\documentclass[reqno, 12pt]{amsart} \usepackage{amsmath,amssymb,amsthm} \usepackage{mathabx} \usepackage{mathrsfs} \usepackage[backref,colorlinks]{hyperref} \usepackage[abbrev,msc-links]{amsrefs} \usepackage{enumerate} \usepackage{enumitem} \usepackage[symbol]{footmisc} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%% Page Layout %\parindent=0pt \linespread{1.3} \normalsize \setlength{\hoffset}{-1.6cm} \addtolength{\textwidth}{4cm} \addtolength{\voffset}{-0.5cm} \addtolength{\textheight}{1cm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newtheorem{theorem}{Theorem} \newtheorem{definition}[theorem]{Definition} \newtheorem{lem}[theorem]{Lemma} \newtheorem{claim}[theorem]{Claim} \newtheorem{subclaim}[theorem]{Subclaim} \newtheorem{fact}[theorem]{Fact} \newtheorem{remark}[theorem]{Remark} \newtheorem*{cl}{\textbf{Container Lemma}} \def\cB{\mathcal{B}} \def\cD{\mathcal{D}} \def\cH{\mathcal{H}} \def\eps{\epsilon} \def\dHH#1{\leavevmode\setbox0=\hbox{#1}\dimen0=\wd0\setbox0=\hbox{.}% \advance\dimen0 by -\wd0% \hbox{#1\raise-0.5ex\hbox to 0pt{\hss.\kern.5\dimen0}}}% \newcommand{\xx}{{\mathcal{X}}} \newcommand{\yy}{{\mathcal{Y}}} \newcommand{\zz}{\mathcal{Z}} \renewcommand{\aa}{\mathcal{A}} \newcommand{\bb}{\mathcal{B}} \newcommand{\cc}{\mathcal{C}} \renewcommand{\thefootnote}{\fnsymbol{footnote}} \newcommand{\lr}{\Leftrightarrow} \begin{document} \title[Note on Ramsey theorem for posets]{Note on Ramsey theorem for posets with linear extensions} \author[A.Arman]{Andrii Arman} \address{Department of mathematics, University of Manitoba, Winnipeg, Manitoba R3T 2N2, Canada} \email{armana@myumanitoba.ca} \thanks{} \author[Vojt\v{e}ch R\"{o}dl]{Vojt\v{e}ch R\"{o}dl} \address{Department of Mathematics and Computer Science, Emory University, Atlanta, GA 30322, USA} \email{rodl@mathcs.emory.edu} %\thanks{The second author was supported by NSF %grant DMS 1301698.?????} \thanks{The second author was supported by NSF grant DMS 1301698} \keywords{Ramsey theorem, posets} \subjclass[2010]{05C55 (primary), 06A07 (secondary)} \date{\today} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%% ABSTRACT %%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{abstract} In this note we consider a Ramsey type result for partially ordered sets. In particular we give an alternative short proof of a theorem for a posets with multiple linear extensions recently obtained by Solecki and Zhao in \cite{SZ}. \end{abstract} \maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%% INTRODUCTION %%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \iffalse \section{Introduction} Ramsey theorem for posets with one linear extension was proved in \cite{NR}, \cite{PTW}, see also \cite{fouche}. Recently, Solecki and Zhao in \cite{SZ} proved Ramsey theorem for posets with multiple linear extensions. Their proof was based on Ramsey theorem for rigid surjection, which was earlier proved by Solecki. In this note, we present a different proof of Solecki-Zhao result. Our proof of this Ramsey type theorem for posets with multiple linear extensions is based on Ramsey type theorem for posets with single linear extension. Our main result is Theorem \ref{thm:main}. \fi \section{Preliminary definitions} A poset is a pair $(X,P^{X})$, where $X$ is a set and $P^{X}$ is a partial order on $X$. We consider partial orders that are strict, i.e. not reflexive. We say that a partial order $L^{X}$ on $X$ \textit{extends} a partial order $P^{X}$ on $X$ if for all $x, y \in X$ $$x P^{X} y \Rightarrow x L^{X} y.$$ If $(X,P^{X})$ is a poset and $U \subset X$ we denote by $P^{X}|_{U}$ the \textit{restriction} of $P^{X}$ onto $U$. Below, we consider collections $\mathcal{L}^{X}_k=(L_1^{X}, L_2^{X}, \dots, L_k^{X})$ ,where each of $L_{i}^{X}$ is a linear order on $X$. \begin{definition} We denote by $PL^{(k)}$ the set consisting of all triplets $(X, P^{X}, \mathcal{L}^{X}_k)$, where $(X,P^{X})$ is a poset and each $L_{i}^{X}$ for $ i \in [k]$ is a linear order that extends $P^{X}$. \end{definition} \begin{definition} Let $\xx , \yy \in PL^{(k)}$, where $\xx=(X, P^{X}, \mathcal{L}^{X}_k)$ and $\yy=(Y, P^{Y}, \mathcal{L}^{Y}_k)$. We write $\xx \subseteq \yy$ if \begin{itemize} \item $X\subseteq Y$ and $P^{Y}|_X$ extends $P^{X}$. \item $L^{Y}_{i}|_X=L^{X}_{i}$ for all $i\in [k]$. \end{itemize} \end{definition} \begin{definition} Let $\xx , \yy \in PL^{(k)}$, where $\xx=(X, P^{X}, \mathcal{L}^{X}_k)$ and $\yy=(Y, P^{Y}, \mathcal{L}^{Y}_k)$. We say that a mapping $\pi : X \to Y$ is order preserving for $\xx$ and $\yy$ if for any $i \in [k]$ and any $x,y \in X$ we have $$ x L_{i}^X y \lr \pi(x)L_{i}^{Y}\pi(y) \; \; \;\text{and} \; \; \; x P^{X}y \lr \pi(x)P^{Y}\pi(y) .$$ \end{definition} \begin{definition} \label{def:isomor} We say that $\pi$ is an isomorphism between $\xx \in PL^{(k)}$ and $\tilde{\xx} \in PL^{(k)}$ if it is order preserving bijection. We say that $\xx \in PL^{(k)}$ is isomorphic to $\tilde{\xx} \in PL^{(k)}$ if there is an isomorphism between $\xx$ and $\tilde{\xx}$. \end{definition} \begin{definition}\label{def:copy} Let $k > 0$ and $\xx, \yy \in PL^{(k)}$. We say that $\tilde{\xx} \in PL^{(k)}$ is a copy of $\xx$ in $\yy$ if $\tilde{\xx} \subseteq \yy$ and $\tilde{\xx}$ is isomorphic to $\xx$. For $\xx, \yy \in PL^{(k)}$ denote by $\binom{\yy}{\xx}$ the set of all copies of $\xx$ in $\yy$. \end{definition} For any $\tilde{\xx} \in \binom{\yy}{\xx}$ there is unique order preserving mapping $\pi: X \to \tilde{X}$. On other hand, any order preserving mapping $\pi : X \to Y$ induces a copy $\tilde{\xx}=\pi(\xx) \in \binom{\yy}{\xx}$. We identify each $\tilde{\xx} \in \binom{\yy}{\xx}$ with corresponding order preserving mapping $\pi$ and will say that $\pi$ is a copy of $\xx$ in $\yy$ instead of saying that $\tilde{\xx}$ is a copy of $\xx$ in $\yy$ with corresponding order preserving mapping $\pi$. The following theorem follows from the result of \cite{NR} (see \cite{Nesetril} and \cite{NR2}). Different proof of Theorem~\ref{thm:one_ext} was also given by Soki\'c~\cite{Sokic1} (using results of \cite{PTW} and \cite{Fouche}). \begin{theorem}\label{thm:one_ext} For any integer $r$ and any $\xx, \yy \in PL^{(1)} $ there is $\zz \in PL^{(1)}$, such that for any $r$-colouring of set $\binom{\zz}{\xx}$ there is $\tilde{\yy}$, a copy of $\yy$ in $\zz$, such that $\binom{\tilde{\yy}}{\xx}$ is monochromatic. \end{theorem} %Ramsey properties of the class of partially ordered sets were considered in \cite{NR} and \cite{PTW}, where all partially ordered sets with P-Ramsey properties were characterised (see also \cite{NR2}). Subsequently some extensions and related results were obtained in \cite{Promel} and \cite{Fouche}, using different methods. %Ramsey properties of the class of partially ordered sets were considered in \cite{NR} and \cite{PTW}, where all partially ordered sets with P-Ramsey properties were characterised (see also \cite{NR2}). Subsequently some extensions and related results were obtained in \cite{Promel} and \cite{Fouche}, using different methods. Next theorem is a product version of the Theorem \ref{thm:one_ext}, that we are going to use in Section \ref{sec:proof}. Proof of this theorem is based on a standard folkloristic product argument. For similar results of this type see e.g. \cite{Promel}. \begin{theorem}\label{thm:product} For any $\xx_i, \yy_{i}\in PL^{(1)}$ with $i\in [k]$ there are $\zz_i \in PL^{(1)}$ with $i \in [k]$, such that for any 2-colouring of set $\binom{\zz_1}{\xx_1} \times \dots \times \binom{\zz_k}{\xx_k}$ there are $\tilde{\yy_i}$, a copies of $\yy_i$ in $\zz_i$ for $i \in [k]$, such that $\binom{\tilde{\yy_1}}{\xx_1} \times \dots \times \binom{\tilde{\yy_k}}{\xx_k}$ is monochromatic. \end{theorem} Based on Theorem \ref{thm:product}, in Section \ref{sec:proof} we provide a proof of the following result, first obtained in \cite{SZ} . \begin{theorem} \label{thm:main} For any integer $k$ any $\mathcal{A}, \bb \in PL^{(k)} $ there is $\cc \in PL^{(k)}$, such that for any colouring $2$-colouring of set $\binom{\cc}{\mathcal{A}}$ there is $\tilde{\bb}$, a copy of $\bb$ in $\cc$, such that $\binom{\tilde{\bb}}{\mathcal{A}}$ is monochromatic. \end{theorem} To distinguish between the objects of $PL^{(1)}$, which play a special role in our proof, and $PL^{(k)}$ for $k\geq 2$, from now on, we use letters $\xx$, $\yy$ and $\zz$ for elements of $PL^{(1)}$ and $\aa$, $\bb$, $\cc$ for elements of $PL^{(k)}$. For the ease of notation we will give a proof of Theorem \ref{thm:main} for case $k=2$. The proof of the general case follows the same lines (and is accessible on arxiv.org). \section{Properties of join and canonical copies} First, we define the join of two elements of $PL^{(1)}$. \begin{definition}\label{def:join} Let $\zz_{i}=(Z_i, P^{Z_i}, L^{Z_i})\in PL^{(1)}$ for $i =1,2$. Define $\cc=\zz_1 \sqcup \zz_2$ by $$\cc=(Z_1 \times Z_2, <_\cc, <_{lx_1}, <_{lx_2}),$$ where $Z_1 \times Z_2$ is Cartesian product of sets $Z_1$ and $Z_2$ , $<_\cc$ is a partial order and $<_{lx_1}, <_{lx_2} $ are linear orders on $Z_1 \times Z_2$ defined by: $$(x_1,x_2)<_\cc(y_1, y_2) \lr x_1P^{Z_1}y_1 \; \; \text{and} \; \; x_2P^{Z_2}y_2 \;, $$ $$(x_1,x_2)<_{lx_1}(y_1, y_2) \lr x_1L^{Z_1}y_1 \; \; \text{or} \; \; \{ x_1=y_1 \; \text{and} \; x_{2}L^{Z_{2}}y_{2} \; \} , $$ $$(x_1,x_2)<_{lx_2}(y_1, y_2) \lr x_2L^{Z_2}y_2 \; \; \text{or} \; \; \{ x_2=y_2 \; \text{and} \; x_{1}L^{Z_{1}}y_{1} \; \} . $$ We say that $\zz_1 \sqcup \zz_2$ is the join of $\zz_1$ and $\zz_2$. \end{definition} Note, that for $\zz_1, \zz_2 \in PL^{(1)}$ we have that $\zz_{1}\sqcup \zz_{2} \in PL^{(2)}$. Indeed, since both $L^{Z_{i}}$ extend $P^{Z_i}$ we infer that both $<_{lx_i}$ also extend $<_\cc$ for $i=1,2$. \begin{claim}\label{claim:1} Let $\zz_{i}=(Z_i, P^{Z_i}, L^{Z_i})\in PL^{(1)}$ for $i =1,2 $ and let $\bb=(Y, P^{Y}, {L}^{Y}_{1}, L^{Y}_2) \in PL^{(2)}$. Set $\cc=\zz_1 \sqcup \zz_2$ and let $\pi_{i} : Y \to Z_{i}$ be a copy of $\yy_{i}=(Y, P^{Y}, L_{i}^{Y})$ in $\zz_{i}$ for $i =1,2$. Then the image of the mapping $\pi : Y \to Z_1 \times Z_2$, defined by $$\pi(y)=(\pi_1(y), \pi_2(y))$$ for each $y \in Y$, is a copy of $\mathcal{B}$ in $\binom{\cc}{\mathcal{B}}$. \end{claim} \begin{remark}\label{remark:1}$ $ \begin{itemize} \item We say that the image of the mapping $\pi$ from Claim \ref{claim:1}, is a \textit{canonical} copy of $\mathcal{\bb}$ in $\cc=\zz_1 \sqcup \zz_2$. \item By $\binom{\cc}{\mathcal{B}}_{can} \subseteq \binom{\cc}{\mathcal{B}}$ we denote a set of all canonical copies of $\mathcal{B}$ in $\cc$. \end{itemize} \end{remark} \begin{proof} We need to verify that $\pi: Y \to Z_1 \times Z_2$ is order preserving for $\aa$ and $\cc$. Indeed, we observe that if $x,y \in Y$, then fact that $\pi_i: Y \to Z_{i}$ preserves $P^{Y}$ for $i =1,2$ combined with definition of $\zz_1 \sqcup \zz_2$ yields $$xP^Yy \lr \begin{array}{l} \pi_1(x) P^{Z_1} \pi_1(y),\\ \pi_2(x) P^{Z_2} \pi_2(y) \end{array} \lr \pi(x)<_\cc \pi(y).$$ Since $\pi_i$ preserves $L^{Y}_i$ for $i=1,2 \;$ , we have $$ xL_i^Yy \lr \pi_i(x)L_i^{Z_i}\pi_i(y) \lr \pi(x)<_{lx_{i}}\pi(y)$$ for $i =1,2$. Hence, $\pi$ preserves $P^{Y}$ and $L^{Y}_i$ for $i =1,2$. \end{proof} For the rest of this section we assume that $\cc=\zz_1 \sqcup \zz_2=(Z_1 \times Z_2, <_\cc, <_{lx_1}, <_{lx_2})$, $\aa=(X, P^{X}, L_1^{X}, L_2^{X})$ and $\bb=(Y, P^{Y}, L_1^{Y}, L_2^{Y})$. \begin{fact}\label{fact:lambda} By construction, $\binom{\cc}{\mathcal{A}}_{can}$ is in 1-1 correspondence with the set $\binom{\zz_1}{\xx_1} \times \binom{\zz_2}{\xx_2}$ and the function $\lambda : (\pi_1(X), \pi_2(X)) \mapsto \pi(X)$ is the bijection between sets $\binom{\zz_1}{\xx_1} \times \binom{\zz_2}{\xx_2}$ and $\binom{\cc}{\mathcal{A}}_{can}$. \end{fact} %For a mapping $\pi$ that is an order preserving mapping for $\bb$ %and $\cc$ denote by $\pi(\bb)$ the copy of $\bb$ in $\cc$ induced by mapping $\pi$. The following Claim states that if $\pi$ is a canonical copy of $\bb$ in $\cc$ and $\tilde{\mathcal{A}}$ is a copy of $\mathcal{A}$ in $\bb$, then $\pi(\tilde{\mathcal{A}})$ is a canonical copy of $\mathcal{A}$ in $\cc$. \begin{claim}\label{claim:2} If $\pi \in \binom{\cc}{\bb}_{can}$ and $\tau \in \binom{\bb}{\aa}$, then $\sigma=\pi \circ \tau \in \binom{\cc}{\aa}_{can}$. \end{claim} \begin{proof} %Let $\bb=(Y, P^{Y}, {L}^{Y}_{1}, {L}^{Y}_{1})$. Since $\pi: Y \to Z_1 \times Z_2$ is a canonical copy, we have that $\pi=(\pi_1, \pi_2),$ where $\pi_i : Y \to Z_i$ are copies of $Y$ in $Z_{i}$ for $i=1,2$. Define $\sigma_{i}=\pi_{i}\circ \tau$ for $i=1,2$. It is sufficient to prove that for $i=1,2$ $\sigma_{i}$ is order preserving for $\xx$ and $\zz_{i}$. Indeed, since $\tau$ preserves $P^{X}, L^{X}_1, L^{X}_2$ and that $\pi_{i}$ preserves $P^{Y}, L^{Y}_i$ for $i=1,2$, we have for any $x,y \in X$ and for $i=1,2$ $$xP^Xy \lr \tau(x)P^{Y}\tau(y) \lr \pi_{i}(\tau(x))P^{Z_{i}}\pi_{i}(\tau(y))\lr \sigma_{i}(x)P^{Z_{i}}\sigma_{i}(y),$$ $$xL^X_{i}y \lr \tau(x)L^{Y}_{i}\tau(y) \lr \pi_{i}(\tau(x))L^{Z_{i}}\pi_{i}(\tau(y))\lr \sigma_{i}(x)L^{Z_{i}}\sigma_{i}(y).$$ Consequently, for $i=1,2$ , $\sigma_{i}$ is order preserving for $\xx$ and $\zz_{i}$, and $\sigma=(\sigma_1, \sigma_2)$ is a canonical copy of $\aa$ in $\cc$. \end{proof} Our final Claim states that if $\tilde \bb$ is a canonical copy of $\bb$ in $\cc$, and $\tilde{\mathcal{A}}$ is a copy of $\mathcal{A}$ in $\tilde \bb$, then $\tilde{\mathcal{A}}$ is a canonical copy of $\mathcal{A}$ in $\cc$. \begin{claim}\label{claim:3} If $\pi \in \binom{\cc}{\bb}_{can}$ and $\sigma \in \binom{\pi(\bb)}{\aa}$, then $\sigma \in \binom{\cc}{\aa}_{can}$. \end{claim} \begin{proof} Since $\pi$ is an isomorphism between $\bb$ and $\pi(\bb)$, the mapping $\pi^{-1}$ exists and is order preserving for $\pi(\bb)$ and $\bb$. Since $\sigma: \aa \to \sigma(\aa) \subseteq \pi(\bb)$ and $\pi^{-1}: \pi(\bb) \to \bb$, the mapping $\tau=\pi^{-1} \circ \sigma: \aa \to \bb$ is well defined. Moreover, $\sigma$ and $\pi^{-1}$ are order preserving, so is also $\tau$. Finally, Claim \ref{claim:2} applied for $\pi$ and $\tau$ gives that $\pi \circ \tau =\sigma$ is a canonical copy of $\mathcal{A}$. \end{proof} \section{Proof of Theorem \ref{thm:main}}\label{sec:proof} Let $\mathcal{A}=(X, P^{X}, L_{1}^{X}, L_{2}^{X})$ and $\bb=(Y,P^{Y}, L_{1}^Y, L_{2}^Y)$ be given. Applying Theorem \ref{thm:product} with $\xx_{i}=(X,P^{X}, L_{i}^X)$ for $i=1,2$ and $\yy_i=(Y,P^{Y}, L_{i}^Y)$ for $i=1,2$ we obtain $\zz_{i}=(Z_{i},P^{Z_{i}}, L_{i}^{Z_{i}})$ for $i=1,2$ . Set $\cc=\zz_1 \sqcup \zz_2$. Let $\chi: \binom{\cc}{\mathcal{A}} \to \{red,blue\}$ be a colouring. Since $\binom{\cc}{\mathcal{A}}_{can} \subseteq \binom{\cc}{\mathcal{A}}$, colouring $\chi$ induces $\{red,blue\}$ colouring of $\binom{\cc}{\mathcal{A}}_{can}$. By Fact \ref{fact:lambda}, $\binom{\zz_1}{\xx_1} \times \binom{\zz_2}{\xx_2}$ and $\binom{\cc}{\mathcal{A}}_{can}$ are in 1-1 correspondence and thus $\chi$ induces a colouring of $\binom{\zz_1}{\xx_1} \times \binom{\zz_2}{\xx_2}$. By a choice of $\zz_1$ and $\zz_2$ (recall that $\zz_i \in PL^{(1)}, \; i=1,2$) there are $\tilde{Y}_i \in \binom{\zz_{i}}{\yy_{i}}$ for $i=1,2$, such that $\binom{\tilde{\yy}_1}{\xx_1} \times \binom{\tilde{\yy}_2}{\xx_2}$ is monochromatic and w.l.o.g we assume that all elements of $\binom{\tilde{\yy}_1}{\xx_1} \times \binom{\tilde{\yy}_2}{\xx_2}$ are red. Let $\pi_{i} : \yy_i \to \zz_i$ be a copy of $\yy_i$ in $\zz_i$, such that $\pi_{i}(\yy_i)=\tilde{\yy_i}$ for $i=1,2$. Then, by Claim \ref{claim:1}, the mapping $\pi : Y \to Z_1 \times Z_2$ defined by $\pi(y)=(\pi_1(y), \pi_2(y))$ is a canonical copy of $\bb$ in $\cc$ (see Remark \ref{remark:1}) i.e. $\pi \in \binom{\cc}{\bb}_{can}$. Let $\sigma \in \binom{\pi(\bb)}{\aa}$, then, by Claim \ref{claim:3}, $\sigma \in \binom{\pi(\bb)}{\aa}_{can}$. Therefore, $\sigma$ is of the form $\sigma(x)=(\sigma_1(x), \sigma_2(x))$, where $\sigma_1 \in \binom{\tilde{\yy_{1}}}{\xx_1}$ and $\sigma_2 \in \binom{\tilde{\yy_{2}}}{\xx_2}$. Since all elements of $\binom{\tilde{\yy_{1}}}{\xx_1} \times \binom{\tilde{\yy_{2}}}{\xx_2}$ are red, we get that the pair $(\sigma_1, \sigma_2)$ and $\sigma$ itself is red. Consequently, every element of $\binom{\pi(\bb)}{\mathcal{A}}$ is colored red. \section{Concluding remarks} We chose to present the argument for $k=2$ for its notational ease. With the concept of join of two posets replaced with join of $k$ posets, as in definition below, the proof follows the line of the argument presented in this note. \begin{definition} Let $\zz_{i}=(Z_i, P^{Z_i}, L^{Z_i})\in PL^{(1)}$ for $i \in [k]$ and set $C=\Pi_{i=1}^{k}Z_i$. Define partial order $<_C$ on set $C$ by $\overline{x}<_{C}\overline{y}$ if $x_{i}P^{Z_{i}}y_{i}$ for all $i \in [k]$. For all $i \in [k]$ define shifted lexicographic orders $<_{lx_i}$ on set $\Pi_{i=1}^{k}Z_i$, by $$\overline{x}<_{lx_i}\overline{y} \lr x_{i+\delta}L^{Z_{i+\delta}}y_{i+\delta},$$ where $\delta$ is the smallest non-negative number $j$, for which $x_{i+j}\neq y_{i+j}$ (with addition mod $k$). Let $\mathcal{L}_{k}^{C}=(<_{lx_1}, <_{lx_2}, \dots, <_{lx_k})$. Then the join of $\zz_1, \dots, \zz_k$ is $$\cc=(C, <_C, \mathcal{L}_{k}^C).$$ \end{definition} During preparation of this paper it was brought to our attention that Theorem \ref{thm:main} also follows from the results of Soki\'c~\cite{Sokic}. Alternative proof of Theorem \ref{thm:main} can be deduced from Theorem 1 in~\cite{Sokic} and follows the same steps as the proof presented in this note. 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