_{s}$ is a uniform equivalence, there exists $m\in\mathbb{N}^{+}$ such that $\Phi_{s^m}(x)=x$ for all $s\in S$. Now, for any $g\in G$, $x  = \Phi_{s^{m k_{1}}_{1}} \Phi_{s^{m k_{2}}_{2}}$ ... $\Phi_{s^{m k_{n}}_{n}}(x) = \Phi_{g^{m}}(x)$ for some $k_{i}\geq 0$ and for all $1\leq i\leq n$. Thus, $x\in P(\Phi^{m})\subset P(\Phi)$.
\end{proof}

\begin{theorem}
Let $\pi:Y\rightarrow X$ be a locally isometric covering map. Suppose that $\Phi\in Act(G,X)$, $\Psi\in Act(G,Y)$ satisfy $\pi\Psi = \Phi\pi$ i.e. $\pi\Psi_{g}(y) = \Phi_{g}\pi(y)$, for all $y\in Y$ and for all $g\in G$. Suppose that there exists $\delta_{0} >0$ such that for each $y\in Y$ and $0<\delta <\delta_{0}$, $\pi : U_{\delta}(y)\rightarrow U_{\delta}(\pi(y))$ is an isometry. Then
\begin{enumerate}
\item[(i)] $\Phi$ is expansive if and only if $\Psi$ is expansive.
\item[(ii)] $\Phi$ has the shadowing property if and only if $\Psi$ has the shadowing property.
\end{enumerate}
\label{3.13}
\end{theorem}
\begin{proof}
(i) Suppose $\Phi$ is expansive with expansive constant $\mathfrak{c}$. We claim that, $0< \gamma<$ min$\lbrace \mathfrak{c}, \delta_{0}\rbrace$ is an expansive constant for $\Psi$. Let $y_{1}, y_{2}\in Y$ such that $p(\Psi_{g}(y_{1}),\Psi_{g}(y_{2}))< \gamma$, for all $g\in G$. Therefore, $\Psi_{g}(y_{2})\in U_{\gamma}(\Psi_{g}(y_{1}))\subset U_{\delta_{0}}(\Psi_{g}(y_{1}))$, for all $g\in G$ which implies that $d(\pi\Psi_{g}(y_{1}), \pi\Psi_{g}(y_{2}))< \gamma$, for all $g\in G$. Hence, $d(\Phi_{g}\pi(y_{1}), \Phi_{g}\pi(y_{2}))< \gamma$, for all $g\in G$. Since $\gamma$ is expansive constant for $\Phi$, we get that $\pi(y_{1}) = \pi(y_{2})$. In particular, for $g=e$ the identity element of $G$, $y_{2}\in U_{\gamma}(y_{1})\subset U_{\delta_{0}}(y_{1})$ and since $\pi$ is an isometry on $U_{\gamma}(y_{1})$, we get that $0= d(\pi(y_{1}),\pi(y_{2})) = p(y_{1}, y_{2})$ or $y_{1} = y_{2}$. Hence, $\Psi$ is expansive with expansive constant $\gamma$.
\medskip

Conversely, suppose that $\Psi$ is expansive with expansive constant $\mathfrak{c}$. Since $\Psi$ is uniformly continuous, we can choose $0< \delta< \delta_{0}$ such that $p(y_{1},y_{2}) < \delta$ implies that $p(\Psi_{s}(y_{1}),\Psi_{s}(y_{2})) < \delta_{0} $, for all $s\in S$. Let $0< \gamma <$ min$\lbrace \mathfrak{c}, \delta\rbrace$ and $x_{1}, x_{2}\in X$ such that $d(\Phi_{g}(x_{1}), \Phi_{g}(x_{2}))< \gamma$, for all $g\in G$. Choose, $y_{1}, y_{2}\in Y$ with $\pi(y_{1})=x_{1}$, $\pi(y_{2})=x_{2}$ and $p(y_{1}, y_{2})= d(x_{1}, x_{2})< \gamma <\delta <\delta_{0}$.
If $d(\Phi_{g}(x_{1}), \Phi_{g}(x_{2}))\leq \gamma$ for all $g\in G$, then $p(\Psi_{s}(y_{1}), \Psi_{s}(y_{2})) = d(\pi\Psi_{s}(y_{1}), \pi\Psi_{s}(y_{2})) = d(\Phi_{s}(x_{1}), \Phi_{s}(x_{2})) \leq \gamma$ for all $s\in G_{1}$.
By induction on $n$, we can show that $p(\Psi_{g}(y_{1}), \Psi_{g}(y_{2})) \leq \gamma$ for all $g\in G_{n}$. Thus $p(\Psi_{g}(y_{1}), \Psi_{g}(y_{2})) \leq \gamma$ for all $g\in G$ and hence by expansivity of $\Psi$, we get that $y_{1} = y_{2}$ or $x_{1} = x_{2}$. Thus, $\Phi$ is expansive with expansivity constant $\gamma$
\medskip

(ii) Suppose that $\Phi$ has the shadowing property. For any $0< \epsilon\leq \delta_{0}$, the shadowing property of $\Phi$ gives $0< \delta <\epsilon$ such that every $\delta$-pseudo orbit can be $\epsilon$-traced. Let $f:G\rightarrow Y$ be a $\delta$-pseudo orbit for $\Psi$ i.e. $p(\Psi_{s}(f(g)), f(sg)) < \delta <\delta_{0}$, for all $s\in S$ and all $g\in G$. Therefore, $f(sg)\in U_{\delta_{0}}(\Psi_{s}f(g))$. Since $\pi$ is an isometry on $U_{\delta_{0}}(\Psi_{s}f(g))$, we get that
$d(\pi \Psi_{s}(f(g)), \pi f(sg)) = p(\Psi_{s}(f(g)), f(sg)) < \delta <\delta_{0}$ or $d(\Phi_{s}\pi f(g), \pi f(sg)) = p(\Psi_{s}(f(g)), f(sg)) < \delta <\delta_{0}$, for all $s\in S$ and all $g\in G$. So, $h:G\rightarrow X$ given by $h(g)= \pi f(g)$ is a $\delta$-pseudo orbit for $\Phi$ and hence, there exists $x\in X$ such that $d(\Phi_{g}(x),\pi f(g))< \epsilon \leq \delta_{0}$ or $d(\Phi_{g} \pi(y), \pi f(g))< \epsilon \leq \delta_{0}$ or $d(\pi \Psi_{g}(y), \pi f(g))< \epsilon \leq \delta_{0}$ for some $y\in Y$ and for all $g\in G$ which implies $p(\Psi_{g}(y), f({g}))< \epsilon$ for all $g\in G$. Hence $f(g)$ is $\epsilon$-traced by $y\in Y$.
\medskip

Conversely, suppose that $\Psi$ has the shadowing property. For $0< \epsilon\leq \delta_{0}$, the shadowing property of $\Psi$ gives $0< \delta <\epsilon$ such that every $\delta$-pseudo orbit of $\Psi$ can be $\epsilon$-traced. Let $f:G\rightarrow X$ be a $\delta$-pseudo orbit for $\Phi$ i.e. $d(\Phi_{s}f(g), f(sg)) < \delta <\delta_{0}$, for all $s\in S$ and all $g\in G$. Since $\pi$ is onto map, there exists $t:G\rightarrow Y$ such that $\pi t(g) = f(g)$ and $d(\Phi_{s}\pi (t(g)), \pi t(sg))< \delta$ or $d(\pi \Psi_{s}(t(g)), \pi t(sg))< \delta$. Since $\pi$ is an isometry on $U_{\delta_{0}}(\Psi_{s} (t(g)))$, we get that $d(\pi \Psi_{s}(t(g)), \pi t(sg)) = p(\Psi_{s}(t(g)), t(sg)) < \delta <\delta_{0}$. So, $t$ is a $\delta$-pseudo orbit of $\Psi$. Then, there exists $y\in Y$ such that $p(\Psi_{g}(y), t(g))< \epsilon \leq \delta_{0}$ which implies that
$d(\pi \Psi_{g}(y), \pi t(g))< \epsilon \leq \delta_{0}$, for some $y\in Y$ or $d(\Phi_{g}\pi (y), \pi t(g))< \epsilon \leq \delta_{0}$ or $d(\Phi_{g}(x), f(g))< \epsilon$, for all $g\in G$. Hence $f$ is $\epsilon$-traced by $x\in X$.
\end{proof}

\section{Gromov-Hausdorff Stability}

In \cite{A3}, authors defined the Gromov-Hausdorff distance between $(X,d_X)$ and $(Y,d_Y)$ by $d_{GH}(X, Y) =$ inf$\lbrace \delta > 0 \mid $ there exists $\delta$-isometries $i: X\rightarrow Y$ and $j: Y\rightarrow X\rbrace$.
\medskip

For a finite symmetric generating set $S$ of $G$ and $\Phi,\Psi\in Act(G, X)$, the map $d_{S}^{X}(\Phi, \Psi) =$ sup$_{(s, x)\in S\times X}d^{X}(\Phi_{s}(x), \Psi_{s}(x))$ is a metric.
\medskip

For non-necessarily continuous map $i:X\rightarrow Y$, $\Phi\in Act(G, X)$ and $\Psi\in Act(G, Y)$, we define
$d_{S}^{Y}(\Psi\circ i,i\circ \Phi) =$ sup$_{(s, x)\in S\times X}d^{Y}(\Psi_{s}(i(x)), i(\Phi_{s}(x)))$.
\medskip

The $G$-Gromov-Hausdorff distance between $\Phi\in Act(G,X)$ and $\Psi\in Act(G,Y)$ with respect to the generating set $S$ is defined as
\begin{center}
$d_{GH}^{S}(\Phi, \Psi) =$ inf$\lbrace \delta > 0\mid$ there exists $\delta-$isometries $i: X\rightarrow Y$ and $j: Y\rightarrow X$ such that $d_{S}^Y(\Psi\circ i, i\circ \Phi) < \delta$ and $d_{S}^X(\Phi\circ j, j\circ \Psi) < \delta \rbrace$.
\end{center}
\medskip

The strong $G$-Gromov-Hausdorff distance between $\Phi\in Act(G, X)$ and $\Psi\in Act(G, Y)$ with respect to the generating set $S$ is defined as
\begin{center}
$\overline{d}_{GH}^{S}(\Phi, \Psi) =$ inf$\lbrace \delta > 0\mid$ there exists $\delta-$isometry $i: Y\rightarrow X$ such that $d_{S}^X(\Phi\circ i, i\circ \Psi)<\delta\rbrace$.
\end{center}
\medskip

A function $d : X\times X\rightarrow \mathbb{R}$  is said to be a pseudo-quasimetric on $X$, if for every triplet $(x, y, z)$ of $X\times X\times X$ following conditions hold
\begin{enumerate}
\item [(i)] $d(x, y) \geq 0$ and $d(x, x) = 0$;
\item [(ii)] $d(x, y) = d(y, x)$;
\item [(iii)] There exists a constant $k\geq 1$ such that $d(x, y)\leq k[d(x, z) + d(z,y)]$, where $k$ is called the coefficient of pseudo-quasi metric.
\end{enumerate}

If only (i) and (iii) hold, then $d$ is called an asymmetric pseudo-quasi metric on $X$.

\begin{theorem}
Let $S$ be a generating set of $G$. For $\Phi\in Act(G, X)$ and $\Psi\in Act(G, Y)$ following statements are true
\begin{enumerate}
\item [(i)] $d_{GH}^{S}(\Phi, \Psi) = $max $\lbrace \overline{d}_{GH}^{S}(\Phi, \Psi), \overline{d}_{GH}^{S}(\Psi, \Phi)\rbrace$ and if $X = Y$, then $d_{GH}^{S}(\Phi, \Psi) \leq d_{S}(\Phi, \Psi)$;
\item [(ii)] $d_{GH}(X, Y) \leq d_{GH}^{S}(\Phi, \Psi)$ and $d_{GH}(X, Y) = d_{GH}^{S}(Id_{X}, Id_{Y})$, where $Id_{X}, Id_{Y}$ denote the trivial actions on $X$ and $Y$ respectively;
\item [(iii)] $d_{GH}^{S}(\Phi, \Psi) \geq 0$ and $d_{GH}^{S}(\Phi, \Phi) = 0$;
\item [(iv)] $d_{GH}^{S}(\Phi, \Psi) = d_{GH}^{S}(\Psi, \Phi)$;
\item [(v)] If $\Lambda \in Act(G, Z)$, then $d_{GH}^{S}(\Phi, \Psi)\leq 2[d_{GH}^{S}(\Phi, \Lambda) + d_{GH}^{S}(\Lambda,\Psi)]$.
\item [(vi)] $\overline{d}_{GH}^{S}(\Phi, \Psi) \geq 0$ and $\overline{d}_{GH}^{S}(\Phi, \Psi) = 0$;
\item [(vii)] If $\Lambda \in Act(G, Z)$, then $\overline{d}_{GH}^{S}(\Phi, \Psi)\leq 2[\overline{d}_{GH}^{S}(\Phi, \Lambda) + \overline{d}_{GH}^{S}(\Lambda,\Psi)]$.
\label{6.1}
\end{enumerate}
\end{theorem}

\begin{proof}
Proof of (i), (ii), (iii), (iv) and (vi) follow from the corresponding definitions. The proof of (vii) is similar to the proof of (v). So, we prove only (v). For fixed $\epsilon > 0$, we can choose $\delta_{1}$-isometries $i : X\rightarrow Z$, $j: Z\rightarrow X$ and $\delta_{2}$-isometries $k: Y\rightarrow Z$, $l : Z\rightarrow Y$ such that $\delta_{1} < d_{GH}^{S}(\Phi, \Lambda) + \epsilon$, $\delta_{2} < d_{GH}^{S}(\Lambda, \Psi) + \epsilon$, $d_{S}^Z(\Lambda\circ i, i\circ \Phi) < \delta_{1}$, $d_{S}^X(j\circ \Lambda, \Phi\circ j) < \delta_{1}$, $d_{S}^Z(\Lambda\circ k, k\circ \Psi) < \delta_{2}$, $d_{S}^Y(l\circ \Lambda, \Psi\circ l) < \delta_{2}$. Clearly, $l\circ i: X\rightarrow Y$ and $j\circ k: Y\rightarrow X$ are $(\delta_{1}+\delta_{2})$-isometries and hence, $2(\delta_{1}+\delta_{2})$-isometries as well. For all $x\in X$, $s\in S$, we have
\begin{align*}
&d^{Y}(\Psi_{s}(l(i(x))), l(i(\Phi_{s}(x))))\\
&\leq d^{Y}(\Psi_{s}(l(i(x))), l(\Lambda_{s}(i(x)))) + d^{Y}(l(\Lambda_{s}(i(x))), l(i(\Phi_{s}(x)))) \\
&< \delta_{2} + d^{Z}(\Lambda_{s}(i(x)), i(\Phi_{s}(x))) + \delta_{2}\\
&< \delta_{1} + 2\delta_{2}< 2(\delta_{1} + \delta_{2}).
\end{align*}

Hence, $d_{S}^Y(\Psi\circ (l\circ i), (l\circ i)\circ \Phi) < 2(\delta_{1} + \delta_{2})$. Similarly, one can prove that $d_{S}^X((j\circ k) \circ \Psi, \Phi\circ (j\circ k)) < 2(\delta_{1}+\delta_{2})$. Therefore, $d_{GH}^{S}(\Phi, \Psi) \leq 2(\delta_{1}+\delta_{2}) < 2(d_{GH}^{S}(\Phi, \Lambda) + d_{GH}^{S}(\Lambda, \Psi) + 2\epsilon)$. Since $\epsilon$ was chosen arbitrarily, we get the result.
\end{proof}

Items (iii), (iv) and (v) imply that $d_{GH}^{S}$ forms a pseudo-quasimetric and items (vi), (vii) imply that the $\overline{d}_{GH}^{S}$ forms an asymmetric pseudo-quasimetric.
\medskip

In this section, without loss of generality we assume that $0 < \epsilon, \delta, \eta, \gamma, \mathfrak{c} < \frac{1}{2}$.

\begin{Definition}
An action $\Phi\in Act(G, X)$ is said to be strongly $GH$-stable (resp. $GH$-stable) with respect to the generating set $S$ of $G$ if for every $\epsilon > 0$ there exists $\delta > 0$ such that for every action $\Psi\in Act(G, Y)$ satisfying $\overline{d}_{GH}^{S}(\Phi, \Psi) < \delta$ (resp. $d_{GH}^{S}(\Phi, \Psi) < \delta$) there exists a continuous $\epsilon$-isometry $k: Y\rightarrow X$ such that $\Phi\circ k = k\circ \Psi$.
\label{6.2}
\end{Definition}

The following result shows that strong $GH$-stability is independent of the generating set.

\begin{theorem}
Let $S$ and $P$ be two finite symmetric generating sets of $G$. If $\Phi\in Act(G, X)$, then $\Phi$ is strongly $GH$-stable with respect to $S$ if and only if $\Phi$ is strongly $GH$-stable with respect to $P$.
\label{6.3}
\end{theorem}
\begin{proof}
Suppose that $\Phi$ is strongly $GH$-stable with respect to the generating set $S$. Then, for every $\epsilon > 0$ there exists $\eta> 0$ such that for each $\Psi\in Act(G, Y)$ satisfying $\overline{d}_{GH}^{S}(\Phi, \Psi) < \eta$ there exists a continuous $\epsilon$-isometry $k: Y\rightarrow X$ such that $\Phi\circ k = k\circ \Psi$. Set $m = $max$_{s\in S}d_{P}(s)$, where $d_{P}$ is the word length metric on $G$ induced by $P$. Choose $\gamma > 0$ such that $m\gamma < \eta$. Fix $\delta > 0$ such that $d^{X}(x_{1}, x_{2}) < \delta$ implies $d^{X}(\Phi_{g}(x_{1}), \Phi_{g}(x_{2})) < \gamma$ for all $g\in G_{m}^{P}$, where $G_{m}^{P}$ is the set of all elements of length at most $m$ with respect to the generating set $P$.
\medskip

For any $s\in S$, we can write $s = p_{1}p_{2}...p_{d_{P}(s)}$, where $d_{P}(s) \leq m$ and $p_{i}\in P$ for all $1\leq i\leq d_{P}(s)$. Let $\Psi\in Act(G, Y)$ with $\overline{d}_{GH}^{P}(\Phi, \Psi) < \delta$. Thus, there exists $\delta$-isometry $i:Y\rightarrow X$ such that $d_{P}^X(\Phi\circ i, i\circ \Psi)< \delta$. Since every $\delta$-isometry is $\eta$-isometry, it is enough to show that $d_{S}^X(\Phi\circ i, i\circ \Psi)< \eta$. Note that for all $y\in Y$, $s\in S$ following holds.
\medskip

$
d^{X}(\Phi_{s}\circ j(y), j\circ \Psi_{s}(y))\\
= d^{X}(\Phi_{p_{1}...p_{d_{P}(s)}}\circ j(y), j\circ \Psi_{p_{1}...p_{d_{P}(s)}}(y)) \\
\leq d^{X}(\Phi_{p_{1}...p_{d_{P}(s)-1}}(\Phi_{p_{d_{P}(s)}}\circ j(y)), \Phi_{p_{1}...p_{d_{P}(s)-1}}(j\circ \Psi_{p_{d_{P}(s)}}(y))) \\
+ d^{X}(\Phi_{p_{1}...p_{d_{P}(s)-2}}(\Phi_{p_{d_{P}(s)-1}}\circ j (\Psi_{p_{d_{P}(s)}}(y))),\Phi_{p_{1}...p_{d_{P}(s)-2}}(j\circ \Psi_{p_{d_{P}(s)-1}} (\Psi_{p_{d_{P}(s)}}(y))) \\
+ ... + d^{X}(\Phi_{p_{1}}\circ j (\Psi_{p_{2}...p_{d_{P}(s)}}(y)), j\circ \Psi_{p_{1}...p_{d_{P}(s)}}(y))< m\gamma< \eta$
\end{proof}

\begin{theorem}
Let $\Phi\in Act(G, X)$ be strongly $GH$-stable. If $\Psi\in Act(G, Z)$ is isometric to $\Phi$ via isometry $k : Z \rightarrow X$, then $\Psi$ is strongly $GH$-stable.
\label{6.5}
\end{theorem}

\begin{proof}
Let $\epsilon > 0$ be fixed and let $\delta > 0$ be given for $\epsilon$ by strong $GH$-stability of $\Phi$. Let $\Lambda\in Act(G, Y)$ such that $\overline{d}_{GH}^{S}(\Psi, \Lambda) < \frac{\delta}{2}$. By Theorem \ref{6.1}, $\overline{d}_{GH}^{S}(\Phi, \Lambda)\leq 2[\overline{d}_{GH}^{S}(\Phi, \Psi)$ $+ \overline{d}_{GH}^{S}(\Psi, \Lambda)] < 2[0 +\frac{\delta}{2}] = \delta$. Since $\Lambda\in Act(G, Y)$, there exists a continuous $\epsilon$-isometry $h: Y\rightarrow X$ such that $\Phi\circ h = h\circ \Lambda$ and hence, $\Psi\circ k^{-1}\circ h = k^{-1}\circ h\circ \Lambda$. Since $k$ is an isometry, one can see that $k^{-1}h: Y\rightarrow X$ is a continuous $\epsilon$-isometry. Hence, $\Psi$ is strongly $GH$-stable.
\end{proof}

\begin{Lemma}
Let $X$ be a metric space in which every closed ball is compact and $\Phi$ be expansive with expnaisve constant $\mathfrak{c}$. Then, for each $x\in X$ and every $\epsilon>0$ there is $n_{\epsilon}(x)\geq 0$ such that sup$_{g\in G_{n_{\epsilon}(x)}}d^{X}(\Phi_{g}(x), \Phi_{g}(x'))$ $\leq \mathfrak{c}$ implies that $d^{X}(x, x')< \epsilon$.
\label{4.5}
\end{Lemma}
\begin{proof}
If not then there exists $\epsilon > 0$ such that for each $n\geq 0$ there exists $x'_{n}\in X$ satisfying sup$_{g\in G_{n}}d^{X}(\Phi_{g}(x), \Phi_{g}(x'_{n}))\leq \mathfrak{c}$ and $d^{X}(x, x'_{n}) \geq \epsilon$. Since closed ball of radius $\mathfrak{c}$ around $x$ is compact, without loss of generality we may assume that $x'_{n}\rightarrow x'$. Since $G = \cup_{n\geq 0}G_{n}$, we get that $d^{X}(\Phi_{g}(x), \Phi_{g}(x'))\leq \mathfrak{c}$ for all $g\in G$ and $d^{X}(x, x')\geq \epsilon$, which contradicts the expansivity of $\Phi$.
\end{proof}

\begin{theorem}
Let $X$ be a metric space in which every closed ball is compact. If $\Phi\in Act(G, X)$ is expansive and has the shadowing property, then it is strongly $GH$-stable.
\end{theorem}

\begin{proof}
Let $\mathfrak{c}$ be an expansivity constant for $\Phi\in Act(G,X)$ and let us fix generating set $S$ of $G$. Let $\epsilon>0$ be given and we choose $0 < \eta < \frac{1}{8}$min$\lbrace \epsilon, \mathfrak{c} \rbrace$. Let $0 < \delta < \eta$ be given for $\eta$ by the shadowing property of $\Phi$. Let $Y$ be a metric space and let $\Psi\in Act(G, Y)$ be such that $\overline{d}_{GH}^{S}(\Phi, \Psi) < \delta$. Thus, we can choose $\delta$-isometry  $i: Y\rightarrow X$ satisfying $d_{S}^{X}(\Phi\circ i, i\circ \Psi) < \delta$.
\medskip

For each $y\in Y$, set $x_{g}^{y} = i(\Psi_{g}(y))$ for all $g\in G$. So, for all $g\in G$, $s\in S$ and each $y\in Y$,
$d^{X}(x_{sg}^{y}, \Phi_{s}(x_{g}^{y}))= d^{X}(i(\Psi_{s}\Psi_{g}(y)), \Phi_{s}(i(\Psi_{g}(y))))= d^{X}(i\circ \Psi_{s}(\Psi_{g}(y)), \Phi_{s}\circ i(\Psi_{g}(y))) < \delta.$
\medskip

Therefore by expansivity and the shadowing property, there exists a unique $\eta$-tracing point $x^{y}$ of $\lbrace x_{g}^{y}\rbrace$ for each $y\in Y$. In particular, $d^{X}(\Phi_{g}(x^{y}), x_{g}^{y}) < \frac{\mathfrak{c}}{2}$ for all $g\in G$. Define $h : Y\rightarrow X$ by $h(y) = x^{y}$ satisfying $d^{X}(\Phi_{g}(h(y)), i(\Psi_{g}(y))) \leq \eta$ for all $y\in Y$, $g\in G$.
\medskip

For $g = e$, we have $d^{X}(h(y), i(y))\leq \eta$ for all $y\in Y$. Then, for all $y_{1}, y_{2}\in Y$, we have
\begin{align*}
&|d^{X}(h(y_{1}), h(y_{2})) - d^{Y}(y_{1}, y_{2})| \\
&\leq  |d^{X}(h(y_{1}), h(y_{2})) - d^{X}(i(y_{1}), i(y_{2}))| + |d^{X}(i(y_{1}), i(y_{2})) - d^{Y}(y_{1}, y_{2})| \\
&\leq  |d^{X}(h(y_{1}), h(y_{2})) - d^{X}(h(y_{1}), i(y_{2}))|+ |d^{X}(h(y_{1}), i(y_{2})) - d^{X}(i(y_{1}), i(y_{2}))| + \delta \\
&\leq d^{X}(h(y_{2}), i(y_{2})) + d^{X}(h(y_{1}), i(y_{1})) + \delta \\
&\leq 2\eta +\delta < \epsilon
\end{align*}

Further, $d_{H}^{X}(h(Y), X)\leq d_{H}^{X}(h(Y), i(Y)) + d_{H}^{X}(i(Y), X))\leq \eta + \delta < \epsilon$. Thus, it follows that $h$ is an $\epsilon$-isometry.
\medskip

For all $y\in Y$, $d^{X}(\Phi_{g}(h(\Psi_{s}(y))), i(\Psi_{g}(\Psi_{s}(y)))) \leq \eta$, $d^{X}(\Phi_{g}(\Phi_{s}(h(y))), i(\Psi_{gs}(y))) \leq \eta$. Therefore, by expansivity of $\Phi$ we get that $\Phi_{s}\circ h = h\circ \Psi_{s}$ for all $s\in S$. In other words, we have $\Phi\circ h = h\circ \Psi$.
\medskip

Let $y\in Y$ and $h(y) = x$. Further, let $\gamma > 0$ and $n_{\gamma}(x)$ is the number in Lemma \ref{4.5} corresponding to $\gamma$. Let $\delta > 0$ be such that $d^Y(x,y)<\delta$ implies that $d^Y(\Psi_g(x),\Psi_g(y))<\frac{\eta}{8}$ for all $g\in n_{\gamma}(x)$. Then, for all $g\in G_{n_{\gamma}(x)}$ and $y'\in Y$ such that $d^{Y}(y, y') < \delta$, we get that
\medskip

$d^{X}(\Phi_{g}(h(y)), \Phi_{g}(h(y')))
\\
=  d^{X}(h(\Psi_{g}(y)), h(\Psi_{g}(y')))
\\
\leq d^{X}(h(\Psi_{g}(y)), i(\Psi_{g}(y'))) + d^{X}(i(\Psi_{g}(y)), i(\Psi_{g}(y')))+ d^{X}(i(\Psi_{g}(y')), h(\Psi_{g}(y')))
\\
\leq 2\eta + \delta + d^{Y}(\Psi_{g}(y),\Psi_{g}(y'))
\\
\leq 3\eta + \frac{\eta}{8}
\\
< \mathfrak{c}$
\medskip

This implies $d^{X}(h(y), h(y')) < \gamma$ and hence, $h$ is a continuous map. Thus, we conclude that $\Phi$ is strongly $GH$-stable.
\end{proof}

\begin{Corollary}
Every expansive action with the shadowing property on a relatively compact metric space is strongly $GH$-stable and hence, $GH$-stable.
\end{Corollary}

\begin{Example}
Let $G = \langle a, b : ba =a^{2}b \rangle$ be a solvable group with the symmetric generating set $S = \lbrace a, a^{-1}, b, b^{-1}\rbrace$ and let $F$ be an expansive homeomorphism with the shadowing property. Let $\mathfrak{c}$ be an expansivity constant for $F$. Further, assume that $F^{-1}$ is a contraction map i.e. there exists $0\leq K < 1$ such that $d(F^{-1}(x), F^{-1}(y)) \leq K d(x, y)$ for all $x, y\in X$. Let $\Phi\in Act(G,X)$ be generated by $\Phi_{a}(x) = x$ and $\Phi_{b}(x) = F(x)$ for all $x\in X$. It is then clear that $\Phi$ is an expansive action. Moreover, we claim that $\Phi$ has the shadowing property.
\medskip

Let $\epsilon>0$ be given and $2\gamma = $min$\lbrace \mathfrak{c}, \epsilon\rbrace$. Further, let $\delta > 0$ be such that every $\delta$-pseudo orbit of $F$ is $\gamma$-traced. Let $f: G\rightarrow X$ be a $\delta$-pseudo orbit of $\Phi$. Now, for each $g\in G$, $\lbrace x_{i} = f(b^{i}g)\rbrace_{i\in \mathbb{Z}}$ is a $\delta$-pseudo orbit of $F$. Let $x_{g}$ be its $\gamma$-tracing point i.e. $d(F^{i}(x_{g}),f(b^{i}g))<\gamma$. Since $d(F^{i}(x_{bg}), f(b^{i}bg)) < \gamma$, $d(F^{i}(F(x_{g})), f(b^{i+1}g)) < \gamma$, we have that $d(F^{i}(x_{bg}), F^{i}(F(x_{g})) < 2\gamma \leq \mathfrak{c}$ for all $i\in \mathbb{Z}$. By expansivity of $F$, we have $x_{bg} = F(x_{g}) = \Phi_{b}(x_{g})$ for all $g\in G$. Since $\Phi_{a}(x) = x$, $d(f(g), f(ag)) < \delta$ for all $g\in G$. Then, $b^{i}a = a^{2^{i}}b^{i}$ for $i \geq 1$ implies that $d(f(a^{2i-k-1}b^{i}g), f(a^{2i-k}b^{i}g)) < \delta$ for $0\leq k\leq 2i$. Hence, $d(f(a^{2i}b^{i}g), f(b^{i}g)) < \delta(2i)$ for all $g\in G$. Since $f(a^{2^{i}}b^{i}g) = f(b^{i}ag)$, we have $d(F^{i}(x_{ag}), f(b^{i}ag)) < \gamma$ and $d(F^{i}(x_{g}), f(b^{i}g)) < \gamma$ for all $g\in G$, $i\geq 0$. Now since, $F^{-1}$ is a contraction map, we have that $d(x_{ag}, x_{g}) \leq K^{i}d(F^{i}(x_{ag}), F^{i}(x_{g})) \leq K^{i}(2\gamma + \delta(2i))$ for all $i\geq 0$. As $i\rightarrow \infty$, right hand side of the last inequality tends to zero, which implies that $x_{ag} = x_{g}$ for all $g\in G$. Thus, we have that $x_{bg} = F(x_{g}) = \Phi_{b}(x_{g})$ and $x_{ag} = \Phi_{a}(x_{g})$ for all $g\in G$. Therefore, $d(\Phi_{g}(x_{e}), f(g)) < \epsilon$ for all $g\in G$ and hence, we conclude that $\Phi$ has the shadowing property.
\label{E4.8}
\end{Example}

\section{Sequential Shadowing Property}

Smale's spectral decomposition theorem (Theorem 3.1.11 \cite{A1}) is one of the celebrated result in topological dynamics. Existence of such a decomposition is still not proved for finitely generated group actions satisfying expansivity and the shadowing property. In this section, we have introduced another variant of the shadowing property, called sequential shadowing property which surprisingly does not need the support of expansive property to spectrally decompose the system.
\medskip

\begin{Definition}
Let $x,y\in X$ and $\delta>0$. We say that a finite sequence $\lbrace x = x_{0}, x_{1},. . ., x_{k} =y\rbrace_{i=0}^{k}$ of elements of $X$ is a weak $\delta$-chain of $\Phi\in Act(G,X)$ from $x$ to $y$ with respect to the generating set $S$  if there is a sequence $\lbrace s_{x_i}\rbrace_{i=0}^{k-1}$ of elements of $S$ such that $d(\Phi_{s_{x_i}}(x_{i}), x_{i+1}) < \delta$ for $0\leq i\leq (k-1)$. We say that $x$ is weakly $\delta$-related to $y$ (written as $x WR^{S}_{\delta} y$) with respect to $S$, if there is weak $\delta$-chains from $x$ to $y$ and from $y$ to $x$. If $x WR^{S}_{\delta} y$ for each $\delta>0$, then $x$ is said to be weakly related to $y$ (written $x WR^{S} y$) with respect to $S$. Action $\Phi\in Act(G,X)$ is said to be weak chain transitive with respect to $S$ if $x WR^{S} y$ for all $x, y\in X$.
\label{5.2}
\end{Definition}

For $\delta > 0$, $s\in S$ and $x\in X$, the sequence $\lbrace x, \Phi_{s}(x), x\rbrace$ is a weak $\delta$-chain of $\Phi\in Act(G,X)$ from $x$ to itself i.e. $x WR^{S} x$ for all $x\in X$. Thus, the set $WCR^{S}(\Phi)=\lbrace x\in X\mid x WR^{S} x\rbrace = X$ is independent of the generating set $S$. It is easy to check that the relation $WR^{S}$ is an equivalence relation on $WCR^{S}(\Phi) = X$. Thus, we can write $X = \cup_{\lambda\in \Lambda}B^{S}_{\lambda}$, where $B^{S}_{\lambda}[x]=\lbrace y : y WR^{S} x\rbrace$ is the equivalence class containing $x$.

\begin{Definition}
A sequence $\lbrace x_{i}\rbrace_{i=-\infty}^{\infty}$ of elements of $X$ is said to be $\delta$-sequential pseudo orbit (SPO) of $\Phi\in Act(G,X)$ with respect to $S$ if there is a sequence $\lbrace s_{x_i}\rbrace_{i=-\infty}^{\infty}$ of elements of $S$ such that $d(\Phi_{s_{x_i}}(x_{i}), x_{i+1}) < \delta$ for all $i\in \mathbb{Z}$. A sequence $\lbrace x_{i}\rbrace_{i=-\infty}^{\infty}$ is $\epsilon$-sequentially traced (ST) if there is $x\in X$ and a sequence $\lbrace g_{i} \rbrace _{i = -\infty}^{\infty}$ with $g_{0} = e$ and $g_i\neq e$ for all $i\neq 0$, such that $d(\Phi_{g_{i}}(x), x_{i}) < \epsilon$ for all $i\in \mathbb{Z}$. Action $\Phi$ is said to have sequential shadowing property (SSP) if for every $\epsilon>0$ there is $\delta>0$ such that every $\delta$-SPO is $\epsilon$-ST by some point in $X$.
\label{5.3}
\end{Definition}

In case of homeomorphisms SSP suggests us that numerically found trajectories with uniformly small one-step forward or backward errors at each step, can be seen through a part of real trajectory. It helps us to understand the larger class of pseudo-trajectories by studying the original trajectories.

\begin{theorem}
Let $\Phi\in Act(G,X)$. Then,
\\
(i) SSP and weak chain transitivity of $\Phi$ are independent of the generating set.
\\
(ii) SSP and weak chain transitivity of $\Phi$ are invariant under uniform conjugacy.
\label{5.4}
\end{theorem}

\begin{proof}
(i) Let $S$ and $S'$ be two generating sets of $G$. Since any two word norms are bi-Lipschitz equivalent, we can choose a constant $C \geq 1$ such that $\frac{|g|_{S'}}{C}\leq |g|_{S} \leq C|g|_{S'}$ for all $g\in G$. Let $x,y\in X$ be such that $d(\Phi_{s}(x), y) < \delta$. If $s = s'_{1}s'_{2}...s'_{C}$ for $s'_{i}\in S'$, $1 \leq i \leq C$ then, $\lbrace x, \Phi_{s'_{C}}(x), \Phi_{s'_{C-1}s'_{C}}(x),..., \Phi_{s'_{2}s'_{C-1}s'_{C}}(x), y\rbrace$ forms a weak $\delta$-chain with respect to $S'$. So, if $\lbrace x_{0}, x_{1}, ..., x_{k}\rbrace$ is weak $\delta$-chain with respect to $S$, then using the same methodology it can be extended to weak $\delta$-chain with respect to $S'$. Similarly, any $\delta$-SPO with respect to $S$ can be extended to $\delta$-SPO with respect to $S'$. Therefore, we can conclude that SSP and weak chain transitivity of $\Phi$ are independent of the generating set.

(ii) The proof of this part is easy to work out.
\end{proof}

\begin{theorem}
Let $X$ be a compact metric space. The trivial action $\Phi\in Act(G, X)$ has SSP if and only if $X$ is totally disconnected.
\label{5.9}
\end{theorem}
\begin{proof}
Suppose that $\Phi$ has SSP and $X$ has at least one non-trivial connected component $C$. Let $x$,$y$ be two distinct points in $C$ and $\epsilon= \frac{d(x, y)}{3}$. Let $\delta$ be given for $\epsilon$ by SSP of $\Phi$. Further, let $z_0 = x,$...$,z_k = y$ be a sequence of points from $x$ to $y$ so that $d(z_{j+1},z_j) < \delta$ for $j = 0,$...$,k-1$. Such a chain exists by the connectedness of $C$. Let $s\in S$ be fixed and
\[
p_{i} =
	\begin{cases}
	\Phi_{s^{i}}(x) & i < 0 \\
	\Phi_{s^{i}}(z_{i}) & 0\leq i\leq k\\
	\Phi_{s^{i}}(y) & i > k
	\end{cases}
\]
Since for $i = 0,$...$,k-1$, $d(\Phi_{s}(p_{i}), p_{i+1}) = d(\Phi_{s}\Phi_{s^{i}}(z_{i}), \Phi_{s^{i+1}}) = d(z_{i}, z_{i+1}) < \delta$, $\lbrace p_{i}\rbrace_{i=-\infty}^{\infty}$ is $\delta$-SPO. Suppose $\lbrace p_i\rbrace$ is $\epsilon$-ST by a point $p\in X$ via the sequence $\lbrace g_{i}\rbrace_{i=-\infty}^{\infty}$. Thus, by definition of $\epsilon$-ST, we get that $d(p, x) < \epsilon$ and $d(\Phi_{g_{k}}(p), \Phi_{s^{p_{k}}}(y)) = d(p, y) < \epsilon$. Then, $0 < \epsilon = \frac{d(x, y)}{3} \leq \frac{d(x, p) + d(p, y)}{3} < \frac{2\epsilon}{3}$, which is absurd. So, $X$ must be totally disconnected.
\medskip

Conversely, if $X$ is totally disconnected, then for each $\epsilon>0$ there exists a finite open cover $U_{0},..., U_{m}$ of $X$ such that $U_{i}\cap U_{j}=\phi$, for $i\neq j$ and $diam(U_{i})<\epsilon$, for each $i$, $j\in \lbrace 0 , 1 ,... , m\rbrace$.
Choose $\delta$ such that $0<\delta <$ min$\lbrace d(U_{i},U_{j}):i\neq j)\rbrace$ and choose a $\delta$-pseudo orbit $\lbrace x_{i}\rbrace_{i=-\infty}^{\infty}$ of $I$. Then, there is $U_{j}$ such that $\lbrace x_{i}\rbrace_{i=-\infty}^{\infty} \subset U_{j}$ and hence any point of $U_{i}$ can $\epsilon$-trace this orbit.
\end{proof}

\begin{theorem}
Let $\pi:Y\rightarrow X$ be a locally isometric covering map. Suppose that $\Phi\in Act(G,X)$, $\Psi\in Act(G,Y)$ satisfy $\pi\Psi = \Phi\pi$ i.e. $\pi\Psi_{g}(y) = \Phi_{g}\pi(y)$, for all $y\in Y$, $g\in G$. Further, suppose that there exists $\delta_{0} >0$ such that for each $y\in Y$ and $0<\delta <\delta_{0}$, $\pi : U_{\delta}(y)\rightarrow U_{\delta}(\pi(y))$ is an isometry. Then, $\Phi$ has SSP if and only if $\Psi$ has SSP.
\label{5.10}
\end{theorem}
\begin{proof}
The proof is similar to the proof of Theorem \ref{3.13}.
\end{proof}
\begin{theorem}
Let $\Phi\in Act(G, X)$. If $x R^S_{\delta} y$, then $x WR^{S}_{\delta} y$.
\label{5.5}
\end{theorem}
\begin{proof}
Let $f$ be $\delta$-pseudo orbit such that $f(g) = x$ and $f(h) = y$ for distinct $g,h\in G$. Assume that $g = g_{1}g_{2}$...$g_{p}$ and $h = h_{1}h_{2}$...$h_{q}$, where $g_{i}, h_{j}\in S$ for all $1\leq i\leq p$, $1\leq j\leq q$. Since $d(\Phi_{s}(f(g)), f(sg)) < \delta$ for all $s\in S$, $g\in G$, the sequence $ x_{0}= x =f(g)$, $x_{1} = f(g_{2}$...$g_{p})$, $x_{2} = f(g_{3}$...$g_{p})$,..., $x_{p-1} = f(g_{p})$, $x_{p} = f(e)$, $x_{p+1} = f(h_{q})$, $x_{p+2} = f(h_{q-1}h(q))$,..., $x_{p+q} = f(h_{1}$...$h_{q})$ forms a weak $\delta$-chain via sequence $\lbrace g_{1}^{-1}, g_{2}^{-1}$,..., $g_{p}^{-1}, h_{q}, h_{q-1}$,...,$h_{1}\rbrace$ of elements of $S$ from $x$ to $y$. Similarly, we can construct a weak $\delta$-chain from $y$ to $x$. Hence, $x WR^S_{\delta} y$.
\end{proof}

\begin{Corollary}
For $\Phi\in Act(G, X)$, transitivity implies chain transitivity implies weak chain transitivity.
\label{5.6}
\end{Corollary}
\begin{proof}
It follows from Theorem \ref{4.7}, Theorem \ref{5.5} together with \textit{Proposition 2.8} \cite{B1}.
\end{proof}

\begin{Lemma}
Let $\Phi\in Act(G, X)$, then $B^{S}_{\lambda}$ is closed and $\Phi$-invariant for each $\lambda \in \Lambda$.
\label{5.7}
\end{Lemma}
\begin{proof}
Let $\epsilon > 0$ and $\lbrace y_{j}\rbrace$ be a sequence in $B^{S}_{\lambda} = [z]$ such that $y_{j}$ converges to $y$.
Choose $m >0$ such that $d(y,y_{j})<\epsilon$ for all $j\geq m$. Since $y_{m}\in B^{S}_{\lambda}$, there exist weak $\epsilon$-chains $\lbrace y_{m} = x_{0}, x_{1}$,...,$x_{k-1}, x_{k} = z\rbrace$ and $\lbrace z = x_{0}, x_{1}$,...,$x_{k'-1}, x_{k'}=y_{m}\rbrace$. Fix $s\in S$, then $\lbrace y, \Phi_{s}(y), y_{m}, x_{1}$,...,$x_{k-1}, x_{k}=z\rbrace$ and $\lbrace z = x_{0}, x_{1}$, ...,$x_{k'-1}, x_{k'} = y_{m}, \Phi_{s}(y_{m}), y\rbrace$ are weak $\epsilon$-chains from $y$ to $z$ and from $z$ to $y$. Hence, $y\in B^{S}_{\lambda}$ and thus $B^{S}_{\lambda}$ is closed for each $\lambda\in \Lambda$.

Now, Let $s'\in S$, $\lambda\in \Lambda$, $\epsilon > 0$. Suppose that $y\in \Phi_{s'}(B^{S}_{\lambda}) = \Phi_{s'}[z]$ for some $z\in X$. Then, $y = \Phi_{s'}(x)$ for some $x\in B^{S}_{\lambda}$. Let $\lbrace x = x_{0}, x_{1}$,...,$x_{k-1}, x_{k} = z\rbrace$ and $\lbrace z = x_{0}, x_{1}$,...,$x_{k'-1}, x_{k'} = x\rbrace$ are weak $\epsilon$-chains from $x$ to $z$ and from $z$ to $x$. Then, $\lbrace y = \Phi_{s'}(x), x = x_{0}, x_{1}$,...,$x_{k-1}, x_{k} = z\rbrace$ and $\lbrace z = x_{0}, x_{1}$,...,$x_{k'-1}, x_{k'} = x, \Phi_{s'}(x) \rbrace$ are weak $\epsilon$-chains from $y$ to $z$ and from $z$ to $y$. Hence, $\Phi_{s'}(B^{S}_{\lambda})\subset B^{S}_{\lambda}$. Since $s'$ and $\lambda$ was chosen arbitrary and $S$ is symmetric, we have  $B^{S}_{\lambda}\subset \Phi_{s'}(B^{S}_{\lambda})$. Thus, $B^{S}_{\lambda}$ is $\Phi$-invariant for every $\lambda\in \Lambda$.
\end{proof}

\begin{theorem}
If $\Phi\in Act(G, X)$ has SSP, then $X = \Omega(\Phi)$.
\label{5.11}
\end{theorem}
\begin{proof}
Let $\epsilon > 0$ and let $\delta>0$ be given for $\epsilon$ by SSP of $\Phi$. Then, there exists weak $\delta$-chain $x_{0}=x, x_{1}$,...,$x_{k} = x$ with sequence $\lbrace s_{x_i}\rbrace_{i=0}^{k-1}$. We can extend this chain to $\delta$-SPO $\lbrace y_{i}\rbrace_{i=-\infty}^{\infty}$, where $y_{ki+j} = x_{j}$ for all $i\in \mathbb{Z}$, $0\leq j\leq k$ with sequence $\lbrace s_{x_i}\rbrace_{i=-\infty}^{\infty}$, where $s_{x_{ki+j}} = s_{x_j}$ for all $i\in \mathbb{Z}$, $0\leq j\leq k$. Thus, by SSP there exists $z\in X$ which $\epsilon$-ST this orbit and so we can choose non-identity element $g_{k}$ such that $d(\Phi_{g_{k}}(z), x)< \epsilon$. Hence, $\Phi_{g_{k}}(U_{\epsilon}(x))\cap U_{\epsilon}(x)\neq \phi$, where $U_{\epsilon}(x)=\lbrace y\in X : d(y,x) <\epsilon \rbrace$. Since $\epsilon$ was chosen arbitrary, we have $x\in \Omega(\Phi)$.
\end{proof}

\begin{Lemma}
If $\Phi$ has SSP, then $B^{S}_{\lambda}$ is open in $\Omega(\Phi)$ for each $\lambda$.
\label{5.8}
\end{Lemma}
\begin{proof}
We have $B^{S}_{\lambda}\subset \Omega(\Phi) = WCR^{S}(\Phi) = X$. Let us fix $x\in B^{S}_{\lambda}$ and $\delta > 0$. Choose $0 <\gamma <\delta$ such that $d(x,y)<\gamma$ implies $d(\Phi_{s}(x),\Phi_{s}(y))<\delta$ for all $s\in S$. For $y\in B^{S}_{\lambda}$ there is a $\delta$-chain $\lbrace x=x_{0},$...,$x_{k} = y\rbrace$ with sequence $\lbrace s_{x_{0}},$...,$s_{x_{k}}\rbrace$ in $\Omega(\Phi)$. It is enough to show that $U_{\gamma}(x)\subset B^{S}_{\lambda}$. Let $x'\in U_{\gamma}(x)$, then $\lbrace x', \Phi_{s_{x_{0}}}(x'), x_{0},$...,$x_{k}=y\rbrace$ is a $\delta$-chain with sequence $\lbrace s_{x_{0}}, s^{-1}_{x_{0}}, s_{x_{0}},$...,$s_{x_{k}}\rbrace$. Since $x, y\in \Omega(\Phi)$ there is a $\delta$-chain from $\lbrace y=y_{0},$...,$y_{l}=x\rbrace$. Hence, for any $s_{y_{l}}\in S$, we can construct a $\delta$-chain $\lbrace y=y_{0},$...,$y_{l}=x,\Phi_{s_{y_{l}}}(x'), x'\rbrace$.
\end{proof}

\begin{theorem}
If $\Phi\in Act(G,X)$ is weak chain transitive having SSP, then it is transitive.
\label{5.12}
\end{theorem}
\begin{proof}
The proof is similar to the proof of Theorem \ref{5.11}.
\end{proof}
\begin{Corollary}
If $\Phi\in Act(G, X)$ has SSP, then
\begin{enumerate}
\item $\Omega(\Phi)= CR(\Phi)= X$.
\item $\Phi$ is weak chain transitive if and only if $\Phi$ is transitive.
\end{enumerate}
\label{5.13}
\end{Corollary}
\begin{proof}
Proof of (1) follows from Theorem \ref{5.11} and Lemma 2.10 \cite{B1} and proof of (2) follows from Theorem \ref{5.12} and Corollary \ref{5.6}.
\end{proof}

\begin{theorem}
Let $X$ be a connected metric space. If $\Phi\in Act(G,X)$ has SSP, then it is transitive. In addition, if $X$ is compact then transitivity of $\Phi$ implies that $\Phi$ has SSP.
\label{5.14}
\end{theorem}
\begin{proof}
By Theorem \ref{5.11}, $X = \Omega(\Phi)$ which is closed and $\Phi$-invariant. We claim that $\Phi$ is weak chain transitive. Let $x$, $y$ be two distinct points in $X$ and $\epsilon > 0$. Choose $0 < \delta < \epsilon$ such that $d(a,b) < \delta$ implies that $d(\Phi_{s}(a), \Phi_{s}(b)) < \epsilon$ for all $s\in S$. By connectedness, choose a sequence of points $z_0 = x,$...$,z_k = y$ so that $d(z_{j+1},z_j) < \delta$ for all $j = 0,$...$,k-1$.
Let $s'\in S$ be fixed and observe that the sequence $z_0 = x, \Phi_{s'}(z_{1}), z_{1}, \Phi_{s'}(z_{2}), z_{2}$...$z_{k-2}, \Phi_{s'}(z_{k-1}), z_{k-1}, \Phi_{s'}(z_{k}), z_k = y$ forms weak $\epsilon$-chain from $x$ to $y$. Similarly, we can form a weak $\epsilon$-chain from $y$ to $x$. Since $\epsilon$ was chosen arbitrary, we get that $x WR y$. Thus, $\Phi$ is weak chain transitive and hence by Theorem \ref{5.12} $\Phi$ is transitive. The additional fact holds because transitivity implies existence of dense orbit.
\end{proof}

\begin{theorem}
Let $X$ be a compact metric space. If $\Phi\in Act(G,X)$ has SSP, then $X$ can be written as finite union of disjoint closed $\Phi$-invariant sets on each of which $\Phi$ is transitive. In addition, if $X$ is connected then the decomposition is trivial.
\label{5.15}
\end{theorem}
\begin{proof} By Theorem \ref{5.11}, Lemma \ref{5.8} and Lemma \ref{5.7}, $X = \cup_{i=1}^{p}B^{S}_{\lambda_{i}} = \Omega(\Phi)$ where $B^{S}_{\lambda_{i}}$ are closed and invariant, for all $1\leq i\leq p$.
For given $0 < \epsilon < min \lbrace d(B^{S}_{\lambda_{i}}, B^{S}_{\lambda_{j}}) : 1\leq i, j\leq p\rbrace$,  choose $ 0 < \delta < \epsilon$ by the SSP of $\Phi|_{\Omega(\Phi)} = \Phi$. Note that every $\delta$-SPO through $\Phi|_{B^{S}_{\lambda_{i}}}$ will be $\epsilon$-ST through a point of $B^{S}_{\lambda_{i}}$, for each $1\leq i\leq p$. Hence $\Phi|_{B^{S}_{\lambda_{i}}}$ has SSP, for each $1\leq i\leq p$. Since $\Phi|_{B^{S}_{\lambda_{i}}}$ is weakly chain transitive with respect to $S$, Theorem \ref{5.12} completes a proof of the first part. The additional part is clear from Theorem \ref{5.14}.
\end{proof}

\begin{Example} Let $G = < a, b : ba =a^{n}b>$ be a finitely generated group, $\lambda > n > 1$ and
\[
A =
\begin{bmatrix}
1 & 0 \\
1 & 1
\end{bmatrix},
B =
\begin{bmatrix}
\lambda & 0 \\
0 & n\lambda
\end{bmatrix}
\]
Define $\Phi : G\times \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ by $\Phi_{a}(x) =Ax$ and $\Phi_{b}(x) = Bx$ for all $x\in \mathbb{R}^{2}$. Note that any open disk of radius $\frac{p}{4}$ centred at point $(p, p)$, for $p > 0$ is moved vertically upward or downward under $\Phi_{a}$ and move only within the first quadrant without touching any axis under $\Phi_{b}$. Thus, under $\Phi$ such disk always move in first or fourth quadrant without touching $y$-axis. Therefore, $\Phi$ cannot be transitive. By Theorem 4.4 \cite{O2} $\Phi$ has the shadowing property and hence by Lemma 2.9 \cite{B1}, $\Phi$ is not chain transitive.
\label{5.17}
\end{Example}
\begin{Example}
Let $G = < a, b : ba =a^{2}b>$ be a finitely generated group and $m>1$ be a fixed integer. Define $\Phi : G\times \mathbb{R}\rightarrow \mathbb{R}$ by $\Phi_{a}(x) = x$ and $\Phi_{b}(x) = mx$ for fixed $m > 1$. By Example \ref{E4.8}, $\Phi$ has the shadowing property. Since $\Phi$ is not transitive, therefore by Lemma 2.9 \cite{B1} it is not chain transitive. Clearly, $\Phi$ is weakly chain transitive. Thus, weak chain transitivity need not imply transitivity or chain transitivity. Further by Theorem \ref{5.12}, $\Phi$ does not have sequential shadowing property. Therefore, we conclude that the shadowing property need not imply sequential shadowing property.
\label{5.16}
\end{Example}

Acknowledgements: The first author is supported by CSIR-Junior Research Fellowship (File No.-09/045(1558)/2018-EMR-I) of  Government of India. The second author is supported by Department of Science and Technology, Government of India, under INSPIRE Fellowship (Resgistration No-IF150210) program.

redpajama_set_name: RedPajamaArXiv


When you are young you take your legs for granted for in running, jumping, walking, swimming, and especially for dancing. They are to do their expected job and most often they do it quite well. But as we get older and end up with sedentary office jobs, for example, our diminished activity and body height means that our blood has to be pumped back up from your legs to your upper body. This by itself should not be a source of problems, but it aggravates some medical conditions that develop elsewhere in your body.
Doctors have statistically noted that most older people have some kind of disorder in the veins of their legs, ranging from mild complaints like chronically tired feeling legs, the presence of aches and pains or swollen ankles, and extending to more serious agonies like deep vein thrombosis, clots, lymphedemia, etc. In most cases like these considerable relief can be obtained by wearing graduated compression support hosiery. These give the needed support to reduce discomfort and assist in the correct circulation of blood. Nowadays these are made to look quite stylish and can even be seen as a fashion-enhancing item, though actually serving as medical appliance that provides needed support and therapy.
There are other conditions that arise that also need this type of support, such as when taking long trips or if there is edema in late pregnancy. These are not usually chronic problems but are more temporary in nature. Whatever the reason that a support stocking is required, the products offered give you the highest quality of compression treatment. Several grades of support strengths are provided and you may choose the type that is perfect for your needs. The various grades and forms of support stocking are explained in detail. The featured brand, Sigvaris, uses a special double-layer covering process that develops a therapeutical 4-way stretch and this makes them considerably more comfortable to wear, without sacrificing a very stylish look. Visit the support hosiery website to see if these will give your legs the holiday that they deserve after so many years of dependable service dancing, walking and running.


This introductory offer is for 6 FREE bottles of fine wine with the purchase of 6 for just $59.88 (that's just $4.99 per bottle).
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Every 3 months you'll receive a superb 12-bottle case selected by our expert team of buyers. You can choose an all-red, all-white or mixed case. There is no obligation to make any further purchases. You can cancel at any time, and the Vintner's Reserve Opener is yours to keep. Your satisfaction is guaranteed.
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Early Galilean Aramaic, the mother tongue of Jesus, is a language that has all but fallen into obscurity. It is perhaps one of the least understood of the ancient Aramaic dialects and is very distinct. Increasingly referred to as "Jewish Palestinian Aramaic" it is a Western dialect of Aramaic whose closest contemporary cousins were Samaritan Aramaic and Christian Palestinian Aramaic (CPA).
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Buying real estate is not only finding a place to call home. The trend of investing in real estate has been rapidly increased over the past 50 years and now, it has become a common investment vehicle. Real estate market has a number of opportunities to make big gains but actually buying and owning real estate is much more difficult and complicated than investing in bonds and stocks.
This is an old method of investment, almost as old as the practice of landownership. A person can buy any property such as a house or a shop and rent it out to a tenant. The owner of that property will be responsible for paying taxes and other maintenance costs of that property. The landlord usually charges enough rent to cover all these costs. Even after paying for these expenses like mortgage and tax the owner can still make a monthly profit out of the rent. If a person buys a rentable property at a good location he can cover the cost of buying that property within a year or two and afterwards he will make a bigger profit out of it. In the case of mortgaged property, one will have to be patient and charge enough rent to cover the expenses until the mortgage has been paid.
It is quite a wild side of real estate investment. Real estate traders are entirely different from buy and rent landlords. Actually, a real estate trader buys a property to hold it for a short period of time, which mostly consists of three to four months, and he sells it whenever there is a chance to earn a profit out of it. This is called flipping properties technique which is based on buying properties that are either undervalued or they are in a very hot market.
Real estate investors get a tool of leverage while investing their money but stock investors do not get this tool. If a person wants to buy stocks he has to pay the full value of the stock at the time he places the buy order. But that is not the case with real estate buying; most conventional mortgages require 5-25% down.?? That means you can own and control the whole property and the equity it holds through paying a small fraction of its total value. Of course you will have to pay the total value eventually but you can control the property the very moment you sign the papers.
You have looked at these benefits of investing in real estate, but you know what? You have just scratched the surface. There are countless variations of real estate investment. You have to be care full about your choice and you must weigh out the cost and benefits of your investment before diving in.


Wasn't aware it changed automatically. I saved it myself and uploaded it to my own host so I know it stays there.
I've created a new banner, that is a bit bigger.
I do as well! That's a great idea!
Sorry that it took so long. The banner is now officially online.


You really can eat in a healthy way if you watch your calories, carbs and fat when visiting fast food restaurants.
Includes items from McDonalds, Burger King, Wendy's and more.
Thanks and we'll return to order again. High quality. Rosa was great. She helped out a ton. Thanks!
Good brand in a competitive cost! It was a pleasure dealing with your company. Courteous and fast.I dealt with Rosy and she was great. Always received answers to my questions quickly.
Logo design looked great. Looks nice, plenty of color options, affordable.


I applaud you and all the hard work that has gone into a great resource for citizens journalists, academics, and really anyone who wants to be informed. I am sure you will get A LOT of suggestions about the site whether it's what to add or questioning Wikipedia as a verification tool (ppl should get over it, it's a legit starting point). My thought or suggestions is a little different. It is not lost on me HOW MUCH work you're already doing. Maybe in the future or as a side project you could add a "learn by doing" tab or something to that effect. In this tab you could post a youtube video or picture or story (really anything that has ALREADY been verified). You could then task your followers or readers to use the tools you've shared with them to verify some aspect of what you shared. Allowing them to practice what you've preached in a safe environment(one where wannabe BrownMosses don't post nonsense online using half assed techniques) and having the ability to learn, gain feedback and results through peer collaboration in forums or from contributors. Essentially, putting the users of Bellingcat in the role you have undertaken as citizen investigator. I think this would drive participation and be a great way to truly learn your techniques. You would be gamifiying the Bellingcat experience. There could be a subscription service almost like your teaching a class or you can use it to just drive traffic. I teach economics and have gamified my class and there is a greater participation and understanding compared to when I was just a lecturer. This could be something that drives major traffic( although I hope you don't need to drive traffic I hope in the brilliance of what you've done it becomes a self-sufficient entity). Thank you for all you have done and continue to do.


As you go through life, you probably hope you will never need to hire a personal injury lawyer. But the fact is that many people do at some point. That's because so many situations call for the help of this kind of legal professional. Here is just a sampling of some of the most common situations in which an attorney of this kind would be helpful.
Dog Bites If you are bitten by someone else's pet and have sustained medical bills as a result, you might want to hire a personal injury lawyer. This is especially important if the owner is not remorseful about the event and has made no attempt to help you afterward. This means he or she has not learned from the incident and will likely not discipline the animal, which indicates it could happen again. For example, if a neighbor lets a dog run around off the leash and continues to let this occur after you are bitten, it is likely that another person will be a victim, as well. A lawsuit gives you a chance to not only get compensation for your medical bills and any time off work, but also help prevent the problem from happening to someone else.
Car Accidents Car accidents often call for the assistance of a personal injury lawyer. If you were injured and have had to use your own insurance or cash to pay for medical expenses, you might be able to change this by hiring an attorney who will go after the at-fault party's insurance provider. If he or she does not have insurance, it might make sense to initiate a lawsuit against the driver in order to collect the money you need to pay for your medical costs. If you are not taken care of by the insurance company or driver responsible for the accident, then you should at least meet with an attorney to go over your options.
Slip and Fall If you go to a store and end up slipping or tripping on something there, resulting in injuries, you should contact a personal injury lawyer. As long as you can show that you were injured and it was not your fault, you might be able to get compensation. Just as with other cases, a lawsuit could ensure that the person at fault learns a lesson and becomes better at preventing trips, slips, and general injuries on the property he or she owns.
The typical personal injury lawyer offers a free or inexpensive consultation. This means that everyone should be able to afford to find out the legal options available to them. Therefore, if there is even a chance you can win your case, you should make some time to meet with an attorney.


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Lauren Comeau: Javi Was Cheating On Me With Kailyn Lowry the Whole Time We Were Together!
by Tyler Johnson at January 20, 2021 10:08 am .
When Javi Marroquin and Lauren Comeau broke up for the second time back in October, fans assumed the split would actually stick this time.
After all, Javi had cheated on Lauren again, and she found out about it in the most painful way imaginable.
During an October episode of Teen Mom 2, Kail revealed that Javi tried to have sex with her in a Wawa parking lot.
For a long time, Lauren was silent on this matter.
Comeau would throw shade on Instagram with memes about toxic people, and everyone assumed she was talking about Javi -- but she refused to comment on the situation directly.
That all changed yesterday.
In a tearful Instagram Live session, Lauren told her side of the story for the first time.
Not only did she confirm that she and Javi have broken up, she implied that Javi had been cheating on her with Kail for the entirety of their three-year relationship.
"It's just really f--king upsetting that two people can still be ok with causing people this much pain," she told the camera.
"It's really f--king upsetting. I've had my fair share of betrayal and pain my whole life but what I learned today, from the kids nonetheless, just makes me feel like the last three years they've just been waiting for this," Comeau continued.
"I've always known people have been against me and Javi, people have wanted to see us break up our whole relationship and now that it's finally here, it's just like, where did those three years go?" she said.
"People can move on with their life, their relationship gets better and I'm just here to pick up the pieces with my kid, stuck here [in Delaware] with no family or friends."
Comeau concluded the video by calling the viewer's attention to "proof" of Kail and Javi's affair.
"If anyone sees Javi driving around in a black Suburban, you'll know why I'm crying today," she said.
Yes, according to Lauren, not only did Javi try to sleep with Kail that faithful day at the Wawa -- the two of them have been hooking up on the down-low for YEARS!
That's a problem, of course, as Javi was engaged to Lauren for much of that time.
And apparently, Kail bought him a new truck as a reward for all that free D!
Yes, Lauren says she was informed by Javi's eldest son that Kail and Javi spent the day together at a car dealership and pics from their Instagram pages (above) seem to corroborate that story.
The response to Lauren's video was a swift and harsh backlash against both Javi and Kail.
"Damn I would hate to be stuck in another city with no friends and family," wrote Briana DeJesus, another ex of Javi's.
"Thank god I dodged that bullet lol f--k datttttt."
In the hours since Lauren's video went live, fans and media outlets have been scrambling to verify Lauren's story.
A source tells The Ashley's Reality Roundup that Kail did not buy Javi a car, he simply purchased her old truck.
"He had a white SUV that he traded into the dealership in order to get the black Suburban Kail had recently traded into the dealership," the source said.
"Lauren feels like Javi driving a car that used to belong to Kail is a slap in the face to her. But what's funny is the white SUV that she and Javi both use to drive was also Kail's car originally," the insider said.
"Lauren used to drive it around, knowing it was Kail's old car and it never bothered her."
Javi Marroquin: I Did NOT Cheat on Lauren Comeau With Kailyn Lowry! [UPDATED]
Start Gallery
Apparently, Lauren has taken all this wheeling and dealing as evidence of an affair.
"She just feels like Javi buying Kail's car is a sign that they're still hooking up, and have been behind her back for years," The Ashley's source said.
Kail has remained silent on the situation, but Javi offered his side of the story in an interview with In Touch.
"While I have tried to avoid a he said/she said situation [by] staying quiet, the accusations Lauren made against me this afternoon on Instagram are false," Javi told the tabloid.
"My only focus right now is creating healthy coparenting relationships with both Lauren and Kail for the sake of my boys and bettering myself."
Kail has not responded to the situation personally, but rep emailed The Ashley with a denial of any wrongdoing:
"Kail is disappointed that a business arrangement over vehicles today was blown out of proportion," the rep wrote.
"All [insinuations] made against her on [Lauren's] live were false."
Needless to say, there's a lot going on here, and it seems that the drama is just beginning.
We'll continue to keep you updated on this ongoing story as more information becomes available.
Tags: Kailyn Lowry, Javi Marroquin, Lauren Comeau
Lauren Comeau: Javi Was Cheating On Me With Kailyn For YEARS!

redpajama_set_name: RedPajamaCommonCrawl


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HOME EVENTS & ENTERTAINING
How to Start an After School Program in My Church
By: Kevin M. Jackson
Difference Between Profit & Non Profit Day Care Centers
Health & Safety Policy for Child Care Settings
What Are the Duties of a Church Trustee Board?
Federal Government Grants for Child Day Care Centers
How to Start a Mother's Morning Out Program
children playing image by L. Shat from Fotolia.com
After school programs assist parents and communities in providing children activities, tutoring and childcare during the hours of the day that have been deemed the most dangerous time for children. Children need places that will occupy their time with meaningful experiences to prevent them from getting involved in gangs or violence, or other kinds of trouble. Community organizations committed to serving children impact the future of our communities, cities and, ultimately, the nation.
Meet with the pastor of the church and determine his willingness to support an after school program. This program will require financial, spiritual and manpower support. It is important that the pastor and the congregation buys into the program for it to be successful.
Brainstorm to determine the focus of the program. Consideration should be given to educational goals, spiritual goals, nutritional goals and physical fitness goals. All are significant components of most after school programs.
Determine what population of children the program will target. Determine how many children the program can effectively support.
Find out from the church administrator what type of insurance the after care program will need to cover incidents and accidents during the duration of the program.
Write down the vision statement, mission statement and program goals. Ensure that the pastor is in agreement then put each in writing on official letterhead to present to potential volunteer or paid staff workers.
Create program rules and regulations for the after care program participants and the adult workers.
Advertise within the church community the need for volunteer or paid workers.
Submit an official background check for each adult worker. Gently inform persons who are not eligible.
Select adult workers who will work with the after school program. Provide the workers with the official vision statement, mission statement and program goals.
Advertise the after care program enrollment to the target audience through word of mouth, church bulletin, radio announcements and the newspaper.
Determine enrollment cut-off date and establish a day to meet with the program participants' parents and guardians to review policies, rules and regulations.
Fun Family Education: After care activities
Based in Virginia, Kevin M. Jackson has been writing professionally since 2003. He is the author of the books "Life Lessons for My Sons" and "When GOD Speaks." Jackson holds a Bachelor of Science in biology from Savannah State University and a Master of Arts in urban education from Norfolk State University.
Copyright © 2019 Leaf Group Ltd., all rights reserved. // Leaf Group Lifestyle


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FYZICAL Therapy & Balance Centers:
FYZICAL Therapy & Balance Centers Is Now an Accredited Orthopedic Residency Program
By: FYZICAL Therapy & Balance Centers | 0 Shares 80 Reads
The accreditation means the physical therapy franchise can officially offer a clinical pathway for Physical Therapists to become certified Orthopedic Clinical Specialists.
SARASOTA, FLA. (PRWEB) February 09, 2021 - FYZICAL Therapy & Balance Centers now offers an orthopedic residency program that has been accredited by the American Board of Physical Therapy Residency and Fellowship Education (ABPTRFE).
The accreditation means that FYZICAL now offers a clinical pathway for its Physical Therapists to undergo advanced training and become certified Orthopedic Clinical Specialists.
Since 1996, only 322 residency programs have earned accreditation from ABPTRFE, putting FYZICAL in an elite category that is even smaller when accounting only for orthopedic residency programs. That means that for Physical Therapists hoping to become certified Orthopedic Clinical Specialists, FYZICAL is among the best and most accessible options.
The accreditation also supports FYZICAL's mission to provide the highest level of care for patients and reduce practice variability across locations.
"By offering an accredited residency program, we are further fulfilling our commitment to excellence across our franchise system," said Dr. Gavin Hamer, FYZICAL's national director of education. "We now have a new resource to attract Physical Therapists who are committed to life-long learning and evidence-based practice. In turn, those Therapists will provide the best possible care for patients, improving their outcomes."
FYZICAL's accreditation as an orthopedic residency was years in the making. In 2016, Dr. Rick Douglass, FYZICAL's president of operations, proposed his initial vision for the residency program and submitted an application to the ABPTRFE. It was two years before the application for candidacy was accepted and another two years - plus lots of education, programming and paperwork - before the board conducted a site visit at FYZICAL's Bonita Springs, Florida location and approved the program, making FYZICAL the 117th physical therapy orthopedic residency in the United States.
"We started with a vision to become the country's best orthopedic residency program, and then committed to the extensive application process," Dr. Douglass said. "It was a serious undertaking. There is so much education to be done, and it's a significant financial commitment. That's why there are so few programs. We had to prove that we have the resources and knowledge to support the program. When the board conducted its site visit, they were impressed by how well equipped we are, and they are eager for us to expand the program across our system."
Currently, FYZICAL's residency program is offered at the franchise's Bonita Springs location. FYZICAL plans to roll out its residency program across its system.
SOURCE FYZICAL Therapy & Balance Centers
Lauren Turner
No Limit Agency
Lauren@nolimitagency.com
$1M+ Avg. Revenue per Location. No medical experience necessary. Manage the manager franchise. Comprehensive training and support. Unique business model to help achieve peak profitability while simultaneously providing exceptional patient care.


What can be said....This design has been probably one of the best selling large dome tents available in the United Kingdom for the last 3 years. As a result Sunncamp decided to make the tent available in a Polyester/Cotton version known as Sunntex. This gives the tent a longer life as well as being cooler as this material can breathe.
The tent has a large front porch area with an groundsheet supplied and two separate zipped sleeping compartments with a ground sheet between the two areas, so if you wish or need this can be made up to a 6 berth dome tent which is tall enough for the average person to stand up in the centre of the tent.


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"The Secret" is the thirteenth episode of the second season of the American comedy television series The Office, and the show's nineteenth episode overall. It was written by Lee Eisenberg and Gene Stupnitsky and directed by Dennie Gordon. The episode first aired on January 19, 2006 on NBC.

The series depicts the everyday lives of office employees in the Scranton, Pennsylvania branch of the fictional Dunder Mifflin Paper Company.  In this episode, Jim Halpert (John Krasinski) is forced into spending time with Michael Scott (Steve Carell) so that Michael will not reveal Jim's feelings for Pam Beesly (Jenna Fischer). Meanwhile, Oscar Martinez (Oscar Nunez) takes a "sick day" and Dwight Schrute (Rainn Wilson) investigates whether he is actually sick.

"The Secret" was written in roughly 26 hours and was the fastest episode written for the series at the time. The title for the episode is purposely vague and refers to Jim's hidden feelings for Pam, Dwight and Angela's relationship, and Oscar's homosexuality. "The Secret" is also the first episode where many of the actors not credited in the opening sequence are credited directly afterwards instead of as guest appearances. The scenes at the Hooters restaurant were filmed relatively early in the day, and a majority of the scenes were improvised by Carell. "The Secret" received largely positive reviews from television critics and was watched by 8.7 million viewers, ranking as the forty-fourth most-watched television episode of the week.

Plot
Jim Halpert (John Krasinski) realizes that Michael Scott (Steve Carell) mistakenly assumed he was making his feelings for Pam Beesly (Jenna Fischer) known to everyone. He tells Michael that he meant this to be a secret between the two of them. Michael concludes from this that he and Jim are friends, which leads to an awkward lunch at Hooters paid for with a corporate credit card. Michael ultimately reveals Jim's secret to everyone, forcing Jim to confess his crush to Pam himself, although he tells her that he got over it three years ago. However, Michael later tells her that he learned of the crush during the recent "booze cruise", leading her to suspect Jim is still infatuated.

Meanwhile, Dwight Schrute (Rainn Wilson) "investigates" Oscar Martinez's (Oscar Nunez) claimed sick day from work, learning that the co-worker is actually taking time off to ice-skate. Dwight blackmails Oscar, saying he owes him a favor in return for not revealing his unauthorized leave-taking. He then watches a movie with Oscar and Oscar's "roommate" Gil (Tom W. Chick).

Production

Writing
"The Secret" was written by Lee Eisenberg and Gene Stupnitsky. The episode was written in roughly 26 hours and was the fastest episode written for the series, at the time. The idea to make the entry the "spring cleaning episode" was "throw[n] in at the last minute" because the writers were "desperate". The subplot was inspired by a friend of Eisenberg's, who suggested the story after he was forced to undergo spring cleaning at his office.

The title for the episode is purposely vague. During the commentary, the cast revealed that there are actually three "secrets" in the episode: Jim's hidden feelings for Pam, Dwight and Angela's relationship, and Oscar's homosexuality. Stupkitsky later joked that there are "actually nine secrets in the episode". The cold opening – wherein Michael tries to tell a joke where the punch line is "What's up dog?" – was inspired by the same prank being pulled on Stupnitsky. The original cold opening for the episode featured Michael playing golf in his office, but it was cut.

Jenna Fischer felt that Krasinski's performance in the episode was "really great" and called "The Secret" her "favorite John Krasinski episode". Krasinski said that Creed Bratton's line, "Which one is Pam?" was his favorite moment "in the entire show, ever". Fischer, in turn, said that her favorite moment was when Jim confesses that he does not have a crush on her, due to the emotion involved. Fischer later admitted that, after the shot was filmed, she cried because it "broke [her] heart".

Filming

"The Secret" was directed by Dennie Gordon, who would go on to direct the season two episode "Boys and Girls". Tom W. Chick portrayed Gil. The cast were particularly impressed with his acting; Fischer called him a "great casting" choice. Lindsey Stoddart, who plays one of the Hooters waitresses, was an improv acquaintance of Martin's.

The shots at the Hooters restaurant were filmed relatively early in the day, and a majority of the scenes were improvised by Carell. In fact, Carell decided to play the penis game – in which two or more people shout the word "penis" at varying levels in an attempt to out do each other – which Krasinski called "the craziest thing I have experienced". The scene where Pam and Kelly discuss wedding plans was particularly difficult to shoot all in one scene, due to the amount of activity going on. John Krasinski later noted that he was impressed that the director and cameramen were able to successfully get the shot. The sequence featuring Michael and Jim talking in the break room was made of two composition shots, because Krasinski was laughing "the entire" time. Michael buying Stanley a peach ice tea and telling him that he will "hate it" was entirely improvised.

Deleted scenes
The Season Two DVD contains a number of deleted scenes from this episode. Notable cut scenes include the cut cold opening of Michael playing with his new putting toy, Dwight expounding on his thoughts on dust bunnies, Michael surveying his "worker bees", Ryan Howard (B. J. Novak) finding an unfinished People Magazine crossword puzzle from 1999 in Michael's office, Michael describing his college fraternity experience, and Michael buying Jim a Hooters T-shirt.

Cultural references
Michael attributes the 1981 song "Our Lips Are Sealed" to The Bangles, when it was really sung by The Go-Go's. Michael and Jim go to Hooters, a company whose waiting staff are primarily young, attractive waitresses usually referred to simply as "Hooter Girls" whose revealing outfits and sex appeal is played up and is a primary component of the company's image. At the restaurant, Michael makes several breast jokes. Near the end of the episode, Michael makes reference to a nonexistent Cinemax movie called More Secrets of a Call Girl.

Reception
"The Secret" originally aired on NBC on January 19, 2006. The episode was watched by 8.7 million viewers. This marked a slight decrease in viewers from a lead-in episode of "My Name is Earl", but more than a follow-up episode of "ER". "The Secret" ranked as the forty-fourth most-watched episode for the week ending January 22.

M. Giant of Television Without Pity awarded the episode an "A". Brendan Babish of DVD Verdict gave the episode a moderately positive review and awarded the entry a "B". He wrote that while it was ""a solid episode" it "lacks any of the belly laughs the show frequently elicits". Michael Sciannamea of AOLTV called the installment "a terrific episode" and wrote that "Michael Scott […] was at his obnoxious best […] in this episode". Furthermore, he highly praised the story, noting that "the Jim/Pam scenario has definitely taken a more interesting turn". During the filming of "The Secret", the cast of the show discovered that Carell had been nominated for a Golden Globe Award. Fischer later noted that it was "fun that ['The Secret' is] the episode that airs after his win".

References

External links
"The Secret" at NBC.com

2006 American television episodes
American LGBT-related television episodes
The Office (American season 2) episodes

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