# Flowers in graph-like spaces

Ann-Kathrin Elm  Research conducted at Universität Hamburg, Department of
Mathematics, Hamburg, Germany Ann-Kathrin Elm: Universität Heidelberg,
Department of Computer Science, Heidelberg, Germany<EMAIL_ADDRESS>heidelberg.de<EMAIL_ADDRESS>and  Hendrik Heine

###### Abstract.

One perspective on tree decompositions is that they display (low-order)
separations of the underlying graph or matroid. The separations displayed by a
tree decomposition are necessarily nested. In 2013, Clark and Whittle proved
the existence of tree decompositions with flowers added in which, up to a
natural equivalence, all low-order separations are displayed. An important
step in that proof is to show that flowers can be extended to maximal flowers.
In this paper we generalise the notion of a flower to pseudoflowers in graph-
like spaces and show for our generalisation, flowers can be extended to
maximal flowers.

###### Key words and phrases:

flower, tree decomposition, graph-like space, separation, maximal flower

###### 2020 Mathematics Subject Classification:

05C40, 05C63 (Primary), 05C05, 05C70 (Secondary)

##  1\. Introduction

Tree decompositions play an important role in the proof of the Graph Minor
Theorem by Robertson and Seymour and have since become a central tool of
structural graph theory. Tangles are a way of fuzzily encoding highly
connected regions of a graph $G$: They are objects that orient every
separation of order less than $k$ (this $k$ is then the _order_ of the tangle)
towards the side where the highly connected region of the graph can be found.
In a tree decomposition of a graph $G$, every edge of the decomposition graph
$T$ induces a separation of $G$. So every $k$-tangle orients every edge of $T$
that induces a separation of $G$ of order less than $k$ and thus points
towards a subtree of $T$ whose edges induce separations of $G$ of order at
least $k$. In this way, every tangle induces a subtree of $T$.

A tree of tangles of $G$ is a tree decomposition such that for any two
distinct tangles of the same order, the induced subtrees of the decomposition
tree are disjoint. Additionally all separations induced by edges of the
decomposition tree have to distinguish two tangles, that is, the two tangles
orient the separation differently. The subtrees of the decomposition tree $T$
induced by the tangles are naturally arranged in a tree-like fashion by $T$,
and usually a tree of tangles is seen as a tool to exhibit the resulting tree-
like structure of the tangles.

From another viewpoint, a tree of tangles also displays the tree-like
structure of the separations of $G$ it induces. As any two separations induced
by a tree of tangles cannot cross, the amount of separations that can be
induced by one tree of tangles is very limited. One way around this
restriction is to declare two separations as equivalent if they distinguish
the same tangles, and to consider a separation as represented by a tree
decomposition if some equivalent separation is induced by the tree
decomposition. This approach is justified by the fact, implied by tangle-tree-
duality theorems, that for any two equivalent separations $r$ and $s$ of which
one is induced by a tree of tangles there is a sequence
$r=p_{0},\ldots,p_{n}=s$ of separations such that the “difference” between to
consecutive separations is less than a separation that is small in the sense
that it does not distinguish tangles.

Even when allowing for this equivalence relation, it is in general not
possible to represent all separations in one tree decomposition, even in the
special case where the tangles to be distinguished all have the same order $k$
and all contain the same tangle of order $k-1$. The problem here is that there
might be separations $r$ and $s$ that _properly cross_ , that is, there are
tangles $T_{1}$, $T_{2}$, $T_{3}$ and $T_{4}$ to be distinguished such that
both $T_{1}$ and $T_{2}$ orient $r$ the same way, $T_{1}$ and $T_{3}$ orient
$s$ the same way, $r$ distinguishes $T_{1}$ from both $T_{3}$ and $T_{4}$ and
$s$ distinguished $T_{1}$ from both $T_{2}$ and $T_{4}$. In this case, any
separations $r^{\prime}$ and $s^{\prime}$ equivalent to $r$ and $s$
respectively also have to cross, and thus cannot be induced by a common tree
decomposition.

The concepts of tangles, tree decompositions and trees of tangles in matroid
theory are closely related to the corresponding concepts in graph theory. In
the context of finite matroids, the problem of representing separations that
properly cross has been solved by adding flowers to the vertices of the
decomposition tree. Flowers are a way of encoding many separations
simultaneously that mostly pairwise cross. Flowers and tree decompositions
with added flowers were introduced for $3$-tangles in [10] and then
generalised to $k$-tangles in [1] and [4]. [4, Theorem 7.1] implies that for a
set $\mathcal{T}$ of tangles of the same order $k$ that all contain the same
tangle of order $k-1$, there is a tree of tangles with flowers such that the
flowers conform with the tree decomposition and such that every separations
distinguishing elements of $\mathcal{T}$ is equivalent to either a separation
induced by the tree decomposition or to a separation displayed by one of the
flowers. A central part of this theorem is the use of flowers that distinguish
a maximal set of elements of $\mathcal{T}$, and the fact that under certain
circumstances for a flower $\Phi$, a separation that is only equivalent to
separations that cross separations displayed by $\Phi$ can be added to $\Phi$.

In this paper, as a first step towards a version of the theorem by Clark and
Whittle for graph-like spaces (a generalisation of infinite graphs), maximal
flowers in graph-like spaces are investigated. We find a definition of
pseudoflower that extends the notion of flower from [4] while being tailored
to taking care of several technical difficulties. A similar extension of
flowers for infinite matroids can also be found in [6]. We then show that
every pseudoflower that is not too small can be extended to another
pseudoflower that is maximal in the sense that further subdividing one of its
petals does not yield another pseudoflower (Theorem 5.29). Finally we show
that under certain assumptions on $\mathcal{T}$, these maximal pseudoflowers
also distinguish maximally many elements of $\mathcal{T}$ (Theorem 5.30).

##  2\. Tools and terminology

###  2.1. Separations, universes and profiles

The definition of a graph-like space is rather involved and for the results of
this paper other than examples we only use a few properties of the separations
of a graph-like space. For this reason, we will mostly work within the
framework of a universe of vertex separations.

A _universe of vertex separations_ on ground set $V$ (thought of as vertex
set) is a tuple $\mathcal{U}=(U,\wedge,\vee,^{*})$ where $U$ consists of pairs
$(A,B)$ with $A\cup B=V$ and is closed under $\wedge$, $\vee$ and ∗;
$\wedge:U\times U\rightarrow U$ is defined via $(A,B)\wedge(C,D)=(A\cap
C,B\cup D)$, $\vee:U\times U\rightarrow U$ is defined via
$(A,B)\vee(C,D)=(A\cup C,B\cap D)$ and ${}^{*}:U\rightarrow U$ is defined via
$(A,B)^{*}=(B,A)$. The elements of $U$ are called _separations_.

The name “universe of vertex separations” derives from a more general
definition of universe, in which $U$ is just some set and there need not be a
ground set $V$, as defined for example in [5]. Universes of vertex separations
are studied in depth under the name of separation systems implemented by set
separations in [3]. The main example for universes of vertex separations arise
from graphs and a generalisation of graphs called graph-like spaces.

The separation $(B,A)$ is the _inverse_ of the separation $(A,B)$, and $(A,B)$
and $(B,A)$ are the _orientations_ of $(A,B)$. Just as with graphs, the
relation $\leq$ on $U$ where $(A,B)\leq(C,D)$ if $A\subseteq C$ and
$D\subseteq B$ is a partial order in which $(A,B)\wedge(C,D)$ is the infimum
of $(A,B)$ and $(C,D)$ and $(A,B)\vee(C,D)$ is the supremum of $(A,B)$ and
$(C,D)$. Two separations $(A,B)$ and $(C,D)$ that have orientations that are
comparable by $\leq$ are called _nested_ , and if two separations are not
nested then they _cross_. A _corner_ of two crossing separations $(A,B)$ and
$(C,D)$ is the supremum or infimum of an orientation of $(A,B)$ and an
orientation of $(C,D)$, so two crossing separations have (up to) $8$ corners.
The _order_ of a separation $(A,B)$ is the size of $A\cap B$ if that is finite
and $\infty$ otherwise. Just as with graphs, the order function is symmetric
and submodular. A $\leq k$-separation is a separation of order at most $k$,
and a $<k$-separation is a separation of order less than $k$.

We also need the notion of a separation system. The more general definition of
abstract separation systems, together with related definitions and several
easy facts about them can be found in [5]. We only need the separation systems
$S_{k}$ that consist, for a universe $(U,\wedge,\vee,^{*})$ of vertex
separations and some $k\in\mathbb{N}$, of all separations of $U$ that have
order less than $k$. The partial order of $U$ induces a partial order of
$S_{k}$. In this context, for two elements $r$ and $s$ of $S_{k}$, the infimum
$r\wedge s$ is always taken in the surrounding universe, not in the partial
order of $S_{k}$. Thus the infimum always exists, but is not always contained
in $S_{k}$. The same holds for $r\vee s$.

In order to obtain our results we want the universe of vertex separations to
behave well with respect to limits of chains of separations that all have the
same order. In this case, we ask that if $k$ is a non-negative integer and
$(A_{i},B_{i})_{i\in I}$ is a chain (with respect to $\leq$) of separations of
order at most $k$, then the supremum of $(A_{i},B_{i})_{i\in I}$ in $U$
exists, has at most order $k$ and is of the form $(A\cup X,B)$ for
$A=\bigcup_{i\in I}A_{i}$, $B=\bigcap_{i\in I}B_{i}$ and some set $X\subseteq
V$. We call a universe of vertex separations whose order function has this
property _limit-closed_. Throughout this paper we fix a limit-closed universe
of vertex separations $\mathcal{U}$ on vertex set $V$. Note that a sub-
universe of $\mathcal{U}(V)$ whose induced order function is still limit-
closed need not be a limit-closed universe of vertex separations.

For a given graph, the universe of separations of that graph clearly is a
limit-closed universe of vertex separations. We will now show that the
separations of a graph-like space form a limit-closed universe of vertex
separations. As this and the example in Section 6 are the only parts of the
paper where we actually work with graph-like spaces, we do not give the
(slightly involved) definition here but refer the interested reader to [2] and
[11]. In a graph-like space, let a pair $(A,B)$ be a separation of that graph-
like space if $A\cup B$ is the vertex set of the graph-like space and every
arc meeting both $A$ and $B$ contains a vertex of $A\cap B$. In light of the
use of pseudo-arcs in [2], one might want to allow for different definitions
of separations in graph-like spaces, for example that additionally every
pseudo-arc meeting both $A$ and $B$ has to meet $A\cap B$. Such other possible
definitions can be captured for example by the notion of separation of a path
space, to be found in [7]. The following proof that the universe of
separations of a graph-like space also is a limit-closed universe of vertex
separations also works for path spaces.

######  Lemma 2.1.

Let $k\in\mathbb{N}$, let $G$ be a graph-like space and let
$(A_{i},B_{i})_{i\in I}$ be a chain of separations of order $\leq k$. Let
$A=\bigcup_{i\in I}A_{i}$ and $B=\bigcap_{i\in I}B_{i}$. Then there exists
$X\subseteq V(\mathcal{P})$ such that $(A\cup X,B)$ is a separation of order
at most $k$.

###### Proof.

Let $((A_{i},B_{i}))_{i\in I}$ be an increasing chain of separations of order
at most $k\in\mathbb{N}$. Let $A$ be the union of all $A_{i}$ and $B$ the
intersection of all $B_{i}$. If there are $k+1$ disjoint arcs
$P_{1},\ldots,P_{k+1}$ that meet both $A$ and $B$, then there is some
sufficiently large $i$ such that all $P_{j}$ meet $A_{i}$. As all $P_{j}$ also
meet $B_{i}$, this implies that all $P_{j}$ meet $A_{i}\cap B_{i}$,
contradicting the fact that $(A_{i},B_{i})$ has order at most $k$. So let
$P_{1},\ldots,P_{l}$ be a maximum set of disjoint arcs meeting both $A$ and
$B$. Without loss of generality assume that the first vertex of each $P_{j}$
is contained in $A$ while its last vertex is contained in $B$. For each $j$
let $x_{j}$ be the supremum of $P_{j}\cap A$ in $P_{j}$, and let $X$ be the
set of all $x_{j}$. By maximality of the family of the $P_{j}$, every vertex
in $P_{j}x_{j}$ is contained in $A+x_{j}$, every vertex in $xP_{j}$ is
contained in $B+x_{j}$, and every vertex of $A\cap B$ has to be contained in
some $P_{j}$ and to be equal to $x_{j}$. In particular $A\cap B\subseteq X$
and $|(A\cap B)\cup X|=|X|=l\leq k$.

Assume for a contradiction that there is an arc $Q$ meeting both $A$ and $B$
but not $X$. Without loss of generality, the first vertex of $Q$ is in $A$ and
the last vertex of $Q$ is in $B$. For each $j$ such that $Q$ meets
$P_{j}x_{j}$ let $q_{j}$ be the supremum in $Q$ of $Q\cap P_{j}x_{j}$, and let
$q$ be the maximum of all $q_{j}$ (or the first vertex of $Q$ if $Q$ meets no
$P_{j}x_{j}$). Similarly, for each $j$ such that $qQ$ meets $x_{j}P_{j}$ let
$r_{j}$ be the infimum in $qQ$ of $qQ\cap x_{j}P_{j}$ and let $r$ be the
minimum of all $r_{j}$ (or the last vertex of $Q$ if $qQ$ meets no
$x_{j}P_{j}$). Then $qQr$ has its first vertex in $A$, its last vertex in $B$
and is internally disjoint from the $P_{j}$. By maximality of the family
$(P_{j})$, either all inner vertices of $qQr$ are contained in $B\setminus A$,
or $qQr$ has an inner vertex and all inner vertices are contained in
$A\setminus B$. Consider the case that all inner vertices of $qQr$ are
contained in $B\setminus A$, and let $i$ be an index such that $q\in A_{i}$.
Then for all $i^{\prime}\geq i$, $q\in A_{i^{\prime}}$ and $r\in
B_{i^{\prime}}$ while all vertices other than $q$ are contained in
$B_{i^{\prime}}\setminus A_{i^{\prime}}$. As $qQr$ has to meet all
$A_{i^{\prime}}\cap B_{i^{\prime}}$ with $i^{\prime}\geq i$, $q$ is contained
in all $B_{i^{\prime}}$ and thus in $A\cap B$, a contradiction. Next consider
the case that $qQr$ has an inner vertex $y$ and all inner vertices are
contained in $A\setminus B$. Then $yQr$ starts in $A$ and ends in $B$ and, by
maximality of the family $(P_{j})$, $r$ is contained in some $P_{j}$. As $Q$
avoids $X$, this last vertex comes later than $x_{j}$ in $P_{j}$. So by the
maximality of the family $(P_{j})$, $P_{j}q$ without its last vertex does not
contain a subarc meeting both $A$ and $B$, hence $x_{j}\in A\setminus B$ and
$r$ is the successor of $x_{j}$ in $P_{j}$. But then $x_{j}\in A_{i}\setminus
B_{i}$ for some index $i$, and $r\in B_{i}$. Thus the edge $x_{j}r$ has one
end vertex in $A_{i}\cap B_{i}$, implying $r\in A_{i}\subseteq A$ and thus
$r\in A\cap B$, contradicting the assumption that $Q$ be disjoint from $X$. So
$Q$ does not exist. ∎

In the situation of Lemma 2.1, for any two separations of the form $(A\cup
X,B)$ and $(A\cup Y,B)$, the infimum $(A\cup(X\cap Y),B)$ is also a separation
of the graph-like space. Thus there is a smallest $X$ for which $(A\cup X,B)$
is a separation, and this $X$ satisfies that $(A\cup X,B)$ is the supremum of
the chain $(A_{i},B_{i})_{i\in I}$ in the universe of separations of the
graph-like space. In the special case of the separations of a graph, the set
$X$ is always empty. In graph-like spaces, that is not necessarily the case.

###  2.2. Cyclic orders

######  Definition 2.2.

[9] A set of triples $Z$ in $S\times S\times S$ is a _cyclic order_ of the set
$S$ if it has the following four properties:

  * •

(cyclic) $\forall a,b,c\in S:(a,b,c)\in Z\Rightarrow(b,c,a)\in Z$

  * •

(antisymmetric) $\forall a,b,c\in S:(a,b,c)\in Z\Rightarrow(c,b,a)\notin Z$

  * •

(linear) $\forall a,b,c\in S\text{ pairwise distinct}:(a,b,c)\notin
Z\Rightarrow(c,b,a)\in Z$

  * •

(transitive) $\forall a,b,c,d\in S:(a,b,c)\in Z\wedge(a,c,d)\in
Z\Rightarrow(a,b,d)\in Z$.

So formally a cyclic order is a different set than its underlying ground set,
but in this paper (except in the appendix) this distinction is not made. Note
that what is called a cyclic order here is sometimes (also in [9]) called a
linear cyclic order or a complete cyclic order, with a cyclic order not
necessarily being linear. The distinction is not made here as all cyclic
orders under consideration are linear.

For two elements $a$ and $b$ of $S$, the set of elements $c\in S$ satisfying
$(a,c,b)\in Z$ is denoted by $\mathopen{]}a,b\mathclose{[}$. The sets
$\mathopen{]}a,b\mathclose{[}+a$, $\mathopen{]}a,b\mathclose{[}+b$ and
$\mathopen{]}a,b\mathclose{[}+a+b$ are denoted by $[a,b\mathclose{[}$,
$\mathopen{]}a,b]$ and $[a,b]$ respectively. When necessary to resolve
ambiguities, we may add the cyclic order in which these are taken as a
subscript. We call a subset $I$ of $S$ is an _interval_ if and only if for all
$s,t\in I$ either $[s,t]\subseteq I$ or $[t,s]\subseteq I$. Clearly, subsets
of $S$ of the form $\mathopen{]}a,b\mathclose{[}$, $\mathopen{]}a,b]$,
$[a,b\mathclose{[}$ or $[a,b]$ are all intervals of $Z$. We call an interval
_non-trivial_ if it is neither the empty set nor all of $S$. Similarly to
linear orders, an element $s\neq a$ is the _successor_ of $a$ in $S$ if
$b\notin[a,s]$ for all for all other elements $b$. Also, an element $p\neq a$
is the _predecessor_ of $a$ if $b\notin[p,a]$ for all other elements $b$. A
_neighbor_ of $a$ is an element that is a predecessor or successor of $a$.

######  Remark 2.3.

[8, Lemma 1.4] The definition of a cyclic order implies that if $Z$ is a
cyclic order on a set $S$ and $(s,s^{\prime},t)$ is an element of $Z$ then
$s$, $s^{\prime}$ and $t$ are pairwise distinct.

We will use cyclically ordered sets of a special kind as the index set of our
$k$-pseudoflowers:

######  Definition 2.4.

Let $I$ be a cyclically ordered set with at least two elements. A cyclically
ordered set $C$ is a _cycle completion_ of $I$ if $I$ is a subset of $C$, the
cyclic order on $I$ is the one induced by $C$, and every non-trivial interval
of $I$ can uniquely be written as $[v,w]\cap I$ for elements $v,w$ of
$C\setminus I$.

Intuitively, the cycle completion of a cyclic order arises from that cyclic
order as follows: Let $I$ be some cyclically ordered set. If one envisions $I$
as boxes arranged in a circle according to the cyclic order (see also Fig. 1),
then it is possible to cut up the circle at two places without cutting through
boxes, thereby dividing the set of boxes into two intervals. If $I$ is finite,
then every one of these “cut points” is between two boxes. So $I$ and the set
of possible cut points form together another cyclically ordered set. If $I$ is
infinite, then not every cut point is between two boxes, but still $I$ and the
set of possible cut points form together a cyclically ordered set, the cycle
completion of $I$.

Cycle completions can be formalised in several ways, and doing so via cuts, as
is done in the appendix, is only one possibility of many.

Figure 1. A set (whose elements are indicated by boxes) which is cyclically
ordered (indicated by the arrangement of the boxes on a circle) and two
“cutting points” dividing the cyclically ordered set into two intervals.

The cycle completion has several properties which we will use throughout the
article and which are intuitively clear. Also, although we did not find the
exact construction of the cycle completion described in the literature,
several very similar constructions are. For these reasons, we will only state
the properties needed in this section, and the proofs can be found in the
appendix, together with a description of how to obtain the cycle completion
from similar constructions.

The first of the facts about cycle completions which we need is that cycle
completions exist and are essentially unique. In order to compare different
cyclic orders, we will use _monotone_ maps, that is maps $f$ between cyclic
orders where $f(v)\in\mathopen{]}f(u),f(w)\mathclose{[}$ implies
$v\in\mathopen{]}u,w\mathclose{[}$, and _isomorphisms of cyclic orders_ ,
bijective maps $f$ where $f(v)\in\mathopen{]}f(u),f(w)\mathclose{[}$ is
equivalent to $v\in\mathopen{]}u,w\mathclose{[}$. Note that if the image of
some monotone map has at least three elements, then the preimage of every
interval is also an interval.

######  Lemma 2.5 (Lemmas 7.13 and 7.18).

For every cyclically ordered set $I$ with at least two elements there is a
cycle completion. If $C$ and $C^{\prime}$ are two cycle completions of $I$,
then there is an isomorphism of cyclic orders $F:C\rightarrow C^{\prime}$ such
that the restriction of $F$ to $I$ is the identity.

The fact that the cycle completion of a cyclically ordered set $I$ is unique
up to isomorphism justifies calling it “the” cycle completion of $I$ and
denoting it as $C(I)$. We will also call the elements of $C(I)\setminus I$
_cutpoints_ , a terminology partly inspired by the term cut (see appendix).

######  Lemma 2.6 (Lemma 7.15).

For a cyclically ordered set $I$ with at least two elements let $v$ and $w$ be
distinct elements of $C(I)$. Then $[v,w]\cap I$ is a non-trivial interval of
$I$.

######  Lemma 2.7 (Lemma 7.18).

Let $I$ and $I^{\prime}$ be cyclically ordered sets with at least two elements
and let $f:I^{\prime}\rightarrow I$ be a surjective monotone map. Then there
is a unique surjective monotone map $F:C(I^{\prime})\rightarrow C(I)$ whose
restriction to $I^{\prime}$ is $f$.

In particular this implies that the isomorphism from Lemma 2.5 is unique.

######  Lemma 2.8 (Lemma 7.16).

Let $I$ and $I^{\prime}$ be cyclically ordered sets with at least two
elements. Let $F:C(I^{\prime})\rightarrow C(I)$ be a surjective monotone map
with $F(I^{\prime})\subseteq I$. Then for every $v\in C(I)\setminus I$ there
is an element $w$ of $C(I^{\prime})$ such that $F^{-1}(v)=\\{w\\}$. This
element $w$ satisfies $w\in C(I^{\prime})\setminus I^{\prime}$. Also
$F(I^{\prime})=I$.

So for every surjective monotone map $F:C(I^{\prime})\rightarrow C(I)$ which
satisfies $F(I^{\prime})=I$ there is an injective monotone map
$\hat{f}:C(I)\setminus I\rightarrow C(I^{\prime})\setminus I^{\prime}$ such
that for all $v\in C(I)\setminus I$ the equation $F^{-1}(v)=\\{\hat{f}(v)\\}$
holds.

######  Lemma 2.9 (Lemma 7.17).

Let $F:C(I^{\prime})\rightarrow C(I)$ be a surjective monotone map with
$F(I^{\prime})=I$, and $\hat{f}:C(I)\setminus I\rightarrow
C(I^{\prime})\setminus I^{\prime}$ the injective monotone map with
$F\circ\hat{f}(v)=v$ for all $v\in C(I)\setminus I$. Assume that $I$ has at
least two elements. Then for all $v,w\in C(I)\setminus I$,
$F^{-1}([v,w])=[\hat{f}(v),\hat{f}(w)]$ and
$F^{-1}(\mathopen{]}v,w\mathclose{[})=\mathopen{]}\hat{f}(v),\hat{f}(w)\mathclose{[}$.

##  3\. $k$-Pseudoflowers

In this section we will define $k$-pseudoflowers and prove a few basic facts
about them. This definition is derived from a definition of $k$-flowers in
matroids in [4]. A $k$-flower is, essentially, a neat way to encode a
collection of separations that mostly cross. The definition we will give is a
lot more complicated than the original one, and there are mainly two reasons
for that: First, a separation of a matroid is a bipartition of the ground set,
and thus it is possible to encode several bipartitions in a partition of the
ground set. But in our context we are working with vertex separations, and
elements of separators have to be contained in several of the sets
corresponding to partition classes. Second, we work in infinite structures and
still want to obtain maximal $k$-pseudoflowers. Thus we have to make limit
processes work, and as a chain of separations of order $k$ may have a limit
separation whose order is less than $k$, we have to allow that separations do
not have order exactly $k$, but at most $k$. Making limit processes work also
requires to put elements of separators between petals, sometimes in addition
to putting them into petals. Also, every finite cyclic order is determined by
its cardinality, but that is not the case for infinite cyclic orders, and thus
we have to allow arbitrary cyclic orders as index sets and cannot work with an
easily described representative.

Recall that we fixed, for the whole paper, a limit-closed universe of vertex
separations on ground set $V$.

######  Definition 3.1.

Let $I$ be a set of size at least $2$ with a cyclic ordering and
$(P_{v})_{v\in C(I)}$ a family of vertex sets. Define
$X=V\backslash\bigcup_{v\in C(I)}P_{v}$ and for all $v,w\in C(I)\setminus I$
let $V(v,w)=\bigcup_{z\in[v,w]}P_{z}\cup X$. Then $(P_{i})_{i\in C(I)}$ is a
_$k$ -pseudoflower_ if

  * •

For every $v\in C(I)\setminus I$ we have $|P_{v}|=\frac{k-|X|}{2}$;

  * •

For all distinct $v,w\in C(I)\setminus I$ the pair $S(v,w)=(V(v,w),V(w,v))$ is
a separation of order at most $k$ and $V(v,w)\cap V(w,v)=P_{v}\cup P_{w}\cup
X$;

  * •

For every $i\in I$ we have $P_{p(i)}\cup P_{s(i)}\subseteq P_{i}$, where
$p(i)$ and $s(i)$ are the predecessor and successor of $i$ in $C(I)$
respectively; and

  * •

for all $i,i^{\prime}\in I$,

$\text{if $|P_{i}|=(k-|X|)/2$ and $P_{i}=P_{i^{\prime}}$ then
$i=i^{\prime}$}.$

A $k$-pseudoflower is _finite_ if the index set is finite. A _$k$ -flower_ is
a $k$-pseudoflower in which all separations $S(v,w)$ are separations with
order exactly $k$ such that neither $V(v,w)\setminus V(w,v)$ nor
$V(w,v)\setminus V(v,w)$ is empty.

For $v\neq w\in C(I)\setminus I$ the set $V(v,w)$ is the _interval set_ of
$[v,w]$ and the separation $S(w,v)$ is the _interval separation_ of $[v,w]$.
For a non-trivial interval $I^{\prime}\subseteq I$ let $v$ and $w$ be the
unique elements of $C(I)\setminus I$ such that $I^{\prime}=I\cap[v,w]$. Then
the _interval set_ $V(I^{\prime})$ of $I^{\prime}$ is $V(v,w)$ and the
_interval separation_ $S(I^{\prime})$ of $I^{\prime}$ is $S(w,v)$. For each
index $i\in I$ the _petal_ of $i$ is its interval set $V(\\{i\\})$ and the
_petal separation_ of $i$ is its interval separation $S(\\{i\\})$. As is usual
for maps defined on power sets, we shorten $V(\\{i\\})$ and $S(\\{i\\})$ to
$V(i)$ and $S(i)$. Note that $V(i)=P_{i}\cup X$ for all $i\in I$. A
$k$-pseudoflower is called a _$k$ -pseudoanemone_ if the sets $P_{v}$ for
$v\in C(I)\setminus I$ are all empty and a _$k$ -pseudodaisy_ otherwise. Note
that the number $(k-|X|)/2$, which is the size of the sets $P_{v}$ with $v\in
C(I)\setminus I$, indicates how far away a $k$-pseudodaisy is from being a
$k$-pseudoanemone. In this sense it plays a very similar role to the
difference between the local connectivity of adjacent petals and the local
connectivity of non-adjacent petals as used, for example, in [1]. Also note
that in a $k$-pseudoanemone, all the interval separations have $X$ as their
separator. In the case where the limit-closed universe comes from a graph $G$,
this means that the set of petals corresponds to a partition of the components
of $G-X$, and that $X$ has exactly $k$ elements.

######  Remark 3.2.

In a graph, $k$-pseudoanemones correspond to partitions of the components of
$G-X$. Also there are no infinite $k$-pseudodaisies in graphs. See Example 6.1
for an example of an infinite $k$-pseudodaisy in a graph-like space.

###  3.1. Basic properties

We start with a lemma that is, among other things, essential for determining
separations of the form $S(I^{\prime})\vee S(I^{\prime\prime})$.

$[a,b]$$[c,d]$$a$$b$$c$$d$ Figure 2. A cyclic order with two intervals. This
figure depicts some of the notation used in Lemmas 3.3 and 5.19.

######  Lemma 3.3.

Let $\Phi$ be a $k$-pseudoflower on index set $I$. Let $a$, $b$, $c$ and $d$
be elements of $C(I)\setminus I$ such that $a$, $b$ and $c$ are pairwise
distinct and $b\in[a,c]$. Also assume that if $d\in\\{a,b,c\\}$, then $d=a$,
and if $d\notin\\{a,b,c\\}$ then $d\in[c,a]$. Then $V(a,b)\cap
V(c,d)=(P_{a}\cap P_{d})\cup(P_{b}\cap P_{c})\cup X$ and $V(a,c)\cap
V(b,d)=V(b,c)\cup(P_{a}\cap P_{d})$.

###### Proof.

Clearly $V(a,b)\cap V(c,d)\subseteq V(a,c)\cap V(c,a)=P_{a}\cup P_{c}\cup X$.
Similarly $V(a,b)\cap V(c,d)$ is a subset of $P_{a}\cup P_{b}\cup X$,
$P_{d}\cup P_{b}\cup X$ and $P_{d}\cup P_{c}\cup X$. Together these
subsetrelations imply

$\begin{split}&V(a,b)\cap V(c,d)\\\ &\quad\subseteq(P_{a}\cup P_{b}\cup
X)\cap(P_{a}\cup P_{c}\cup X)\cap(P_{d}\cup P_{b}\cup X)\cap(P_{d}\cup
P_{c}\cup X)\\\ &\quad=(P_{a}\cap P_{d})\cup(P_{b}\cap P_{c})\cup
X.\end{split}$

As also $(P_{a}\cap P_{d})\cup(P_{b}\cap P_{c})\cup X\subseteq V(a,b)\cap
V(c,d)$, these two sets are equal. Thus

$\begin{split}V(a,c)\cap V(b,d)&=V(b,c)\cup(V(a,b)\cap V(c,d))\\\
&=V(b,c)\cup(P_{a}\cap P_{d})\cup(P_{b}\cap P_{c})\cup X\\\
&=V(b,c)\cup(P_{a}\cap P_{d}).\qed\end{split}$

######  Corollary 3.4.

If $P_{a}\cap P_{d}=\emptyset$ then $S(b,c)=S(a,c)\wedge S(b,d)$.

$w$$v$$p^{\prime}$$s^{\prime}$$p$$s$$i^{\prime}$$i$ Figure 3. Some notation
from the proof of Lemma 3.5.

The next lemma and the previous corollary will often be used together.

######  Lemma 3.5.

Let $(P_{i})_{i\in C(I)}$ be a $k$-flower on index set $I$. Then $P_{v}\cap
P_{w}=\emptyset$ for all $v\not=w\in C(I)\setminus I$.

###### Proof.

By definition of a $k$-flower, the set $P_{v}\cup P_{w}\cup X$, which equals
$V(v,w)\cap V(w,v)$, has exactly $k$ many elements. Also $P_{v}$ and $P_{w}$
both have exactly $(k-|X|)/2$ many elements, so they must be disjoint. ∎

######  Lemma 3.6.

For all $x\in\bigcup_{v\in C(I)\setminus I}P_{v}$ the set $\\{w\in C(I):x\in
P_{w}\\}$ is an interval of $C(I)$.

###### Proof.

Assume otherwise. Then there are pairwise distinct $a,b,c,d\in C(I)$ such that
$b\in[a,c]$, $d\in[c,a]$, $x\in P_{b}\cap P_{d}$ and $x\notin P_{a}\cup
P_{c}$. If $a\in I$ then let $a^{\prime}$ be the predecessor of $a$ in $C(I)$,
otherwise let $a^{\prime}=a$. Similarly, if $c\in I$ then let $c^{\prime}$ be
the predecessor of $c$ in $C(I)$, otherwise let $c^{\prime}=c$. Then $x\notin
P{a^{\prime}}\cup P_{c^{\prime}}$, and $a^{\prime},b,c^{\prime},d$ are
distinct elements of $C(I)\setminus I$. Now

$x\in P_{b}\cap P_{d}\subseteq V(a^{\prime},c^{\prime})\cap
V(c^{\prime},a^{\prime})=P_{a^{\prime}}\cup P_{c^{\prime}}\cup X,$

a contradiction. ∎

###  3.2. Concatenations of $k$-pseudoflowers

A separation $(A,B)$ is _displayed_ by a $k$-pseudoflower if it is an interval
separation of that $k$-pseudoflower. We are now going to define a pre-order on
the set of $k$-pseudoflowers which has the property that if $\Phi\leq\Psi$
then all separations displayed by $\Phi$ are also displayed by $\Psi$.

######  Definition 3.7.

A $k$-pseudoflower $\Phi^{\prime}$ _extends_ another $k$-pseudoflower $\Phi$,
written $\Phi\leq\Phi^{\prime}$, if the sets $X$ and $X^{\prime}$ coincide and
there is a surjective map $F:C(I^{\prime})\rightarrow C(I)$ respecting the
cyclic ordering such that $F(I^{\prime})=I$ and $V(v,w)=X\cup\bigcup_{z\in
F^{-1}([v,w])}P^{\prime}_{z}$ for all $v,w\in C(I)\setminus I$. If
$\Phi^{\prime}$ extends $\Phi$, then $\Phi$ is a _concatenation_ of
$\Phi^{\prime}$.

Note that if $\Phi\leq\Psi\leq\Phi$ for two $k$-pseudoflowers $\Phi$ and
$\Psi$ then property (• ‣ 3.1) $\Phi$ and $\Psi$ are equal up to renaming of
the index sets.

Given a map $F:C(I^{\prime})\rightarrow C(I)$ that witnesses that
$\Phi\leq\Phi^{\prime}$, by Lemma 2.8 there is for every $v\in C(I)\setminus
I$ a unique $w\in C(I^{\prime})$ with $F(w)=v$. Furthermore $w\in
C(I^{\prime})\setminus I^{\prime}$, and $P_{v}=P_{w}$ by definition of
$\Phi\leq\Phi^{\prime}$. This allows us to identify
$F^{-1}(C(I^{\prime})\setminus I^{\prime})$ with $C(I^{\prime})\setminus
I^{\prime}$, and thus view $\Phi$ as a substructure of $\Phi^{\prime}$. (See
Section 5.2 for a more detailed explanation of how this identification can be
done for a whole chain of $k$-pseudoflowers.) Note that the witness of
$\Phi\leq\Phi^{\prime}$ might not be unique, and that the identification of
cutpoints depends on the witness. Conversely, given a finite set $D$ of
cutpoints of a $k$-pseudoflower $\Phi^{\prime}$ on index set $I^{\prime}$, we
can take the subflower $\Phi^{\prime}(D)$ of $\Phi^{\prime}$ whose cutpoints
correspond to $D$ as follows: Let $I$ be a cyclic order with $|D|$ many
elements, and let $\hat{f}:C(I)\setminus I\rightarrow D$ be an isomorphism of
cyclic orders. Extend the inverse of $\hat{f}$ to a surjective monotone map
$F:C(I^{\prime})\rightarrow C(I)$ with $F(I^{\prime})\subseteq I$. For $v\in
C(I)\setminus I$ there is a unique $d\in D$ with $F^{-1}(v)=d$, define
$P_{v}=P_{d}$. For $i\in I$ let $p$ and $s$ be the predecessor and successor
of $i$ in $C(I)$ respectively, and define $P_{i}=\bigcup_{z\in
F^{-1}([p,s])}P_{z}$. Then $(P_{z})_{z\in C(I)}$ is a $k$-pseudoflower, and
$F$ witnesses $(P_{z})_{z\in C(I)}\leq\Phi^{\prime}$. Note that we need this
construction only for finite $D$, but it works for all sets of cutpoints $D$
that can be expressed as $C(I)\setminus I$ for some cyclically ordered set
$I$.

######  Lemma 3.8.

Let $\Phi$ be a $k$-pseudoflower that extends a $k$-flower with three petals.
Then for every $z\in C(I)$ there is $v\in C(I)\setminus I$ such that
$P_{z}\cap P_{v}=\emptyset$. Furthermore, for every $i\in I$ and all non-empty
$Y\subseteq P_{i}$ the set $\\{z\in C(I):Y\subseteq P_{z}\\}$ is a non-trivial
interval of $C(I)$.

###### Proof.

As $\Phi$ extends a $k$-flower with three petals, $C(I)\setminus I$ contains
three distinct elements $v_{1}$, $v_{2}$ and $v_{3}$ such that $P_{v_{1}}$,
$P_{v_{2}}$ and $P_{v_{3}}$ are pairwise disjoint. By renaming if necessary
assume that $z\in[v_{1},v_{2}]$ and $v_{3}\in[v_{2},v_{1}]$. Then

$P_{z}\cap P_{v_{3}}\subseteq V(v_{1},v_{2})\cap V(v_{2},v_{1})=P_{v_{1}}\cup
P_{v_{2}}\cup X.$

As every $P_{y}$ with $y\in C(I)$ is disjoint from $X$, this implies that
$P_{z}\cap P_{v_{3}}=\emptyset$.

Let $Y\subseteq P_{i}$ and for every $y\in P_{i}$ let $C_{y}$ be the interval
of those $z\in C(I)$ with $y\in P_{z}$. By the previous paragraph there exists
$v\in C(I)\setminus I$ such that $P_{i}\cap P_{v}=\emptyset$. As all $C_{y}$
contain $i$ but none contains $v$, the intersection of the $C_{y}$ is a non-
trivial interval of $C(I)$. ∎

We already stated that given two $k$-pseudoflowers $\Phi\leq\Psi$, there might
be several maps witnessing the relation. It may not be obvious from the
definition, but if $\Phi$ is a sufficiently meaningful $k$-pseudoflower, then
two witnessing maps can only differ in very restricted ways. The rest of this
section contains an analysis of the exact ways in which they can differ.

For $k$-pseudoflowers $\Phi\leq\Phi^{\prime}$ such that $\Phi$ extends a
$k$-flower with three petals let $I^{\prime}_{N}$ be the set of those
$i^{\prime}\in I^{\prime}$ such that there are witnesses of
$\Phi\leq\Phi^{\prime}$ that map $i^{\prime}$ to different elements of $I$.

######  Lemma 3.9.

Let $i^{\prime}\in I_{N}^{\prime}$. Then the set of images of $i^{\prime}$
under witnesses of $\Phi\leq\Phi^{\prime}$ contains only two elements of $I$,
and those have a common neighbor $v(i^{\prime})$ in $C(I)$ that satisfies
$P_{i^{\prime}}^{\prime}=P_{v(i^{\prime})}$.

###### Proof.

In order to show that $P^{\prime}_{i^{\prime}}=P_{v}$ for some $v\in
C(I)\setminus I$, let $i_{1}\neq i_{2}$ be possible images of $i^{\prime}$
under witnesses of $\Phi\leq\Phi^{\prime}$. Then in particular
$P^{\prime}_{i^{\prime}}\subseteq P_{i_{1}}$ and
$P^{\prime}_{i^{\prime}}\subseteq P_{i_{2}}$. By Lemma 3.8 there is $v_{1}\in
C(I)\setminus I$ such that $P_{v_{1}}\cap P_{i_{1}}=\emptyset$. Without loss
of generality $v_{1}\in[i_{1},i_{2}]$, and as $i_{1}\neq i_{2}$ there is
$v_{2}\in[i_{2},i_{1}]\setminus I$. So

$P^{\prime}_{i^{\prime}}\subseteq P_{i_{1}}\cap P_{i_{2}}\subseteq
V(v_{1},v_{2})\cap V(v_{2},v_{1})=P_{v_{1}}\cup P_{v_{2}}\cup X,$

implying that $P^{\prime}_{i^{\prime}}\subseteq P_{v_{2}}$ and thus by first
and third condition of $k$-pseudoflowers that
$|P^{\prime}_{i^{\prime}}|=|P_{v_{2}}|=(k-|X|)/2$.

If there is $i\in I$ with $P_{i}=P^{\prime}_{i^{\prime}}$, then every witness
$F$ of $\Phi\leq\Phi^{\prime}$ has to map some $i_{F}^{\prime}$ to $i$, and
$P^{\prime}_{i_{F}^{\prime}}=P_{i}=P^{\prime}_{i^{\prime}}$ then implies by (•
‣ 3.1) that $i_{F^{\prime}}=i^{\prime}$. Thus in this case every witness of
$\Phi\leq\Phi^{\prime}$ maps $i^{\prime}$ to $i$, contradicting $i^{\prime}\in
I^{\prime}_{N}$.

Let $F$ be some witness of $\Phi\leq\Phi^{\prime}$, then
$P^{\prime}_{i^{\prime}}\subseteq P_{F(i^{\prime})}$. By Lemma 3.8 there is
$z_{1}\in C(I)\setminus I$ such that $P_{F(i^{\prime})}\cap
P_{z_{1}}=\emptyset$, and the set $C:=\\{z\in
C(I):P^{\prime}_{i^{\prime}}\subseteq P_{z}\\}$ is a non-trivial interval of
$C(I)$. There is $z_{1}^{\prime}\in C(I^{\prime})\setminus I^{\prime}$ with
$F(z_{1}^{\prime})=z_{1}$.

Assume that there is $i\in C\cap I$ that is (in the linear order of the
interval $C$) neither the biggest nor the smallest element of $C\cap I$. Then
the two neighbors $s$ and $p$ of $i$ in $C(I)$ are also contained in $C$, thus
$P_{s}=P_{p}=P^{\prime}_{i^{\prime}}\neq P_{i}$. Hence there is $x\in P_{i}$
with $x\notin P_{z}$ for all other $z\in C(I)$, and there is $z_{2}\in
C(I^{\prime})$ with $x\in P_{z_{2}}$. Now $F(z_{2})=i$. If
$i^{\prime}\in\mathopen{]}z_{1}^{\prime},z_{2}\mathclose{[}$ then
$F(i^{\prime})\in\mathopen{]}z_{1},i]$ and otherwise
$i^{\prime}\in\mathopen{]}z_{2},z_{1}^{\prime}\mathclose{[}$, implying
$F(i^{\prime})\in[i,z_{1}]$. Both $\mathopen{]}z_{1},i]\cap C$ and
$[i,z_{1}\mathclose{[}\cap C$ are intervals of $C$.

As $F$ was chosen as an arbitrary witness of $\Phi\leq\Phi^{\prime}$, this
implies that $C\cap I$ contains an interval of at most two elements that
contains all images of $i^{\prime}$ under witnesses of
$\Phi\leq\Phi^{\prime}$. As $i^{\prime}\in i_{N}^{\prime}$ this implies that
there are exactly two such images, and that they are neighbors in $I$ and
hence have a common neighbor $v(i^{\prime})$ in $C(I)$. Then
$v(i^{\prime})\notin I$, and

$P^{\prime}_{i^{\prime}}\subseteq V(v_{1},v(i^{\prime}))\cap
V(v(i^{\prime}),v_{1})=X\cup P_{v_{1}}\cup P_{v(i^{\prime})}.$

As $P^{\prime}_{i^{\prime}}$ is disjoint from $X$ and $P_{v_{1}}$ and is at
least as big as $P_{v(i^{\prime})}$, this implies
$P^{\prime}_{i^{\prime}}=P_{v(i^{\prime})}$. ∎

######  Lemma 3.10.

For every $i\in I$ there is some $i^{\prime}\in I^{\prime}\setminus
I^{\prime}_{N}$ such that every witness of $\Phi\leq\Phi^{\prime}$ maps
$i^{\prime}$ to $i$.

###### Proof.

Let $I^{\prime}_{i}$ be the set of those elements of $I^{\prime}$ that are
mapped to $i$ by some witness of $\Phi\leq\Phi^{\prime}$. If $I^{\prime}_{i}$
contains only one element, then the lemma holds, so assume otherwise. Assume
for a contradiction that all elements of $I^{\prime}_{i}$ are contained in
$I^{\prime}_{N}$. Denote the neighbors of $i$ in $C(I)$ by $s$ and $p$, then
by Lemma 3.9 every element $i^{\prime}\in I^{\prime}_{i}$ satisfies either
$P_{i^{\prime}}=P_{s}$ or $P_{i^{\prime}}=P_{p}$, hence there are exactly two
such elements $i_{1}$ and $i_{2}$. Again by Lemma 3.9, $P_{i_{1}}$ and
$P_{i_{2}}$ have both the same size as all sets $P_{v}$ with $v\in
C(I)\setminus I$. But if $v$ denotes the common neighbour of $i_{1}$ and
$i_{2}$ in $C(I^{\prime})$, then $P_{v}\subseteq P_{i_{1}}\cap P_{i_{2}}$,
implying $P_{i_{1}}=P_{i_{2}}$ and thus contradicting property (• ‣ 3.1). ∎

######  Lemma 3.11.

For every $v\in C(I)\setminus I$ that is not of the form $v(i^{\prime})$ for
some $i^{\prime}\in I^{\prime}_{N}$ there is a unique $v^{\prime}\in
C(I^{\prime})\setminus I^{\prime}$ that is mapped to $v$ by all witnesses of
$\Phi\leq\Phi^{\prime}$.

###### Proof.

As $\Phi$ extends a $k$-flower with three petals, there is $i\in I$ such that
$P_{v}\cap P_{i}=\emptyset$ and $v$ is not a neighbor of $i$. Denote the
predecessor and successor of $i$ in $C(I)$ by $p$ and $s$ respectively. In
particular $p\neq v\neq s$. Let $i^{\prime}$ be an element of $I^{\prime}$
that is mapped to $i$ by every witness that $\Phi\leq\Phi^{\prime}$ and denote
its successor in $C(I^{\prime})$ by $s^{\prime}$.

Assume that there are witnesses $F_{1}$ and $F_{2}$ of $\Phi\leq\Phi^{\prime}$
such that $F_{1}^{-1}(v)=v_{1}$ and $F_{2}^{-1}(v)=v_{2}$ and
$v_{1}\in\mathopen{]}s,v_{2}\mathclose{[}$. Then for $j\in\\{1,2\\}$,
$F_{j}^{-1}(s)\in[s^{\prime},v_{j}]$ and thus $V(s^{\prime},v_{2})\subseteq
V(p,v)$ and $V(v_{1},s^{\prime})\subseteq V(v,s)$. Hence every element of
$V(v_{1},v_{2})$ is contained in $V(i)\cup V(s,v)$ and in $V(v,s)$. But
$V(v,s)\cap(V(i)\cup V(s,v))=V(i)\cup P_{v}$ and $V(v_{1},v_{2})\cap(V(i)\cup
P_{v})\subseteq V(v_{1},v_{2})\cap V(v_{2},v_{1})=X\cup P_{v}$, hence every
element of $V(v_{1},v_{2})$ is contained in $X\cup P_{v}$. So
$P_{\hat{\imath}}=P_{v}$ for all $\hat{\imath}\in[v_{1},v_{2}]$ and in
particular there can be at most one such $\hat{\imath}$.

As $v_{1}\neq v_{2}$, some such $\hat{\imath}$ exists. Then
$F_{2}(\hat{\imath})\in[s,v]$ and, because $P_{\hat{\imath}}\cap
P_{i}=P_{v}\cap P_{i}=\emptyset$, $F_{1}(\hat{\imath})\in[v,p]$. So
$\hat{\imath}\in I^{\prime}_{N}$ and $v(\hat{\imath})=v$, contradicting the
assumptions about $v$. ∎

Let $F$ be a witness that $\Phi\leq\Phi^{\prime}$ and $f$ the restriction of
$F$ to $I^{\prime}$. Let $f^{\prime}:I^{\prime}\rightarrow I$ be obtained from
$f$ by choosing, for every $i^{\prime}\in I^{\prime}_{N}$, some neighbor of
$v(i^{\prime})$ in $C(I)$ as the image of $f^{\prime}$ and mapping every other
element of $I^{\prime}$ to its image under $f$. Let
$F^{\prime}:C(I^{\prime})\rightarrow C(I)$ be the unique surjective map
respecting the cyclic order whose restriction to $I^{\prime}$ is $f^{\prime}$
(see Lemma 2.7).

######  Lemma 3.12.

The map $F^{\prime}$ is well-defined.

###### Proof.

It suffices to show that $f^{\prime}$ is surjective and respects the cyclic
order. As for every $i\in I$ there is some $i^{\prime}\in I^{\prime}$ that is
mapped to $i$ by all witnesses of $\Phi\leq\Phi^{\prime}$ and is in particular
not contained in $I_{N}^{\prime}$, $f^{\prime}$ is indeed surjective.

In order to show that $f^{\prime}$ respects the cyclic order, consider
elements $i_{1}$, $i_{2}$ and $i_{3}$ of $I^{\prime}$ such that
$f^{\prime}(i_{2})\in\mathopen{]}f^{\prime}(i_{1}),f^{\prime}(i_{3})\mathclose{[}$.
In order to show that $i_{2}\in\mathopen{]}i_{1},i_{3}\mathclose{[}$ it
suffices to consider the case that $f^{\prime}$ differs from $f$ in at most
three elements of $I$, and for that it is enough to show that $f^{\prime}$
respects the cyclic order if it maps only one element of $I^{\prime}$
differently than $f$. So assume that $f=f^{\prime}$ except that
$f^{\prime}(i_{1})$ is the successor of $f(i_{1})$, the other cases are
symmetric. Let $i_{1}^{\prime}$ be an element of $I^{\prime}$ that is mapped
to $f^{\prime}(i_{1})$ by all witnesses of $\Phi\leq\Phi^{\prime}$. In
particular $f(i_{1}^{\prime})=f^{\prime}(i_{1}^{\prime})$, and $f(i_{1})$ is
the predecessor of $f(i_{1}^{\prime})$. If $f(i_{1})\neq f(i_{3})$, then
$i_{2}\in\mathopen{]}i_{1},i_{3}\mathclose{[}$ because $f$ respects the cyclic
order. Otherwise $f^{-1}(f(i_{1}))$ is an interval of $I^{\prime}$ containing
both $i_{1}$ and $i_{3}$ but neither $i_{2}$ nor $i_{1}^{\prime}$, and
$f^{\prime-1}(f^{\prime}(i_{1}))$ is an interval of $I^{\prime}$ containing
both $i_{1}$ and $i_{1}^{\prime}$ but neither $i_{2}$ nor $i_{3}$. Together
with $i_{2}\in\mathopen{]}i_{1},i_{3}\mathclose{[}$ this implies that
$i_{1}\in\mathopen{]}i_{3},i_{1}^{\prime}\mathclose{[}$ and hence that
$i_{1}\in\mathopen{]}i_{3},i_{2}\mathclose{[}$. ∎

######  Lemma 3.13.

The map $F^{\prime}$ is a witness of $\Phi\leq\Phi^{\prime}$.

###### Proof.

Let $v$ and $w$ be distinct elements of $C(I)\setminus I$. If
$F^{\prime-1}([v,w])$ differs from $F^{-1}([v,w])$, then it does so in
elements $i$ of $I_{N}^{\prime}$ with $v(i)=v$ or $v(i)=w$ and one of their
neighbors. Thus

$X\cup\bigcup_{z\in F^{-1}([v,w])}P_{z}^{\prime}=X\cup\bigcup_{z\in
F^{\prime-1}([v,w])}P_{z}^{\prime}.\qed$

###  3.3. Tangles and profiles in $k$-pseudoflowers

The theory presented in this paper will be presented for two very similar
objects: Profiles and tangles. The main results hold for the more general
object, namely profiles. Thus they also hold for tangles, and in this case the
proof for tangles is shorter. As the term tangle is more widely known than the
term profile, the main results will be presented for both tangles and
profiles, with the main differences to be found in this subsection. In the
rest of the paper, we will simply work with a set $\mathcal{P}$ of
$k+1$-profiles, and as every $k+1$-tangle is also a $k+1$-profile the reader
is free to think of a set of tangles.

The following definitions are again to be found in [5]. Given a separation
systems $S_{k}$ of a universe of vertex separations, a _consistent
orientation_ of $S_{k}$ is a set $O\subseteq S_{k}$ such that for all $s\in
S_{k}$, exactly one element of the set $\\{s,s^{*}\\}$ is contained in $O$,
and such that for two elements $r,s\in O$ if $r^{*}\leq s$ then
$\\{r,r^{*}\\}=\\{s,s^{*}\\}$. A _$k$ -profile_ of $S_{k}$ is a consistent
orientation $P$ such that for all elements $r$ and $s$ of $P$ $(r\vee
s)^{*}\notin P$ holds. A $k$-tangle of $S_{k}$ is a consistent orientation $T$
of $S_{k}$ such that for all elements $(A_{1},B_{1})$, $(A_{2},B_{2})$ and
$(A_{3},B_{3})$ of $T$, $(A_{1},B_{1})\vee(A_{2},B_{2})\vee(A_{3},B_{3})$ is
not cosmall, which for universes of vertex separations means that $A_{1}\cup
A_{2}\cup A_{3}\neq V$.

Given a graph, a $k$-profile of that graph is defined to be a profile of
$S_{k}$ in the universe of vertex separations of $G$. So it is obvious what
the definition of a $k$-profile of a graph-like space or a limit-closed
universe of vertex separations should be: A $k$-profile of the universe of all
vertex separations. Similarly, a $k$-tangle of a limit-closed universe of
vertex separations is a tangle of $S_{k}$.

Readers interested in the technical details might note that if the universe is
the universe of vertex separations of a graph, then this definition of a
$k$-tangle is not exactly the same as the definition of a $k$-tangle of a
graph. The two definitions are very close, however, and every $k$-tangle of
the graph is also a $k$-tangle of the universe.

For finite $k$-flowers, a $(k+1)$-tangle points towards a unique petal $i$ of
the flower in the sense that the $(k+1)$-tangle contains $S(i)$. A slightly
weaker form of this property also holds in infinite $k$-pseudoflowers. Given a
$k$-pseudoflower $\Psi$ on index set $I$ a $k+1$-profile $P$ _is located by
$\Psi$_ if there is some $v\in C(I)\setminus I$ such that either $\forall w\in
C(I)\setminus I-v:S(v,w)\in P$ or $\forall w\in C(I)\setminus I-v:S(w,v)\in
P$.

For tangles we always have this property.

######  Lemma 3.14.

Let $\Phi$ be a $k$-pseudoflower and let $T$ be a $k+1$-tangle. Then $T$ is
located by $\Phi$.

###### Proof.

Let $I$ be the index set of $\Phi$ and let $u\in C(I)\setminus I$. Let
$V^{\prime}$ be the set of all elements $w$ of $(C(I)\setminus I)-u$ such that
$S(u,w)\in T$. If $V^{\prime}$ is empty or $(C(I)\setminus I)-u$, then $u$
witnesses that $T$ is located by $\Phi$, so assume otherwise. As $T$ is
consistent, $V^{\prime}$ is an interval of $C(I)\setminus I$ and thus of the
form $\mathopen{]}u,v\mathclose{[}\cap C(I)\setminus I$ or
$\mathopen{]}u,v]\cap C(I)\setminus I$ for some $v\in C(I)\setminus I$. First
consider the case that $V^{\prime}$ is of the form
$\mathopen{]}u,v\mathclose{[}\cap C(I)\setminus I$ for some $v\in
C(I)\setminus I$. Then $S(u,v)$ is not an element of $T$ and thus $S(v,u)\in
T$. Hence for all $w\in\mathopen{]}v,u\mathclose{[}$ we have $S(v,w)\in T$ by
consistency of $T$. Also for all $w\in\mathopen{]}u,v\mathclose{[}$ we have
that $S(u,w)\in T$ so because also $S(v,u)\in T$ the tangle property implies
$S(v,w)\in T$. So in this case we are done. Now consider the case that
$V^{\prime}$ is of the form $\mathopen{]}u,v]$ for some $v\in C(I)\setminus
I-u$. Then $S(u,v)\in T$, so also $S(w,v)\in T$ for all
$w\in[u,v\mathclose{[}$ by consistency. Furthermore for all
$w\in\mathopen{]}v,u\mathclose{[}$ we have $S(u,w)\notin T$, so $S(w,u)\in T$
and thus $S(w,v)\in T$ by the tangle property. So in this case we are done as
well. ∎

For general $k+1$-profiles, the statement holds with an additional assumption
that the $k$-pseudoflower is sufficiently large.

$S(t,s)$$\in P$$s$$t$$x$$y$$v$ Figure 4. $S(t,s)$ is an interval separation of
a $k$-pseudoflower and is contained in a $k+1$-profile $P$. See also Lemma
3.15.

######  Lemma 3.15.

Let $\Phi$ be a $k$-pseudoflower which extends a $k$-flower with at least four
petals and let $P$ be a $k+1$-profile. Then $P$ is located by $\Psi$.

###### Proof.

Let $I$ be the index set of $\Psi$. Let $s$, $t$, $x$ and $y$ be elements of
$C(I)\setminus I$ such that $\Phi(\\{s,t,x,y\\})$ is a $k$-flower with four
petals and $t\in[s,x]$ and $y\in[x,s]$. Then $P$ contains an orientation of
$S(s,x)$ and an orientation of $S(t,y)$, and by Corollaries 3.4 and 3.5 $P$
also contains the supremum of those two orientations. So $P$ contains the
inverse of a petal separation of $\Phi(\\{s,t,x,y\\})$, without loss of
generality assume that $P$ contains $S(t,s)$.

In the interval $[s,t]$ let $v$ be the supremum of
$\\{w\in[s,t\mathclose{[}\setminus I\colon S(t,w)\in P\\}$. There are three
cases: $v=t$, $S(v,x)\in P$ and $S(x,v)\in P$. In the first case we have
$\forall w\in C(I)\setminus I-v:S(v,w)\in P$ and are done.

Consider the second case: $v\neq t$ and $S(v,x)\in P$. Then $S(v,w)=S(v,x)\vee
S(t,w)$ for all $w\in[s,v\mathclose{[}$ by Corollaries 3.4 and 3.5 and thus
$S(v,w)\in P$. Hence $\forall w\in C(I)\setminus I-v:S(v,w)\in P$.

In the last case, $v\neq t$ and $S(x,v)\in P$. Again $S(t,v)=S(t,s)\vee
S(x,v)$ and thus $S(t,v)\in P$. If $t$ is the successor of $v$ in
$C(I)\setminus I$, then $S(t,v)\in P$ implies that $\forall w\in C(I)\setminus
I-v:S(w,v)\in P$. Otherwise let $w\in\mathopen{]}v,t\mathclose{[}\setminus I$.
By the definition of $v$, $P$ does not contain $S(t,w)$, so $S(t,w)=S(t,s)\vee
S(y,w)$ implies that $S(y,w)$ is not an element of $P$. Thus $P$ contains
$S(w,y)$, and $S(w,v)=S(w,y)\vee S(x,v)$ implies that $S(w,v)\in P$. As this
is true for all $w\in\mathopen{]}v,t\mathclose{[}\setminus I$, $\forall w\in
C(I)\setminus I-v:S(w,v)\in P$ holds. ∎

For the remainder of this article, instead of differentiating between tangles
and profiles, we will always work more generally with profiles located by the
relevant flowers. Since, as noted above, Lemma 3.14 also works for tangles of
graphs, our approach encompasses this case as well.

We want to have $k$-pseudoflowers that distinguish as many elements of
$\mathcal{P}$ as possible in the following sense: We call two $\leq
k$-separations _equivalent (with respect to $\mathcal{P}$_) if they are are
contained in the same elements of $\mathcal{P}$. An equivalence class $E$ of
$\leq k$-separations is _displayed_ by a $k$-pseudoflower $\Phi$ if an element
of $E$ is displayed by $\Phi$. We are interested in $k$-pseudoflowers
displaying as many equivalence classes as possible. For that, we also consider
the following relation on $k$-pseudoflowers. Note that it is not a partial
order as it is not antisymmetric.

######  Definition 3.16.

Let $\Phi$ and $\Phi^{\prime}$ be two $k$-pseudoflowers. Define
$\Phi\preccurlyeq\Phi^{\prime}$ if and only if every relevant separation
displayed by $\Phi$ is equivalent to a separation displayed by
$\Phi^{\prime}$. If $\Phi\preccurlyeq\Phi^{\prime}\preccurlyeq\Phi$, then the
$k$-pseudoflowers are _equivalent_.

##  4\. Extending flowers

Two separations _properly cross_ if all of their corner separations are
contained in profiles of $\mathcal{P}$.

Let $\Psi$ be a $k$-pseudoflower and $i$ a petal of $\Psi$ with predecessor
$p$ and successor $s$. A separation $(C,D)$ properly crossing $S(i)$ is
_anchored_ at $v\in C(I)\setminus I$ if $(C,D)\wedge S(i)=S(v,p)$ and
$(D,C)\wedge S(i)=S(s,v)$.

######  Lemma 4.1.

Let $\Phi$ be a $k$-pseudoflower distinguishing at least three elements of
$\mathcal{P}$ located by $\Phi$. Let $(C,D)$ be a separation of order $k$
which properly crosses some petal separation $S(i)$ of $\Phi$. Then there is a
separation $(C^{\prime},D^{\prime})$ properly crossing $S(i)$ such that

  * •

$(C,D)\vee S(i)=(C^{\prime},D^{\prime})\vee S(i)$ and $(D,C)\vee
S(i)=(D^{\prime},C^{\prime})\vee S(i)$.

  * •

$(C^{\prime},D^{\prime})$ or its inverse is anchored at some $v\in
C(I)\setminus I$.

###### Proof.

Let $P_{1}\in\mathcal{P}$ be a profile which contains both $(C,D)$ and $S(i)$.
If $\Phi$ distinguishes two profiles which contain both $(C,D)$ and $S(i)^{*}$
then let $P_{2}\in\mathcal{P}$ be a profile which contains both $(D,C)$ and
$S(i)^{*}$. Then some profile $P_{3}\in\mathcal{P}$ which contains both
$(C,D)$ and $S(i)^{*}$ is distinguished from $P_{2}$ by $\Phi$. If $\Phi$ does
not distinguish any profiles which contain both $(C,D)$ and $S(i)^{*}$, then
let $P_{3}$ be some such profile. As $\Phi$ distinguishes at least three
profiles, it distinguishes some profile $P_{2}$ from both $P_{1}$ and $P_{3}$.
Because $P_{2}$ is distinguished from $P_{1}$, it contains $S(i)^{*}$. Because
$P_{2}$ is also distinguished from $P_{3}$, it does not contain $(C,D)$, so it
contains $(D,C)$. In both cases, $P_{2}$ contains $S(i)^{*}$ and $(D,C)$,
while $P_{3}$ contains $S(i)^{*}$ and $(C,D)$. Furthermore $\Phi$
distinguishes $P_{2}$ and $P_{3}$.

Denote the predecessor of $i$ in $C(I)$ by $p$ and the successor by $s$. As
$\Phi$ distinguishes $P_{2}$ and $P_{3}$, there is a cutpoint $v\in
C(I)\setminus I$ such that $S(v,s)$ distinguishes $P_{2}$ and $P_{3}$. Because
both $P_{2}$ and $P_{3}$ contain $S(i)^{*}$, $S(v,p)$ also distinguishes
$P_{2}$ and $P_{3}$, and $P_{2}$ contains $S(v,s)$ if and only if it contains
$S(v,p)$.

From here on consider the case that $P_{3}$ contains $S(v,p)$, the other case
is symmetric. Define $(C^{\prime\prime},D^{\prime\prime})=(C,D)\vee S(v,p)$.
Because $P_{3}$ contains both $(C,D)$ and $S(v,p)$ and $P_{2}$ contains both
$S(p,v)$ and $(D,C)$, the separation $(C^{\prime\prime},D^{\prime\prime})$ has
order $k$ and distinguishes $P_{2}$ and $P_{3}$. Then
$(C^{\prime\prime},D^{\prime\prime})\vee S(i)=(C,D)\vee S(v,p)\vee
S(i)=(C,D)\vee S(i)$ and in particular
$(C^{\prime\prime},D^{\prime\prime})\vee S(i)$ is the same as $(C,D)\wedge
S(i)$. Let $P_{4}$ be a profile that contains both $S(i)$ and $(D,C)$. Then
$(D^{\prime\prime},C^{\prime\prime})\leq(C,D)$ implies that
$(D^{\prime\prime},C^{\prime\prime})$ is contained in $P_{4}$. The profile
$P_{3}$ contains both $(C^{\prime\prime},D^{\prime\prime})$ and $S(i)^{*}$ and
the profile $P_{4}$ contains both $(D^{\prime\prime},C^{\prime\prime})$ and
$S(i)$, so $(D^{\prime\prime},C^{\prime\prime})\vee S(i)$ is a separation of
order $k$. Also $(D^{\prime\prime},C^{\prime\prime})\vee S(i)\leq(D,C)\vee
S(i)$. Because $V(v,p)\cup V(s,p)=V$, we have $D\cup V(s,p)=(D\cap V(p,v))\cup
V(s,p)$. The set $D\cup V(s,p)$ is the left side of $(D,C)\vee S(i)$ and
$(D\cap V(p,v))\cup V(s,p)$ is the left side of the separation
$(D^{\prime\prime},C^{\prime\prime})\vee S(i)$. Also, the right side of
$(D,C)\vee S(i)$ is a subset of the right side of
$(D^{\prime\prime},C^{\prime\prime})\vee S(i)$. Because these two separations
have the same order, they are equal.

Define $(D^{\prime},C^{\prime})=(D^{\prime\prime},C^{\prime\prime})\vee
S(s,v)$. Just as in the previous paragraph $(C^{\prime},D^{\prime})$ is a
separation of order $k$ distinguishing $P_{2}$ and $P_{3}$. Also
$(D^{\prime},C^{\prime})\vee S(i)=(D^{\prime\prime},C^{\prime\prime})\vee
S(i)$ and $(C^{\prime},D^{\prime})\vee
S(i)=(C^{\prime\prime},D^{\prime\prime})\vee S(i)$. Furthermore $P_{1}$
contains $(C^{\prime},D^{\prime})$ and $S(i)$, and $P_{3}$ contains
$(D^{\prime},C^{\prime})$ and $S(i)^{*}$, so $(C^{\prime},D^{\prime})\wedge
S(i)$ has order $k$. Because $S(v,p)\leq(C^{\prime},D^{\prime})\leq S(v,s)$,
also

$(V(v,p)\cup P_{s},V(p,v))=S(v,s)\wedge S(i)\geq(C^{\prime},D^{\prime})\wedge
S(i)\geq S(v,p)\wedge S(i)=S(v,p),$

which, as $(C^{\prime},D^{\prime})\wedge S(i)$ has the same order as $S(v,p)$,
implies that $(C^{\prime},D^{\prime})\wedge S(i)=S(v,p)$. Similarly
$(D^{\prime},C^{\prime})\wedge S(i)=S(s,v)$. ∎

######  Lemma 4.2.

Let $\Psi$ be a $k$-pseudoflower and $i$ a petal of $\Psi$ with predecessor
$p$ and successor $s$. If a $k$-separation $(C,D)$ properly crossing $S(i)$ is
anchored at $v\in C(I)\setminus I$, then $C\cap P_{s}=\emptyset$, $D\cap
P_{p}=\emptyset$,$X\subseteq C\cap D$ and $(C\cap D)\setminus V(p,s)=P_{v}$.

###### Proof.

Since $P_{s}$ is contained in the separator of $S(i)$, but does not meet
$V(v,p)$ it cannot meet $C$. Similarly $(C,D)\wedge S(i)=S(s,v)$ implies that
$P_{p}$ does not meet $D$. The third statement is immediate from the fact that
$X$ occurs in the separators of both $S(s,v)$ and $S(v,p)$.

For the last equation, note that $(C,D)\wedge S(i)=S(v,p)$ implies both $C\cap
V(s,p)=V(v,p)$ and $D\cup V(p,s)=V(p,v)$. Taking the intersection gives
$(C\cap D\cap V(s,p))\cup(C\cap V(s,p)\cap V(p,s))=V(v,p)\cap V(p,v)$.
Deleting $V(p,s)$ on both sides and simplifying gives the result. ∎

######  Lemma 4.3.

Let $\Phi$ be a $k$-pseudoflower distinguishing at least three elements of
$\mathcal{P}$ located by $\Phi$. Let $(C,D)$ be a separation of order $k$
which properly crosses some petal separation $S(i)$ of $\Phi$ and is anchored
at $v\in C(I)\setminus I$. Let $p$ be and $s$ be the predecessor and successor
of $i$ in $C(I)$ respectively. Then there is an extension $\Phi^{\prime}$ of
$\Phi$, witnessed by $F$, such that

  * •

For all $i^{\prime}\in I\setminus\\{i\\}$ there is only one element contained
in $F^{-1}(i^{\prime})$.

  * •

$F^{-1}(i)$ contains exactly three elements $i_{1}$, $m$ and $i_{2}$. By
switching the names of $i_{1}$, $m$ and $i_{2}$ if necessary we can assume
that $m$ is the successor of $i_{1}$ and the predecessor of $i_{2}$.

  * •

$P_{m}=(C\cap D)\setminus V(s,p)$.

  * •

The interval set of $i_{1}$ is $C\cap V(i)$ and the interval set of $i_{2}$ is
$D\cap V(i)$.

  * •

$(C,D)$ is an interval separation of $\Phi^{\prime}$.

###### Proof.

Let $\Phi^{\prime}$ be obtained from $\Phi$ by replacing $i\in I$ with
$i_{1}$, $m$ and $i_{2}$ in order and setting $P_{m}=(C\cap D)\setminus
V(s,p)$, $P_{i_{1}}=C\cap P_{i}$ and $P_{i_{2}}=D\cap P_{i}$.

We will now prove that $\Phi^{\prime}$ is a $k$-pseudoflower. The third
condition is trivial. By Lemma 4.2 we know that $C\cap D$ consists of exactly
$P_{m}$, $X$ and $P_{v}$, so $|P_{m}|=k-|P_{v}|-|X|=\frac{k-|X|}{2}$.

Assume for a contradiction that (• ‣ 3.1) fails. We may assume without loss of
generality that $|P_{i_{1}}|=\frac{k-|X|}{2}$. It follows that
$P_{m}=P_{i_{1}}$ and thus $P_{i_{1}}\subseteq P_{i_{2}}$. But then $(D,C)\vee
S(i)^{*}$ is cotrivial with witness $(C,D)$ and therefore is contained in
every profile, contradicting the fact that $S(i)$ properly crosses $(C,D)$.

To show that $\Phi^{\prime}$ is a $k$-pseudoflower, it thus remains to show
the second condition. Let $x,y\in C(I)$ be arbitrary. Because $\Phi$ was a
$k$-pseudoflower, w.l.o.g. $y=m$. Furthermore, since $P_{i_{1}}$ and
$P_{i_{2}}$ meet only in $P_{m}$, we have $V(x,m)\cap V(m,x)=P_{x}\cup
P_{m}\cup X$.

Thus it is enough to show that $(V(x,m),V(m,x))$ is a separation. Without loss
of generality $x$ occurs in the interval in the interval from $m$ to $v$ in
$C(I)$, the other case is analogous. Note that, as $(C,D)$ is anchored at $v$,
$S(v,p)\leq(C,D)\leq S(v,s)$ and thus $V(v,p)\subseteq C\subseteq V(v,s)$. So

$V(x,p)\cup C=V(x,p)\cup(C\cap V(i))=V(x,m)$

and similarly $V(p,x)\cap D=V(m,x)$. Hence $S(v,p)\vee(C,D)=S(x,m)$, implying
that $S(x,m)$ is indeed a separation.

Thus $\Phi^{\prime}$ is a $k$-pseudoflower. By construction, $(C,D)$ appears
as the interval separation $S(m,v)$. ∎

$i$$i_{1}$$i_{2}$$v$$p$$m$$s$$(C,D)$ Figure 5. The the index $i$ is replaced
by the indices $i_{1}$ and $i_{2}$, creating a new cutvertex $m$. This allows
us to display the separation $(C,D)$, see also Lemma 4.3.

##  5\. Maximal $k$-pseudoflowers

Let $(\Phi_{j})_{j\in J}$ be a $\leq$-chain of $k$-pseudoflowers which
distinguish at least three elements of $\mathcal{P}$ that are located by the
$\Phi_{j}$. In this section we will prove that there is a $k$-pseudoflower
which is an upper bound of the chain $(\Phi_{j})_{j\in J}$. The proof works
for both $k$-pseudodaisies and $k$-pseudoanemones, but can be shortened a lot
if the $\Phi_{j}$ are $k$-pseudoanemones as follows: The careful construction
of the sets $P_{v}$ with $v\in V_{N}$ in Section 5.3 becomes redundant as all
those sets are necessarily empty. For the same reason, in Section 5.5 there is
no need to derive the sets $P_{v}$ with $v\in C(I\setminus
I_{N})\setminus(I\setminus I_{N})$ from their counterparts in $C(I)\setminus
I_{N}$. Furthermore, it is possible to carry the following stronger version of
(• ‣ 3.1) through the limit process: That no petal should be empty. This
stronger version then makes all witnesses of comparability of
$k$-pseudoflowers unique, making the search for a compatible choice of
witnesses unnecessary. Also, the definition of $I^{\prime}$ can be made
easier, as it simply consists of those $i\in I$ with $P_{i}\neq\emptyset$, and
$C(I^{\prime})$ is more obviously related to $C$. Lastly, in $k$-anemones the
condition of Corollary 3.4 that $P_{a}\cap P_{d}=\emptyset$ always holds,
which means that the proof in Section 5.4 that all interval separations of the
newly constructed $k$-pseudoflower are indeed interval separations can be
streamlined.

###  5.1. Finding compatible witnesses

The first step in constructing an upper bound of a chain of $k$-pseudoflowers
is to construct the index set. For that we would like to take the inverse
limit of the index sets, and for that we need compatible witnesses of the
comparability of the $k$-pseudoflowers in the chain. Phrased more formally, we
need for every $j\leq l\in J$ a witness $F_{lj}$ of $\Phi_{j}\leq\Phi_{l}$
such that for $j\leq l\leq m\in J$ the concatenation of $F_{lj}$ with $F_{ml}$
is $F_{mj}$.

In Section 3.2 we showed that, given sufficiently large $k$-pseudoflowers
$\Phi\leq\Phi^{\prime}$ on index sets $I$ and $I^{\prime}$, only few elements
$i$ of $I^{\prime}$ have several possible images under witnesses of
$\Phi\leq\Phi^{\prime}$, and that picking some possible image for every such
$i$ gives again a witness of $\Phi\leq\Phi^{\prime}$. So defining the $F_{lj}$
amounts to finding a way of making all these choices in a way that is
compatible over the whole chain of $k$-pseudoflowers. For that, we first show
how these choices interact in a chain of only three $k$-pseudoflowers.

So for the next two lemmas let $\Phi_{1}\leq\Phi_{2}\leq\Phi_{3}$ be
$k$-pseudoflowers on index sets $I_{1}$, $I_{2}$ and $I_{3}$ respectively such
that $\Phi_{1}$ extends a $k$-flower with three petals. For some $i_{3}\in
I_{3}$ we are now interested in possible images of $i_{3}$ under witnesses of
$\Phi_{2}\leq\Phi_{3}$, their images under witnesses of $\Phi_{1}\leq\Phi_{2}$
and how those relate to witnesses of $i_{3}$ under witnesses of
$\Phi_{1}\leq\Phi_{3}$.

######  Lemma 5.1.

If $i_{3}$ has two possible images $i_{1}$ and $i_{1}^{\prime}$ under
$\Phi_{1}\leq\Phi_{3}$, and there is $i_{2}\in I_{2}$ whose possible witnesses
under $\Phi_{1}\leq\Phi_{2}$ are also $i_{1}$ and $i_{1}^{\prime}$, then all
witnesses of $\Phi_{2}\leq\Phi_{3}$ map $i_{3}$ to $i_{2}$.

###### Proof.

By Lemma 3.10 there is some $i_{3}^{\prime}$ whose unique image under
$\Phi_{2}\leq\Phi_{3}$ is $i_{2}$. Then $i_{3}^{\prime}$ has $i_{1}$ and
$i_{1}^{\prime}$ as possible images under $\Phi_{1}\leq\Phi_{3}$ and
$P_{i_{3}}=P_{i_{3}^{\prime}}$ by Lemma 3.9, so $i_{3}=i_{3}^{\prime}$. ∎

######  Lemma 5.2.

If $i_{3}$ has two possible images $i_{1}$ and $i_{1}^{\prime}$ under
$\Phi_{1}\leq\Phi_{3}$, $i_{1}^{\prime}$ successor of $i_{1}$ in $I_{1}$, and
$i_{3}$ has two possible images $i_{2}$ and $i_{2}^{\prime}$ under
$\Phi_{2}\leq\Phi_{3}$, $i_{2}^{\prime}$ successor of $i_{2}$ in $I_{2}$, then
every witness of $\Phi_{1}\leq\Phi_{2}$ maps $i_{2}$ to $i_{1}$ and
$i_{2}^{\prime}$ to $i_{1}^{\prime}$.

$i_{3}$$j$$\hat{\imath}_{3}$$i_{2}^{\prime}$$i_{2}$$\hat{\imath}_{2}$$i_{1}^{\prime}$$i_{1}$$\hat{\imath}_{1}$
Figure 6. Three compatible $k$-pseudoflowers
$\Phi_{1}\leq\Phi_{2}\leq\Phi_{3}$. The outer circle depicts $\Phi_{3}$ and
three elements of its index set, the inner circle depicts $\Phi_{1}$.

###### Proof.

See Fig. 6 for a depiction of the notation used in this proof. As a
concatenation of witnesses is again a witness, all witnesses of
$\Phi_{1}\leq\Phi_{2}$ map $i_{2}$ to one of $i_{1}$ and $i_{1}^{\prime}$, and
similarly for $i_{2}^{\prime}$. If one element of $I_{2}$ had both $i_{1}$ and
$i_{1}^{\prime}$ as possible images under $\Phi_{1}\leq\Phi_{2}$, then by the
previous lemma $i_{3}$ could not have both $i_{2}$ and $i_{2}^{\prime}$ as
possible images under $\Phi_{2}\leq\Phi_{3}$. Hence no element of $I_{2}$ has
both $i_{1}$ and $i_{1}^{\prime}$ as possible images under witnesses of
$\Phi_{1}\leq\Phi_{2}$, and in particular both $i_{2}$ and $i_{2}^{\prime}$
have a unique image under such witnesses. Let $\hat{\imath}_{1}$ be an element
of $I_{1}$ that is neither $i_{1}$ nor $i_{1}^{\prime}$, and let
$\hat{\imath}_{3}$ be an element of $I_{3}$ that is mapped to
$\hat{\imath}_{1}$ by all witnesses of $\Phi_{1}\leq\Phi_{3}$.

Assume for a contradiction that both $i_{2}$ and $i_{2}^{\prime}$ have the
same image in $I_{1}$ under witnesses of $\Phi_{1}\leq\Phi_{2}$, and that this
image is $i_{1}^{\prime}$ (the other case is symmetric). Let $j$ be an element
of $I_{3}$ that is mapped to $i_{2}$ by all witnesses of
$\Phi_{2}\leq\Phi_{3}$. As no witness of $\Phi_{2}\leq\Phi_{3}$ maps
$\hat{\imath}_{3}$ to $i_{2}$ or $i_{2}^{\prime}$ and there is some witness
that maps $i_{3}$ to $i_{2}^{\prime}$, we have
$j\in\mathopen{]}\hat{i_{3}},i_{3}\mathclose{[}$. Also, by Lemma 3.12 (and
because $j\neq i_{3}$) there is a witness of $\Phi_{1}\leq\Phi_{3}$ that maps
$j$ to $i_{1}^{\prime}$, $i_{3}$ to $i_{1}$ and $\hat{i_{3}}$ to
$\hat{i_{1}}$. This implies
$j\in\mathopen{]}i_{3},\hat{\imath}_{3}\mathclose{[}$, a contradiction.

So $i_{2}$ and $i_{2}^{\prime}$ have distinct unique images under witnesses of
$\Phi_{1}\leq\Phi_{2}$. If $\hat{\imath}_{2}$ denotes the image of
$\hat{\imath}_{3}$ under some witness of $\Phi_{2}\leq\Phi_{3}$, then
$\hat{\imath}_{2}\in\mathopen{]}i_{2}^{\prime},i_{2}\mathclose{[}$ and
$\hat{\imath}_{1}\in\mathopen{]}i_{1}^{\prime},i_{1}\mathclose{[}$ implies
that indeed all witnesses of $\Phi_{1}\leq\Phi_{2}$ map $i_{2}$ to $i_{1}$ and
$i_{2}^{\prime}$ to $i_{1}^{\prime}$. ∎

Now we want to use these two lemmas to define the maps $F_{lj}$. For all
indices $j,l\in J$ with $j\leq l$ define a map $f_{lj}:I_{l}\rightarrow I_{j}$
as follows: For $i\in I_{l}$, if there are indices
$j=j_{0}<j_{1}<\ldots<j_{n}=l$, $n\geq 1$, in $J$ such that for $i_{n}=i$ and
for every $m\leq n$ every witness of $\Phi_{j_{m-1}}\leq\Phi_{j_{m}}$ maps
$i_{m}$ to the same element $i_{m-1}$ of $I_{m-1}$, then let $f_{lj}(i)=i_{0}$
(first case). Otherwise there are two possible images $i_{j}$ and
$i_{j}^{\prime}$ of $i$ under witnesses of $\Phi_{j}\leq\Phi_{l}$ such that
$i_{j}^{\prime}$ is the successor of $i_{j}$. In this case let
$f_{lj}(i)=i_{j}^{\prime}$ (second case).

######  Lemma 5.3.

For all $j\leq l\in J$ and all $i\in I_{j}$, $f_{lj}(i)$ is well defined.

###### Proof.

In order to show that $f_{lj}(i)$ is well-defined, it suffices to consider the
case where $f_{lj}(i)$ is defined via the first case. Assume that
$j=j_{0}\leq\ldots j_{n}=l$ in $J$ and $j=l_{0}\leq\ldots\leq l_{m}=l$ are two
chains of elements of $J$ via which $f_{lj}(i)$ could be defined. We will show
by induction on $n+m$ that the value of $f_{lj}(i)$ is the same in both cases.
If one of $n$ and $m$ is $0$, then $j=l$ and the claim holds, so assume
otherwise. Furthermore, if $j_{n-1}=l_{m-1}$, then it suffices to apply the
induction hypothesis to the chains $j_{0}\leq\ldots\leq j_{n-1}$ and
$l_{0}\leq\ldots\leq l_{m-1}$. So assume that $j_{n-1}<l_{m-1}$, the other
case is symmetric. Let $r$ be an integer such that $l_{r-1}\leq j_{n-1}\leq
l_{r}$, and $i_{r}$ the image of $i$ in $I_{l_{r}}$ under the chain of the
$l_{s}$. If $i_{r}$ has several images under witnesses of
$\Phi_{j_{n-1}}\leq\Phi_{l_{r}}$, then $i$ also has several images under
witnesses of $\Phi_{j_{n-1}}\leq\Phi_{l}$. So $i_{r}$ has a unique image
$i_{s}$ under witnesses of $\Phi_{j_{n-1}}\leq\Phi_{l_{r}}$, and that image is
also the unique image of $i$ under witnesses of $\Phi_{j_{n-1}}\leq\Phi_{l}$.
Then $j_{0}\leq\ldots\leq j_{n-1}\leq l_{r}$ and $l_{0}\leq\ldots\leq l_{r}$
are chains on which, because $n+r\leq n+m-1$, the induction hypothesis can be
applied. Thus the lemma holds. ∎

######  Lemma 5.4.

For all $j\leq l\leq m$ in $J$ and $i\in I_{m}$, $f_{lj}\circ
f_{ml}(i)=f_{mj}(i)$.

###### Proof.

First consider the case that $f_{mj}(i)$ is defined via the first case. Let
$j_{0}\leq\ldots j_{n}$ be a sequence via which $f_{mj}(i)$ could have been
defined, with $j_{r-1}\leq l\leq j_{r}$. If $i_{r}$ has two possible images
$i_{l},i_{l}^{\prime}$ under witnesses of $\Phi_{l}\leq\Phi_{j_{r}}$, then
$f_{ml}(i)$ is one of those. Also, every witness of $\Phi_{r-1}\leq\Phi_{l}$
maps both $i_{l}$ and $i_{l}^{\prime}$ to $i_{r-1}$, implying that both
$f_{lj}(i_{l})$ and $f_{lj}(i_{l}^{\prime})$ are defined via the first case
and are equal to $f_{mj}(i)$. So in this case $f_{lj}\circ
f_{ml}(i)=f_{mj}(i)$. If $i_{r}$ has a unique image $i_{l}$ under witnesses of
$\Phi_{j}\leq\Phi_{r}$, then all witnesses of $\Phi_{r-1}\leq\Phi_{l}$ map
$i_{l}$ to $i_{r-1}$, and thus $f_{ml}(i)$ is defined via the first case and
also $f_{lj}(i_{l})$ is defined via the first case. As $f_{mj}(i)$ is well
defined, this implies that $f_{lj}\circ f_{ml}(i)=f_{mj}(i)$.

So assume that $f_{mj}(i)$ is defined via the second case, and that $i_{j}$
and $i_{j}^{\prime}$ are the two possible images of $i$ under witnesses of
$\Phi_{j}\leq\Phi_{m}$. If there is $i_{l}\in I_{l}$ such that both $i_{l}$
and $i_{l}^{\prime}$ are images of witnesses of $\Phi_{j}\leq\Phi_{l}$, then
by Lemma 5.1 every witness of $\Phi_{l}\leq\Phi_{m}$ maps $i$ to $i_{l}$. Thus
$f_{mj}(i)=f_{ml}(i_{l})=f_{ml}\circ f_{lj}(i)$. So assume that there is no
$i_{l}\in I_{l}$ that has $i_{j}$ and $i_{j}^{\prime}$ as possible images
under witnesses of $\Phi_{j}\leq\Phi_{l}$. By renaming assume that
$i_{j}^{\prime}$ is the successor of $i_{j}$. Then by Lemma 5.2 there are
$i_{l}$ and $i_{l}^{\prime}$ in $I_{l}$ such that every witness of
$\Phi_{j}\leq\Phi_{l}$ maps $i_{l}$ to $i_{j}$ and $i_{l}^{\prime}$ to
$i_{j}^{\prime}$. Also, $i_{l}^{\prime}$ is the successor of $i_{l}$, and
every witness of $\Phi_{l}\leq\Phi_{m}$ maps $i$ to $i_{l}$ or
$i_{l}^{\prime}$. Then $f_{ml}(i)$ cannot be defined via the first case,
because that would imply that $f_{mj}(i)$ would also be defined via the first
case. So $f_{mj}(i)=i_{j}^{\prime}=f_{lj}(i_{l}^{\prime})$ and
$f_{ml}(i)=i_{l}^{\prime}$. ∎

For $j\leq l\in J$ let $F_{lj}$ be the unique extension of $f_{lj}$ to a
surjective map $C(I_{l})\rightarrow C(I_{j})$ respecting the cyclic order with
$F_{lj}(I_{l})=I_{j}$, which exists by Lemma 2.7.

######  Lemma 5.5.

The maps $F_{lj}$ are witnesses of $\Phi_{j}\leq\Phi_{l}$ and are compatible
in the sense that for $j\leq l\leq m\in J$ the concatenation of $F_{lj}$ and
$F_{ml}$ is $F_{mj}$.

###### Proof.

As $f_{lj}$ can be obtained from the restriction of any witness of
$\Phi_{j}\leq\Phi_{l}$ to $I_{l}$ by changing the images of some elements of
$I_{j}$ to another image they can have under witnesses of
$\Phi_{j}\leq\Phi_{l}$, by Lemma 3.13 $F_{lj}$ is also a witness of
$\Phi_{j}\leq\Phi_{l}$. Furthermore the restriction of $F_{lj}\circ F_{ml}$ to
$I_{m}$ is equal to $f_{mj}$, and $F_{lj}\circ F_{ml}$ is surjective, respects
the cyclic order and satisfies $F_{mj}(I_{m})=I_{j}$. Thus, by uniqueness of
the extensions, $F_{mj}=F_{lj}\circ F_{ml}$. ∎

######  Corollary 5.6.

There are witnesses $F_{lj}$ of $\Phi_{j}\leq\Phi_{l}$, one for all pairs of
indices $j\leq l\in J$, such that for all $j\leq l\leq m\in J$ the
concatenation of $F_{lj}$ and $F_{ml}$ is $F_{mj}$.∎

###  5.2. Inverse limits of $k$-pseudoflowers

As we showed in the last section, there is a compatible family of witnesses
$F_{lj}$ of $\Phi_{j}\leq\Phi_{l}$, one for every pair of indices $j\leq l\in
J$. For the following construction we fix one such family. Then there is an
inverse limit $I$ of the $I_{j}$ with projections $(\pi_{j})_{j\in J}$.
Because all $F_{lj}$ respect the cyclic order, the inverse limit also has a
cyclic order which is respected by the projections. Further, because all the
$F_{lj}$ are surjective, so are the projections. We will construct a partition
$\Psi:=(P_{i})_{i\in C(I)}$ that is an upper bound of the chain
$(\Phi_{j})_{j\in J}$ and is a $k$-pseudoflower except that it may violate
property (• ‣ 3.1). So far we can only define $P_{i}$ with $i\in I$: It is the
intersection of all sets $P_{\pi_{j}(i)}$ with $j\in J$.

$I_{j}$$I_{l}$$I_{m}$$C(I_{j})$$C(I_{l})$$C(I_{m})$$f_{lj}$$f_{ml}$$f_{mj}$$F_{lj}$$F_{ml}$$F_{mj}$

$I_{j}$$I_{l}$$I$$C(I_{j})$$C(I_{l})$$C(I)$$\pi_{j}$$f_{lj}$$\pi_{l}$$\Pi_{j}$$F_{lj}$$\Pi_{l}$

Figure 7. To the left: The maps $(F_{lj})_{j\leq l\in J}$ extend the
compatible maps $(f_{lj})_{j\leq l\in J}$ and are therefore compatible
witnesses that the $k$-pseudoflowers are comparable. To the right: The maps
$(\Pi_{j})_{j\in J}$ defined via the projections $(\pi_{j})_{j\in J}$ are
compatible with the maps $(F_{lj})_{j\leq l\in J}$.

We will now relate $C(I)$ to the cyclically ordered sets $C(I_{j})$. By Lemma
2.7 every projection $\pi_{j}:I\rightarrow I_{j}$ can be extended uniquely to
a surjective map $\Pi_{j}:C(I)\rightarrow C(I_{j})$ that respects the cyclic
order. For $j\leq l\in J$, the restriction of $F_{lj}\circ\Pi_{l}$ to $I_{l}$
is $\pi_{j}$, and hence $F_{lj}\circ\Pi_{l}=\Pi_{j}$. By Lemma 2.8, the
cutpoints of the $C(I_{j})$ can be identfied with each other and with
cutpoints of $C(I)$ as follows: Given $j\leq l\leq n\in J$ and $v\in
C(I_{j})\setminus I_{j}$, let $u$ be the unique element of $C(I_{l})\setminus
I_{l}$ with $F_{lj}(u)=v$ and let $u^{\prime}$ be the unique element of
$C(I_{n})\setminus I_{n}$ with $F_{nl}(u^{\prime})=u$. Then $u^{\prime}$ is
also the unique element of $C(I_{n})\setminus I_{n}$ with
$F(nj)(u^{\prime})=v$. Also, if $w$ is the unique element of $C(I)\setminus I$
with $\Pi_{l}(w)=u$, then $w$ is the unique element of $C(I)\setminus I$ with
$\Pi_{j}(w)=v$. So it is well-defined to identify $v$ with $w$ and with all
cutpoints that get mapped to $v$ via some $F_{lj}$. We do this identification
for all cutpoints of all $C(I_{j})$ and denote the set of cutpoints of $C(I)$
that are identified with a cutpoint of some $C(I_{j})$ by $V_{F}$. Then for
$v\in V_{F}$, $P_{v}$ is the same for all $j\in J$ where $v$ is a cutpoint, so
we take that vertex set also to be $P_{v}$ for $\Psi$. Also note that, for
$v,w\in V_{F}$, $V(v,w)$ does not depend on the $k$-pseudoflower $\Phi_{j}$
with respect to which it is defined. But $V(v,w)$ taken in $\Psi$ is not yet
defined, and when it is we will first have to show that it is equal to
$V(v,w)$ taken in some $\Phi_{j}$.

In order to properly distinguish here, for cutpoints $v$ and $w$ of $C(I)$ we
introduce the notation

$\displaystyle V^{\prime}(v,w)$ $\displaystyle=X\cup P_{v}\cup
P_{w}\cup\bigcup_{z\in[v,w]\cap(I\cup V_{F})}P_{z}$ and
$\displaystyle\hat{V}(v,w)$ $\displaystyle=X\cup\bigcup_{z\in[v,w]\cap
C(I)}P_{z}$

while the notation $V(v,w)$ is reserved for the case $v,w\in V_{F}$ and the
value taken in some $\Phi_{j}$. Note that up to now, in many cases
$V^{\prime}$ and $\hat{V}$ are not yet well-defined because $P_{z}$ is not yet
defined for cutpoints $z\notin V_{F}$. Those sets $P_{z}$ are going to be
defined later.

For $v,w\in V_{F}$, we can see that $V^{\prime}(v,w)$ and $V(v,w)$ are the
same:

######  Lemma 5.7.

For all distinct $v,w\in V_{F}$ we have $V(v,w)\setminus
X=\bigcup_{z\in[v,w]}P_{z}$ where the interval is taken in $I\cup V_{F}$.

###### Proof.

Let us first show that $V(v,w)\setminus X$ is a subset of
$\bigcup_{z\in[v,w]}P_{z}$. For this, let $u\in V(v,w)\setminus X$. If also
$u\in V(w,v)$ then $u\in P_{v}\cup P_{w}$ and we are done, so assume that
$u\notin V(w,v)$. If there is $z\in V_{F}$ such that $u\in P_{z}$, then
$u\notin V(w,v)$ implies that $z\notin[w,v]$ and thus $z\in[v,w]$. If there is
no $z\in V_{F}$ such that $u\in P_{z}$, then for all $j\in J$ there is a
unique $i_{j}\in I_{j}$ such that $u\in P_{i_{j}}$. In this case also
$F_{lj}(i_{l})=i_{j}$ for all $j\leq l\in J$. So there is $i\in I$ such that
$\Pi_{j}(i)=i_{j}$ for all $j\in J$, and $u\in P_{i}$. Again $u\notin V(v,w)$
implies $i\in[v,w]$.

To prove the other inclusion, let $j\in J$ be sufficiently large such that
$C(I_{j})$ contains $v$, $w$ and $z$ if $z\in V_{F}$. Then $\Phi_{j}$
witnesses that $P_{z}\subseteq V(v,w)$. Furthermore all $P_{z}$ are disjoint
from $X$. ∎

######  Corollary 5.8.

$X=V(G)\setminus\bigcup_{z\in I\cup V_{F}}P_{z}$.∎

######  Corollary 5.9.

For all distinct $v,w\in V_{F}$ we have $V(v,w)=V^{\prime}(v,w)$.∎

######  Corollary 5.10.

For all distinct $v,w\in V_{F}$ the pair $(V^{\prime}(v,w),V^{\prime}(w,v))$
is a separation of order at most $k$ with separator $P_{v}\cup P_{w}\cup X$.∎

###  5.3. Completing the index set

Let $V_{N}$ be the set of cutpoints of $C(I)$ that are not contained in
$V_{F}$. In this subsection we will define the values $P_{z}$ with $z\in
V_{N}$.

######  Lemma 5.11.

Every $v\in V_{N}$ has a unique neighbor in $C(I)$, which is an element of
$I$.

###### Proof.

As $v\in V_{N}$, no $\Pi_{j}$ maps $v$ to a cutpoint. Let $i$ be the element
of $I$ with $\pi_{j}(i)=\Pi_{j}(v)$ for all $j\in J$. If it exists, let $z$ be
a neighbor of $v$ in $C(I)$. For all $j\in J$, $\Pi_{j}(z)$ is $\Pi_{j}(v)$ or
one of its neighbors in $C(I_{j})$. Also no cutpoint can be a neighbor of
another cutpoint by Lemma 2.6, so $z\in I$ and $z=i$. Hence if $v$ has a
neighbor in $C(I)$ then that neighbor is $i$. As $i$ is the only element of
$I$ with $\Pi_{j}(i)=\Pi_{j}(v)$ for all $j\in J$, $i$ is indeed a neighbor of
$v$ in $C(I)$. ∎

Now we want to define $P_{v}$ for $v\in V_{N}$. By Lemma 5.11 there is a
unique $i\in I$ which is a neighbor of $v$ in $C$. For every $j\in J$ let
$u_{j}$ be the predecessor and $w_{j}$ the successor of $\Pi_{j}(i)$ in
$C(I_{j})$. Let $z\in V_{F}$ be a cutpoint such that for all sufficiently
large $j\in J$ both $S(z,w_{j})$ and $S(z,u_{j})$ distinguish two elements of
$\mathcal{P}$. In particular, for all sufficiently large indices $j$, $P_{z}$
is disjoint from $P_{u_{j}}$ and $P_{w_{j}}$ and thus $P_{z}$ is disjoint from
$V(u_{j},w_{j})$.

$\Pi_{j}(i)$$i$$(Y,Z)$$S(u_{j},z)$$z$$u_{j}$$v$$w_{j}$

$\Pi_{j}(i)$$i$$(Y^{1},Z^{1})$$(Z^{2},Y^{2})$$S(z_{2},z_{1})$$z_{2}$$z_{1}$$u_{j}$$v$$w_{j}$

Figure 8. To the left: In the case that $v\in V_{N}$ is the predecessor of its
neighbor $i$ in $C(I)$, the set $P_{v}$ is defined via the limit of the
separations $S(z,u_{j})$ for some suitable $z\in V_{F}$.
To the right: $P_{v}$ does not depend on the choice of $z$, see also Lemma
5.15.

If $v$ is the predecessor of $i$, then let $(Y,Z)$ be the supremum of the set
$\\{S(z,u)\colon u\in\mathopen{]}z,v\mathclose{[}\cap V_{F}\\}$ and otherwise
let $(Y,Z)$ be the infimum of the set $\\{S(z,w)\colon
w\in\mathopen{]}v,z\mathclose{[}\cap V_{F}\\}$. In both cases $(Y,Z)$ has
order at most $k$ because the order function is limit-closed. Define
$P_{v}:=(Y\cap Z)\setminus(P_{z}\cup X)$. As $S(z,u_{j})\leq(Y,Z)\leq
S(z,w_{j})$ for all sufficiently large $j\in J$, $Y\cap Z$ contains $P_{z}$.
Also $(Y,Z)$ distinguishes two elements of $\mathcal{P}$ so it has order $k$.
Hence $P_{v}$ has $(k-|X|)/2$ many elements.

######  Lemma 5.12.

$P_{v}\subseteq V(u_{j},w_{j})$ for all $j\in J$.

###### Proof.

It suffices to show the claim for all $j\in J$ such that $z$ is a cutpoint of
$C(I_{j})$. As $S(z,u_{j})\leq(Y,Z)\leq S(z,w_{j})$ by definition, $Y\cap Z$
is contained in $V(z,w_{j})\cap V(u_{j},z)$, which equals $V(u_{j},w_{j})\cup
P_{z}$ by Lemma 3.3. As $P_{v}$ is a subset of $Y\cap Z$ and is disjoint from
$P_{z}$, the lemma holds. ∎

######  Corollary 5.13.

$P_{v}\subseteq P_{i}$.

###### Proof.

As every $\Phi_{j}$ is a $k$-pseudoflower, for sufficiently large $j\in J$ we
have

$\displaystyle V(u_{j},w_{j})=X\cup P_{\Pi_{j}(i)}\cup P_{u_{j}}\cup
P_{w_{j}}=X\cup P_{\Pi_{j}(i)}.$

Because $P_{v}$ is disjoint from $X$, this implies
$P_{v}\subseteq\bigcap_{j\in J}P_{\Pi_{j}(i)}=P_{i}$. ∎

######  Corollary 5.14.

For $i^{\prime}\in I$, if $u$ is a neighbor of $i^{\prime}$ in $C(I)$ then
$P_{u}\subseteq P_{i^{\prime}}$.

###### Proof.

If $u\in V_{N}$, then $i^{\prime}$ is the unique neighbor of $u$ and
$P_{u}\subseteq P_{i^{\prime}}$ by Corollary 5.13. If $u\in V_{F}$, then
$P_{u}\subseteq P_{\Pi_{j}(i^{\prime})}$ for all $j\in J$ for which $u$ is a
cutpoint, also implying $P_{u}\subseteq P_{i^{\prime}}$. ∎

######  Lemma 5.15.

The set $P_{v}$ does not depend on the choice of $z$.

###### Proof.

Let $z_{1}$ and $z_{2}$ be two possible choices for $z$, assume
$z_{1}\in[v,z_{2}]$. Denote the sets defined by $z_{n}$ by a superscript index
$n$, e.g. $(Y^{1},Z^{1})$ is the separation $(Y,Z)$ defined by $z_{1}$. This
proof is for the case that $v$ is the predecessor of $i$, the proof of the
other case is symmetric. Let $l$ be sufficiently large that both
$S(z_{1},w_{l})$ and $S(z_{2},u_{l})$ are defined in $\Phi_{l}$ and
distinguish two elements of $\mathcal{P}$.

As $S(z_{1},u_{j})\geq S(z_{2},u_{j})$ for all $j\geq l$, also
$(Y^{1},Z^{1})\geq(Y^{2},Z^{2})$. Furthermore for all $j\geq l$,

$\displaystyle S(z_{1},u_{l})\vee(Y^{2},Z^{2})\geq S(z_{1},u_{l})\vee
S(z_{2},u_{j})=S(z_{1},u_{j}).$

So $(Y^{1},Z^{1})\leq S(z_{1},u_{l})\vee(Y^{2},Z^{2})\leq(Y^{1},Z^{1})$ and
thus the two separations are equal. In particular $P_{v}^{1}\subseteq
Z^{1}\subseteq Z^{2}$ and $P_{v}^{1}\subseteq Y^{2}\cup V(z_{1},u_{l})$.
Because also $P_{v}^{1}\subseteq V(u_{l},w_{l})$, together we have

$\displaystyle P_{v}^{1}\subseteq V(u_{l},w_{l})\cap(Y^{2}\cup
V(z_{1},u_{l}))=(V(u_{l},w_{l})\cap Y^{2})\cup(X\cup P_{u_{l}})\subseteq
Y^{2}$

where the equality holds by Lemma 3.3. Hence $P_{v}^{1}\subseteq Y^{2}\cap
Z^{2}$. Also $P_{v}^{1}\subseteq V(u_{j},w_{j})$ for all sufficiently large
$j\in J$ implies that $P_{v}^{1}\cap V(z_{1},z_{2})=\emptyset$ and thus
$P_{v}^{1}=P_{v}^{2}$ as those two sets have the same size. ∎

######  Lemma 5.16.

$(Y,Z)=(V^{\prime}(z,v),V^{\prime}(v,z))$.

###### Proof.

We are going to show the lemma in the case that $v$ is the predecessor of $i$,
the other case is symmetric.

We have $V^{\prime}(z,v)=P_{v}\cup\bigcup_{t\in[z,v]\cap V_{F}}V(z,t)\subseteq
Y$. Assume there is $u\in Y\setminus V^{\prime}(z,v)$, then $u\notin
V(z,u_{j})$ for all sufficiently large $j\in J$. So $u\in Z$ and thus $u\in
P_{z}\cup P_{v}\cup X$, a contradiction. So $Y=V^{\prime}(v,z)$.

We have $V^{\prime}(v,z)\subseteq V^{\prime}(u_{j},z)=V(u_{j},z)$ for all
sufficiently large $j\in J$ and thus $V^{\prime}(v,z)\subseteq Z$. Let $u\in
Z\setminus(X\cup P_{v})$ and let $w\in I\cup V_{F}$ be some element such that
$u\in P_{w}$. If $w\in[v,z]$, then also $u\in V^{\prime}(v,z)$. Otherwise
$w\in[z,v]$, so $w\in[z,u_{j}]$ for some $j\in J$ and thus $u\in
V(z,u_{j})\subseteq Y$. In this case we have $u\in Y\cap Z=P_{v}\cup P_{z}\cup
X\subseteq V^{\prime}(v,z)$. Thus $Z\subseteq V^{\prime}(v,z)$, and these two
sets are equal. ∎

######  Corollary 5.17.

$(V^{\prime}(z,v)\cup P_{v},V^{\prime}(v,z)\cup P_{v})$ is a separation with
separator $P_{v}\cup P_{z}\cup X$.

###  5.4. Interval separations are indeed $k$-separations

The goal of this section is to show that the interval separations of the limit
we are constructing have the properties required in the definition of
pseudoflowers.

######  Lemma 5.18.

For all distinct $v,w\in V_{F}\cup V_{N}$ we have
$V^{\prime}(v,w)=\hat{V}(v,w)$.

###### Proof.

$V^{\prime}(v,w)\subseteq\hat{V}(v,w)$ is clear by definition of
$V^{\prime}(v,w)$. In order to show the reverse inclusion, let $t$ be an
element of $[v,w]$. and $u\in P_{t}$. If $t\in\\{v,w\\}$ or $t\in I\cup
V_{F}$, then clearly $P_{t}\subseteq V^{\prime}(v,w)\cup P_{v}\cup P_{w}$.
Otherwise $t$ has a neighbor $i\in I$, and $i\in[v,w]$. Then by Corollary
5.13, $P_{v}\subseteq P_{i}\subseteq V^{\prime}(v,w)$. ∎

######  Lemma 5.19.

(The cyclic order is illustrated in Fig. 2). Let $a$, $b$, $c$ and $d$ be
elements of $V_{F}\cup V_{N}$ such that

  * •

$b\in[a,c]$ and $d\in[c,a]$

  * •

$(V^{\prime}(a,c),V^{\prime}(c,a))$ is a separation with separator $X\cup
P_{a}\cup P_{c}$

  * •

$(V^{\prime}(b,d),V^{\prime}(d,b))$ is a separation with separator $P_{b}\cup
P_{d}\cup X$.

  * •

$P_{a}\cap P_{d}=\emptyset$.

Then $(V^{\prime}(b,c),V^{\prime}(c,b))$ is a separation with separator
$P_{b}\cup P_{c}\cup X$.

###### Proof.

We show that

$\displaystyle(V^{\prime}(b,c),V^{\prime}(c,b))=(V^{\prime}(a,c),V^{\prime}(c,a))\wedge(V^{\prime}(b,d),V^{\prime}(d,b)).$

$V^{\prime}(c,b)=V^{\prime}(c,a)\cup V^{\prime}(d,b)$ and
$V^{\prime}(b,c)\subseteq V^{\prime}(a,c)\cap V^{\prime}(b,d)$ clearly hold.
Let $u\in V^{\prime}(c,b)\setminus(P_{b}\cup P_{c}\cup X)$. Because $P_{a}\cap
P_{d}$ is empty, either $u\in V^{\prime}(c,a)\setminus(P_{a}\cup P_{c}\cup X)$
or $u\in V^{\prime}(d,b)\setminus(P_{b}\cup P_{d}\cup X)$. In the first case,
because $(V^{\prime}(a,c),V^{\prime}(c,a))$ is a separation with separator
$P_{a}\cup P_{c}\cup X$, $u$ is an element of $V^{\prime}(c,a)\setminus
V^{\prime}(a,c)$. In the second case we similarly get $u\in
V^{\prime}(d,b)\setminus V^{\prime}(b,d)$. Hence in both cases $u\notin
V^{\prime}(a,c)\cap V^{\prime}(b,d)$, showing that $V^{\prime}(a,c)\cap
V^{\prime}(b,d)=V^{\prime}(b,c)$. Furthermore, $V^{\prime}(b,c)\cap
V^{\prime}(c,b)\subseteq P_{b}\cup P_{c}\cup X$, and so the inclusion must be
an equality. ∎

######  Lemma 5.20.

For all distinct $v,w\in V_{F}\cup V_{N}$ the pair
$(\hat{V}(v,w),\hat{V}(w,v))$ is a separation with separator $X\cup P_{v}\cup
P_{w}$.

###### Proof.

By Lemma 5.18 it suffices to show that $(V^{\prime}(v,w),V^{\prime}(w,v))$ is
a separation with the correct separator for all distinct elements $v$ and $w$
of $V_{F}\cup V_{N}$. If both $v$ and $w$ are contained in $V_{F}$ then this
is true by Corollary 5.10. Consider first the case that exactly one is
contained in $V_{F}$, by switching the names we may assume $w\in V_{F}$. If
$w$ is a suitable candidate for $z$ in the definition of $P_{v}$, then
$(V^{\prime}(w,v),V^{\prime}(v,w))$ is a separation with separator $P_{v}\cup
P_{w}\cup X$ by Corollary 5.17. So assume otherwise. We will consider the case
that for all sufficiently large $j\in J$, $S(w,u_{j})$ does not distinguish
elements of $\mathcal{P}$, the case where $S(w,w_{j})$ does not distinguish
elements of $\mathcal{P}$ is symmetric.

Let $z\in V_{F}$ from which $P_{v}$ might have been defined. Then by choice of
$z$ there is $t\in V_{F}$ such that $S(t,z)$ distinguishes elements of
$\mathcal{P}$ and such that $t\in[v,z]$ if $w\in[z,v]$ and $t\in[z,v]$ if
$w\in[v,z]$. Also $(V^{\prime}(z,v),V^{\prime}(v,z))$ is a separation with
separator $P_{z}\cup P_{v}\cup X$ by Corollary 5.17 and we already saw that
$(V^{\prime}(w,t),V^{\prime}(t,w))$ is a separation with separator $P_{w}\cup
P_{t}\cup X$. Furthermore $S(t,z)=(V^{\prime}(t,z),V^{\prime}(z,t))$, and
because $S(t,z)$ distinguishes elements of $\mathcal{P}$ and thus has
connectivity $k$ this implies $P_{t}\cap P_{z}=\emptyset$. Now we can apply
Lemma 5.19. If $t\in[v,z]$ then we apply Lemma 5.19 for $a=z$, $b=w$, $c=v$
and $d=t$. Otherwise we apply the lemma for $a=t$, $b=v$, $c=w$ and $d=z$. In
both cases $(V^{\prime}(w,v),V^{\prime}(v,w))$ is a separation with separator
$P_{v}\cup P_{w}\cup X$.

Now assume that both $v$ and $w$ are contained in $V_{N}$. Because every
$\Phi_{j}$ distinguishes at least three elements of $\mathcal{P}$, by swapping
the names of $v$ and $w$ if necessary we may assume that there are $t$ and $u$
in $V_{F}$ such that $t\in[v,w]$ and $u\in[t,w]$ and $S(t,u)$ distinguishes
elements of $\mathcal{P}$. We already showed that
$(V^{\prime}(v,u),V^{\prime}(u,v))$ and $(V^{\prime}(t,w),V^{\prime}(w,t))$
are $k$-separations with separator $P_{v}\cup P_{u}\cup X$ and $P_{t}\cup
P_{w}\cup X$ respectively. Furthermore $P_{t}\cap P_{u}=\emptyset$ because
$S(t,u)$ distinguishes elements of $\mathcal{P}$, thus we may apply Lemma 5.19
to $a=u$, $b=w$, $c=v$ and $t=d$ and we are done. ∎

###  5.5. Deletion of redundant petals

$\Phi$ is now nearly a $k$-pseudoflower, the only property missing is (• ‣
3.1). In order to fix that, we are going to delete troublesome elements from
$I$.

Let $\mathcal{Y}$ be the set of subsets of $V$ of size $(k-|X|)/2$ that are of
the form $P_{i}$ for some $i\in I$. In order to ensure that (• ‣ 3.1) holds,
it suffices to delete, for every $Y\in\mathcal{Y}$, all but one index $i\in I$
with $P_{i}=Y$. For that, we first define one element we want to keep: If
there is $j\in J$ and $i_{j}\in I_{j}$ whose petal is $Y$, then we want to
define $i_{Y}$ to be an element of $I$ with $\pi_{j}(i_{Y})=i_{j}$. If that
does not exist, then we pick $i_{Y}$ arbitrarily with $P_{i_{Y}}=Y$.

Actually if $i_{Y}$ is defined via the first case, then $i_{Y}$ is uniquely
determined:

######  Lemma 5.21.

Let $Y\in\mathcal{Y}$ and $i_{j}$ be an index element of some $\Phi_{j}$ whose
petal is $Y$. Then there is a unique $i\in I$ with $\pi_{j}(i)=i_{j}$, and if
$P_{i_{l}}=Y$ for some $l\in J$ and $i_{l}\in I_{l}$, then $\pi_{l}(i)=i_{l}$.

###### Proof.

Let $l\in J$ with $j\leq l$. By Lemma 3.10 there is some $i_{l}\in I_{l}$ that
is mapped to $i_{j}$ by all witnesses of $\Phi_{j}\leq\Phi_{l}$, in particular
$F_{lj}(i_{l})=i_{j}$. Thus $P_{i_{l}}$ is contained in $P_{i_{j}}$ and has
size at least $(k-|X|)/2$. But $P_{i_{j}}$ also has size $(k-|X|)/2$, so
$P_{i_{l}}=P_{i_{j}}=Y$. By (• ‣ 3.1), $i_{l}$ is the only element of $I_{l}$
with $P_{i_{l}}=Y$, and thus $i_{l}$ is the only element of $I_{l}$ with
$F_{lj}(i_{l})=i_{j}$.

So if $i$ and $i^{\prime}$ are elements of $I$ with
$\pi_{j}(i)=\pi_{j}(i^{\prime})=i_{j}$, then $\pi_{l}(i)=\pi_{l}(i^{\prime})$
for all $l\geq j$, implying that $i=i^{\prime}$. By construction of $I$ there
is some $i\in I$ with $\pi_{j}(i)=i_{j}$, thus there is a unique such element.
∎

The previous lemma implies that if there is some index $i_{j}$ of some
$\Phi_{j}$ with $P_{i_{j}}=Y$, then there is $i\in I$ such that all indices of
any $\Phi_{j}$ whose petal is $Y$ are the image of $i$ under $\pi_{j}$.

Obtain $I^{\prime}$ from $I$ by deleting, for every element $Y$ of
$\mathcal{Y}$, all $i\in I-i_{Y}$ with $P_{i}=Y$. Obtain $C$ from $C(I)$ by
deleting $I\setminus I^{\prime}$ and, for all $i\in I\setminus I^{\prime}$,
the predecessor of $i$ in $C(I)$.

The following lemma implies that the elements of $I^{\prime}$ are close to
being “dense” in $C(I)$:

######  Lemma 5.22.

Let $w$ and $w^{\prime}$ be distinct elements of $V_{F}$ such that for some
$Y\in\mathcal{Y}$, $P_{z}=Y$ for all $z\in[w,w^{\prime}]\setminus I$. Then for
some $i^{\prime}\in[w,w^{\prime}]\cap I$ there is $j\in J$ with
$P_{\pi_{j}(i^{\prime})}=Y$ or there is $y\in P_{i^{\prime}}$ such that
$y\notin P_{i^{\prime\prime}}$ for all $i^{\prime\prime}\in I-i^{\prime}$.

###### Proof.

Let $j\in J$ be large enough that $C(I_{j})$ contains both $w$ and
$w^{\prime}$. Then there is $i_{j}\in[w,w^{\prime}]\cap I_{j}$, where the
interval is taken in $C(I_{j})$, and some $i^{\prime}\in I$ with
$\pi_{j}(i^{\prime})=i_{j}$. For the predecessor $p$ and the successor $s$ of
$i_{j}$ in $C(I_{j})$, $P_{s}=P_{p}=Y$. If $P_{i_{j}}=Y$, then the lemma
holds, so assume otherwise. Let $y$ be an element of $P_{i_{j}}$ that is not
contained in $Y$. Then $s$ and $p$ are contained in $[w,w^{\prime}]$ and
$V(p,s)$ does not depend on whether it is taken in $(P_{z})_{z\in C(I)}$ or in
$\Phi_{j}$. Hence there is $z\in[w,w^{\prime}]$ such that $y\in P_{z}$, and
$z$ has to be contained in $I$. So $y\in P_{z}$ for some
$z\in[w,w^{\prime}]\cap I$, implying that the predecessor $p^{\prime}$ and
successor $s^{\prime}$ of $z$ in $C(I)$ both have $Y$ as vertex set. Then the
separating set of the separation $S(z)$ is $Y$, implying that $y$ is only
contained in $P_{z}$ and in no $P_{i^{\prime\prime}}$ with
$i^{\prime\prime}\in I-z$. ∎

So every element of $C(I)\setminus C$ is close to some element of $C$:

######  Lemma 5.23.

Let $z\in C(I)\setminus C$. Then there is $v\in C\setminus I^{\prime}$ such
that $[z,v]$ is finite, no element of $[z,v\mathclose{[}$ is contained in $C$,
and $P_{z^{\prime}}=P_{v}$ for all $z^{\prime}\in[z,v\mathclose{[}$.

###### Proof.

If $z\in I$, then let $i_{0}=z$, otherwise let $i_{0}$ be the successor of
$z$. Recursively, if $i_{l}$ is defined and has a successor in $I$ that is not
contained in $I^{\prime}$, then define $i_{l+1}$ to be that successor.

Assume for a contradiction that $i_{2}$ is defined. For any distinct
$i,i^{\prime}\in I$, some projection $\pi_{j}$ maps $i$ and $i^{\prime}$ to
different images. So there is $w\in V_{F}$ such that $w\in[i,i^{\prime}]$.
Thus the common neighbor $w_{0}$ of $i_{0}$ and $i_{1}$, and the common
neighbor $w_{1}$ of $i_{1}$ and $i_{2}$, are both contained in $V_{F}$. Now
the previous lemma implies that $i_{1}$ is contained in $I^{\prime}$, a
contradiction.

If $i_{1}$ is defined, then let $v$ be the successor of $i_{1}$ in $C(I)$.
Otherwise let $v$ be the successor of $i_{0}$ in $C(I)$. Thus, if $v$ is the
predecessor of some element of $I$, then that is also contained in
$I^{\prime}$. For $i\in I\setminus I^{\prime}$, and the predecessor $p$ and
successor $s$ of $i$ in $C(I)$, $P_{p}=P_{i}=P_{s}$. In particular
$P_{z}=P_{i_{0}}=P_{v}$, and if $i_{1}$ is defined then also
$P_{v}=P_{i_{1}}=P_{w_{0}}$ for the common neighbor $w_{0}$ of $i_{0}$ and
$i_{1}$ in $C(I)$. ∎

######  Lemma 5.24.

$C$ is a cycle completion of $I^{\prime}$.

###### Proof.

In order to show that the identity on $I^{\prime}$ can be extended uniquely to
a bijective monotone map from $C$ to $C(I^{\prime})$ it suffices to show that
all non-trivial intervals of $I^{\prime}$ can be uniquely written as
$[v,w]\cap I^{\prime}$ for elements $v$ and $w$ of $C\setminus I^{\prime}$. In
order to do so, let $I^{\prime\prime}$ be a non-trivial interval of
$I^{\prime}$, and let $i\in I^{\prime}\setminus I^{\prime\prime}$. Let
$\hat{I}$ be the set of all $\hat{\imath}$ for which there are $i_{1},i_{2}\in
I^{\prime\prime}$ such that the interval $[i_{1},i_{2}]$, taken in $I$,
contains $\hat{\imath}$ but not $i$. Then $\hat{I}$ is an interval of $I$ that
contains $I^{\prime\prime}$ as a subset and that does not contain $i$. So
$\hat{I}$ can be written as $[v,w]\cap I$ for unique elements $v$ and $w$ of
$C(I)\setminus I$. Also, $I^{\prime\prime}=\hat{I}\cap I^{\prime}=[v,w]\cap
I^{\prime}$.

In order to find $v$ and $w$ that are contained in $C$, first consider $v$. If
$v$ is the predecessor of some $i^{\prime}\in I$, then $i^{\prime}$ is
contained in $I^{\prime\prime}$ and thus in $I^{\prime}$. So $v$ cannot be the
predecessor of an element of $I\setminus I^{\prime}$, implying that $v$ is
contained in $C$. But $w$ could indeed be contained in $C(I)\setminus C$. If
it is, then apply Lemma 5.23 to $w$ and obtain $w^{\prime}$. Otherwise let
$w=w^{\prime}$. Then $\hat{I}=[v,w]\cap I^{\prime}=[v,w^{\prime}]\cap
I^{\prime}$.

In order to show that $v$ and $w$ are unique, it suffices to show for any
distinct $v,w\in C\setminus I^{\prime}$ that $[v,w]\cap I^{\prime}$ is non-
empty. As in the proof of Lemma 5.23, if $[v,w]$ contains enough elements (at
least 7) then it contains two elements of $V_{F}$ and thus an element of
$I^{\prime}$. Otherwise, $[v,w]$ is finite and $v$ is the predecessor in
$C(I)$ of some $i\in[v,w]\cap I$. Because $v\in C$, this implies that $i\in
I^{\prime}$. ∎

Let $\tilde{F}:C(I^{\prime})\rightarrow C$ be a bijective monotone map whose
restriction to $I^{\prime}$ is the identity. This map then identifies
$C(I^{\prime})$ with $C$. Denote the family $(P_{z})_{z\in C(I^{\prime})}$
where each $P_{z}$ is equal to $P_{\tilde{F}(z)}$, by $\Phi^{\prime}$. As
$\Phi$ satisfies all properties of a $k$-pseudoflower with the possible
exception of (• ‣ 3.1), also $\Phi^{\prime}$ satisfies all properties of a
$k$-pseudoflower with the possible exception of (• ‣ 3.1).

######  Lemma 5.25.

For all elements $v$ and $w$ of $C\setminus I^{\prime}$ we have

$\bigcup_{z\in[v,w]_{C}}P_{z}=\bigcup_{z\in[v,w]_{C(I)}}P_{z}$

where the intervals are taken in $C$ and $C(I)$, respectively.

###### Proof.

It is clear that the left-hand side is a subset of the right-hand side. In
order to show the other direction, let $u\in[v,w]_{C(I)}$. If $u\in C$ then
$u\in[v,w]_{C}$, so assume otherwise. Apply Lemma 5.23 to $u$ and obtain
$z^{\prime}\in C\setminus I^{\prime}$. Then
$w\notin[u,z^{\prime}\mathclose{[}$, as $w\in C$, so $z^{\prime}\in[v,w]$. As
$P_{u}=P_{z^{\prime}}$, the lemma holds. ∎

######  Lemma 5.26.

$\Phi^{\prime}$ is a $k$-pseudoflower such that $\Phi_{j}\leq\Phi^{\prime}$
for all $j\in J$.

###### Proof.

Recall that by $\tilde{F}$, $C(I^{\prime})$ is identified with $C$. For all
$v,w\in C(I^{\prime})\setminus I^{\prime}$,

$\displaystyle\bigcup_{z\in[v,w]_{C(I^{\prime})}}P_{z}$
$\displaystyle=\bigcup_{z\in[v,w]_{C}}P_{z}$ by Lemma 5.25
$\displaystyle=V^{\prime}(v,w)\setminus X$ by Lemma 5.18

so by Lemma 5.20 $S(v,w)$ taken in $\Phi^{\prime}$ is a separation with
separator $P_{v}\cup P_{w}\cup X$ and also $X=V\setminus\bigcup_{z\in
C(I^{\prime})}P_{z}$. As every $P_{v}$ with $v\in C(I)\setminus I$ has size
$(k-|X|)/2$, this is also true for every $v\in C(I^{\prime})\setminus
I^{\prime}$. Hence every $S(v,w)$ has order at most $k$. If some $i\in
I^{\prime}$ has a neighbor $v$ in $C(I^{\prime})$, then $P_{v}\subseteq P_{i}$
by Corollary 5.14. And by definition of $I^{\prime}$, $\Phi^{\prime}$
satisfies (• ‣ 3.1). Thus $\Phi^{\prime}$ is a $k$-pseudoflower.

In order to show that $\Phi_{j}\leq\Phi^{\prime}$ for all $j\in J$, we first
show that the restriction of $\pi_{j}$ to $I^{\prime}$ is surjective. For
that, let $i_{j}\in I_{j}$ and let $p$ and $s$ be the predecessor and
successor of $i_{j}$ in $C(I_{j})$. Then $p$ and $s$ are contained in $V_{F}$,
and thus by Lemma 5.22 there is $i\in[v,w]\cap I^{\prime}$, and
$\pi_{j}(i)=i_{j}$. So the restriction of $\pi_{j}$ to $I^{\prime}$ with
codomain $I_{j}$ is a surjective monotone map, and there is a unique monotone
surjective extension $F_{j}:C(I^{\prime})\rightarrow C(I_{j})$ with
$F_{j}(I^{\prime})=(I_{j})$.

Let $v$ and $w$ be distinct elements of $C(I_{j})\setminus I_{j}$. If $v\in
C$, then let $v^{\prime}=v$. Otherwise there is, by Lemma 5.23, an element
$v^{\prime}\in C\setminus I^{\prime}$ such that $[v,v^{\prime}]\cap
C=v^{\prime}$ and $P_{v}=P_{v^{\prime}}$. Define $w^{\prime}$ similarly. Then

$V(v,w)=\hat{V}(v,w)=\hat{V}(v^{\prime},w^{\prime})=V(\tilde{F}^{-1}(v^{\prime}),\tilde{F}^{-1}(w^{\prime}))$

so $F_{j}$ witnesses that $\Phi_{j}\leq\Phi^{\prime}$. ∎

So in this section it was shown that if $(\Phi_{j})_{j\in J}$ is a
$\leq$-chain of $k$-pseudoflowers which distinguish at least three elements of
$\mathcal{P}$ that they locate, then there is a $k$-pseudoflower which is an
upper bound of the chain $(\Phi_{j})_{j\in J}$. In particular, if $\Psi$ is a
$k$-pseudoflower which distinguishes at least three elements of $\mathcal{P}$
that it locates, then in the set of $k$-pseudoflowers $\Phi^{\prime}$ with
$\Psi\leq\Phi^{\prime}$ every $\leq$-chain has an upper bound. Thus the
following theorem follows by Zorn’s Lemma:

######  Theorem 5.27.

Let $\Phi$ be a $k$-pseudoflower which distinguishes at least three elements
of $\mathcal{P}$ that it locates. Then there is a $\leq$-maximal
$k$-pseudoflower $\Psi$ such that $\Phi\leq\Psi$.

In this theorem, the condition that $\Phi$ distinguishes at least three
elements of $\mathcal{P}$ that it locates can be weakened to $\Phi$
distinguishing any three profiles or tangles that it locates. As every
$k$-pseudoflower locates every $k$-tangle, and every $k$-pseudoflower
extending a $k$-flower with four petals locates every $k$-profile, the theorem
can be specialized to the following versions:

######  Theorem 5.28.

Let $\Phi$ be a $k$-pseudoflower which distinguishes at least three
$k$-profiles and extends a $k$-flower with four petals. Then there is a
$\leq$-maximal $k$-pseudoflower $\Psi$ such that $\Phi\leq\Psi$.

######  Theorem 5.29.

Let $\Phi$ be a $k$-pseudoflower which distinguishes at least three
$k$-tangles. Then there is a $\leq$-maximal $k$-pseudoflower $\Psi$ such that
$\Phi\leq\Psi$.

###  5.6. Existence of $\preccurlyeq$-maximal $k$-pseudoflowers

Call an element $P$ of $\mathcal{P}$ _closed_ if whenever $(S_{j})_{j\in J}$
is a chain of elements of $P$, then its supremum (which exists because their
order is bounded) is contained in $\mathcal{P}$. If $\mathcal{P}$ only
contains closed elements, then $\leq$-maximal $k$-pseudodaisies are also
$\preccurlyeq$-maximal, and thus there are $\preccurlyeq$-maximal
$k$-pseudodaisies.

$v$$t$$w$$y$$S(v,t)$$\in P_{3}$$S(t,w)$$\in P_{4}$$W$ Figure 9. Some of the
notation used in the proof of Theorem 5.30.

######  Theorem 5.30.

Let $\Phi$ be a $k$-pseudoflower that is $\leq$-maximal and distinguishes
three elements of $\mathcal{P}$ that it locates. If all elements of
$\mathcal{P}$ are closed, then $\Phi$ is also $\preccurlyeq$-maximal.

###### Proof.

See Fig. 9 for a depiction of some of the notation. Let $\Phi$ be a
$k$-pseudoflower that distinguishes three elements of $\mathcal{P}$ that it
locates and which is not $\preccurlyeq$-maximal among all $k$-pseudoflowers,
as witnessed by $\Phi^{\prime}$. Let $P_{1}$ and $P_{2}$ be two elements of
$\mathcal{P}$ which are distinguished by $\Phi^{\prime}$ but not by $\Phi$.

By Lemma 3.14 there is a cutpoint $v\in C(I)\setminus I$ such that either
$S(v,w)\in P_{1}$ for all $w\in C(I)\setminus I-v$ or $S(w,v)\in P_{1}$ for
all $w\in C(I)\setminus I-v$. We will assume that $S(v,w)\in P_{1}$ for all
$w\in C(I)\setminus I-v$, the other case is symmetric. Because $\Phi$
distinguishes at least three elements of $\mathcal{P}$ there is an interval
separation $S(v,w)$ of $\Phi$ whose inverse is contained in two elements
$P_{3}$ and $P_{4}$ of $\mathcal{P}$ which are distinguished and located by
$\Phi$. As $\Phi$ distinguishes and locates $P_{3}$ and $P_{4}$, and
distinguishes them from $P_{1}$, there is $t\in\mathopen{]}v,w\mathclose{[}$
such that $S(v,t)$ distinguishes $P_{3}$ and $P_{4}$. By swapping the names of
$P_{3}$ and $P_{4}$ if necessary, we may assume that $S(t,v)\in P_{3}$. Again
by the profile property also $S(t,w)$ distinguishes $P_{3}$ and $P_{4}$ with
$S(w,t)\in P_{4}$.

The $k$-pseudoflower $\Phi^{\prime}$ distinguishes $P_{1}$, $P_{2}$, $P_{3}$
and $P_{4}$ pairwise, so there is an interval separation $(C,D)$ of
$\Phi^{\prime}$ which is contained in $P_{3}$ but not $P_{4}$ and which
distinguishes $P_{1}$ and $P_{2}$. By swapping the names of $P_{1}$ and
$P_{2}$ if necessary we may assume that $(C,D)$ is contained in $P_{1}$ and
$P_{3}$ and that its inverse is contained in $P_{2}$ and $P_{4}$.

If $v$ has a predecessor $v^{\prime}$ in $C(I)\setminus I$, then $(C,D)$
properly crosses $S(v,v^{\prime})$ and by Lemmas 4.1 and 4.3 there is an
extension $\Phi^{\prime\prime}$ of $\Phi$ which distinguishes $P_{1}$ and
$P_{2}$. Assume for a contradiction that $v$ has no predecessor in
$C(I)\setminus I$. Let $W$ be the interval
$\mathopen{]}w,v\mathclose{[}\setminus I$. If $\Phi$ is a $k$-pseudoanemone,
then $S(t,v)$ is the supremum of the separations $S(t,x)$ with $x\in W$.
Because $P_{1}$ is closed and contains all $S(t,x)$ with $x\in W$ but not
$S(t,v)$, $\Phi$ cannot be a $k$-pseudoanemone and thus has to be a
$k$-pseudodaisy.

For all $x\in W$ let $S_{x}$ be the separation $((C,D)\vee S(t,x))\wedge
S(t,v)$. All three $k$-separations $(C,D)$, $S(t,x)$ and $S(t,v)$ are
contained in $P_{3}$, not contained in $P_{4}$ and have order at most $k$, so
$S_{x}$ has order at most $k$. Denote the unique supremum of $(S_{x})_{x\in
W}$, which exists by limit-closedness, by $(A,B)$. As the union of all sets
$V(v,x)$ with $x\in W$ is the whole ground set $V$, there is some $y\in W$
such that $V(v,y)$ contains $A\cap B$. But all profiles are closed, so $(A,B)$
is still contained in $P_{1}$ and thus properly crosses $S(v,y)$. Hence
applying Lemma 4.2 to the concatenation of $\Phi$ on vertices $v$, $t$ and $y$
shows that $A\cap B$ contains a vertex not in $V(v,y)$, a contradiction to the
choice of $y$. ∎

##  6\. Example of an infinite $k$-pseudoflower

Recall that the relevant definitions for graph-like spaces can be found for
example in [2].

######  Example 6.1.

Let $a$ and $n$ be natural numbers and define $k=a+2n$. Let $G$ be a
$k$-connected graph and $(P_{i})_{1\leq i\leq n}$ a family of pairwise
disjoint arcs. Let $V$ be a vertex set of $G$

Figure 10. An example of a $k$-pseudoflower with $|X|=2$ and $d=3$.

of size $a$ which does not contain any first or last vertices of some $P_{i}$.
Then we can define a $k$-connected graph-like space $G^{\prime}$ as follows:
Let $(G_{j})_{j\in\mathbb{Z}}$ be a family of disjoint copies of $G$, $X$ a
set of size $a$ disjoint from all $G_{i}$ and $Y=\\{y_{1},...,y_{n}\\}$ a set
disjoint from all $G_{i}$ and from $X$. For all suitable $i$ and $j$ identify
the last vertex of $P_{i}$ in $G_{j}$ with the first vertex of $P_{i}$ in
$G_{j+1}$ and add all edges between the copy of $V$ in $G_{j}$ and $X$.

Let the topology of $G^{\prime}-Y$ be the simplex topology of
$G^{\prime}_{Y}$. Define sets $A(i,j,l,\epsilon)$ for $1\leq i\leq n$, $j\leq
l\in\mathbb{Z}$ and $1>\epsilon>0$ as follows: Let $A(i,j,l,\epsilon)\cap V$
consist of the vertices of copies of $P_{i}$ in graphs $G_{m}$ where $m<j$ or
$m>l$. Add the inner points of edges with both end vertices in
$A(i,j,l,\epsilon)$. For edges $e$ of which only the starting vertex is in
$A(i,j,l,\epsilon)$ add $\iota_{e}([0,\epsilon\mathclose{[})$ and for edges
$e$ of which only the end vertex is contained in $A(i,j,l,\epsilon)$ add
$\iota_{e}(\mathopen{]}1-\epsilon,1])$. The open sets of $G-Y$ and the sets of
the form $A(i,j,l,\epsilon)$ form a base of the topology of $G^{\prime}$.

######  Claim 6.2.

$G^{\prime}$ is a graph-like space.

###### Proof.

For its being a graph-like space we have to show that for each two distinct
vertices $v,w$ in $V(G^{\prime})$ there are two open sets of $G^{\prime}$
partitioning $V(G^{\prime})$ such that $v$ and $w$ are not contained in the
same open set. If one of these vertices, $v$ say, is not contained in $Y$,
then there are two disjoint open sets $U,V$ of $G^{\prime}-Y$ such that $V\cap
V(G^{\prime})=\\{v\\}$ and $U\cap V(G^{\prime})=V(G^{\prime})\backslash Y-v$.
Also for each $i$ there is some $A(i,j,l,\epsilon)$ which does not contain
$v$, let $W$ be the union of $U$ and such a $A(i,j,l,\epsilon)$ for all $i$.

So consider the case that both $v$ and $w$ are contained in $Y$ and that
$v=y_{i}$ for some $i$. Pick some suitable $j$ and $l$ and define $U$ to be
the union of all $A(m,j,l,\frac{1}{2})$ where $m\neq i$. Let $V$ be the union
of $A(i,j,l,\frac{1}{2})$ and all $v^{0.5}$ where $v\in V(G^{\prime}-U)$. Then
$U$ and $V$ are disjoint open sets, one of them contains $v$ and the other $w$
and every vertex is contained in one of these sets. ∎

In order to show that $G^{\prime}$ is $k$-connected, note that for each
$P_{i}$ the copies of it induce a pseudo-cycle containing $y_{i}$. Then
$G^{\prime}$ clearly has a $k$-pseudoflower with $(k-|X|)/2=n$ whose petal
sets are of the form $V(G_{i})\cup X$.

## References

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##  7\. Appendix

The goal of this appendix is to show that cycle completions exist, are unique
and have the properties we outlined in Section 2.2. While there is a lot of
prior work in this area, some of which is referenced here, none of it matches
our requirements exactly, so we include these proofs for completeness.

This section starts with a short collection of basics about cyclic orders and
their connection to linear orders. In this section we often consider distinct
linear or cyclic orders on the same ground set. Because of this, and in
contrast to the rest of the paper, in this section cyclic orders and linear
orders are not implicit but introduced more formally as relations on the
ground set. So a cyclic order of a set $S$ is a set $Z\subseteq S\times
S\times S$ that is cyclic, antisymmetric, linear and transitive.

The notation of $\leq$ is kept for a linear order, but the linear order in
question is added as an index where necessary, for example in $s\leq_{L}t$ for
a linear order $L$. Similarly, for intervals of cyclic orders, the cyclic
order in question may be indicated by an index.

######  Definition 7.1.

[9, Definition 1.1] Given two linear orders $A$ and $B$ on disjoint ground
sets, the linear order $A\oplus B$ is the linear order defined on the union of
the ground sets of $A$ and $B$ by letting $x\leq y$ if $x\leq_{A}y$ or
$x\leq_{B}y$ or $x\in A$ and $y\in B$.

######  Definition 7.2.

[9, Lemma 1.11, Definition 2.1, Theorem 2.3] Given a linear order $L$ on a set
$S$, the cyclic order $Z$ _induced by $L$_ consists of those triples
$(s,s^{\prime},t)$ of elements of $S$ such that in $L$ one of the equations
$s<s^{\prime}<t$, $s^{\prime}<t<s$ and $t<s<s^{\prime}$ holds. Given a cyclic
order $Z$ on a set $S$, a _cut of $Z$_ is a linear order $L$ on $S$ such that
$Z$ is the cyclic order induced by $L$.

######  Lemma 7.3.

[8, Theorem 3.1] For every cyclic order $Z$ on set $S$ and every $s$ in $S$
there is a cut of $Z$ whose smallest element is $s$.

######  Remark 7.4.

By Lemma 7.3, every non-trivial interval of a cyclic order $Z$ is also an
interval of a cut $L$ of $Z$. Also, such an interval inherits a linear order
from every cut of which it is an interval, and that linear order does not
depend on the chosen cut.

######  Observation 7.5.

Let $Z$ be a cyclic order on a set $S$, $L$ a cut of $Z$ and $s$, $s^{\prime}$
and $t$ elements of $S$ such that $(s,s^{\prime},t)\in Z$. If $s\leq t$ in
$L$, then $s<s^{\prime}<t$ in $L$.∎

######  Definition 7.6.

Given a cyclic order $Z$ on ground set $S$, the cyclic order

$\\{(t,s,r)\colon(r,s,t)\in Z\\}$

on $S$ is the _mirror_ of $Z$.

Monotone maps have the property that preimages of intervals are again
intervals. Maps with that property are close to being monotone:

######  Lemma 7.7.

Let $Z$ be a cyclic order on a set $S$ and $Z^{\prime}$ a cyclic order on a
set $S^{\prime}$. Let $f:S\rightarrow S^{\prime}$ be a map such that for all
intervals $I$ of $Z^{\prime}$ the set $f^{-1}(I)$ is an interval of $Z$. Then
$f$ is a monotone map or a mirror of a monotone map.

###### Proof.

If for all elements $r$, $s$ and $t$ of $S$ the implication

$(f(r),f(s),f(t))\in Z^{\prime}\Rightarrow(t,s,r)\in Z$

holds, then $f$ is a mirror of a monotone map. So assume that there are
elements $r$, $s$ and $t$ of $S$ such that $(f(r),f(s),f(t))\in Z^{\prime}$
and $(r,s,t)\in Z$. We will start with an observation that we will refer to
later in this proof. For this we consider any $u\in S$ such that
$(f(r),f(u),f(t))\in Z^{\prime}$. Then $f^{-1}([f(t),f(r)])$ is an interval of
$S$ which contains $t$ and $r$ but not $s$, so $[t,r]$ is a subset of
$f^{-1}([f(t),f(r)])$. As also $u\notin f^{-1}([f(t),f(r)])$, $u\notin[t,r]$
and thus $(r,u,t)\in Z$.

Now let $r^{\prime}$, $s^{\prime}$ and $t^{\prime}$ be elements of $S$ such
that $(f(r^{\prime}),f(s^{\prime}),f(t^{\prime}))\in Z^{\prime}$. In order to
show that $(r^{\prime},s^{\prime},t^{\prime})\in Z$, first consider the case
that the number $n$ of elements in
$\\{f(r^{\prime}),f(s^{\prime}),f(t^{\prime})\\}$ which are not contained in
$\\{f(r),f(s),f(t)\\}$ is zero. Assume, by renaming if necessary, that
$f(r^{\prime})=f(r)$, $f(s^{\prime})=f(s)$ and $f(t^{\prime})=f(t)$. Then by
three applications of our observation, $(r,s^{\prime},t)\in Z$ and thus
$(s^{\prime},t^{\prime},r)\in Z$ and hence
$(r^{\prime},s^{\prime},t^{\prime})\in Z$.

Next consider the case that $n=1$. Assume, again by renaming if necessary,
that $(f(r),f(r^{\prime}),f(s))\in Z^{\prime}$ (see also the left cyclic order
of Fig. 11). By our observation, $(r^{\prime},s,t)\in Z$ and
$(r^{\prime},t,r)\in Z$ and hence also $(r^{\prime},s,r)\in Z$. Then there are
three cases: Either $f(s^{\prime})=f(s)$ and $f(t^{\prime})=f(t)$ or
$f(s^{\prime})=f(t)$ and $f(t^{\prime})=f(r)$ or $f(s^{\prime})=f(s)$ and
$f(t^{\prime})=f(r)$. In all three cases, by the case $n=0$ also
$(r^{\prime},s^{\prime},t^{\prime})\in Z$.

Next consider the case that $n=2$, and that one of the intervals
$\mathopen{]}f(r),f(s)\mathclose{[}$, $\mathopen{]}f(s),f(t)\mathclose{[}$,
and $\mathopen{]}f(t),f(r)\mathclose{[}$ contains both elements of
$\\{f(r^{\prime}),f(s^{\prime}),f(t^{\prime})\\}$ which are not contained in
$\\{f(r),f(s),f(t)\\}$. Assume, by renaming if necessary, that both
$f(r^{\prime})$ and $f(s^{\prime})$ are contained in
$\mathopen{]}f(r),f(s)\mathclose{[}$. In that case, the fact that
$(f(r^{\prime}),f(s^{\prime}),f(t^{\prime}))\in Z^{\prime}$ implies that
$f(r^{\prime})\in\mathopen{]}f(r),f(s^{\prime})\mathclose{[}$ (see also the
middle cyclic order in Fig. 11). Also, by the case $n=1$, $(r^{\prime},s,t)$,
$(s^{\prime},s,t)$ and $(t,r,r^{\prime})$ are all contained in $Z$. As
$f^{-1}([f(t),f(r^{\prime})])$ contains $t$ and $r^{\prime}$ but neither $s$
or $s^{\prime}$, $(r^{\prime},s,t)\in Z$ implies $[t,r^{\prime}]\subseteq
f^{-1}([f(t),f(r^{\prime})])$. Thus $s^{\prime}$ is not contained in
$[t,r^{\prime}]$ and hence $(r^{\prime},s^{\prime},t)\in Z$. Because
$(s^{\prime},s,t)$ and $(t,r,r^{\prime})$ are contained in $Z$, also
$(r^{\prime},s^{\prime},s)$ and $(r^{\prime},s^{\prime},r)$ are contained in
$Z$. By the case $n=0$, also $(r^{\prime},s^{\prime},t^{\prime})\in Z$.

Next consider the case that $n=2$ and none of the intervals
$\mathopen{]}f(r),f(s)\mathclose{[}$, $\mathopen{]}f(s),f(t)\mathclose{[}$ or
$\mathopen{]}f(t),f(s)\mathclose{[}$ contains two elements of
$\\{f(r^{\prime}),f(s^{\prime}),f(t^{\prime})\\}$. Assume, by renaming if
necessary, that $f(t^{\prime})\in\\{f(r),f(s),f(t)\\}$ and that there is
$u\in\\{r,s,t\\}$ such that $(f(r^{\prime}),f(u),f(s^{\prime}))\in Z^{\prime}$
(see also the right cyclic order of Fig. 11). In this case
$(t^{\prime},r^{\prime},u)$ and $(u,s^{\prime},t^{\prime})$ are both contained
in $Z^{\prime}$ by the case $n=1$ and thus
$(r^{\prime},s^{\prime},t^{\prime})$ is also contained in $Z^{\prime}$.

The only case left is the case $n=3$. Assume, by renaming if necessary, that
$(f(r^{\prime}),f(s),f(s^{\prime}))\in Z^{\prime}$. Then
$(s,s^{\prime},t^{\prime})$ and $(s,t^{\prime},r^{\prime})$ are contained in
$Z$ by the case $n=2$ and thus $(r^{\prime},s^{\prime},t^{\prime})\in Z$. ∎

$f(r)$$f(r^{\prime})$$f(s)$$f(t)$

$f(r)$$f(s)$$f(t)$$f(r^{\prime})$$f(s^{\prime})$

$f(r^{\prime})$$f(s^{\prime})$$f(t^{\prime})$$f(u)$

Figure 11. Three of the cases in the proof of Lemma 7.7

######  Definition 7.8.

[8, Remark 2.3] Given a cyclic order $Z$ of a set $S$ and a subset
$S^{\prime}$ of $S$, the set of triples in $Z$ which only contain elements of
$S^{\prime}$ is a cyclic order on $S^{\prime}$, the _induced cyclic order_ on
$S^{\prime}$.

######  Theorem 7.9.

[9, Theorem 3.6] Let $Z$ be a cyclic order on set $S$ and let $K$ and $L$ be
distinct cuts of $Z$. Then there are non-empty disjoint subsets $A$ and $B$ of
$S$ such that $A\cup B=S$,
$K\mathbin{\upharpoonright}A=L\mathbin{\upharpoonright}A$,
$K\mathbin{\upharpoonright}B=L\mathbin{\upharpoonright}B$,
$K=K\mathbin{\upharpoonright}A\oplus K\mathbin{\upharpoonright}B$ and
$L=K\mathbin{\upharpoonright}B\oplus K\mathbin{\upharpoonright}A$.

One example of a construction similar to the cycle completion is the
following: The cycle completion of a cyclically ordered set $I$ can be
obtained from the Dedekind completion of one of its cuts $L$ by adding as many
elements to the ground set as necessary such that every element of the
original ground set has a predecessor and a successor not in the ground set
and then identifying the new smallest and biggest elements. Also constructing
the pseudo-line as in [2] from $L$, contracting all inner points of an edge to
one point and then again identifying the new smallest and biggest elements
yields the cycle completion. Third, the restriction of the cycle completion to
the set of cuts is already described in [9].

We will now give a precise construction which gives a linear order $D(L)$
starting from a linear order $L$ and is very similar to both the Dedekind
completion of $L$ and the pseudo-line $L(L)$ as in [2, Definition 4.1].
Similarly to the Dedekind completion, $D(L)$ consists of initial segments of
$L$ and of the elements of $L$ itself. But here an element $l$ of $L$ is not
identified with an initial segment of $L$. The construction of $D(L)$ can be
obtained from the pseudo-line $L(L)$ by replacing all the intervals
$(0,1)\times\\{l\\}$ by just $l$. The topology of the pseudo-line is not
needed in the context of this paper.

######  Example 7.10.

(See also [2]) Let $L$ be a linear order on a set $S$ and let $V(L)$ be the
set of _initial segments of $L$_, i.e. subsets $S^{\prime}$ of $S$ which
satisfy that if $s$ is an element of $S^{\prime}$ and $t$ is an element of $S$
with $t<s$ then also $t\in S^{\prime}$. The subset relation is a natural
linear order on $V(L)$. Define a linear order on the disjoint union of $S$ and
$V(L)$ by letting $x\leq y$ if either both $x$ and $y$ are contained in $S$
and $x\leq y$ in $L$ or both are contained in $V(L)$ and $x\leq y$ in $V(L)$
or $x\in y$ or $y\in S\setminus x$. Denote the resulting linear order on
$S\cup V(L)$ by $D(L)$. The smallest element of $D(L)$ is the empty set and
the biggest element of $D(L)$ is $S$. Denote $S\cup(V(L)\setminus\\{S\\})$ by
$V^{\prime}(L)$, the restriction of $D(L)$ to $V^{\prime}(L)$ by
$D^{\prime}(L)$ and the cyclic order induced by $D^{\prime}(L)$ by $Z(L)$. For
every element $s$ of $S$, the set $\\{t\in S\colon t<s\\}$ is the predecessor
and the set $\\{t\in S\colon t\leq s\\}$ is the successor of $s$ in $D(L)$.

Every subset of $D(L)$ has a supremum and an infimum in $D(L)$, which can be
seen as follows: Given a subset $V^{\prime}$ of $V(L)$, the set $\bigcup
V^{\prime}$ is an initial segment of $S$ and is the supremum of $V^{\prime}$
both in $D(L)\mathbin{\upharpoonright}V(L)$ and in $D(L)$. Similarly the set
$\bigcap V^{\prime}$ is the infimum of $V^{\prime}$ in
$D(L)\mathbin{\upharpoonright}V(L)$ and $D(L)$. So in order to show that every
subset of $S\cup V(L)$ has a supremum and an infimum in $D(L)$, it suffices to
consider subsets $S^{\prime}$ of $S$, and by symmetry it suffices to show that
$S^{\prime}$ has a supremum in $D(L)$. The set $\\{s\in S|\exists\,t\in
S^{\prime}:s\leq t\\}$, denoted by $S^{\prime\prime}$, is an initial segment
of $S$ which is an upper bound of $S^{\prime}$. Also, no proper subset of
$S^{\prime\prime}$ is an upper bound of $S^{\prime}$. So if $S^{\prime}$ has
an upper bound in $D(L)$ which is less than $S^{\prime\prime}$, then that
upper bound is contained in $S$. In particular, as $D(L)$ contains between any
two elements of $S$ at least one element of $V(L)$, there is at most one upper
bound of $S^{\prime}$ which is less than $S^{\prime\prime}$. Thus $S^{\prime}$
has a supremum in $D(L)$.

######  Lemma 7.11.

Let $L$ and $K$ be cuts of a cyclic order $Z$ on a set $S$ with at least two
elements such that $K=(L\mathbin{\upharpoonright}(S\setminus
S^{\prime}))\oplus(L\mathbin{\upharpoonright}S^{\prime})$ for an initial
segment $S^{\prime}$ of $L$. Then the map

$V^{\prime}(L)\rightarrow V^{\prime}(K),\quad x\mapsto\begin{cases}x&x\in S\\\
x\cup(S\setminus S^{\prime})&x\in V^{\prime}(L)\setminus S,\ x\subsetneq
S^{\prime}\\\ x\setminus S^{\prime}&x\in V^{\prime}(L)\setminus S,\
S^{\prime}\subseteq x\end{cases}$

is the unique isomorphism of $Z(L)$ and $Z(K)$ which preserves $S$.

###### Proof.

The map $F_{1}:V^{\prime}(L)\rightarrow
V^{\prime}(L\mathbin{\upharpoonright}S^{\prime})\cup
V^{\prime}(L\mathbin{\upharpoonright}(S\setminus S^{\prime}))$ which maps
elements of $S$ to themselves, initial segments which are properly contained
in $S^{\prime}$ to themselves and initial segments $I$ containing $S^{\prime}$
to $I\setminus S^{\prime}$ is an isomorphism of the linear orders
$D^{\prime}(L)$ and $D^{\prime}(L\mathbin{\upharpoonright}S^{\prime})\oplus
D^{\prime}(L\mathbin{\upharpoonright}(S\setminus S^{\prime}))$. Similarly the
map $F_{2}:V^{\prime}(K)\rightarrow
V^{\prime}(L\mathbin{\upharpoonright}S^{\prime})\cup
V^{\prime}(L\mathbin{\upharpoonright}(S\setminus S^{\prime}))$ which maps
every elements of $S$ to themselves, initial segments properly contained in
$S\setminus S^{\prime}$ to themselves and initial segments $I$ containing
$S\setminus S^{\prime}$ to $I\cap S^{\prime}$ is an isomorphism of the linear
orders $D^{\prime}(K)$ and $D^{\prime}(L\mathbin{\upharpoonright}(S\setminus
S^{\prime}))\oplus(L\mathbin{\upharpoonright}S^{\prime})$. Thus the map given
in the lemma, which equals $F_{2}^{-1}\circ F_{1}$, is an isomorphism of
$Z(L)$ and $Z(K)$.

Let $F$ and $G$ be two isomorphisms of $Z(L)$ and $Z(K)$ which preserve $S$.
Assume for a contradiction that there is $v\in V^{\prime}(L)$ such that $F(v)$
is less than $G(v)$ in $D^{\prime}(K)$. As $F$ and $G$ both are bijective and
preserve $S$, $F(v)$ and $G(v)$ are both contained in $V^{\prime}(K)\setminus
S$. Thus there are elements $s\in G(v)\setminus F(v)$ and $t\in S\setminus
G(v)$. Then $(F(t),F(v),F(s))$ equals $(t,F(v),s)$ and is thus contained in
$Z(K)$. Because $F$ is monotone, this implies that $(t,v,s)\in Z(L)$. But
similarly $(s,G(v),t)\in Z(K)$ and thus $(s,v,t)\in Z(L)$, a contradiction. ∎

######  Corollary 7.12.

Let $Z$ be a cyclic order on set $S$ and let $L$ and $K$ be cuts of $Z$. Then
there is a unique isomorphism of $Z(L)$ and $Z(K)$ which preserves $S$.

###### Proof.

Let $S^{\prime}\subseteq S$ such that
$K=(L\mathbin{\upharpoonright}(S\setminus
S^{\prime}))\oplus(L\mathbin{\upharpoonright}S^{\prime})$ and such that
$S^{\prime}$ is an initial segment of $L$. Such a set exists by Theorem 7.9.
Then the statement follows from Lemma 7.11. ∎

######  Lemma 7.13.

Let $Z$ be a cyclic order on a non-empty set $S$ and let $\mathcal{V}$ be the
set of cuts of $Z$. For every cut $L\in\mathcal{V}$ denote the map
$V^{\prime}(L)\rightarrow S\cup\mathcal{V}$ which maps every element of $S$ to
itself and every initial segment $S^{\prime}$ to
$(L\mathbin{\upharpoonright}(S\setminus
S^{\prime}))\oplus(L\mathbin{\upharpoonright}S^{\prime})$ by $\eta_{L}$. Also
denote $\\{(\eta_{L}(a),\eta_{L}(b),\eta_{L}(c))\colon(a,b,c)\in Z(L)\\}$ by
$T_{L}$. Then $T_{L}$ is a cyclic order on $S\cup\mathcal{V}$ which does not
depend on the choice of $L$.

###### Proof.

By Theorem 7.9 the maps $\eta_{L}$ are surjective, so they are bijections
between $V^{\prime}(L)$ and $S\cup\mathcal{V}$. Thus every $T_{L}$ arises from
$Z(L)$ by renaming the elements of $V^{\prime}(L)$ and thus is a cyclic order
on $S\cup\mathcal{V}$, and $\eta_{L}$ is an isomorphism of $Z(L)$ and $T_{L}$.
Let $L$ and $K$ be elements of $\mathcal{V}$ and let $S^{\prime}$ be an
initial segment of $L$ such that $K=(L\mathbin{\upharpoonright}(S\setminus
S^{\prime}))\oplus(L\mathbin{\upharpoonright}S^{\prime})$ (such a segment
exists by Theorem 7.9). Denote the unique isomorphism of $Z(L)$ and $Z(K)$
preserving $S$, which exists by Lemma 7.11, by $F$. Then $\eta_{K}\circ
F(s)=\eta_{L}(s)$ for all $s\in S$. Also, for all initial segments $I$ of $L$
which are properly contained in $S^{\prime}$,

$\displaystyle\eta_{K}\circ F(I)$ $\displaystyle=\eta_{K}(I\cup(S\setminus
S^{\prime}))=(K\mathbin{\upharpoonright}(S^{\prime}\setminus
I))\oplus(K\mathbin{\upharpoonright}(I\cup(S\setminus S^{\prime})))$
$\displaystyle=(L\mathbin{\upharpoonright}(S^{\prime}\setminus
I))\oplus(L\mathbin{\upharpoonright}(S\setminus
S^{\prime}))\oplus(L\mathbin{\upharpoonright}I)$
$\displaystyle=(L\mathbin{\upharpoonright}(S\setminus
I))\oplus(L\mathbin{\upharpoonright}I)=\eta_{L}(I),$

and similarly for all initial segments $I$ of $L$ which contain $S^{\prime}$

$\displaystyle\eta_{K}\circ F(I)$ $\displaystyle=\eta_{K}(I\setminus
S^{\prime})=(K\mathbin{\upharpoonright}(S\setminus(I\setminus
S^{\prime})))\oplus(K\mathbin{\upharpoonright}(I\setminus S^{\prime}))$
$\displaystyle=(L\mathbin{\upharpoonright}(S\setminus
I)))\oplus(L\mathbin{\upharpoonright}S^{\prime})\oplus(L\mathbin{\upharpoonright}(I\setminus
S^{\prime}))$ $\displaystyle=(L\mathbin{\upharpoonright}(S\setminus
I))\oplus(L\mathbin{\upharpoonright}I)=\eta_{L}(I).$

So $\eta_{K}\circ F=\eta_{L}$ and thus $\eta_{K}\circ F\circ\eta_{L}^{-1}$ is
the identity. But $\eta_{K}\circ F\circ\eta_{L}^{-1}$ is also a composition of
isomorphisms of cyclic orders and thus the identity is an isomorphism of
$T_{L}$ and $T_{K}$, so $T_{L}=T_{K}$. ∎

Given a cyclic order $Z$ on set $S$ and a cut $L$ of $Z$, the previous lemma
shows that $T_{L}$ only depends on $Z$ and not on $L$. $T_{L}$ will be a cycle
completion of $Z$. From now on, denote $T_{L}$ by $\mathcal{Z}(Z)$ and its
ground set by $\mathcal{S}(Z)$.

The next lemma shows that $\mathcal{Z}(Z)$ is really a cycle completion of a
cyclic order $Z$. As a result, cycle completions of cyclic orders exists.

######  Lemma 7.14.

Let $Z$ be a cyclic order on set $S$. Then for every non-trivial interval $I$
of $S$ there are unique elements $v$ and $w$ of $\mathcal{S}(Z)\setminus S$
such that $I=[v,w]\cap S$.

###### Proof.

Let $L$ be a cut of $Z$ such that $I$ is an interval of $L$ and such that some
element of $S$ is bigger than all elements of $I$ in $L$. By construction of