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\title[The chain covering number]{The chain covering number of a poset with no infinite antichains}
\alttitle{Le nombre de chaînes recouvrant un ensemble ordonné sans antichaînes infinies}
\author{\firstname{Uri} \lastname{Abraham}\IsCorresp}
\address{Math \& CS Dept., Ben-Gurion University, Beer-Sheva, 84105 Israel}
\email{abraham@math.bgu.ac.il}
\email[U. Abraham]{uriabr@gmail.com}
\author{\firstname{Maurice} \lastname{Pouzet}}
\address{ICJ, Mathématiques, Université Claude-Bernard Lyon1, 43 bd. 11 Novembre 1918, 69622 Villeurbanne Cedex, France}
\address{Mathematics \& Statistics Department, University of Calgary, Calgary, Alberta, Canada T2N 1N4}
\email[M. Pouzet]{pouzet@univ-lyon1.fr}
\thanks{This research benefited from the Abkasian seminar running through the Covid epidemic and was supported by LABEX MILYON (ANR-10-LABX-0070) of Université de Lyon within the program ``Investissements d'Avenir (ANR-11-IDEX-0007)'' operated by the French National Research Agency (ANR)}
\CDRGrant[LABEX MILYON]{ANR-10-LABX-0070}
\CDRGrant[ANR]{ANR-11-IDEX-0007}
\subjclass{03E05, 06A07}
%\keywords{posets, graphs, well quasi ordering, better quasi ordering}
\begin{abstract}
The chain covering number $\Cov(P)$ of a poset $P$ is the least number of chains needed to cover $P$. For an uncountable cardinal $\nu$, we give a list of posets of cardinality and covering number $\nu$ such that for every poset $P$ with no infinite antichain, $\Cov(P)\geq \nu$ if and only if $P$ embeds a member of the list. This list has two elements if $\nu$ is a successor cardinal, namely $[\nu]^2$ and its dual, and four elements if $\nu$ is a limit cardinal with $\cf(\nu)$ weakly compact. For $\nu= \aleph_1$, a list was given by the first author; his construction was extended by F. Dorais to every infinite successor cardinal $\nu$.
\end{abstract}
\begin{altabstract}
Le nombre de recouvrement par chaînes d'un ensemble ordonné $P$ (poset), noté $\Cov(P)$, est le plus petit nombre de chaînes nécessaires pour recouvrir $P$. Pour un cardinal donné $\nu$, on donne une liste de posets $Q$ de nombre de recouvrement par chaînes $\nu$ telle que pour tout poset $P$ sans antichaîne infinie, $\Cov(P)\geq \nu$ si et seulement si $P$ contient une copie d'un membre de la liste. Cette liste est constituée de posets de cardinal $\nu$, elle a deux éléments si $\nu$ est un cardinal successeur, à savoir $[\nu]^2$ et son dual, et quatre éléments si $\nu$ est un cardinal limite avec $\cf(\nu)$ faiblement compact. Pour $\nu= \aleph_1$, une liste a été donnée par le premier auteur; sa construction a été étendue par F. Dorais à tout cardinal successeur infini $\nu$.
\end{altabstract}
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\begin{document}
\maketitle
\section{Introduction}
Let $P:= (V, \leq)$ be an ordered set (poset). The \emph{chain covering number} $\Cov(P)$ of $P$ is the least number of chains needed to cover $V$.
A famous result due to R.P. Dilworth~\cite{dilworth} asserts that for every poset $P$ and integer $n$, $\Cov(P)\geq n$ if and only if $P$ contains an antichain of cardinality $n$.
If $n$ is replaced by $\aleph_0$, then $\Cov(P)\geq \aleph_{0}$ if and only if $P$ contains either an infinite antichain or the lexicographical sum of antichains of finite unbounded cardinality indexed by the chain of the integers or its reverse (see~\cite[Lemma~4.1, p.~491]{milner-pouzet2}).
M. Perles~\cite{perles} and independently E. S. Wolk~\cite{wolk} proved that the assumption that every antichain of the poset is finite does not suffice to get a cover of the poset with a countable number of chains (or with any prescribed number $\nu$ of chains). Their counter-example is the direct product $\nu \times \nu$ of an uncountable cardinal $\nu$ by itself. It turns out that the subset $[\nu]^2:=\{(\alpha, \beta): \alpha<\beta< \nu\}$ has the same properties. This is the main protagonist in this article.
According to F. Dorais~\cite{dorais}, F. Galvin conjectured in 1968 that $\Cov(P)\geq \aleph_{1}$ if and only if $V$ contains a subset $X$ of cardinality $\aleph_1$ such that $\Cov (P{\restriction X})= \aleph_1$. It is not known if the truth of this conjecture is consistent with ZFC. The first author proved in~\cite{abraham} that this conjecture holds with the additional hypothesis that $P$ has no infinite antichain. He introduced \emph{Perles posets} having cardinality and chain covering number $\aleph_1$ and proved that if a poset $P$ has no infinite antichain, $\Cov(P)\geq \aleph_{1}$ if and only if some poset induced by $P$ is isomorphic to a Perles poset. This result was extended by F. Dorais~\cite{dorais2} to infinite successor cardinals.
It was conjectured by R. Rado~\cite{rado} that Galvin's conjecture holds for interval orders (a poset $P$ is an
\emph{interval order} if its vertices can be represented by intervals of a chain, a vertex $x$ being strictly before $y$ in the ordering if the interval associated to $x$ is to the left of the interval associated to $y$, these intervals being disjoint). Todor\v{c}evi\'c~\cite {todorcevic1} proved that Rado's conjecture is consistent with the existence of a supercompact cardinal. In~\cite{todorcevic2, todorcevic3}, Todor\v{c}evi\'c proved that large cardinals are indeed necessary to establish the consistency of Rado’s conjecture. He also proved that the truth of Rado's conjecture implies that Galvin's conjecture holds for $2$-dimensional posets (see~\cite{dorais}).
In this paper, we prove Galvin's conjecture for posets with no infinite antichain and for each cardinal $\nu$ we give an explicit list of posets of cardinality and chain covering number $\nu$.
In order to present our result, we need few \emph{notations}. Let $P:= (V, \leq)$ be a poset. If $X$ is a subset of $V$, the poset \emph{induced} by $P$ on $X$ is $P{\restriction X}:= (X, \leq \cap X\times X)$. The term {\em subposet} always refers here to induced subposet. If $P:= (V, \leq)$ is a poset, two elements $x, y$ of $V$ are \emph{comparable} if $x\leq y$ or $y\leq x$; otherwise they are \emph{incomparable}. We write $x \bot y$ to say that they are incomparable. For each subset $A$ of $P$, we write $(\Inc A)$ or $A^\Inc$ for the set of vertices of $P$ that are incomparable to every element of $A$. Also, for $p\in P$, $p^\Inc$ denotes the set of vertices of $P$ that are incomparable to $p$.\linebreak We define $\downarrownogap A:= \{x\in P: x\leq a\; \text {for some } a\in A\}$; similarly we have $\uparrownogap A$. If $A= \{a\}$, we denote these sets by $\downarrownogap a$ and $\uparrownogap a$. If $a\leq b$ in $P$, we write $[a,b]_P:= \{ z\in P: a\leq z\leq b\}$ and $[a, b)$ is similarly defined. The \emph{dual} of $P$ is the poset on the same domain equipped with the opposite order, we denote it by $P^*$. We recall that a subset $I$ of $P$ is an \emph{initial segment} of $P$ if $x\leq y$ and $y\in I$ imply $x\in I$; a subset of $P$ is a \emph{final segment} if this is an initial segment of $P^*$, the dual of $P$. Initial and final segments being ordered by set inclusion, if $A$ is a subset of $P$, $\downarrownogap A$ is the least initial segment of $P$ containing $A$ and $\uparrownogap A$ is the least final segment of $P$ containing $A$. A subset $A$ of $P$ is
\emph{up-directed} if every pair $x,y$ of elements of $A$ has some upper bound $z$ in $A$. An \emph{ideal} of $P$ is any up-directed nonempty initial segment $J$ of $P$.
The \emph{dominance} relation is defined on subsets of $P$: $B$ dominates $A$, written $A\nearrow B$, when $\forall a\in A\, \forall b\in B (a\leq_P b)$.
A poset $P$ is \emph{embeddable} in a poset $P'$, and we set $P\leq P'$, if $P$ is isomorphic to some subposet of~$P'$. Two posets $P$ and $P'$ are \emph{equimorphic} if each is isomorphic to a subposet of the other. Clearly, if $P\leq P'$ then $\Cov(P)\leq\Cov(P')$, while $\Cov(P)=\Cov(P^*)$. A subset $A$ of $P$ is \emph{cofinal} in $P$ if every element of $P$ is majorized by some element $a\in A$, i.e, $\downarrownogap A = P$. The \emph{cofinality} of $P$ is the least cardinality of a cofinal subset of $P$, we denote it by $\cf (P)$. Accordingly, if $\nu$ is an ordinal, we denote by $\cf (\nu)$ the cofinality of $\nu$. For a set $X$ of ordinals, $\sup X$ (often notated $\bigvee X$) is the supremum of~$X$.
\begin{defi}\label{Perles}
Let $\nu$ be an infinite ordinal, we set $[\nu]^2: =\{(\alpha, \beta):
\alpha<\beta<\nu\}$, and order that set by setting $(\alpha, \beta)\leq (\alpha', \beta')$ if $\alpha\leq \alpha'$ and $\beta\leq \beta'$. As a subset of the direct product $\nu\times \nu$, this poset is the upper half of $\nu\times \nu$ ordered componentwise.
\end{defi}
Our main result asserts:
\begin{theo}\label{thm:covering}
Let $\nu$ be an uncountable cardinal and $P$ be a poset with no infinite antichain.
\begin{enumerate}
\item\label{theo2.1} If $\nu$ is a successor cardinal, i.e., $\nu= \kappa^+$, then $\Cov(P)\geq \nu= \kappa^+$ if and only if $P$ or $P^*$ contains a copy of $[\kappa^+]^2$.
\item\label{theo2.2} If $\nu$ is an uncountable limit cardinal, then $\Cov(P)\geq \nu$ if and only if $P$ or $P^*$ contains a poset $Q$ of the form $\sum_{a\in C}Q_{a}$, where $C$ is a chain of cardinality $\cf (\nu)$, $Q_{a} \simeq [\kappa_{a}^+]^2$ and $(\kappa_{a}^+)_{a\,\in\,C}$ is a family of distinct cardinals that satisfies $\sup \{\kappa_{a}^+: a\in C\} = \nu$.
\end{enumerate}
\end{theo}
\paragraph{Comments about item~\meqref{theo2.1}} In~\cite{abraham}, the first author introduced a collection $\bfP(\omega_1, \omega_1)$ of posets of size $\aleph_1$ and proved that these posets cannot be covered by less that $\aleph_1$ chains (Proposition~1.15) and furthermore that a poset with no infinite antichain can be covered by less than $\aleph_1$ chains if and only if it does not embed a member of $\bfP(\omega_1, \omega_1)$ (Theorem~2.1). In~\cite{dorais2} F.~Dorais extended this result to successor cardinals. For such a cardinal, say $\kappa^+$, he introduced posets of \emph{Perles type} and showed that a poset with no infinite antichain can be covered by at most $\kappa$ chains if and only if it does not embed a Perles type poset of cardinality $\kappa^+$. Item~\eqref{theo2.1} is equivalent to the fact that each of these posets embeds a copy of $[\kappa^+]^2$ or its dual. This fact was not stated in~\cite{abraham} nor in~\cite{dorais2}. Since this is a fact of independent interest, we give a proof in Subsection~\ref{subsection:perlestype}.
\paragraph{Comments about item~\meqref{theo2.2}}
Let $\lambda:= \cf(\nu)$. Suppose that $\lambda$ is \emph{weakly compact}, that is every chain of cardinality $\lambda$ contains either a chain of order type $\lambda$ or of its dual. Then, in item~\eqref{theo2.2}, we may replace the chain $C$ by $\lambda$ or its dual. If we compare the posets $Q$ by embeddability, we get only four nonequimorphic posets, namely $\sum_{\alpha\,\in\,\lambda}[\kappa^+_{\alpha}]^2$, $\sum_{\alpha\,\in\,\lambda^*}[\kappa^+_{\alpha}]^2$, and their duals, where the sequence $(\kappa_\alpha)_{\alpha\,<\,\lambda}$ is increasing and with limit $\nu$. Two such families $(\kappa_{\alpha}^+)_{\alpha\,\in\,\lambda}$ give equimorphic posets.
If $\lambda$ is not weakly compact, then among the posets we obtain in item~\eqref{theo2.2}, some cannot be expressed as a well-ordered or reversely well ordered chain of type $\lambda$. For an example, suppose that $C$ is the chain $\R$ of real numbers. Let $c$ be a bijective map from $\R$ onto $\kappa:= 2^{\aleph_0}$ and $f$ be a strictly increasing map from $\kappa$ in the collection of successor cardinals. If $A$ is any subchain of $C$ set $Q_{A}:= \sum _{a\,\in\,A}[f(c(a))]^2$. Then $\Cov(Q_{A})=\sup \{f(c(a)): a\in A\}$. Since a well-ordered or reverse well-ordered subchain $A$ of $\R$ is at most countable, and the cofinality of $2^{\aleph_0}$ is uncountable, $\Cov(Q_{A})<\Cov(Q_{C})$.
\begin{enonce}{Problem}
Let $\nu$ be an uncountable limit cardinal such that $\lambda:=\cf(\nu)$ is not weakly compact,
find the least number $G(\nu)$ of posets $Q$ needed in item~\eqref{theo2.2}?
\end{enonce}
An easy consequence of Item~\eqref{theo2.1} of Theorem~\ref{thm:covering} is about the notion of indivisibility. A poset $P$ is \emph{indivisible} if for every partition of its domain in two parts, $P$ embeds into a restriction of $P$ to one of these parts; more generally, $P$ is $<\nu$-\emph{indivisible} for some cardinal $\nu$ if for every partition of $P$ into less than $\nu$ parts $P$ embeds in a restriction of $P$ to some part. \cite {fraisse} contains several results about the indivisibility of posets and relations.
\begin{coro}
If $\nu$ is an infinite successor cardinal, then $[\nu]^2$ is $<\nu$-indivisible.
\end{coro}
Indeed, let $(A_{\alpha})_{\alpha\,<\,\kappa}$ where $\kappa^+= \nu$ be a partition (or a covering) of $[\nu]^2$. If the covering number of each part is at most $\kappa$ then the covering number of $[\nu]^2$ is at most $\kappa$. But this is impossible, hence the covering number of some part is at least $\nu$. From Item 1 of Theorem~\ref{thm:covering}, this part embeds~$[\nu]^2$.
Using Lemma~\ref{lem:thirdcofinalchain} the conclusion extends to regular $\nu$. We do not know if $[\nu]^2$ is $<\cf(\nu)$-indivisible. Extension of these results to $[\nu]^n$ need to be considered. This paper was motivated by a question of Abraham, Bonnet, Kubis~\cite{abk} about two well studied notions in the theory of ordered sets, namely well quasi orders (w.q.o) and better quasi orders (b.q.o), (\cite{fraisse} for reference). They ask whether any w.q.o is a countable union of b.q.o subposets. A positive answer would imply that any $\omega$-indivisible w.q.o is b.q.o.
We present two proofs of Theorem~\ref {thm:covering}. They have a common part, which is a reduction to a special case (Theorem~\ref{thm:covering2}). With that reduction, Item~\eqref{theo2.2} reduces to Item~\eqref{theo2.1}. Our first proof of Item~\eqref{theo2.1} is based on the definition of the class $\calP(\kappa)$ of posets. We prove that any poset in that class embeds $[\nu]^2$ or its dual (Theorem~\ref{Tm1.3}). Then we obtain the conclusion of Item 1 by proving that a poset not coverable by $\nu$ chains embeds a poset in $\calP(\nu^+)$ (Theorem~\ref{Tm1.3}). This latter fact was known for $\aleph_1$-Perles posets~\cite{abraham}. Our second proof relies on a characterization of posets containing $[\nu]^2$ in terms of chains of ideals (Lemma~\ref{lem:thirdcofinalchain}) and on a proof that a poset not coverable by $\nu$ chains contains such a chain of ideals. This latter proof relies on the notion of purity introduced in~\cite{assous-pouzet}. The reason for including two proofs of the same theorem is that each proof touches on some ideas that we found interesting and which could be useful for other investigations in this field. The reader may choose to read just one proof: read the common part of the two proofs, and then continue with either section.
\section{Basic tools}\label{Basic}
Some fundamental tools that are used in the paper are the Erd\"os--Dushnik--Miller partition theorem~\cite{dushnik-miller}, a description of cofinal subsets of posets with no infinite antichains due to the second author~\cite{pouzet} (see~\cite{milner-prikry, milner-pouzet1, fraisse}) and the notion of purity with~\cite[Theorem~24]{assous-pouzet}.
%theoA ?
\begin{theo}[{Erd\"os--Dushnik--Miller}]
For every infinite cardinal $\lambda$, every graph with $\lambda$ vertices contains either an infinite independent set or a complete subgraph on $\lambda$ vertices. (In arrow notation this is written $\lambda\rightarrow (\aleph_0, \lambda)^2$.)
\end{theo}
A poset with no infinite antichains is a union of finitely many up-directed subsets (R.~Bonnet~\cite{Bonnet}, see also~\cite[4.7.2 p.~124]{fraisse}).
The cofinality theorem of Pouzet (see~\cite{fraisse}) states:
\begin{theo}\label{cofinalthm}
Every poset $P$ with no infinite antichain has a cofinal subset $A$ which decomposes in a finite union $A_1 \cup \dots \cup A_k$ with no comparabilities between elements of $A_i$ and $A_j$ for every $i\not =j$, and such that the $<_P$ order on each $A_i$ is isomorphic to a finite product $\alpha_{1,i}\times \dots \times \alpha_{n_i, i}$ of regular distinct cardinals.
\end{theo}
A consequence (in fact an equivalent statement) is this:
\begin{theo}[Pouzet]\label{cofinalthm2}
Every up-directed poset $P$ with no infinite antichain has a cofinal subset $A$ which is order isomorphic to a finite product $\alpha_{1}\times \dots \times \alpha_{n}$ of regular distinct ordinals.
\end{theo}
\begin{defi}[Purity]\label{def:pure}
A poset $P$ is \emph{pure} if every proper initial segment $I$ of $P$ is strictly bounded above (that is, some $x\in P\setminus I$ dominates $I$).
\end{defi}
This condition amounts to the fact that every non cofinal subset of $P$ is strictly bounded above (indeed, if a subset $A$ of $P$ is not cofinal, then $\downarrownogap A\not = P$ hence from purity, $\downarrownogap A$, and thus $A$, is strictly bounded above. The converse is immediate).
A pure poset is necessarily up-directed: Given $x,y\in P$ there is $z$ such that $x,y\leq z$. If $x$ and $y$ are comparable, this is evident. So we assume that $x\bot y$ and then $x\in I = P\setminus \uparrownogap y$. As $I\neq P$ is a proper initial segment there is $z \in\, \uparrownogap y$ a bound of $I$. Hence $x, y \leq z$. Hence a pure poset with infinite cofinality has no maximal element. Indeed, if $x\in P$, then for some $y\in P$, $y\not\leq x$ (or else the cofinality of $P$ is $1$) and there is $z$ with $x,y\leq z$. Necessarily $x< z$.
We extract the following result from~\cite[Theorem~24]{assous-pouzet}.
\begin{theo}\label{theo:pure}
Let $P$ be a poset with uncountable cofinality $\nu$. Then $P$ is pure if and only if $P$ is a lexicographical sum $\sum_{a\,\in\,\nu }P_{a}$ with arbitrary components $P_a$.
\end{theo}
\begin{proof}
To prove that the lexicographical sum of posets is pure does not require that the sum is over a regular $\nu$. It suffices for the index ordering to be any chain with no maximum. If $P = \sum_{a\,\in\,K}P_{a}$ where $K$ is a chain with no maximum, then $P$ is clearly pure, because any proper initial segment occupies only a bounded in $K$ set of indexes.
For the other direction, assume that $P$ is pure with uncountable cofinality $\nu$. Let $\{ b_\alpha \mid \alpha <\nu\}$ be cofinal in $P$. We are going to construct by induction on $\alpha<\nu$ a strictly increasing sequence $(x_\alpha)_{\alpha\,<\,\nu}$ such that $b_\alpha\leq x_{\alpha}$ (thereby assuring that the constructed sequence is also cofinal). The final segment $S_\alpha= \{ p\in P\mid \forall \zeta < \alpha\, (x_\zeta\leq p)\}$ is defined to satisfy for every $\alpha<\nu$ that
\begin{equation}\label{eq:property}
P\setminus S_\alpha\nearrow S_{\alpha+1}.
\end{equation}
The definition of $S_\alpha$ implies that: (1) $\alpha<\alpha'\rightarrow S_{\alpha'}\subseteq S_\alpha$. (2) If $\delta<\nu$ is a limit then $S_\delta = \bigcap \{ S_\zeta\mid \zeta < \delta\}$. (3) $S_{\alpha+1} = \{ p\in P \mid x_\alpha\leq p\} = (\uparrownogap x_\alpha)$. (4) $S_0=P$ and $S_\nu = \emptyset$.
Start by defining $x_0\in P$ as any member of $P$ such that $b_0\leq x_0$.
\begin{enumerate}
\item Suppose that $x_\alpha$ is defined, then find $x_{\alpha+1}$ that strictly dominate the proper initial segment $P\setminus \uparrownogap x_\alpha$. In fact, we add the requirement that $b_{\alpha+1}\leq x_{\alpha+1}$ (which is possible because a pure poset is up-directed). Since $P\setminus \uparrownogap x_\alpha\nearrow x_{\alpha+1}$, we get that $P\setminus S_{\alpha+1} \nearrow S_{\alpha+2}$.
\item Suppose now that $\alpha< \nu$ is a limit ordinal and $(x_\zeta\mid \zeta< \alpha)$ is defined. Since $\alpha< \nu=\cf(P)$, $I=\downarrownogap \{x_\zeta\mid \zeta<\alpha\}\neq P$, and hence $I$ is strictly bounded in $P$. That is, $J = P\setminus S_\alpha$ is a proper initial segment of $P$. Define $x_\alpha$ that strictly dominates $P\setminus S_\alpha$, and suppose additionally that $x_\alpha$ dominates $b_\alpha$. So $P\setminus S_\alpha \nearrow x_\alpha\nearrow S_{\alpha+1}$. Thus~\eqref{eq:property} holds also when $\alpha$ is a limit ordinal.
\end{enumerate}
This ends the definition of the strictly increasing and cofinal sequence $(x_\alpha)_{\alpha\,<\,\nu}$.
Define the rings $R_\alpha = S_{\alpha}\setminus S_{\alpha+1}$ for $\alpha<\nu$. We get a partition of $P$, $P=\bigcup_{\alpha\,<\,\nu} R_\alpha$. It follows from the definitions that $\alpha+2\leq \beta<\nu\rightarrow R_\alpha \nearrow R_\beta$. (Indeed, $R_\alpha = S_\alpha\setminus S_{\alpha+1}\subseteq P\setminus S_{\alpha+1}\nearrow S_{\alpha+2}$, and $R_\beta\subseteq S_\beta\subseteq S_{\alpha+2}$. Thus $R_\alpha\nearrow R_\beta$.) Redefine the rings by setting for every limit ordinal $\delta<\nu$, $\calR_\delta =
\bigcup_{n\,\in\,\omega}R_{\delta + n}$, then we get a coarser partition $P=\bigcup_{\lim \delta\,<\,\nu} {\calR}_\delta$ such that if $\delta_1<\delta_2$ then ${\calR}_{\delta_1}\nearrow {\calR}_{\delta_2}$. Hence $P=\sum_\delta {\calR}_\delta$ as required.
\end{proof}
\subsection{\texorpdfstring{$\calP(\kappa)$}{calP(kappa)}, a class of \texorpdfstring{$\kappa$}{kappa}-Perles posets}\label{subsection:perlestype}
As we saw, for any cardinal $\nu$, $[\nu]^2$ is not a union of less than $\nu$ chains. This leads to the following definition.
\begin{defi}\label{Perles0}
For an uncountable cardinal $\nu$, a poset $P$ with only finite antichains is said to be of $\nu$-Perles type if it is not a union of less than $\nu$ chains.
\end{defi}
Thus, $[\nu]^2$ and its dual are examples of $\nu$-Perles type posets. For regular $\kappa$, we define a class of poset containing only $\kappa$-Perles types which will be found out to be the class of all $\kappa$-Perles types.
\begin{defi}\label{Perles1}
For a regular uncountable cardinal $\kappa$, a poset is in the class $\calP(\kappa)$ if it contains a subposet $P$ whose universe, $|P|$, is $\kappa$ (or a subset of $\kappa$) so that $P$ carries both the partial ordering $<_P$\goodbreak\noindent and the ordinal ordering $<$, and there is a function $F: |P|=\kappa\to \kappa$ whose range is $\kappa$, and is such that the following holds with respect to the the incomparability relation $\Inc$ on $P$:
\begin{enumerate}
\item\label{def11.1} For every $\alpha\in \kappa$, $|F^{-1}\{\alpha\}|=\kappa$,
\item\label{def11.2} If $\alpha_1<\alpha_2<\kappa$, $\xi_1\in F^{-1}\{\alpha_1\}$ and $\xi_2\in F^{-1}\{\alpha_2\}$ are such that $\xi_1>\xi_2$, then ${\xi_1}\ \Inc\ {\xi_2}$.
\end{enumerate}
\end{defi}
\begin{rema}\label{R1.0}
If $P=(\kappa, <, <_P) \in \calP(\kappa)$ and $Q$ is a subposet of $P$ such that for every $\alpha\in F`` Q$, $F^{-1}\{\alpha\}$ has cardinality $\kappa$, then $Q\in \calP(\kappa)$ as well. (Collapse the universe of $Q$ so that it becomes $\kappa$, and likewise collapse $F'' Q$.)
\end{rema}
\begin{enonce}[plain]{Claim}\label{R1.2}
Let $\kappa$ be a regular uncountable cardinal.
\begin{enumerate}
\item\label{claim13.1} If $P\in \calP(\kappa)$ then $P^\ast\in \calP(\kappa)$ as well, where $P^\ast$ denotes the dual ordering of $P$.
\item\label{claim13.2} $P$ in $\calP(\kappa)$ is not covered by $<\kappa$ chains.
\item\label{claim13.3} $[\kappa]^2$ (and its dual) are in the $\calP(\kappa)$ family of posets.
\end{enumerate}
\end{enonce}
The following notes indicate the proofs.
\begin{enumerate}
\item Item~\eqref{claim13.1} is obvious since the definition of $\calP(\kappa)$ does not mention the ordering of $P$, only the comparability relation is involved.
\item For item~\eqref{claim13.2}, let $F:P=\kappa \to \kappa$ be a witness for $P\in\calP(\kappa)$. Suppose that $P$ \emph{is} covered by $\kappa_0 <\kappa$ chains $\{P_i\mid i<\kappa_0\}$. Then for any $\alpha<\kappa$, $F^{-1}\{\alpha\}$ has an intersection of size $\kappa$ with (at least) one of the chains $C_{i(\alpha)}$ (by regularity of $\kappa$). For a set $S$ of size $\kappa$ there is a fixed chain $C_{i}$ such that for any $\alpha\in S$, $i=i(\alpha)$. This is impossible: take $\alpha_1<\alpha_2$ in $S$, take any $\xi_2\in C_i\cap F^{-1}\{\alpha_2\}$, and then take $\xi_1\in C_i\cap F^{-1}\{\alpha_1\}$ such that $\xi_1>\xi_2$. Then $\xi_1\ \Inc\ \xi_2$ shows that they cannot be in the same chain.
\item For Item~\eqref{claim13.3}, prove that the poset $[\kappa]^2$ is in $\calP(\kappa)$: namely there exist an isomorphic copy $P$ of $[\kappa]^2$ whose universe is $\kappa$ and a function $F$ that satisfy Definition~\ref{Perles1}. Let $<_R$ be the right-order lexicographical ordering on $[\kappa]^2$ defined by $\langle{\alpha,\beta}\rangle<\langle{\alpha',\beta'}\rangle$ iff $\beta<\beta'$ or $\beta=\beta'\, \wedge\,
\alpha<\alpha'$. The order-type of $[\kappa]^2$ in $<_R$ is $\kappa$ in its ordinal ordering $<$. Let $p:(\kappa, <) \to ([\kappa]^2, <_R)$ be the order-isomorphism. We notate $p(\xi) = (\alpha_\xi,\beta_\xi)$. Thus the poset $[\kappa]^2$ induces an isomorphic copy $P = (\kappa, <_P)$ on $\kappa$ by setting $\xi \leq_P \xi'$ iff $\alpha_\xi\leq \alpha_{\xi'} \wedge \beta_\xi\leq\beta_{\xi'}$. Define $F(\xi)=\alpha_\xi$. The function $F$ is used to shows that $[\kappa]^2 \in \calP(\kappa)$. Define a closed unbounded subset $C\subset \kappa$ such that every $\delta\in C$ has the following property.
\begin{enumerate}
\item[] If $\delta\in C$, then $\xi<\delta\rightarrow \alpha_\xi,\beta_\xi <\delta$ and, for every $\alpha<\beta$ in $\delta$, there is $\xi<\delta$ such that $(\alpha,\beta)=(\alpha_\xi,\beta_\xi)$.
\end{enumerate}
Now define a subset $X\subset \kappa$ such that for every $\delta\in C$, $X\cap[\delta,\delta')$ is a singleton, and for every $\alpha<\kappa$, $\{\xi\in X\mid \alpha=\alpha_\xi\}$ is unbounded in $\kappa$. Then $P\restriction X$ and $F:P\restriction X\to \kappa$ show that $[\kappa]^2\in \calP(\kappa)$.
\end{enumerate}
\begin{theo}\label{Tm1.3}
Let $\kappa$ be a regular uncountable cardinal. If $P$ has only finite antichains ($P$ is a FAC poset) then $P$ embeds $[\kappa]^2$ iff $P$ is $\calP(\kappa)$.
\end{theo}
\begin{proof}
We noted above that $[\kappa]^2$ is in $\calP(\kappa)$.
Suppose that $P\in \calP(\kappa)$ where $P=(\kappa, <_P)$ and $F: P=\kappa\to \kappa$ is a surjection such that conditions~\eqref{def11.1} and~\eqref{def11.2} of Definition~\ref{Perles1} hold. We have to prove that $P$ or its dual contain a copy of $[\kappa]^2$. Given $\alpha\in\kappa$ apply the Dushnik--Miller partition theorem, $\kappa\rightarrow (\kappa,\omega)^2$, to the set $[F^{-1}\{\alpha\}]^2$ and to the coloring that gives to $\{\xi_1,\xi_2\}\in [F^{-1}\{\alpha\}]^2$ one color if $\xi_1$ and $\xi_2$ are comparable in $<_P$ and another color if they are not. Since there is no infinite antichain, we get a homogeneous set $C_\alpha\subseteq F^{-1}\{\alpha\}$ for the comparability colour which is a\linebreak $<_P$-chain of cardinality $\kappa$.
An element $\xi\in C_\alpha$ is \emph{exceptional} (in $C_\alpha$) if the sets $\{\eta\in C_\alpha\mid \eta \leq_P \xi\}$ and $\{\eta\in C_\alpha\mid \xi\leq_P \eta\}$ are both of cardinality $\kappa$.
\begin{enonce}[plain]{Claim}
If $\xi_1\!\in\! C_{\alpha_1}$ and $\xi_2\!\in\! C_{\alpha_2}$ are both exceptional in their chains (where $\alpha_1\neq\alpha_2$) then $\xi_1\ \Inc\ \xi_2$.
\end{enonce}
\begin{proof}
Suppose on the contrary that $\xi_1$ and $\xi_2$ are comparable.
For example, suppose that $\alpha_1<\alpha_2$ and $\xi_1\leq_P \xi_2$ (any other pattern works as well). Since $\xi_1$ is exceptional in $C_{\alpha_1}$, there are $\kappa$ members $\eta$ of $C_{\alpha_1}$ such that $\eta \leq_P \xi_1$. Take such $\eta\in C_{\alpha_1}$ that is above $\xi_2$. Then $\eta\ \Inc\ \xi_2$. But this contradicts $\eta \leq_P \xi_1 \leq_P \xi_2$.
\end{proof}
This claim and the finite antichain condition of $P$ imply that there is only a finite number of chains with an exceptional member. Throwing out this finite number of chains $C_\alpha$, we may assume that no chain contains an exceptional member.
We say that $\xi\in C_\alpha$ is of type I if $|\{\eta\in C_\alpha\mid \eta \leq_P \xi\}|<\kappa$. And we say that $C_\alpha$ itself is of type I if it contains $\kappa$ members that are all of type I. In this case, and since $\kappa$ is regular, the set of members of $C_\alpha$ of type I contains a subset $D_\alpha\subset C_\alpha$ of order-type $\kappa$ such that for $\xi_1<\xi_2$ in $D_\alpha$, $\xi_1 <_P \xi_2$.
If $C_\alpha$ is not of type I then it has $<\kappa$ members of type I, and for each of the remaining members $\xi$ that are not of type I, $|\{\eta\in C_\alpha\mid \eta \leq_P \xi\}| =\kappa$. But $\xi$ is not exceptional and hence $|\{\eta\in C_\alpha\mid \xi \leq_P \eta\}|<\kappa$. In this case, we can find $D_\alpha\subseteq C_\alpha$ of cardinality $\kappa$ such that for all $\xi_1, \xi_2\in D_\alpha$, $\xi_1<\xi_2\rightarrow
\xi_2<_P \xi_1$.
Fixing such sets $D_\alpha$, there is a set $R\subseteq \kappa$ of cardinality $\kappa$ such that for every $\alpha\in R$, either $<_P$ is strictly increasing on $D_\alpha$, or else for every $\alpha\in R$ $<_P$ is strictly decreasing. In the first case we will get an embedding of $[\kappa]^2$ into $P$, and in the second case an embedding into $P^\ast$. By symmetry of the argument we may suppose that for every $\alpha\in R$, $<_P$ is increasing over $D_\alpha$.
\begin{enonce}[plain]{Claim}\label{Cl1}
If $\alpha_1<\alpha_2$ are in $R$, and $\zeta_i\in D_{\alpha_i}$ for $i=1,2$ are $\leq_P$-comparable, then $\zeta_1 <_P \zeta_2$.
\end{enonce}
\begin{proof}
If it is not the case that $\zeta_1 <_P \zeta_2$, then $\zeta_2 <_P \zeta_1$ since
$\zeta_1, \zeta_2$ are comparable. Pick $\zeta'_1\in D_{\alpha_1}$ such that $\zeta_1, \zeta_2 < \zeta_1'$ (namely as ordinals). Then $\zeta_2 \bot \zeta_1'$, but $\zeta_2 <_P \zeta_1 <_P \zeta_1'$ contradicts incomparability.
\end{proof}
By Remark~\ref{R1.0}, we may assume that $P=\kappa=\bigcup_{\alpha\,\in\,R} D_\alpha$ and $F:\kappa\to \kappa$ is a surjection which satisfies definition~\ref{Perles1}.
Thus we have the following situation. The universe of $P$ is $\kappa$ and a surjection~$F:\kappa \to \kappa$ exists such that:
\begin{enumerate}\Penumi
\item\label{P1} For every $\alpha\in \kappa$, $D_\alpha=F^{-1}\{\alpha\}$ is unbounded in $\kappa$, and, for every $\xi_1<\xi_2$ in $D_\alpha$, $\xi_1 <_P \xi_2$.
\item\label{P2} If $\alpha_1<\alpha_2<\kappa$, $\xi_1\in D_{\alpha_1}$, $\xi_2\in D_{\alpha_2}$ and $\xi_1, \xi_2$ are comparable, then $\xi_1 <_P \xi_2$ (see Claim~\ref{Cl1}).
\end{enumerate}
On the set $\calD =\{D_\alpha\mid \alpha<\kappa\}$ define a partial ordering relation $\prec$ by $D_\alpha\prec D_\beta$ iff $\forall x\in D_\alpha\, \exists y\in D_\beta\ x <_P y$. We have just seen that
\begin{equation}\label{EQ1}
\text{if } \alpha<\beta<\kappa\ \text{then }\neg \left(D_\beta\prec D_\alpha\right).
\end{equation}
(In fact, by Claim~\ref{Cl1}, if $b\in D_\beta$ and $a\in D_\alpha$ then it is not the case that $b<_P a$.)
\begin{enonce}[plain]{Claim}
Relation $\prec$ is a well-quasi-ordering on $\calD$.
\end{enonce}
\begin{proof}
Let $\langle D_{\alpha_0},D_{\alpha_1},\,\ldots\rangle$ be an infinite sequence from $\calD$. We may assume that the ordinals $\alpha_k$ are increasing with $k\in\omega$. We have to show that $D_{\alpha_i}\prec D_{\alpha_j}$ for some $ix(i,j)$ for every $j>i$. But then $\{ x_i\mid i\in\omega\}$ is an infinite antichain in $P$ (by Claim~\ref{Cl1} and since $x_i \not< x_j$ when $i< j$).
\end{proof}
Again we use the Erd\"os--Dushnik--Miller partition theorem $\kappa\rightarrow (\kappa,\omega)$, but now on the set of chains $\calD$ which forms a w.q.o. If $D_\alpha, D_\beta$ are two chains in $\calD$, color the pair in the comparability-incomparability colours of the $\prec$ ordering. Since there is no infinite set of incomparable members of a wqo, we have a chain of cardinality $\kappa$. In fact, it follows from~\eqref{EQ1} that the order-type of this\goodbreak\noindent chain is $\kappa$. So for some set $S\subseteq \kappa$ of order-type $\kappa$, if $\alpha_1<\alpha_2$ are in $S$, then $D_{\alpha_1}\prec D_{\alpha_2}$. We may assume that $S=\kappa$, and to the two properties~\eqref{P1} and~\eqref{P2} we add~\eqref{P3}:
\begin{enumerate}\Penumi
\setcounter{enumi}{2}
\item\label{P3} If $\alpha_1<\alpha_2<\kappa$ then $D_{\alpha_1}\prec D_{\alpha_2}$
\end{enumerate}
The following claim concludes the proof of Theorem~\ref{Tm1.3}.
\begin{enonce}[plain]{Claim}\label{Clm:embedding}
Properties~\eqref{P1}, \eqref{P2}, \eqref{P3} entail that $[\kappa]^2\hookrightarrow P$.
\end{enonce}
\begin{proof}
Define an order embedding $G:[\kappa]^2 \to P$, so that $G(\alpha,\beta)\in D_\alpha$ for every $\alpha<\beta<\kappa$. By induction on $\beta<\kappa$ we define $G \restriction [\beta]^2$. Suppose that $G$ is defined on $[\beta_0]^2$ for some $\beta_0<\kappa$. We then define $G(\alpha,\beta_0)$ for $\alpha<\beta_0$ by induction on $\alpha$. Suppose $\alpha_0<\beta_0$ is such that for every $\alpha<\alpha_0$, $G(\alpha,\beta_0)$ is defined. Let $E$ denote the domain of the function $G$ at this stage of the construction, that is $E= [\beta_0]^2 \cup \alpha_0\times \{ \beta_0\}$. $E$ contains no pair that is above $(\alpha_0,\beta_0)$ in the $[\kappa]^2$ ordering, and so $E$ is partitioned into a set $E_<$ consisting of those members of $E$ that are below $(\alpha_0,\beta_0)$, and a set $E_{\bot}$ consisting of those members of $E$ that are incomparable with $(\alpha_0,\beta_0)$. For $G(\alpha_0,\beta_0)$ we have to pick $p\in D_{\alpha_0}$ that is above all members of $G '' E_<$ and is incomparable with all members of $G'' E_{Inc}$. Observe that the cardinality of $E$ is $<\kappa$ and that for $\alpha<\beta<\kappa$:
\begin{enumerate}
\item $(\alpha,\beta)\in E_<$ iff $\alpha<\alpha_0 \wedge \beta\leq \beta_0$ or $\alpha = \alpha_0 < \beta<\beta_0$, and
\item $(\alpha,\beta)\in E_\bot$ iff $\alpha_0 < \alpha < \beta<\beta_0$.
\end{enumerate}
Recall that if $\alpha<\alpha_0$ then for every $\xi\in D_\alpha$, $\xi <_P \zeta$ for some $\zeta\in D_{\alpha_0}$. And if $\alpha_0<\alpha$ and $\zeta\in D_{\alpha_0}$ then $\zeta$ is incomparable with $\zeta' \in D_\alpha$ whenever $\zeta>\zeta'$. Thus $\zeta\in D_{\alpha_0}$ that is high enough can be found to satisfy the requirements for $\zeta = G(\alpha_0,\beta_0)$. This ends the proof of Theorem~\ref{Tm1.3}.
\end{proof}\let\qed\relax
\end{proof}
Another interesting characterization of posets $P$ such that $P$ of the dual $P^*$ embeds $[\kappa]^2$ is in the following theorem. Recall that an ideal $I$ in a poset $P$ is an up-directed initial segment of $P$.
\begin{theo}\label{lem:thirdcofinalchain}
Let $\kappa$ be a regular cardinal and $P$ be a poset. Then $P$ or its dual contains $[\kappa]^2$ if and only if $P$ or its dual contains a poset that is the union of a strictly increasing chain $(J_{\alpha})_{\alpha\,<\,\kappa}$ of unbounded ideals, each containing a cofinal chain of order-type $\kappa$.
\end{theo}
\begin{proof}
Consider $[\kappa]^2$ and for every $\alpha<\kappa$ define $I_\alpha = \{(\tau,\beta)\mid \tau\leq \alpha$ and $\tau<\beta< \kappa\}$. Then $I_\alpha$ is a proper ideal of $[\kappa]^2$, and $[\kappa]^2$ is the union of these ideals. In each $I_\alpha$ the chain $(\alpha,\beta)$, $\alpha<\beta<\kappa$ is cofinal.
Now suppose that $P$ contains a poset that satisfies the condition. For every ideal $J_\alpha$ let $C_\alpha\subset J_\alpha$ be a cofinal subset of order-type $\kappa$. Note that if $\alpha_1<\alpha_2<\kappa$ then for every $x\in J_{\alpha_1}\subset J_{\alpha_2}$ there is $y\in C_{\alpha_2}$ such that $x<_P y$ (because $J_{\alpha_1}\subset J_{\alpha_2}$). Also, for every $\alpha<\kappa$ there is a bounded set of points in $C_{\alpha}$ that are in $J_{\alpha'}$ for some $\alpha'<\alpha$, and hence by trimming $C_\alpha$ we may assume for every $\alpha' < \alpha$ that if $x\in J_{\alpha'}$ and $y\in C_\alpha$ are comparable then the only possible relation is $x<_P y$. Also, since $C_\alpha$ is not bounded in $P$, if $\alpha<\beta<\kappa$ then for every $y\in C_\beta$ there is a cofinal segment of members of $C_\alpha$ that are incompatible with $y$.
We define an embedding from $[\lambda]^2$ in $P$ such that $f(\alpha, \beta)\in J_{\alpha}$ for every $\alpha<\beta<\lambda$. Order $[\lambda]^2$ lexicographically according to the second difference, that is $(\alpha' \beta')\prec (\alpha, \beta)$ if either $\beta' <\beta$ or $\beta= \beta'$ and $\alpha'<\alpha$. The order-type of $[\kappa]^2$ under $\prec$ is $\kappa$. By induction on $\beta\in\kappa$ define $f(\alpha,\beta)\in C_\alpha$ for every $\alpha<\beta$. Suppose that $\beta<\kappa$ and that $f$ is defined on all of $[\beta]^2=\{(\zeta,\xi)\mid \zeta<\xi <\beta\}$. We define $f(\alpha,\beta)\in C_\alpha$ by induction on $\alpha<\beta$.
For that, suffices to show that $J_{\alpha}$ contains some $x_{\alpha, \beta}$ larger than $f(\alpha, \beta')$ for all $\alpha <\beta'<\beta$ and larger than $f(\alpha', \beta)$ for all $\alpha'<\beta$, and yet $x_{\alpha,\beta}$ has to be incompatible to every $f(\alpha^*,\beta^*)$ when $\alpha<\alpha^*<\beta^*<\beta$. Since the number of these restrictions is less than $\kappa$, such $x_{\alpha,\beta}$ can be found and defining $f(\alpha,\beta)=x_{\alpha,\beta}$ works. One checks that the map $f$ is an embedding.
\end{proof}
\goodbreak
So the following statements about a FAC poset $P$ and regular cardinal $\kappa$ are equivalent: (1) $P$ contains a poset in the class $\calP(\kappa)$, (2) $P$ contains $[\kappa]^2$ or the dual of $[\kappa]^2$, (3) $P$ or $P^*$ contains a subposet $Q$ that is the union of an increasing chain of length $\kappa$ of ideals, unbounded in $Q$ and each containing a cofinal chain of order-type $\kappa$.
\section{Proofs of Theorem~\ref{thm:covering}}
The ``if'' directions of both items~\eqref{theo2.1} and~\eqref{theo2.2} of the theorem are easily obtained, since $\Cov([\kappa^+]^2=\kappa^+$ for any infinite cardinal $\kappa$.
\begin{enumerate}
\item In case $\nu = \kappa^+$ is a successor cardinal and $Q:=[\kappa^+]^2$ or its inverse is a subposet of $P$, then $\Cov(P)\geq \kappa^+$.
\item Suppose that $\nu$ is a limit uncountable cardinal and $P$ or its inverse $P^*$ contains a subposet $Q= \sum_{a\,\in\,C}Q_{a}$, where $C$ is a chain of cardinality $\cf (\nu)$, $Q_{a}:= [\kappa_{a}^+]^2$ and $(\kappa_{a}^+)_{\alpha\,\in\,C}$ is a family of cardinals such that $\sup \{\kappa_{a}^+: a\in C\} = \nu$. Then $\Cov(Q)= \nu$, because otherwise $\Cov(Q)=\kappa<\nu$ despite the fact that $P$ contains a copy of $[\kappa_\alpha^+]^2$ for some $\kappa_\alpha^+>\kappa$.
\end{enumerate}
Thus the main burden of the proof of Theorem~\ref{thm:covering} is in the ``only if'' direction.
\ignore{ Thus, in view of the inductive scheme we assume the following about an uncountable cardinal $\nu_0$:
\begin{equation}
\text{Theorem~\ref{thm:covering} holds for every uncountable cardinal $\nu < \nu_0$.}
\end{equation}
We are given a poset $P_0$ with $\Cov(P_0)\geq \nu_0$, and we have to prove the theorem for $P_0$ and $\nu_0$. If there is some $p\in P_0$ such that $\Cov(p^\bot)\geq \nu_0$, then induction can be applied to $p^\bot$ because the antichain-rank of $p^\bot$ is smaller than that of $P_0$. This would conclude the proof of Theorem~\ref{thm:covering}. So we may assume about $P_0$ that $\forall p\in P_0\ \Cov(p^\bot) < \nu_0$. Thus to prove Theorem~\ref{thm:covering}, it suffices to prove the following seemingly weaker version. }
It suffices to prove the seemingly weaker version of Theorem~\ref{thm:covering} in which an additional assumption is made (equation~\eqref{Eq:ind}).
\begin{theo}[{Theorem~\ref{thm:covering}*}]
Let $P$ be a poset with no infinite antichain and let $\nu$ be an uncountable cardinal such that $\Cov(P)\geq \nu$ and
\begin{equation}\label{Eq:ind}
\forall p\in P\ \Cov\left(p^\bot\right) < \nu.
\end{equation}
\begin{enumerate}
\item\label{theo2'.1} If $\nu$ is a successor cardinal, i.e., $\nu= \kappa^+$, then $P$ or its dual $P^*$ contains a copy of $[\kappa^+]^2$.
\item\label{theo2'.2} If $\nu$ is a limit cardinal then $P$ or its dual $P^*$ contains a poset $Q$ of the form $\sum_{a\,\in\,C}Q_{a}$, where $C$ is a chain of cardinality $\cf (\nu)$, $Q_{a} \cong [\kappa_{a}^+]^2$ and $(\kappa_{a}^+)_{a\in C}$ is a family of distinct successor cardinals, such that $\sup \{\kappa_{a}^+: a\in C\} = \nu$.
\end{enumerate}
\end{theo}
\begin{lemm}\label{lem:two-versions}
For every uncountable cardinal $\nu$, if Theorem~\ref{thm:covering}* holds for $\nu$ then Theorem~\ref{thm:covering} also holds for $\nu$, so that these two forms are equivalent.
\end{lemm}
Assume the statement of Theorem~\ref{thm:covering}*, and let $P$ be a FAC poset such that $\Cov(P)\geq \nu$. The antichains of $P$ form a well-founded relation under the inverse inclusion relation, and we let $P_0$ be a subposet of $P$ with the least height of its antichains such that still $\Cov(P_0)\geq \nu$. Then for every $p\in P_0$, the antichain height of $P_0\cap p^\bot$ is smaller than the antichain height of $P_0$, and hence $\Cov(P_0\cap p^\bot)<\nu$. Thus Theorem~\ref{thm:covering}* can be applied to $P_0$. So $P_0$ contains subposets as required by item~\eqref{theo2'.1} or~\eqref{theo2'.2}, and these are subsets of $P$ which prove that Theorem~\ref{thm:covering} follows.
Our plan is to first define an equivalence relation $\sim$ on $P$, and get some information on the covering number of the equivalence classes of $\sim$ (in Lemma~\ref{Lem:A}). Then, we formulate a reduction theorem, Theorem~\ref{thm:covering2}, which seems to be weaker than Theorem~\ref{thm:covering}* in two aspect: it deals only with the case that $\nu=\kappa^+$ is a successor cardinal, and it strengthen assumption~\eqref{Eq:ind} of Theorem~\ref{thm:covering}* to $\forall p\in P\ \Cov(P\setminus \uparrownogap p) < \nu$. With the aid of the analysis of the equivalent classes of $\sim$ it turns out that the reduction theorem suffices to yield the full Theorem~\ref{thm:covering} for every uncountable cardinal $\nu$.
\begin{defi}
For any poset $P$, an equivalence relation $\sim$ is defined over $P$ by $a\sim b$ iff $a=b$ or there is a finite sequence $x_1,\,\ldots,\,x_n$ such that $a=x_1$, $b= x_n$, and $x_i \Inc x_{i+1}$ for every $1\leq i < n$. We say that this sequence joins $a$ to $b$ in $\sim$. Borrowing the language of graph theory, equivalence classes of $\sim$ are also called \emph{connected components} or just \emph{components}.
\end{defi}
Let $A$ and $B$ be equivalence classes of $\sim$. Recall that $A \nearrow B$ is a shorthand for $\forall a\in A\, \forall b\in B\, (a\leq_P b)$. We claim that the equivalence classes of $P$ are linearly ordered under the $\nearrow$ reflexive and transitive relation. Indeed, if $A\neq B$ are distinct equivalence classes, if $a\in A$ and $b\in B$, then $a$ and $b$ are comparable in $<_P$, and if $a<_P b$ for example, then for every $a'\in A$ and $b'\in B$, $a'<_P b'$. This can be proved by induction on the length of the $\Inc$ paths that lead from $a$ to $a'$ and from $b$ to $b'$. It follows immediately that any poset $P$ is the sum of its components along their linear ordering. This result which belongs to the folklore of the theory of ordered sets (see~\cite[I.1.2]{pouzet78} and~\cite[2.7]{abraham}) is used here in connection with the covering numbers of its components in the following.
\begin{lemm}\label{Lem:A}
\label{lem:covering3}
Any poset $P$ is the lexicographical sum, $P:= \sum_{i\,\in\,D} P_i$, of its equivalence classes $P_i$, indexed by the chain $D$ that reflects the $\nearrow$ ordering on the equivalence classes.
Suppose that $\Cov(P_i)=\rho_i$ for every $i\in D$, and let $\kappa := \sup\{\rho_i \mid i\in D\}$ be the supremum of these cardinals. Then
\begin{equation}\label{eq:kappa}
\Cov(P) = \kappa.
\end{equation}
Moreover, there are two exclusive possibilities:
\begin{enumerate}
\item\label{lemm23.1} Either $\Cov(P_i)<\kappa$ for every $i\in D$, and then $\kappa$ is a limit cardinal, and for some $D'\subset D$ of cardinality $\cf(\kappa)$ and (distinct) successor cardinals $\kappa^+_i$, $\Cov(P_i)\geq \kappa_i^+$ for every $i\in D'$, and
\[
\kappa = \sup \left\{\kappa_i^+ \mid i\in D'\right\}.
\]
Or else,
\item\label{lemm23.2} There exists an equivalence class $P_{i_0}$ in $P$ with a maximal cardinality $\Cov(P_{i_0})=\Cov(P)$ (i.e. $\kappa = \Cov(P_{i_0})$).
\end{enumerate}
\end{lemm}
\begin{proof}
For every $i\in D$, let $(L_\xi^i\mid \xi < \rho_i)$ be an enumeration of chains that cover $P_i$. For every given $\xi < \kappa$, the union of the chains, $L_\xi = \bigcup \{ L^i_\xi\mid i\in D \wedge \xi < \rho_i\}$, is a chain of $P$, and $P$ is covered by $\calL = \{ L_\xi\mid \xi < \kappa\}$. Hence $\Cov(P)\leq |\calL | \leq \kappa$. But as $\rho_i\leq \Cov(P)$ for every $i\in D$, $\Cov(P)=|\calL | = \kappa$, and thus~\eqref{eq:kappa} holds.
In case this supremum is a maximum, i.e. for some $i_0\in D$, $\Cov(P_i)\leq\rho_{i_0} = \Cov(P_{i_0})$ for all~$i$, then $\Cov(P)=\rho_{i_0}$. Otherwise $\kappa$ is a limit cardinal, and for any $\mu<\kappa$ there is an equivalence class $P_i$ with $\Cov(P_i)>\mu$. In this case the set $D'$ can easily be chosen so that the first possibility holds.
\end{proof}
The following simple lemma turns out to be very useful in our proof (as was~\cite[Lemma~2.9]{abraham}).
\begin{lemm}\label{lem:metric}
Let $P$ be a poset and $x, y$ be
such that $x\sim y$. Suppose that
\begin{equation}\label{assumption-on-xy}
x=x_0,\,\ldots,\,x_n=y \quad\text{is a path that joins $x$ to $y$.}
\end{equation}
Then
\begin{equation}\label{Eq:Inc}
\left[x,y\right]_P \subseteq x_1^\bot \cup \dots \cup x_{n-1}^\bot.
\end{equation}
\end{lemm}
\begin{proof}
Suppose that $z\in [x,y]$, that is $x\leq_P z\leq_P y$. If $z=x_0$, then $z\bot x_1$ evidently. Assume that $x_0 <_P z \leq_P x_n$. Noting that $z\not \leq_P x_0$, let $0< m \leq n$ be the first such that $z \leq_P x_m$ (thus $m \geq 1$). So $z \not \leq_P x_{m-1}$, and since $x_{m-1}\leq_P z$ is impossible (for $x_{m-1}\bot x_m$), $x_{m-1} \bot z$ follows. (This proof includes the case that $x\bot y$. In that case $[x,y] = \{x,y \}$ and $x_1^\bot \cup \dots \cup x_{n-1}^\bot= x_1^\bot\cup x_0^\bot$, so that $\{x,y\}\subseteq x_1^\bot\cup x_0^\bot$ is evident.)
\end{proof}
\begin{coro}\label{cor:interval}
Let $\nu$ be an infinite cardinal, and $P$ be a poset such that $\Cov(x^\Inc) < \nu$ for every $x\in P$. If $Q$ is a component of $P$, then $\Cov([x, y]_P) < \nu$ for every $x,y\in Q$ such that $x <_P y$.
\end{coro}
\begin{proof}
Since $ [x, y]_P \subseteq x_1^\bot \cup \dots \cup x_{n-1}^\bot$, $\Cov([x, y]_P) \leq \Cov(x_1^\bot)+ \dots + \Cov (x_n^\bot)$ and since $\Cov(x^\bot) <
\nu$ for each $x\in P$, the corollary follows immediately.
\end{proof}
Lemma~\ref{Lem:A} and Corollary~\ref{cor:interval} are used next.
\begin{lemm}\label{lem:cov}
Let $\nu$ be an uncountable cardinal and $P$ be a poset such that $\Cov (P)\geq \nu$, but
\begin{equation}\label{Eq:cov-p-bot-small}
\forall p\in P\ \Cov\left(p^\bot\right)<\nu.
\end{equation}
Then either
\begin{enumerate}
\item\label{lemm26.1}
each connected component of $P$ has covering number $<\nu$, and in this case $\Cov(P)=\nu = \sup \{\Cov(X) \mid X \text{ is a component of } P\}$, or else
\item\label{lemm26.2} some connected component $Q$ of $P$ contains a subposet $R$ such that $\Cov (R)\geq \nu$, and $R$ or its dual has the property that $\Cov(R\setminus \uparrownogap x)< \nu$ for every $x\in R$.
\end{enumerate}
\end{lemm}
\begin{proof}
Decompose $P$ into its $\sim$-equivalence classes. By Lemma~\ref{Lem:A}, $\Cov(P) =\sup \{\Cov(X) \mid X \text{ is a component of } P\}$.
If each component $X$ has covering number $<\nu$, then we are in Lemma~\ref{lem:cov}$\MK$\eqref{lemm26.1}, and $\nu =\Cov(P)$ is a limit cardinal. Otherwise, there exists $Q$, a connected component of $P$ such that $\Cov(P) = \Cov (Q)\geq \nu$.
\begin{enumerate}
\item If for every $x\in Q$, $\Cov(Q\cap \uparrownogap x)<\nu$ then combining this condition with assumption~\eqref{Eq:cov-p-bot-small} we get that $\Cov(Q\setminus \downarrownogap x)<\nu$ for every $x\in Q$. In this case $R = Q^*$ satisfies Lemma~\ref{lem:cov}$\MK$\eqref{lemm26.2}.
\item Otherwise, there is some $x_0\in Q$ such that $\Cov(Q\cap \uparrownogap x_0)\geq \nu$. In this case, we apply Corollary~\ref {cor:interval} to $Q$ and get that $\Cov[x_0,y]_P<\nu$ for every $y\in Q$ such that $x_0<_P y$. Thus, the subposet $R = Q\cap (\uparrownogap x_0)$ satisfies $\Cov (R\cap (\downarrownogap x))< \nu$ for every $x\in R$. Combining this property of $R$ with assumption~\eqref{Eq:cov-p-bot-small} we get that $\Cov(R\setminus \uparrownogap x)<\nu$ for every $x\in R$. Now condition~\eqref{lemm26.2} of our lemma holds for $R$ (which is a subposet of $P$).
\end{enumerate}
In both cases, Lemma~\ref{lem:cov}$\MK$\eqref{lemm26.2} holds for $P$ or its inverse.
\end{proof}
Here is the reduction theorem which we intend to prove in the next section.
\begin{theo}\label{thm:covering2}
Let $\nu=\kappa^+$ be any uncountable successor cardinal, and $P$ be a poset with no infinite antichain, satisfying
\begin{equation}\label{eq:assumptions}
\Cov (P)\geq \nu\ \wedge\ \Cov(P\setminus \uparrownogap x)< \nu\ \text{for every } x\in P.
\end{equation}
Then $P$ contains a copy of $[\nu]^2$.
Evidently, in~\eqref{eq:assumptions}, we can replace $\Cov(P\setminus \uparrownogap x)< \nu$ with $\Cov(P\setminus \downarrownogap x)< \nu$ and conclude that $P$ contains a copy of $[\nu]^2$ inverse.
\end{theo}
In this section we prove that Theorem~\ref{thm:covering2} implies Theorem~\ref{thm:covering}, and for the proof of this implication some notations are needed. For any uncountable successor cardinal $\kappa^+$ let $\T$\ref{thm:covering2}$(\kappa^+)$ denote the statement of Theorem~\ref{thm:covering2} for the specific cardinal $\nu = \kappa^+$, and for a limit uncountable cardinal $\nu$, let $\T$\ref{thm:covering2}$(<\nu)$ be the statement that for every successor uncountable cardinal $\kappa^+< \nu$, $\T$\ref{thm:covering2}$(\kappa^+)$. Similar notations such as
$\T$\ref{thm:covering}$(\nu)$ and $\T$\ref{thm:covering}$^*(\nu)$ will be used. Recall that we proved
\begin{equation}\label{Eq:cov-cov*}
\T\ref{thm:covering}(\nu) \iff \T\ref{thm:covering}^*(\nu)
\end{equation}
for every uncountable cardinal $\nu$ (this is Lemma~\ref{lem:two-versions}).
The proof of Theorem~\ref{thm:covering2} will occupy the next section, and we conclude this section with a deduction of Theorem~\ref{thm:covering} from Theorem~\ref{thm:covering2}. This deduction is spread over two theorems, the first for the successor case and the second for the limit case.
\begin{theo}[Successor case]\label{thm:successor}
Let $\kappa^+$ be any uncountable successor cardinal. Then
\begin{equation}\label{Eq-thm-succ}
\T\ref{thm:covering2}\left(\kappa^+\right)\rightarrow \T\ref{thm:covering}\left(\kappa^+\right).
\end{equation}
\end{theo}
\begin{proof}
We are going to prove that
\T\ref{thm:covering2}$(\kappa^+)\rightarrow \T\ref{thm:covering}^*(\kappa^+)$, and then use equivalence~\eqref{Eq:cov-cov*} in order to conclude~\eqref{Eq-thm-succ}. Assume \T\ref{thm:covering2}$(\kappa^+)$, and let $P$ be a FAC poset such that $\Cov(P)\geq \kappa^+$ and for every $p\in P$, $\Cov(p^\bot) \leq \kappa$. We have to prove that $P$ or $P^*$ contains a copy of $[\kappa^+]^2$.
Lemma~\ref{lem:cov} yields two possibilities for $P$:
\begin{enumerate}
\item For every component $X$ of $P$, $\Cov(X)\leq \kappa$. In this case, the chains that cover the different components can be attached to form a covering of $P$ with $\leq \kappa$ chains in contradiction to $\Cov(P)\geq \kappa^+$. So this case is void.
\item There is a subposet $Q$ of $P$ such that $\Cov(Q)\geq \kappa^+$ and $Q$ or $Q^*$ satisfies the condition that
\begin{equation}
\forall x\in Q\ \left(\Cov\left(Q\setminus \uparrownogap x\right)\leq \kappa\right).
\end{equation}
Then we can use \T\ref{thm:covering2}$(\kappa^+)$ and conclude that $Q$ or $Q^*$ contains a copy of $[\kappa^+]^2$. Evidently, it follows that $P$ or $P^*$ contains such a copy.
\end{enumerate}
\end{proof}
\begin{theo}[Limit case]\label{thm:limit}
Let $\nu$ be any uncountable limit cardinal. Then
\T\ref{thm:covering2}$(<\nu)$ implies \T\ref{thm:covering}$(\nu)$.
\end{theo}
\begin{proof}
Suppose \T\ref{thm:covering2}$(<\nu)$, namely that Theorem~\ref{thm:covering2} holds for every uncountable successor cardinal $\kappa^+<\nu$. Then, by the previous theorem (the successor case),
\begin{equation}\label{Eq:all-successors}
\mbox{\T~\ref{thm:covering}$\left(\kappa^+\right)$ holds for every uncountable successor cardinal $\kappa^+<\nu$.}
\end{equation}
Given a FAC poset $P$ such that $\Cov(P)\geq \nu$, our aim is to prove that $P$ or $P^*$ contains a sum $\Sigma_{a\in C}[\kappa_a^+]$ as in item 2 of Theorem~\ref{thm:covering}.
For every connected component $X$ of $P$ define $\rho(X) = \Cov(X)$. Then (by Lemma~\ref{Lem:A})
\begin{equation}
\Cov(P)= \sup\left\{\rho(X)\mid X \text{ is a component of } P\right\}.
\end{equation}
Apply Lemma~\ref{lem:cov} to $P$, and consider its two cases.
\begin{enumerate}
\item The first case is when $\rho(X)<\nu$ for every component $X$ of $P$. In this case, take an increasing sequence of successor cardinals $(\kappa_i^+\mid i<\cf(\kappa))$ with supremum $\nu$, and for each $\kappa_i$ take a specific component $X_i$ with covering number $\rho(X_i)\geq \kappa_i^+$. This procedure yields a linear ordering $D$ (not necessarily well-ordered) of cardinality $\cf(\nu)$ of components such that $P$ contains a sum of components, $\sum_{i\,\in\,D}X_i$, where $\Cov(X_i)\geq \kappa_i^+$.
Since Theorem~\ref{thm:covering} holds for uncountable successor cardinals smaller than $\nu$, $X_i$ embeds either $[\kappa^+_{i}]^2$ or its dual. It follows that for some $C\subseteq D$ of cardinality $\cf(\nu)$, $P$ or $P^*$ contains a sum of the form $\sum_{i\,\in\,C} [\kappa^+_i]^2$ where $\sup\{\kappa_i^+\mid i\in C\}=\nu$.
\item The second case is that $P$ contains a subposet $Q$ such that $\Cov(Q) \geq \nu$ and $Q$ or $Q^*$ satisfies
\begin{equation}\label{eq:assumptions2}
\Cov\left(Q\setminus \uparrownogap x\right)< \nu\ \text{for every } x\in Q.
\end{equation}
\end{enumerate}
The following lemma applies to $Q$ and concludes the proof of Theorem~\ref{thm:limit}.
\begin{lemm}
Suppose that $\nu$ is an uncountable limit cardinal such that \T\ref{thm:covering2}$(<\nu)$ holds. If $Q$ satisfies equation~\eqref{eq:assumptions2}, then $Q$ contains a sum $\Sigma_{a\,\in\,C}[\kappa_a^+]^2$ where $C$ is a linear order of order-type $\cf(\nu)$ and $\sup\{ \kappa_a \mid a\in C\}=\nu$.
\end{lemm}
\begin{proof}
%\renewcommand{\qedsymbol}{}
Suppose that assumption~\eqref{eq:assumptions2} holds for a limit uncountable cardinal $\nu$ (which could be regular or singular). We must conclude that $Q$ contains a sum $\sum_{a \in C} [\kappa^+_a]^2$ as required.
We prove first for any $X\subseteq Q$ that
\begin{equation}\label{fact1}
|X|<\cf(\nu) \ \Rightarrow\ X \text{ is bounded in } Q.
\end{equation}
By assumption~\eqref{eq:assumptions2}, $\Cov (Q \setminus \uparrownogap x)<\nu$ for every $x$, and hence the covering number of $Y = \bigcup_{x\in X} {(Q\setminus \uparrownogap x)}$ is less than $\nu$. So there exists some $x_0\in Q\setminus Y$, which is necessarily an upper-bound of $X$.
Let $(\kappa_{\alpha}^+)_{\alpha\,<\,\cf(\nu)}$ be an increasing sequence of successor cardinals, all less than $\nu$, such that $\sup \{\kappa_{\alpha}^+: \alpha<\cf(\nu)\} = \nu$.
We construct an increasing sequence $(x_{\alpha})_{\alpha\,<\,\cf(\nu)}$ in $Q$ such that $\Cov ([x_{\alpha}, x_{\alpha+1})) \geq \kappa^{+}_{\alpha}$ for every $\alpha <\cf(\nu)$.
Equation~\eqref{fact1} shows that for limit ordinals $\delta<\cf(\nu)$, if the sequence $(x_\alpha)_{\alpha\,<\,\delta}$ is defined, then an upper-bound $x_\delta$ can be obtained (and for $\delta = 0$, $x_0\in Q$ is arbitrarily chosen). Suppose that $x_\alpha\in Q$ is defined. Since $\Cov(Q)\geq \nu$ but $\Cov (Q \setminus \uparrownogap x_\alpha) < \nu$, $\Cov (\uparrownogap x_\alpha) \geq \nu > \kappa_{\alpha}^{++}$. We concluded in~\eqref{Eq:all-successors} that $\T\ref{thm:covering}(\kappa^+)$ holds for any successor cardinals $\kappa^+<\nu$, and thus $\uparrownogap x_\alpha$ contains a copy of $[\kappa_{\alpha}^{++}]^2$ or its reversed dual. Since $[\kappa_\alpha^{++}]^2$ contains a \emph{bounded} copy of $[\kappa_\alpha^+]^2$, there is some element $x_{\delta+1}\in Q$ such that a copy of $[\kappa_{\alpha}^{+}]^2$ or of its reverse is contained in the interval $[x_\delta, x_{\delta+1}]$. (The inverse of any $[\kappa]^2$ is certainly bounded.)
By picking a subsequence, we may assume that all intervals $[x_\delta, x_{\delta+1}]$ contain copies of the corresponding $[\kappa^{+}_{\alpha}]^2$ or else that all of them contain copies of the dual of these posets.
\end{proof}
\end{proof}
This ends the proof of Theorem~\ref{thm:limit} (the limit case), and together with Theorem~\ref{thm:successor} we get that
\begin{equation}
\forall \kappa>\omega \left(\T\ref{thm:covering2}\left(\kappa^+\right)\right)\rightarrow \forall
\nu>\omega \left(\T\ref{thm:covering}(\nu)\right).
\end{equation}
Thus it remains to prove that Theorem~\ref{thm:covering2} holds for all uncountable successor cardinals.
\subsection{Proofs of Theorem~\ref{thm:covering2}}
\begin{lemm}\label{Lem3.7}
Let $\nu = \kappa^+$ be a successor cardinal, and suppose that $P$ is an up-directed FAC poset such that $\Cov(P)\geq \nu$, but $\Cov (\ngp)< \nu$ for every $p\in P$. If $P$ does not contain a copy of $[\kappa^+]^2$, then there is a well-ordered chain with order-type some regular $\lambda\geq \nu$ that is cofinal in $P$.
\end{lemm}
\begin{proof}
Since $\Cov (P\setminus \uparrownogap x)<\nu$ for every $x\in P$ and $\Cov (P) \geq \nu$, then $P$ is up-directed. Indeed, we proved in~\eqref{fact1} that every
set of cardinality $<\cf(\nu)$ is bounded, and thus certainly $P$ is up-directed. According to Theorem~\ref{cofinalthm2}, $P$ has a cofinal subset $A$ isomorphic to a product $\alpha_{1}\times \dots \times \alpha_{n}$ of pairwise distinct regular cardinals enumerated (for clarity) in increasing order.
We claim that, assuming $P$ does not contain a copy of $[\kappa^+]^2$, each cardinal $\alpha_{j}$, for $1\leq j\leq n$, is at least $\nu$. For the proof, suppose on the contrary that $\alpha_{1}<\nu$. Since $\nu=\kappa^+$ is a successor cardinal and $\alpha_{1}$ is a cardinal, $\alpha_{1} \leq \kappa$. Let $A_1:= \{x_{\alpha}: \alpha<\alpha_{1}\}$ be the image in $A$ of the chain $\{(z,0,\,\ldots,\,0)\in \alpha_{1}\times \dots \times \alpha_{n}\mid z<\alpha_{1}\}$. Since $\Cov(P\setminus \uparrownogap x_{\alpha}) \leq \kappa$ and $\alpha_{1}\leq \kappa$, $\Cov(\bigcup_{\alpha<\alpha_{1}} (P\setminus \uparrownogap x_{\alpha})) \leq \kappa$. But this union is $P$ since the chain $A_1$ is unbounded. This contradicts $\Cov (P) \geq \kappa^+$. Thus the product $\alpha_{1}\times \dots \times \alpha_{n}$ consists of cardinals that are $\geq \kappa^+$. If this product consists of two or more cardinals, then $P$ contains a copy of $[\kappa^+]^2$. Thus, if we assume that $P$ does not contain a copy of $[\kappa^+]^2$, then $P$ has a cofinal chain of order-type some regular $\lambda\geq \nu$.
\end{proof}
From this point of the article the two proofs of Theorem~\ref {thm:covering2} split. For one proof jump to section~\ref{sec:secondProof}, and for the other proof continue reading.
\subsubsection{Completion of proof of Theorem~\ref {thm:covering2}}
Let's restate Theorem~\ref{thm:covering2} for convenience.
\begin{theo*}[Theorem~\ref{thm:covering2}]\label{Lem3.2a}
Assume that $P$ has only finite antichains, and that $\nu = \kappa^+$ is an uncountable successor cardinal such that $\Cov(P)\geq \kappa^+$ but $\Cov (\ngp)\leq \kappa$ for every $p\in P$. Then $P$ contains a subposet in $\calP(\kappa^+)$ and hence $P$ contains a copy of $[\kappa^+]^2$. (And, correspondingly, if for every $p\in P$, $\Cov (\nots{p}) \leq \kappa$, then $P^*$ contains a copy of $[\kappa^+]^2$.)
\end{theo*}
\begin{proof}
With Lemma~\ref{Lem3.7} we may further assume that $P$ has a cofinal chain $A=\{ x_\xi\mid \xi < \lambda\}$ of order-type $\lambda\geq \kappa^+$ ($\lambda$ is regular). Let $\gamma_0\leq \lambda$ be the first ordinal such that the initial segment $I=\bigcup \{\downarrownogap x_\xi\mid \xi < \gamma_0\}$ has covering number $\geq \kappa^+$. Then redefine $P:= I$, $\lambda :=\cf\, \gamma_0$, so that the assumptions of our theorem remain for the new $P$ and $\lambda$. The set $\{x_\xi\mid \xi < \lambda\}$ of order-type $\lambda\geq \kappa^+$ is cofinal in $P$, and we gain a useful property that if $Q$ is a subposet of $P$ of cofinality $\cf(Q)<\lambda$ then:
\begin{equation}\label{eq:on-Q}
\Cov(Q)\leq \kappa.
\end{equation}
For every $\zeta<\lambda$ define
\begin{equation}\label{DefLayer}
D_\zeta= \left(\downarrownogap x_\zeta\right)\setminus \bigcup_{\xi\,<\,\zeta}\left(\downarrownogap x_\xi\right).
\end{equation}
$D_\zeta$ is said to be the $\zeta$ \emph{layer} of $P$. Surely $x_\zeta\in D_\zeta$ so that $D_\zeta\neq \emptyset$.
The following is obvious.
\begin{enumerate}
\item $x_\zeta\in D_{\zeta}$ is the maximum of $D_\zeta$, and $P=\bigcup_{\zeta\,<\,\lambda}D_\zeta$ is a partition of $P$ into its layers.
\item If $\zeta'<\zeta<\lambda$, $x\in D_{\zeta'}$, $y\in D_\zeta$ then $\neg(y\leq_P x)$ so that if $x$ and $y$ are comparable, then $x< y$. Thus any $<_P$ relation that exists between members of $P$ at different layers is in accordance with the layers' indexes.
\item Since $D_\zeta\subseteq (\downarrownogap x_\zeta)$, $D_\zeta$ is a union of $\kappa$ chains.
\end{enumerate}
\begin{defi}
Define $F: P \to \lambda$ by $F(p)=\xi$ iff $\xi<\lambda$ is the first ordinal such that $\neg (x_\xi <_P p)$.
\end{defi}
\begin{enonce}[plain]{Claim}\label{CLcc}\leavevmode
\begin{enumerate}
\item\label{claim33.1} For every $\xi<\lambda$, $F^{-1}\{\xi\}\subseteq \{x_\xi \} \cup (P\ \setminus\ \uparrownogap x_\xi)$ and hence $F^{-1}\{\xi\}$ is a union of $\kappa$-chains. If $x_\zeta < p$ then $F(p)>\zeta$. If $p\in D_\zeta$ then $F(p)\leq \zeta$.
\item\label{claim33.2} If $x<_P y$ then $F(x)\leq F(y)$.
\item\label{claim33.3} If $\zeta'<\zeta<\lambda$, $x\in D_{\zeta'}$, $y\in D_\zeta$ and $F(x)>F(y)$, then $x\ \Inc\ y$.
\item\label{claim33.4} For every $\xi<\lambda$, $F(x_\xi)=\xi$.
\end{enumerate}
\end{enonce}
\begin{proof}
All items follow directly the definition of $F$. To prove Item~\eqref{claim33.3} for example, suppose that $x$ is comparable to $y$. Then $x<_P y$ follows and Item~\eqref{claim33.2} implies that $F(x)\leq F(y)$, which contradicts assumption $F(x)>F(y)$. Item~\eqref{claim33.3} will lead to the conclusion that $P$ contains a poset in $\calP(\kappa^+)$.
\end{proof}
For every $\xi\in\lambda$ define $\gamma(\xi)\leq \lambda$ by
\[
\gamma(\xi) = \sup \left\{\mu\in\lambda \mid{ \exists x}\in D_\mu\, (F(x) =\xi) \right\}.
\]
Since $F(x_\xi)=\xi$, $\xi\leq \gamma(\xi)$.
Define a partition of $\lambda$ into two sets, $B$ and $U=\lambda\setminus B$ (Bounded and Unbounded): $\beta\in B$ iff $\gamma(\beta)<\lambda$ (i.e. $F^{-1}\{\beta\}$ intersects only a bounded set of layers). It follows immediately that
\begin{equation}\label{Eq3}
p\in D_\mu \wedge F(p)=\xi\in B \rightarrow \mu\leq \gamma(\xi).
\end{equation}
Since $U=\lambda \setminus B$, $\beta\in U$ iff $\gamma(\beta)=\lambda$, i.e. the set of layers that intersect $F^{-1}$ is unbounded in $\lambda$. Define $\CalB =F^{-1}B$ and $\U=F^{-1}U$, so that $P=\CalB\cup \U$.
We prove next that $\Cov(\CalB)\leq \kappa$ and conclude that $\Cov(\U)>\kappa$. Then we will prove that $\U$ contains a copy of $[\kappa^+]^2$ (thereby proving Theorem~\ref{thm:covering2}).
\begin{lemm}\label{Claim:cov}
$\Cov(\B)\leq \kappa$.
\end{lemm}
\begin{proof}
Consider the function $\gamma\restriction B$ defined over $B$. Cardinal $\lambda$ is uncountable and regular, and hence the set $C=\{\delta\in\lambda\mid \forall \beta\in B\cap\delta\, (\gamma(\beta) < \delta)\}$ is closed unbounded in $\lambda$. For every $\delta\in C$ let $\delta'\in C$ denote the first ordinal in $C$ above $\delta$, and let $I_\delta=[\delta,\delta')$ be the ordinal interval $\delta'\setminus\delta$. Then $\lambda=\bigcup_{\delta\,\in\,C}I_\delta$ is a partition of $\lambda$ into slices of cardinality $<\lambda$ each. Define, for $\delta\in C$,
\begin{equation}
\CalB(\delta):= \bigcup_{\xi\,\in\,I_\delta}D_\xi\cap F^{-1}B.
\end{equation}
Then $\CalB(\delta)$ is a union of $\kappa$ chains because $\bigcup_{\xi\,\in\,I_\delta}(\downarrownogap x_\xi)$ is a subset of $P$ with a cofinal subset of cardinality $|I_\delta|<\lambda$, and hence is a union of $\kappa$ chains by~\eqref{eq:on-Q}. Thus its subset, $\CalB(\delta)$, is also a union of $\kappa$ chains.
\goodbreak
\begin{enonce}[plain]{Claim}\label{Cl:comp}\leavevmode
\begin{enumerate}
\item\label{claim35.1} Suppose that $p\in \CalB(\delta)$ (where $\delta\in C$). Then $p<_P x_{\delta'}$ and $F(p)\geq \delta$.
\item\label{claim35.2} If $\delta_1<\delta_2$ are in the club set $C$ defined above, then for every $p_1\in \CalB(\delta_1)$ and $p_2\in\CalB(p_2)$, $p_1<_P p_2$.
\end{enumerate}
\end{enonce}
\begin{proof}
Suppose that $p\in \CalB(\delta)$. Then $p\in D_\mu$ for some $\mu\in I_\delta$, and hence $p\leq x_\mu \kappa$. It remains to prove that $\U =F^{-1} U$ contains a copy of $[\kappa^+]^2$ in order to conclude the proof of Theorem~\ref{thm:covering2}.
We prove first that $|U|\geq \kappa^+$. If not, if $|U|\leq \kappa$ then $\U = \bigcup_{\zeta\in U} F^{-1} \{\zeta\}$ is covered by $\kappa$ chains since every $F^{-1}\{\zeta\}$ is so covered (by item~\eqref{claim33.1} of Claim~\ref{CLcc}).
To prove that $\U$ contains a copy of $[\kappa^+]^2$, let $U_0\subseteq U$ be a subset of order-type $\kappa^+$. For $\alpha\in U_0$, $\gamma(\alpha)=\lambda$, i.e. $S_\alpha = \{\mu\in\lambda \mid \exists x\in D_\mu\, F(x)=\alpha\}$ is a subset of $\lambda$ of cardinality $\lambda$. It is not difficult to find $R_\alpha\subset S_\alpha$ that are pairwise disjoint and of order-type $\kappa^+$ such that $R=\bigcup_{\alpha\,\in\,U_0}R_\alpha$ is also of order-type $\kappa^+$. We identify $R \simeq \kappa^+$ with $\kappa^+$. For every $\mu\in R$ there is some $\alpha\in U_0$ and $p(\mu)\in D_\mu$ such that $F(p(\mu))=\alpha$. Define $P_0=\{ p(\mu)\mid \mu \in R\}\subset P$, and then the injection $p:R\to P_0$ induces a partial ordering on $R\simeq \kappa^+$ which is isomorphic to $P_0$. That is, for $\zeta_1,\zeta_2\in \kappa^+$, we define $\zeta_1<_P \zeta_2$ iff $p(\zeta_1)<_P p(\zeta_2)$. The function $\bfF:R\to U_0$ defined by $\bfF(\mu)=F(p(\mu))$ shows that $P\in \calP(\kappa^+)$ (use item~\eqref{claim33.3} of Claim~\ref{CLcc}). Thus $[\kappa^+]^2$ is embeddable in $R$. This completes the proof of Theorem~\ref{thm:covering2}.
\end{proof}
\subsubsection{Second proof of Theorem~\ref{thm:covering2}}\label{sec:secondProof}
Recall that in Lemma~\ref{Lem3.7} we proved that if $P$ is a FAC poset such that $\Cov(P)\geq \nu =\kappa^+$, but for every $p\in P$, $\Cov (\ngp)\leq \kappa$, then there is a chain $A$ cofinal in $P$ with well ordered order type some regular $\lambda\geq \nu$ (or else $[\nu]^2$ is embeddable into $P$, and in that case the theorem follows immediately). Hence we may assume in the following discussion that $A=\{x_\xi\mid\xi<\lambda\}$ is an increasing and cofinal in $P$ sequence.
We use the conclusion of Lemma~\ref{Lem3.7} in proving
\begin{lemm}\label{lem:cofinalchain}
Assume that $\nu = \kappa^+$ is a successor cardinal and $P$ is a FAC poset such
that $\Cov(P)\geq \nu$ but for every $p\in P$, $\Cov(P\setminus \uparrownogap p) \leq \kappa$. Then $P$ is impure.
\end{lemm}
\begin{proof}
Suppose for the sake of a contradiction that $P$ is pure. According to Theorem~\ref{theo:pure}, $P$ is a lexicographical sum $\sum_{a\,\in\,K }P_{a}$ where $K$ is a chain with cofinality $\cf(P)=\lambda$. For each $a\in K$, $\Cov(P_{a}) \leq \kappa$, because $P_a$ is bounded by any member of $P_b$ for $a<_K b$, and $\Cov(\downarrownogap p)<\nu$ for any $p\in P$ since $\downarrownogap p \subseteq P\setminus \uparrownogap p$. As $\Cov(P_{a}) \leq \kappa$ for every $a\in K$, and since $P$ is a sum of the $P_a$ convex subposets, we get (by union of chains) that $\Cov(P) \leq \kappa$. This contradicts our hypothesis that $\Cov(P)\geq \nu=\kappa^+$.
\end{proof}
\begin{lemm}\label{Lem:3.17}
Under the assumptions of Lemma~\ref{lem:cofinalchain}, $P$ contains a proper initial segment $J$ that is unbounded and has a cofinal sequence of order-type $\lambda$.
\end{lemm}
\begin{proof}
The previous lemma says that $P$ is impure and hence contains a proper initial segment $I$ that is unbounded in $P$. Let $\cf(I)=\lambda_0$ be the cofinality of $I$, and let $\{a_\alpha\mid\alpha<\lambda_0\}$ be a cofinal subset of $I$. Define a function $f:\lambda\to \lambda_0$ thus: since $x_\alpha\in A$ is not an upper-bound of $I$, there is some $a_\beta$ that is not dominated by $x_\alpha$; define $f(\alpha)=\beta$ for the least $\beta<\lambda_0$ such that $\neg(a_\beta\leq x_\alpha)$. Note that if $\alpha_1 < \alpha_2<\lambda$ then either $a_{f(\alpha_1)}\bot a_{f(\alpha_2)}$ or $a_{f(\alpha_1)} \leq_P a_{f(\alpha_2)}$.
Apply the Erd\"os--Dushnik--Miller theorem to the partition of $\lambda$ that gives one color to the pair $\alpha_1<\alpha_2<\lambda$ if $a_{f(\alpha_1)}\bot a_{f(\alpha_2)}$ and the other color if not. Since $P$ satisfies the FAC, there is a set $X\subset\lambda$ of cardinality $\lambda$ that is homogeneous for the comparability color. The set $K=\{a_{f(\alpha)}\mid \alpha\in X\}$ is a chain of $P$ and the function $x_\alpha\leadsto a_{f(\alpha)}$ is $<_P$ monotonic. Moreover, $K$ is unbounded in $P$ because any $p\in P$ is bounded by some $x_\alpha$ and hence $\neg(a_{f(\alpha)}\leq_P p$. Thus the ideal $J =\downarrownogap K$ has $K$ as a cofinal sequence of order-type $\lambda$.
\end{proof}
Note that if $x\in P$ then $P':= \uparrownogap x$ has the same properties of $P$, i.e. $\Cov(P')\geq \kappa^+$ and for every $p\in P'$, $\Cov(p^\bot)\leq \kappa$. Hence Lemma~\ref{Lem:3.17} applies to $P'$ and there is a proper initial segment of $P'$ that is unbounded (even in $P$) and has a cofinal sequence of order-type $\lambda$.
Let $\frkI_{P}(\lambda)$ be the set of unbounded ideals of $P$ containing a cofinal chain of type $\lambda$. By hypothesis, $P\in \frkI_P(\lambda)$. Let $\frkI_P(\lambda)^{-}:= \frkI_P(\lambda)\setminus \{P\}$. We order $\frkI_P(\lambda)^{-}$ by set inclusion. We saw in Lemma~\ref{Lem:3.17} that $\frkI_P(\lambda)^{-}$ is not empty, and the following lemma shows more with a similar proof.
\begin{lemm}\label{claim:cof}
$\cf(\frkI_P(\lambda)^{-})\geq \lambda$.
\end{lemm}
\begin{proof}
Let $(J_{\alpha})_{\alpha\,<\,\mu}$ with $\mu:=\cf(\frkI_P(\lambda)^{-})$
be a sequence cofinal in $\frkI_P(\lambda)^{-}$. For each $\alpha<\mu$, pick $x_{\alpha} \in A\setminus J_{\alpha}$. If $\mu<\lambda$ then, since $\lambda$ is a regular cardinal, there is some $x\in A$ majorizing all the $x_{\alpha}'s$. In order to get a contradiction, it suffices to show that there is some $J$ such that $x\in J\in \frkI_P(\lambda)^{-}$. According to our hypothesis, the final segment $\uparrownogap x$ is impure. Hence, it contains a proper initial segment $I$ which is unbounded (in $\uparrownogap x$). Let $A_x:= \uparrownogap x\cap A$. For each $y\in A_x$, we may pick some $ c_y\in I\setminus \downarrownogap y$ since $y$ does not majorize $I$. Apply Erd\"os--Dushnik--Miller's Theorem to the sequence $(c_y)_{y\,\in\,A_x}$. Since $P$ has no infinite antichain, there is an increasing subsequence $(c_y)_{y\,\in\,D}$ with $D$ cofinal in $A_x$. The values of this sequence cannot be bounded in $P$. (otherwise, if $a$ is an upper bound, pick $y\geq a$ in $D$ and get a contradiction with $c_y$) hence they generate an unbounded ideal; since it is included in $\downarrownogap I$ it is distinct of $P$.
\end{proof}
\begin{enonce}[plain]{Claim}\label{claim:infinite antichain}
$\frkI_P(\lambda)^{-}$ has no infinite antichain.
\end{enonce}
\begin{proof}
This relies on the fact that $\lambda$ is uncountable and
regular. Indeed, suppose the sake of a contradiction that $\frkI_P(\lambda)^{-}$ contains an infinite antichain $(J_n)_{n\,<\,\omega}$. For each $n\not =m<\omega$ there is some $x_{n,m}\in J_n \setminus J_m$. Since $J_n$ contains a cofinal chain of uncountable regular type, it contains an element $x_n$ majorizing all the $x_{n, m}$. Then $\{x_n: n<\omega\}$ is an infinite antichain in $P$, which is impossible.
\end{proof}
With this claim, we complete the proof of Theorem~\ref{thm:covering2} as follows. We pick a well founded cofinal subset in $\frkI_P(\lambda)^{-}$. It is well quasi-ordered by Claim~\ref{claim:infinite antichain} and by Claim~\ref{claim:cof}, and has cardinality at least $\lambda$. By Erd\"os--Dushnik--Miller's theorem, it contains a chain of type $\lambda$. Let $(J_\alpha\mid \alpha<\kappa^+)$ enumerate an increasing sequence of ideals in $\frkI_P(\lambda)^{-}$, and let $I_\alpha\subset J_\alpha$ be a cofinal subset of order-type $\lambda$. We may assume that if $\alpha_1<\alpha_2$ then no member of $I_{\alpha_2}$ is below one of $I_{\alpha_1}$.
To deduce the proof of Theorem~\ref{thm:covering2} we embed a copy of $[\kappa^+]^2$ into $\bigcup\{ I_\alpha\mid \alpha<\kappa^+\}$. Following the same procedure as in Claim~\ref{Clm:embedding}, we define inductively an embedding of $[\beta]^2$ into $\bigcup_{\alpha\,<\,\beta}I_\alpha$ in such a way that the image of $\{(x,y)\in [\beta]^2\mid x=\alpha\}$ is in $I_\alpha$.
%\section*{Declaration of interests}
%The authors do not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and have declared no affiliations other than their research organizations.
\subsection*{Acknowledgements}
The authors would like to thank Professor James Cummings for the extensive discussions that helped to clarify several arguments in the proofs.
\ignore{\section{Apendix: Proof of the Erd\"os--Dushnik--Miller's theorem}
\begin{theo}[Erd\"os--Dushnik--Miller] For every infiniate cardinal
$\kappa$, $\kappa\rightarrow(\omega,\kappa)^2$.
\end{theo}
\begin{proof}
Let $\kappa$ be an infinite cardinal and $c:[A]^2\to \{0,1\}$ be a coloring of some set $A$ of cardinality $\kappa$ in two colors. Assume that there is no countable infinite subset of $A$ that is homogeneous for the color $0$. We have to get a homogeneous subset of $A$ for color $1$ of cardinality $\kappa$. The case $\kappa=\aleph_0$ is just Ramsey's theorem.
Suppose first that for every $B\subseteq A$ of cardinality $\kappa$ there is some $b\in B$ such that
\begin{equation}\label{Eq:B'}
B^b=\{ x\in B\mid c(b,x)=0 \}.
\end{equation}
Then there surely is an $\omega$-sequence in $A$ that is homogeneous in color $0$. (Define inductively a finite $0$-homogeneous sequence $b_0,\ldots,b_{n-1}$ together with a subset $B_{n-1}$ of cardinality $\kappa$ such that $c\restriction \{b_0,\ldots, b_{n-1}\}\times B_{n-1}$ is constantly $0$.)
So we may assume that there is a subset $B$ of $A$ of cardinality $\kappa$ such that for every $b\in B$ the set $B^b$ as defined above has cardinality $< \kappa$. For notational simplicity we may assume that $B=A$, and
\begin{equation}\label{Eq:A}
A^a = \{ x\in A\mid c(a,x)=0\}
\end{equation}
has cardinality $<\kappa$ for every $a\in A$. Although a proof can be given uniformly for the regular $\kappa$ and singular $\kappa$ cases a clearer presentation of the ideas is when we first do the regular case. So assume that $\kappa$ is a regular infinite cardinal. Not only for every $a\in A$ the set $A^a$ has cardinality $<\kappa$, but for every subset $E\subset A$ of cardinality $<\kappa$, the set
\begin{equation}
A^E = \{ x\in A\mid \exists b\in E\, (c(b, x)=0)\} = \bigcup\{A^a \mid a\in E \}
\end{equation}
has cardinality $<\kappa$ (by regularity of $\kappa$). Now by induction on $\alpha<\kappa$ an increasing $1$-homogeneous sequence $E_\alpha = \{ a_\xi\mid \xi < \alpha\}$ is defined. Given the sequence $E_\alpha$, then any $a_\alpha\in A\setminus A^{E_\alpha}$ is taken in order to define the $1$-homogeneous sequence $(a_\xi \mid \xi \leq \alpha)$.
Suppose now that $\kappa$ is a singular cardinal; $\cf \kappa < \kappa$. The first case is exactly as in the regular case. So we may assume that for every $a\in A$, the set $A^a$ as in~\eqref{Eq:A} has cardinality $<\kappa$.
Let $(\kappa_i \mid i<\cf \kappa)$ be an increasing sequence of regular cardinals with limit $\kappa$, and such that $k_i>\cf \kappa$ for every $i$. Let $J_i\subset A$ be arbitrarily chosen that are of cardinality $\kappa_i$. By the inductive assumption on $\kappa_i$ we may assume that $J_i$ is $1$-homogeneous.
\begin{claim}
For every $J_i$ there are a subset $K_i\subset J_i$, also of cardinality $\kappa_i$ and a super-set $L_i\supset K_i$ of cardinality $\lambda _i <\kappa$ such that if $x\in K_i$ and $y\in A\setminus L_i$ then $c(x,y)=1$.
\end{claim} Proof. For every $a\in J_i$ the cardinality of $A^a$ is $\leq \kappa_{j(a)}$ for some $j(a)<\cf \kappa$. Since $\kappa_i$ is regular, for a subset $K_i\subset J_i$ we have that $j(a)\leq \lambda<\kappa$ for some fixed $\lambda$ and for all $a\in K_i$. Then $L_i = \bigcup\{ A^a\mid a\in K_i\}$ has cardinality $\leq \lambda_i =\kappa_i\cdot \lambda <\kappa$, and hence $L_i$ has the required properties.
Finally, by diluting the sequence cofinal in $\kappa$, we may assume that $\kappa_i\leq \lambda_i <\kappa_{i+1}$. We redefine $M_j = K_j\setminus \bigcup \{ L_i\mid i< j\}$, and then the union $\bigcup \{ M_j\mid j<\cf \kappa\}$ is $1$-homogeneous and of cardinality $\kappa$.
\end{proof}}
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