# An inverse Fraïssé limit for finite posets and duality for posets and
lattices

Szymon Głąb, Michał Pawlikowski

###### Abstract

We consider a category of all finite partial orderings with quotient maps as
arrows and construct a Fraïssé sequence in this category. Then we use commonly
known relations between partial orders and lattices to construct a sequence of
lattices associated with it. Each of these two sequences has a limit object –
an inverse limit, which is an object of our interest as well.

In the first chapter there are some preliminaries considering partial orders,
lattices, topology, inverse limits, category theory and Fraïssé theory, which
are used later. In the second chapter there are our results considering a
Fraïssé sequence in category of finite posets with quotient maps and
properties of inverse limit of this sequence. In the third chapter we
investigate connections between posets and order ideals corresponding to them,
getting an inductive sequence made of these ideals; then we study properties
of the inverse limit of this sequence.

Key words and phrases. partially ordered set, Fraïssé sequence, inverse
Fraïssé limit, inverse limit, order ideals, distributive lattice

2020 Mathematics Subject Classification.  Primary: 06A06; Secondary: 18B35,
06D05

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Definitions and facts concerning posets
    2. 2.2 Topological definitions and facts concerning the inverse limits
    3. 2.3 Definitions and facts from the category theory
  3. 3 Category of finite posets
    1. 3.1 Directedness and the amalgamation property
    2. 3.2 A universal projective model of profinite poset
    3. 3.3 Properties of the inverse Fraïssé limit $\mathbb{P}$
  4. 4 The inverse system of order ideals
    1. 4.1 Induced quotient maps
    2. 4.2 Properties of $\mathbb{O}(P)$
  5. 5 The category of finite lattices with order and join-irreducible preserving quotient maps
    1. 5.1 $\mathcal{O}(P_{n})$ is a Fraïssé sequence in $\mathfrak{L}$
    2. 5.2 A different approach to the inverse limit of $\langle\\{\mathcal{O}(P_{n})\\}_{n\in\mathbb{N}},\\{\hat{p}^{n}_{k}\\}_{k<n}\rangle$.

##  1 Introduction

Fraïssé theory has its beginning in the Georg Cantor’s work, where Cantor has
shown that any countable, dense linear ordering without endpoints is
isomorphic to $(\mathbb{Q},\leq)$. In the 1950s Roland Fraïssé expanded on the
idea of Cantor and formulated it in the model theory framework [5], which
marked the beginning of a theory named after him. Some examples of Fraïssé
limits are well-known objects, such as $(\mathbb{Q},\leq)$ – a limit of all
linear orderings; the random graph – a limit of all finite graphs; and some
are completely new, like for example a limit of all finite partially ordered
sets, called the random poset, as well as many more countable,
ultrahomogeneous structures. In late 1970s James Schmerl in [10] characterized
all possible countable ultrahomogeneous posets. Up to isomorphism, there are
only countably many ultrahomogeneous countable posets, each of one of four
types. Since ultrahomogeneous countable structures are Fraïssé limits of a
finite structures family of the same type (this family is called age of the
structure), Schmerl characterized all possible Fraïssé limits for finite
posets.

Due to Birkhoff Theorem (Theorem 2.10) there is duality between finite posets
and finite distributive lattices. Therefore one could expect the similar
characterization to that of Schmerl for lattices. However Abogatma and Truss
in [1] proved that there are continuum many pairwise non-isomorphic
ultrahomogeneous lattices. On the other hand if one demand families of finite
lattices to be varieties, then by result [4] of Day and Ježek there are only
three possible non-isomorphic Fraïssé limits.

In XXI century new variants of Fraïssé theory were introduced by Sławomir
Solecki, Trevor Irwin [7] and Wiesław Kubiś [9, 8], and then further developed
by other mathematicians. Limit objects in these theories can be uncountable
and have found remarkable applications in topology and functional analysis.

In this article we work in generalised Fraïssé theory formulated by Wiesław
Kubiś [9] which uses category-theoretic approach. In this approach classical
Fraïssé limit for posets is a direct limit of finite posets with embeddings.
Here we are studying finite posets with quotients. Then the limit object is an
uncountable poset with a natural topological structure. It turns out that this
is the Cantor space with the extremely simple order structure, namely there
are two kind of points: isolated or comparable with only one other point. The
motivation for our investigation comes from [6], where the authors studied
inverse Fraïssé limit for graphs.

As we have mentioned finite lattices are dual to finite posets. Therefore we
are also interested in finding dual limit to that of posets.

The paper is organized as follows. In Section 2 we present basic definitions
and facts on posets, lattices and inverse limits, and gives an introduction to
the categorical Fraïssé theory. In Section 3 we build the model of the inverse
Fraïssé limit for finite posets $\mathbb{P}$, and its properties. In Section 4
using duality between posets and lattices, having
$\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$ an
inductive sequence consisting of finite posets with quotient maps and its
limit
$P=\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$,
we investigate an inverse limit $\mathbb{O}(P)$ of order ideals
$\mathcal{O}(P_{n})$. We show that $\mathbb{O}(P)$ is a $\sigma$-complete
atomic lattice, and in Theorem 4.16 we show the connection between
$\mathbb{P}$ and $\mathbb{O}(\mathbb{P})$. In Section 5 we define category of
finite lattice in which $\mathbb{O}(\mathbb{P})$ is an inverse Fraïssé limit,
and present a different way of looking on order ideals $\mathcal{O}(P_{n})$
and induced quotients $\hat{p}^{n}_{k}$.

Definitions and facts concerning posets and lattices are taken from Davey and
Priestley monograph [3]. Awodey’s book [2] and Kubiś’s notes [8] were used to
gather the information on category theory and Fraïssé theory. Some topological
facts are taken from Srivastava’s monograph [11].

##  2 Preliminaries

We will use $\mathbb{N}$ to denote the set of natural numbers starting from
$1$ and we will use $\omega$ to denote the set of natural numbers starting
with $0$.

###  2.1 Definitions and facts concerning posets

In this section we present terminology and notation that we are using in the
sequel. Most of them come from [3].

Let $A$ be a set and $\leq$ be a binary relation on $A$ that is reflexive,
transitive and antisymmetric. We say that structure $(A,\leq)$ is a _poset_.
For $x,y\in A$ if neither $x\leq y$ nor $y\leq x$, then we say that $x$ and
$y$ are incomparable, in symbols $x\parallel y$. We say that a poset
$(A,\leq)$ is a _chain_ , if for all $x,y\in A$, $x\leq y$ or $y\leq x$. If
$A=\\{1,\dots,n\\}$ is considered with the natural order, then we denote such
chain by $\mathbf{n}$. We say that a poset $(A,\leq)$ is an _antichain_ , if
for all $x,y\in A$, $x\leq y\Leftrightarrow x=y$. If $A=\\{1,\dots,n\\}$ is
ordered by equality, then we denote such antichain by $\overline{\mathbf{n}}$.

Let $(A;\leq_{A}),(B;\leq_{B})$ be posets and $\phi$ a mapping
$\phi:A\rightarrow B$. If for all $x,y\in A$ such that $x\leq_{A}y$,
$\phi(x)\leq_{B}\phi(y)$ then we call $\phi$ an _order homomorphism_ (we will
often abbreviate it to _homomorphism_). If $\phi(x)\leq_{B}\phi(y)\iff
x\leq_{A}y$ for all $x,y\in A$, then we say that $\phi$ is an _order-
embedding_. Furthermore, if it is onto, we say that it is an _order-
isomorphism_. We say that a homomorphism $\phi$ is a _quotient map_ if it is
onto and for every $p,r\in B$ such that $p\leq_{B}r$ there are $x,y\in A$ such
that $\phi(x)=p,\phi(y)=r$ and $x\leq_{A}y$.

Let $(A;\leq_{A}),(B;\leq_{B})$ be disjoint posets. We define the _linear sum_
$A\oplus B$ by taking the following order relation on $A\cup B:x\leq y$ if and
only if one of the following conditions is met

  1. 1.

$x,y\in A$ and $x\leq_{A}y$

  2. 2.

$x,y\in B$ and $x\leq_{B}y$

  3. 3.

$x\in A$ and $y\in B$

The _disjoint union_ $A\dot{\cup}B$ is the ordered set $A\cup B$ formed by
defining $x\leq y$ if and only if one of the following conditions is met

  1. 1.

$x,y\in A$ and $x\leq_{A}y$

  2. 2.

$x,y\in B$ and $x\leq_{B}y$

If $A$ and $B$ are not disjoint, we can define their disjoint copies, for
example $A\times\\{0\\}$ and $B\times\\{1\\}$, endow them with order
structures isomorphic to the original ones, and then take their linear sum and
disjoint union. Having this in mind we will write, for example,
$\mathbf{2}\oplus\mathbf{2}=\mathbf{4}$ and
$\overline{\mathbf{2}}\dot{\cup}\overline{\mathbf{2}}=\overline{\mathbf{4}}$
which is true up to isomoprhism.

Let $P$ be a poset and $Q\subset P$. We say that $Q$ is a _down-set_ if for
every $q\in Q$ and for every $p\in P$ such that $p\leq q$ we have that $p\in
Q$. By $\downarrow\\!\\!p$ we shall denote the following set $\\{m\in P:m\leq
p\\}$. We say that it is a _principal_ down-set. It is easy to see that
$\downarrow\\!\\!p$ is a down-set for every $p\in P$. By $\mathcal{O}(P)$ we
denote a partial ordering of all down-sets of $P$ with $\subset$ as an order
relation. Set $\mathcal{O}(P)$ is called an _order ideal_. Let $S\subset P$.
An element $x\in P$ is called an _upper bound_ of $S$ if for every $y\in S$ we
have $y\leq x$. Similarly $x$ is the _lower bound_ of $S$ if $y\geq x$ for
every $y\in S$. The _supremum_ of $S$ is the least upper bound of $S$ and we
denote it by $\sup(S)$. Similarly the _infimum_ of $S$ is the greatest lower
bound of $S$ and we denote it by $\inf(S)$. Supremum and infimum of $S$ may
not exist even for finite sets. However, if they exist, they must be unique.

Let $P$ be a poset. We say that $P$ is a _lattice_ if for every $x,y\in P$
$\sup(\\{x,y\\})\text{ and }\inf(\\{x,y\\})$ exist. Usually we write
$\sup(\\{x,y\\})$, $\inf(\\{x,y\\})$ as $x\vee y$, $x\wedge y$ respectively.

Supremum and infimum are associative, i.e. $x\lor(y\lor z)=(x\lor y)\lor z$
and $x\land(y\land z)=(x\land y)\land z$, and commutative. If a lattice has
the minimal element, then we call it as zero element and denote it by 0.
Similarly the maximal element is denoted by 1. Note that every finite lattice
has both minimal and maximal elements.

Let $L$ be a lattice. We say that $L$ is _distributive_ if for all $x,y,z\in
L$ the following is true:

$x\wedge(y\vee z)=(x\wedge y)\vee(x\wedge z)$ (1)

Let $P$ be a poset. It is easy to see that $\mathcal{O}(P)$ is a lattice with
$\sup\\{x,y\\}=x\cup y$ and $\inf\\{x,y\\}=x\cap y$.

######  Lemma 2.1.

[3, p. 46] Let $\\{A_{i}\\}_{i\in I}$ be a family of sets contained in a
lattice $L$, such that for every $i\in I$, $\sup(A_{i})$ exists and
$\sup(\bigcup_{i\in I}A_{i})$ exists. Then $\bigvee_{i\in
I}\sup(A_{i}):=\sup\\{\sup(A_{i}):i\in I\\}$ exists and $\sup(\bigcup_{i\in
I}A_{i})=\bigvee_{i\in I}\sup(A_{i})$.

Let $L$ be a lattice. An element $x\in L$ is _join-irreducible_ if the
following conditions are met

  1. (i)

$x\neq 0$ (in case $L$ has a zero)

  2. (ii)

$x=a\vee b$ implies $x=a$ or $x=b$ for all $a,b\in L$.

We denote the set of all join-irreducible elements of $L$ by $\mathcal{J}(L)$.
This set inherits $L$’s order relation.

The simple example of elements which are join-irreducible and those which are
not is given after the Definition 2.6. Now we present a relation between
$\leq$ order on $P$ and $\subset$ on $\mathcal{O}(P)$.

######  Lemma 2.2.

[3, p. 21] Let $P$ be a poset and $x,y\in P$. The following statements are
equivalent:

  1. _(i)_

$x\leq y$

  2. _(ii)_

$\downarrow\\!\\!x\subset\downarrow\\!\\!y$;

  3. _(iii)_

For every $Q\in\mathcal{O}(P)$ if $y\in Q$ then $x\in Q$.

Let $L$ and $K$ be lattices. A map $f:L\to K$ is said to be a _lattice
homomorphism_ if $f$ is join-preserving and meet-preserving, that is, for all
$a,b\in L$,

$f(a\vee b)=f(a)\vee f(b)\text{ and }f(a\wedge b)=f(a)\wedge f(b).$

A bijective homomorphism is a lattice isomorphism.

######  Lemma 2.3.

[3, p. 44] Let $L$ an $K$ be lattices and $f:L\to K$ a map. Then $f$ is a
lattice isomorphism if and only if it is an order-isomorphism.

######  Lemma 2.4.

Let $P$ be a finite poset and $Q\subset P$ be a down-set. Then
$Q=\bigcup_{k=1}^{n}\downarrow\\!\\!x_{k}$ where $x_{1},\ldots,x_{n}$ are
pairwise distinct maximal elements of $Q$ (note that $Q$ has maximal elements,
as it is a finite poset). Furthermore it is the only representation of $Q$ as
a union of pairwise incomparable principal down-sets. We shall call the set
$\\{\downarrow\\!\\!x_{k}:k\leq n\\}$ a canonical decomposition of $Q$.

###### Proof.

Let $Q$ be a down-set and $x_{1},\ldots,x_{n}$ be a list of all maximal
elements of $Q$. It is easy to see that
$\bigcup_{k=1}^{n}\downarrow\\!\\!x_{k}\subset Q$ because
$x_{1},\ldots,x_{n}\in Q$ and $Q$ is a down-set. Now if we take $q\in Q$ such
that $q$ is not maximal then, as $Q$ is finite, we will find maximal element
$x_{i}$ such that $q\leq x_{i}$, so
$q\in\bigcup_{k=1}^{n}\downarrow\\!\\!x_{k}$. So
$Q=\bigcup_{k=1}^{n}\downarrow\\!\\!x_{k}$. What follows from Lemma 2.2 and
maximality of $x_{l}$’s is that
$\downarrow\\!\\!x_{i}\parallel\downarrow\\!\\!x_{j}$ for $i\neq j$. Now we
prove that it is the only pairwise incomparable representation of $Q$.

Assume that
$Q=\bigcup_{k=1}^{n}\downarrow\\!\\!x_{k}=\bigcup_{k=1}^{m}\downarrow\\!\\!y_{k}$,
with $\downarrow\\!\\!y_{i}\parallel\downarrow\\!\\!y_{j}\text{ for }i\neq j$.
Clearly $n\leq m$ because otherwise we have that for some
$\downarrow\\!\\!x_{i},\downarrow\\!\\!x_{j}$ we find $\downarrow\\!\\!y_{p}$
such that
$\downarrow\\!\\!x_{i}\subset\downarrow\\!\\!y_{p},\downarrow\\!\\!x_{j}\subset\downarrow\\!\\!y_{p}$
so $x_{i},x_{j}$ would not be distinct. The fact that we will find such
$\downarrow\\!\\!y_{p}$ is because, if
$\downarrow\\!\\!x_{k}\subset\bigcup_{k=1}^{m}\downarrow\\!\\!y_{k}$, then
$x_{k}\in\bigcup_{k=1}^{m}\downarrow\\!\\!y_{k}$, thus
$x_{k}\in\downarrow\\!\\!y_{p}$ for some $p\in\\{1,\dots,m\\}$, so $x_{k}\leq
y_{p}$. It follows that $\downarrow\\!\\!x_{k}\subset\downarrow\\!\\!y_{p}$.

Now it is easy to see that for every $\downarrow\\!\\!x_{k}$ we find
$\downarrow\\!\\!y_{\ell}$ such that
$\downarrow\\!\\!x_{k}\subset\downarrow\\!\\!y_{\ell}$. On the other hand for
every $\downarrow\\!\\!y_{\ell}$ we find $\downarrow\\!\\!x_{p}$ such that
$\downarrow\\!\\!y_{\ell}\subset\downarrow\\!\\!x_{p}$. So
$\downarrow\\!\\!x_{k}\subset\downarrow\\!\\!y_{\ell}\subset\downarrow\\!\\!x_{p}$.
Clearly $k=p$, since $x_{k}$ and $x_{p}$ are comparable. Thus
$y_{\ell}=x_{k}$, and consequently
$\\{x_{1},\dots,x_{n}\\}\subset\\{y_{1},\dots,y_{m}\\}$. Suppose that $m>n$.
Then there is $\ell$ with $y_{\ell}\notin\\{x_{1},\dots,x_{n}\\}$. But then by
maximality of $x_{i}$’s, $y_{\ell}\leq x_{j}$ for some $j$. This contradicts
the assumption that $y_{k}$’s are pairwise incomparable. So $m=n$ and
$\\{y_{1},\ldots,y_{n}\\}=\\{x_{1},\ldots,x_{n}\\}$. ∎

From the above proof we extract the following fact.

######  Lemma 2.5.

Let $P$ be a finite poset, $Q\subset P$ be a down-set. Let
$\\{\downarrow\\!\\!x_{k}:k\leq n\\}$ be the canonical decomposition of $Q$
and let $\\{\downarrow\\!\\!y_{k}:k\leq m\\}$ be such that
$Q=\bigcup_{k=1}^{m}\downarrow\\!\\!y_{k}$. Then
$\\{\downarrow\\!\\!x_{k}:k\leq n\\}\subset\\{\downarrow\\!\\!y_{k}:k\leq
m\\}$.

######  Definition 2.6.

[3, p. 113] Let $L$ be a lattice with the least element $0$ and $a\in L$. We
say that $a$ is an _atom_ in $L$, if $0<a$ and if $x<a$, then $x=0$. A lattice
is called _atomic_ , if for every $x\in L,x\neq 0$ we can find an atom $a$
such that $a\leq x$.

To illustrate the notion of atom, let us consider the power set
$\mathcal{P}(\\{0,1\\})$ with the inclusion as a an order relation. This is a
lattice with the least element $\emptyset$. Then $\\{0\\}$ and $\\{1\\}$ are
atoms while $\\{0,1\\}$ is not. Note also that $\\{0\\}$ and $\\{1\\}$ are
join-irreducible while $\\{0,1\\}$ is not.

######  Lemma 2.7.

[3, p. 117] Let $L$ be a distributive lattice and let $x\in L$, with $x\neq 0$
in case $L$ has a zero. Then the following are equivalent:

  1. (i)

$x$ is join-irreducible

  2. (ii)

if $a,b\in L$ and $x\leq a\vee b$, then $x\leq a$ or $x\leq b$

  3. (iii)

for any $k\in\mathbb{N}$, if $a_{1},\dots,a_{k}\in L$ and $x\leq a_{1}\vee
a_{2}\vee\dots\vee a_{k}$, then $x\leq a_{i}$ for some $i\in\\{1,\dots,k\\}$.

######  Lemma 2.8.

[3, p. 116] Let $P$ be a poset. Then
$\mathcal{J}(\mathcal{O}(P))=\\{\downarrow\\!\\!x:x\in P\\}$

######  Lemma 2.9.

[3, p. 116] Let $P$ be a poset and $\phi:P\to\mathcal{J}(\mathcal{O}(P))$ be
defined as $\phi(x)=\downarrow\\!\\!x$. Then $\phi$ is an order-isomorphism.
In other words $\phi$ is an order-embedding of $P$ into $\mathcal{O}(P)$.

######  Theorem 2.10.

[3, p. 118][Birkhoff’s Theorem] Let $L$ be a finite, distributive lattice.
Then the map $\eta:L\to\mathcal{O}(\mathcal{J}(L))$ defined by

$\eta(a):=\\{x\in\mathcal{J}(L):x\leq a\\}$

is an isomorphism of $L$ onto $\mathcal{O}(L)$.

###  2.2 Topological definitions and facts concerning the inverse limits

In this section we present the definition of the inverse limit and the notion
of profinite poset. Let $X$ be a topological space. We say that $X$ is _$0$
-dimensional_, if it has a base of clopens.

Let $P,Q$ be posets with topology $\tau,\sigma$ respectively. We say that
$f:P\rightarrow Q$ is a _topological isomorphism_ if it is a homeomorphism and
an order-isomorphism.

Let $P_{n}$ be non-empty, finite sets with the discrete topology. Let $E_{n}$
be relations on $P_{n}$. We equip $P_{1}\times P_{2}\times\dots$ with the
product topology. Sets of the form

$\\{x_{1}\\}\times\\{x_{2}\\}\times\dots\times\\{x_{n}\\}\times P_{n+1}\times
P_{n+2}\times\dots$

form a base for topology on $P_{1}\times P_{2}\times\dots$. Moreover,
$P_{1}\times P_{2}\times\dots$ is $0$-dimensional, compact and metrizable.

Now assume we have a family of onto homomorphisms
$\\{p_{k}^{m}:m,k\in\mathbb{N},k<m\\}$ where $p_{k}^{m}:P_{m}\rightarrow
P_{k},m\in\mathbb{N},k<m$, such that $p_{\ell}^{k}\circ
p_{k}^{n}=p_{\ell}^{n}$ for $\ell<k<n$. An inverse limit of $P_{n}$ with
$p_{k}^{m}$ is a set

$\varprojlim\left<\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{m}_{k}\\}_{k<m}\right>:=\left\\{(x_{1},x_{2},\dots)\in\prod_{i\in\mathbb{N}}P_{i}:p_{n}^{n+1}(x_{n+1})=x_{n}\text{
for each }n\in\mathbb{N}\right\\}$

We define relation $E$ on inverse limit as follows:

$(x_{n})E(y_{n})\iff x_{n}E_{n}y_{n}\text{ for each }n\in\mathbb{N}$

If relations $E_{n}$ are partial orders, then $E$ is also a partial ordering.
Then the inverse limit with this ordering (or structure topologically
isomorphic to it) is called a _profinite poset_.

######  Lemma 2.11.

$P:=\varprojlim\left<\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{m}_{k}\\}_{k<m}\right>$
is closed.

###### Proof.

We will show that $(\prod_{i\in\mathbb{N}}P_{i})\setminus P$ is open. Let
$(x_{n})\in(\prod_{i\in\mathbb{N}}P_{i})\setminus P$ and $n$ be the smallest
index such that $p_{n}^{n+1}(x_{n+1})\neq x_{n}$. Then see that
$(x_{n})\in\\{x_{1}\\}\times\\{x_{2}\\}\times\dots\times\\{x_{n+1}\\}\times
P_{n+2}\times\dots\subset(\prod_{i\in\mathbb{N}}P_{i})\setminus P$, so
$(\prod_{i\in\mathbb{N}}P_{i})\setminus P$ is open, thus $P$ is closed. ∎

######  Lemma 2.12.

Relation $E$ is a closed set in $P\times P$.

###### Proof.

We will show that $P^{2}\setminus E$ is an open set. Take
$((x_{n}),(y_{n}))\in P^{2}\setminus E$. Let $n$ be the smallest index such
that $\neg(x_{n}E_{n}y_{n}$).

Then
$((x_{n}),(y_{n}))\in(P\cap\\{x_{1}\\}\times\\{x_{2}\\}\times\dots\times\\{x_{n}\\}\times
P_{n+1}\times\dots)\times(P\cap\\{y_{1}\\}\times\\{y_{2}\\}\times\dots\times\\{y_{n}\\}\times
P_{n+1}\times\dots)\subset P^{2}\setminus E$ and
$(P\cap\\{x_{1}\\}\times\\{x_{2}\\}\times\dots\times\\{x_{n}\\}\times
P_{n+1}\times\dots)\times(P\cap\\{y_{1}\\}\times\\{y_{2}\\}\times\dots\times\\{y_{n}\\}\times
P_{n+1}\times\dots)$ is open in $P^{2}$, thus $P^{2}\setminus E$ is open,
which makes $E$ closed. ∎

Lemma 2.11 proves that our inverse limit is closed and it is a sub-space of
compact space, thus it is compact. See that sets
$P\cap(\\{x_{1}\\}\times\dots\times\\{x_{n}\\}\times P_{n+1}\times\dots)$ form
a base for the inverse limit, thus it is $0$-dimensional and it is also
metrizable.

###  2.3 Definitions and facts from the category theory

In this section we present definitions and facts concerning category theory
and Fraïssé sequences. They come from [2], [8] and [9].

A _category_ is a structure of the form $\mathfrak{K}=\langle
V,A,\operatorname{dom},\operatorname{cod},\circ\rangle$ where $V$ is a class
of objects, $A$ is a class of arrows, $\operatorname{dom}:A\rightarrow V$ and
$\operatorname{cod}:A\rightarrow V$ are the domain and codomain functions, and
$\circ$ is a partial binary operation on arrows, such that the following
conditions are satisfied:

  1. 1.

$f\circ g$ is defined whenever $\operatorname{cod}(g)=\operatorname{dom}(f)$
and $\circ$ is associative, that is, $(f\circ g)\circ h=f\circ(g\circ h)$
whenever $\operatorname{cod}(h)=\operatorname{dom}(g)$ and
$\operatorname{cod}(g)=\operatorname{dom}(f)$

  2. 2.

For each object $a\in V$ there is an arrow $\operatorname{id}_{a}$ with
$\operatorname{dom}(\operatorname{id}_{a})=\operatorname{cod}(\operatorname{id}_{a})=a$
and such that $\operatorname{id}_{a}\circ g=g,f\circ\operatorname{id}_{a}=f$
holds for every $f,g\in A$ with
$\operatorname{cod}(g)=\operatorname{dom}(f)=a$.

  3. 3.

For every objects $a,b\in V$ the class

$\mathfrak{K}(a,b):=\\{f\in
A:\operatorname{dom}(f)=a\wedge\operatorname{cod}(f)=b\\}$

is a set.

Let $\mathfrak{K}$ be a category. We say that $a$ is an _initial object_ , if
for every object $b$ in $\mathfrak{K}$ there is a unique arrow $f:a\rightarrow
b$. $a$ is a _terminal object_ if for every object $b$ in $\mathfrak{K}$ there
is a unique arrow $f:b\rightarrow a$.

Let $\mathfrak{K}$ be a category. We say that $\mathfrak{K}$ is _directed_ if
for every $a,b\in\mathfrak{K}$ there exists $c\in\mathfrak{K}$ such that both
sets $\mathfrak{K}(a,c),\mathfrak{K}(b,c)$ are not empty.

Let $\mathfrak{K}$ be a category. We say that $\mathfrak{K}$ has the
_amalgamation property_ if for every $a,b,c\in\mathfrak{K}$ and for every
morphisms $f\in\mathfrak{K}(a,b),g\in\mathfrak{K}(a,c)$ there exists
$d\in\mathfrak{K}$ and morphisms
$f^{{}^{\prime}}\in\mathfrak{K}(b,d),g^{{}^{\prime}}\in\mathfrak{K}(c,d)$ such
that $f^{{}^{\prime}}\circ f=g^{{}^{\prime}}\circ g$. In other words the
diagram

${b}$${a}$${d}$${c}$$\scriptstyle{f^{{}^{\prime}}}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{g^{{}^{\prime}}}$
(2)

commutes.

Let $\mathfrak{K}$ be a category. A pair of the form
$\langle\\{u_{n}\\},\\{u_{k}^{n}\\}_{k<n}\rangle$ such that
$\\{u_{n}:n\in\mathbb{N}\\}\subset\mathfrak{K}$,
$u_{k}^{n}\in\mathfrak{K}(u_{k},u_{n})$ for every $n,k\in\mathbb{N}$, where
$k<n$, and $u_{k}^{n}\circ u_{\ell}^{k}=u_{\ell}^{n}$ is called an _inductive
sequence_ in $\mathfrak{K}$ and we denote it by $\overrightarrow{u}$. 111An
inductive sequence can be enumerated by any ordinal. The general definition
sounds as follows. Let $\mathfrak{K}$ be a category and $\delta>0$ an ordinal.
A pair of the form
$\langle\\{a_{\xi}\\}_{\xi<\delta},\\{a_{\xi}^{\eta}\\}_{\xi<\eta<\delta}\rangle$
such that $\\{a_{\xi}:\xi<\delta\\}\subset\mathfrak{K}$ and
$a_{\eta}^{\varrho}\circ a_{\xi}^{\eta}=a_{\xi}^{\varrho}$ is called an
_inductive sequence_ in $\mathfrak{K}$ and we denote it by
$\overrightarrow{a}$. The ordinal $\delta$ is the _length_ of
$\overrightarrow{a}$.

Let $\overrightarrow{u}$ be an inductive sequence in category $\mathfrak{K}$.
We say that it has the _extension property_ (E) if for every arrows $f:a\to
u_{n},g:a\to b$ where $n\in\mathbb{N}$, there exists $m>n$ and
$h\in\mathfrak{K}(b,u_{m})$ such that the diagram

${u_{n}}$${u_{m}}$${a}$${b}$$\scriptstyle{u_{n}^{m}}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{h}$

commutes.

Let $\mathfrak{K}$ be a fixed category and let $\overrightarrow{u}$ be an
inductive sequence in that category. We say that $\overrightarrow{u}$ is a
_Fraïssé sequence_ if the following conditions are met

  1. (U)

for every $x\in\mathfrak{K}$, there exists $n\in\mathbb{N}$ such that
$\mathfrak{K}(x,u_{n})\neq\emptyset$ ;

  2. (A)

for every $k\in\mathbb{N}$ and for every arrow $f\in\mathfrak{K}(u_{k},y)$,
where $y\in\mathfrak{K}$, there exist $\ell>k$ and
$g\in\mathfrak{K}(y,u_{\ell})$ such that $u^{\ell}_{k}=g\circ f$.

Let $\mathfrak{K}$ be a fixed category and $\mathcal{F}$ a family of arrows.
By $\operatorname{Dom}(\mathcal{F})$ we denote the set
$\\{\operatorname{dom}(f)\in\mathfrak{K}:f\in\mathcal{F}\\}$. We say that
$\mathcal{F}$ is _dominating_ in $\mathfrak{K}$ (or that $\mathfrak{K}$ is
_dominated_ by $\mathcal{F}$) if the following conditions are met

  1. (D1)

for every $x\in\mathfrak{K}$ there is $a\in\operatorname{Dom}(\mathcal{F})$
such that $\mathfrak{K}(x,a)\neq\emptyset$ ;

  2. (D2)

given $a\in\operatorname{Dom}(\mathcal{F})$ and $f\in\mathfrak{K}(a,y)$, there
exist $g\in\mathfrak{K}(y,b)$ such that $g\circ f\in\mathcal{F}$.

######  Lemma 2.13.

Let $\mathfrak{K}$ be a category with an initial object $a$. If $\mathfrak{K}$
has the amalgamation property, then $\mathfrak{K}$ is directed.

######  Lemma 2.14.

Let $\overrightarrow{u}$ be an inductive sequence in directed category
$\mathfrak{K}$. If $\overrightarrow{u}$ has _(A)_ property, then it has _(U)_
property.

######  Lemma 2.15.

Let $\overrightarrow{u}$ be a Fraïssé sequence in category $\mathfrak{K}$
which has the amalgamation property. Then $\overrightarrow{u}$ has _(E)_
property.

######  Theorem 2.16 (Existence of Fraïssé sequence).

Let $\mathfrak{K}$ be a directed category with amalgamation property. Assume
that $\mathfrak{K}$ is dominated by a countable family of arrows
$\mathcal{F}$. Then $\mathfrak{K}$ has a Fraïssé sequence.

##  3 Category of finite posets

Let $V$ be a class of all finite posets (we identify all isomorphic posets
with each other). Then $V$ is a countable set. Formally, one can define $V$ as
a set of all finite posets on $\mathbb{N}$. Now let $A$ be the class of all
quotient mappings between finite posets; we say that $f$ is an arrow from $a$
to $b$, provided $f$ is quotient mapping from $b$ onto $a$. Then
$\operatorname{dom}f$ is the range of $f$ and $\operatorname{cod}f$ is the
domain of $f$. The composition $g\bullet f$ of two arrows $f$ from $a$ to $b$
and $g$ from $b$ to $c$ is $f\circ g:c\to a$, where $\circ$ is the usual
composition for mappings. The composition of two quotient mapping is a
quotient mapping, identity is a quotient mapping and class of arrows from $a$
to $b$ is a subset of $a^{b}$. Therefore $\mathfrak{K}:=\langle
V,A,\operatorname{dom},\operatorname{cod},\bullet\rangle$ is a category.

In the sequel we will use $\leq$ for poset relation if it is clear what is its
domain. We will use subscripts, say $\leq_{A}$, $\leq_{B}$, to indicate
relations domains if more than one poset is currently considered.

Let $(A,\leq)$ be a poset and let $x,y\in A$. If there is no $z$ less than
$x$, no $z$ greater than $y$ and $x<y$, then we say pair $(x,y)$ is a
_$\mathbf{2}$ -component_ in $A$.

As $\mathfrak{K}$ is a countable category, if we prove that it it directed and
has the amalgamation property, then from Theorem 2.16 we get, that there is a
Fraïssé sequence in that category. In the next section we are going to prove
that our category indeed has these two properties.

###  3.1 Directedness and the amalgamation property

Let $E$ be a relation on $X$. We say that $x$ is $E$-isolated, if for every
$y\neq x$ neither $xEy$ nor $yEx$.

######  Lemma 3.1.

Let $(A,\leq_{A})$ be a finite poset without $\leq_{A}$-isolated points. There
is a quotient map $\varphi:\dot{\bigcup}_{i<n}\mathbf{2}\rightarrow A$ for
some $n\in\mathbb{N}$.

###### Proof.

Let $(x_{0},y_{0}),\ldots,(x_{m-1},y_{m-1})$ be an enumeration of all pairs
$(a,b)\in A\times A$ with $a<_{A}b$. By $(B,\leq_{B})$ we denote the set
$\\{0,1,\ldots,2m-1\\}$ with a partial order given by
$x\leq_{B}y\Leftrightarrow(x=y\vee\exists_{i<m}(x=2i\wedge y=2i+1))$. We
define $f:B\rightarrow A$ as follows

$f(2i)=x_{i},f(2i+1)=y_{i}\text{ for }i<m$ (3)

Clearly $f$ is a quotient map. Since $B$ and $\dot{\bigcup}_{i<m}\mathbf{2}$
are isomorphic, the desired mapping $\varphi$ is a composition of $f$ with the
isomorphism. ∎

######  Lemma 3.2.

Let $(C,\leq_{C})$ be a poset. There is a quotient map
$\chi:\dot{\bigcup}_{i<p}\mathbf{2}\rightarrow C$ for some $p\in\mathbb{N}$.

###### Proof.

Let $I$ be a set of all isolated points in $C$, and let $A=C\setminus I$. Then
there is a quotient map $\phi:\dot{\bigcup}_{j<n}\mathbf{2}\rightarrow A$ for
some $n\in\mathbb{N}$. Let $i$ be the cardinality of $I$. We define
$\psi:\dot{\bigcup}_{j<i}\mathbf{2}\rightarrow I$ so that we map $k$-th
$\mathbf{2}$-component of $\dot{\bigcup}_{j<i}\mathbf{2}$ onto $k$-th point of
$I$. Finally we define $\chi:\dot{\bigcup}_{j<i}\mathbf{2}\text{
}\dot{\cup}\text{ }\dot{\bigcup}_{j<n}\mathbf{2}\rightarrow C$ as
$\chi:=\phi\cup\psi$. ∎

######  Theorem 3.3.

Category of finite posets is directed and has the amalgamation property.

###### Proof.

Firstly we show that this category has the amalgamation property. Let $A,B$
and $C$ be posets and let $f:B\to A$ and $g:C\to A$ be quotient maps. By Lemma
3.2 there are $B^{{}^{\prime}}\text{ and }C^{{}^{\prime}}$, disjoint unions of
$\mathbf{2}$’s, and $f^{{}^{\prime}}:B^{{}^{\prime}}\to B\text{ and
}g^{{}^{\prime}}:C^{{}^{\prime}}\to C$ quotient mappings. If we find a poset
$D$ and quotients $q:D\to B^{{}^{\prime}}\text{ and }p:D\to C^{{}^{\prime}}$
such that the diagram

${B}$${B^{{}^{\prime}}}$${A}$${D}$${C}$${C^{{}^{\prime}}}$$\scriptstyle{f}$$\scriptstyle{f^{{}^{\prime}}}$$\scriptstyle{q}$$\scriptstyle{p}$$\scriptstyle{g}$$\scriptstyle{g^{{}^{\prime}}}$
(4)

commutes, then we are done. Let $D:=B^{\prime}\dot{\cup}C^{\prime}$. We define
$q:D\rightarrow B^{\prime}$ as follows: if $z\in B^{\prime}$ then $q(z)=z$. If
$z\in C^{\prime}$ then let $v\in C^{\prime}$ be such that $z\leq_{D}v$ or
$v\leq_{D}z$. We define $q$ at $z$ and $v$ at the same time, so we may assume
that $z<_{D}v$.

  1. (i)

If $g\circ g^{\prime}(z)=g\circ g^{\prime}(v)$, then because $f$ is onto we
find $x\in B^{\prime}$ such that $f\circ f^{\prime}(x)=g\circ
g^{\prime}(z)=g\circ g^{\prime}(v)$. We put $q(z)=q(v)=x$.

  2. (ii)

If $g\circ g^{\prime}(z)\neq g\circ g^{\prime}(v)$, then $g\circ
g^{\prime}(z)<_{A}g\circ g^{\prime}(v)$. Because $f$ is a quotient map we find
$x,y\in B$ such that $x<_{B}y$ and $f\circ f^{\prime}(x)=g\circ
g^{\prime}(z),f\circ f^{\prime}(y)=g\circ g^{\prime}(v)$. We put
$q(z)=x,q(v)=y$.

Defining $p$ in a similar fashion makes that $f\circ f^{\prime}\circ q=g\circ
g^{\prime}\circ p$ and both $q,p$ are quotient maps. So the functions
$f^{\prime}\circ q$ and $g^{\prime}\circ p$ witness the amalgamation property
in considered category. It is clear that $\mathbf{1}$, that is the one-element
poset, is an initial object of our category. By Lemma 2.13 the category is
directed. ∎

From Theorem 3.3 and Theorem 2.16 we get that there is a Fraïssé sequence in
the category of finite posets. Now we are going to provide its direct
construction.

###  3.2 A universal projective model of profinite poset

The construction in this section is an adaptation of that presented in [6] for
graphs.

Let $\mathbb{P}$ be a set $\\{0,1,2,3\\}^{\omega}$ with a partial order $x\leq
y\Leftrightarrow x=y$ or one of the following conditions is met:

  1. (i)ω

$\exists_{n}\left[\left(\forall_{k<n}\text{
}x(k)=y(k)\right)\wedge\left(x(n)=2,y(n)=3\right)\wedge\left(\forall_{k>n}\text{
}x(k)=y(k)\in\\{0,1\\}\right)\right]$

  2. (ii)ω

$\left(x(0)=0,y(0)=1\right)\wedge\left(\forall_{k\geq 1}\text{
}x(k)=y(k)\in\\{0,1\\}\right)$

Note that this is a partial order. Let $P_{n}:=\\{0,1,2,3\\}^{n}$ for
$n\in\mathbb{N}$. We define a partial ordering $x\leq_{n}y\Leftrightarrow x=y$
or one of the following conditions is met:

  1. (i)n

$\exists_{l<n}\left[\left(\forall_{k<l}\text{
}x(k)=y(k)\right)\wedge\left(x(l)=2,y(l)=3\right)\wedge\left(\forall_{k\in(l,n)}\text{
}x(k)=y(k)\in\\{0,1\\}\right)\right]$

  2. (ii)n

$\left(x(0)=0,y(0)=1\right)\wedge\left(\forall_{k\in[1,n)}\text{
}x(k)=y(k)\in\\{0,1\\}\right)$

The fact that two distinct $x,y\in P_{n}$ are in the order implies that they
are equal at all but one coordinate. Moreover, there are no isolated points
with respect to the order – for any $x\in P_{n}$ we find distinct $y\in P_{n}$
comparable with $x$ by changing one appropriate coordinate of $x$. Therefore
$P_{n}$ is isomorphic to $\dot{\bigcup}_{i<2\cdot 4^{n-1}}\mathbf{2}$. We
define $p_{n}:P_{n+1}\to P_{n}$ such that
$p_{n}(x):=x_{|_{n}}=(x(0),\dots,x(n-1))$ for $n\in\mathbb{N}$. Let us observe
the following:

######  Lemma 3.4.

$p_{n}$ is a quotient map for all $n\in\mathbb{N}$.

###### Proof.

Firstly we show that it is a homomorphism. Let $x\leq_{n+1}y,x\neq y$. We
consider three cases.

  1. (i)

If $x_{|_{n}}=y_{|_{n}}$ then $p_{n}(x)=x_{|_{n}}\leq_{n}y_{|_{n}}=p_{n}(y)$

  2. (ii)

If $x(0)=0,y(0)=1$ and $\forall_{k\in[1,n+1)}\text{ }x(k)=y(k)\in\\{0,1\\}$
then $p_{n}(x)=x_{|_{n}}\leq_{n}y_{|_{n}}=p_{n}(y)$

  3. (iii)

If $x(l)=2,y(l)=3\text{ for some }l<n+1$ then $\forall_{k<l}\text{
}x(k)=y(k)\text{ and }\forall_{k\in(l,n+1)}\text{ }x(k)=y(k)\in\\{0,1\\}$, so
$p_{n}(x)=x_{|_{n}}\leq_{n}y_{|_{n}}=p_{n}(y)$.

Thus $p_{n}$ is a homomorphism. Since $p_{n}$ is onto, we only need to show
that it is a quotient map. Let $a\leq_{n}b$ and $a,b\in P_{n}$. We define
$a^{\prime}:=(a(0),\ldots,a(n-1),0),b^{\prime}:=(b(0),\ldots,b(n-1),0)$. Then
$a^{\prime}\leq_{n+1}b^{\prime}$ and
$p_{n}(a^{\prime})=a,p_{n}(b^{\prime})=b$. ∎

If $n>k$ then $p_{k}^{n}=p_{k}\circ p_{k+1}\circ\ldots\circ p_{n-1}$ is the
restriction to the first $k$ coordinates. Clearly $p_{k}^{n+1}=p_{k}^{n}\circ
p_{n}$. We consider the sequence of posets

${P_{1}}$${P_{2}}$${P_{3}}$${\ldots}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\scriptstyle{p_{3}}$
(5)

Note that, if we show that
$\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p_{k}^{n}\\}_{k<n}\rangle$ is a
Fraïssé sequence in category of finite posets, then the inverse limit
$\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p_{k}^{n}\\}_{k<n}\rangle$
is the inverse Fraïssé limit in this category. To show that $\mathbb{P}$ is
topologically isomorphic to the inverse limit of $P_{n}$’s we must show the
following.

######  Lemma 3.5.

$x\leq y$ if and only if $p_{n}(x)\leq_{n}p_{n}(y)\text{ for all
}n\in\mathbb{N}$.

###### Proof.

The only if part of equivalence is immediate. So let us assume that
$p_{n}(x)\leq_{n}p_{n}(y)\text{ for all }n\in\mathbb{N}$. If
$p_{n}(x)=p_{n}(y)\text{ for all }n\in\mathbb{N}$ then $x=y$. Otherwise we
choose $m:=\min\\{k\in\mathbb{N}:p_{k}(x)\neq p_{k}(y)\\}$ and consider the
following cases:

  1. (i)

If $m>1$ then for all $j<m-1$ we have $x(j)=y(j)$, $x(m-1)=2,y(m-1)=3$ and for
all $j>m-1$ we have $x(j)=y(j)\in\\{0,1\\}$.

  2. (ii)

If $m=1$ then $x(0)=0,y(0)=1$ and for all $j\geq 1$ we have
$x(j)=y(j)\in\\{0,1\\}$ or $x(0)=2,y(0)=3$ and for all $j\geq 1$ we have
$x(j)=y(j)\in\\{0,1\\}$.

In both cases $x\leq y$. ∎

######  Theorem 3.6.

Let
$\phi:\mathbb{P}\to\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$
be defined as follows

$\phi(x):=(x_{|1},x_{|2},\dots)$

Then $\phi$ is topological isomorphism.

###### Proof.

Firstly, we will show that $\phi$ is one-to-one. Take
$x=(x_{0},x_{1},\dots),y=(y_{0},y_{1},\dots)\in\mathbb{P}$ and assume
$\phi(x)=\phi(y)$. Then $x_{|n}=y_{|n}$ for each $n\in\mathbb{N}$, which means
that $x=y$ and consequently $\phi$ is one-to-one.

Now we show that $\phi$ is onto. Let
$\mathbf{a}=(a_{1},a_{2},\dots)\in\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$.
Then $p_{n}(a_{n+1})=a_{n+1}|_{n}=a_{n}$. Let $x_{n}=a_{n+1}(n)$ and
$x=(x_{0},x_{1},\dots)$. Then $\mathbf{a}=(x_{|1},x_{|2},\dots)$ and
$\phi(x)=\mathbf{a}$.

Now we will show that $\phi$ is continuous. Take any basic set

$(\\{x_{|1}\\}\times\\{x_{|2}\\}\times\dots\times\\{x_{|n+1}\\}\times
P_{n+1}\times\dots)\cap\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle.$

Then

$\phi^{-1}((\\{x_{|1}\\}\times\\{x_{|2}\\}\times\dots\times\\{x_{|n+1}\\}\times
P_{n+1}\times\dots)\cap\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle)=$
$\\{x_{0}\\}\times\\{x_{1}\\}\times\dots\times\\{x_{n}\\}\times\\{0,1,2,3\\}\times\dots,$

which means $\phi$ is continuous. Since $\mathbb{P}$ is compact, $\phi$ is a
homeomorphism. By Lemma 3.5 $\phi$ is a topological isomorphism. ∎

Our next step is to show that sequence (5) is a Fraïssé sequence.

######  Proposition 3.7.

A sequence (5) is a Fraïssé sequence in the category of finite posets with
quotient maps.

###### Proof.

From Lemma 2.14 we get that, if our sequence has (A) property, then it also
has (U) property as $\mathfrak{K}$ is directed. So we only need to show that
our sequence has (A) property.

Let $H$ be a finite poset and let $p:H\rightarrow P_{k}$ be a quotient map.
Note that for every $x\in P_{k}$ there is only one $y\in P_{k}$ such that
$x\leq_{k}y$ and $x\neq y$. Since $P_{n}$ is isomorphic to
$\dot{\bigcup}_{i<2\cdot 4^{n-1}}\mathbf{2}$ for all $n\in\mathbb{N}$, then
the cardinality of $\\{(x,y)\in P_{k}\times P_{k}:x<_{k}y\\}$ equals $2\cdot
4^{k-1}$. Let $D_{i}:=\\{x_{i},y_{i}\\}$ be such that $x_{i}<_{k}y_{i}$ for
$i=1,\dots,2\cdot 4^{k-1}$. Let $m>k$. Then for every $(x,y)$ with $x<_{k}y$
we have the following:

  1. 1.

We have at least $2^{m-k-1}$ pairs $(u,v)$ with $u<_{m}v$ and such that
$p_{k}^{m}(u)=p_{k}^{m}(v)=x$. Indeed if we take pairs $(u,v)$ of the form
$u=x\hat{\;\;}2\hat{\;\;}t$ and $v=x\hat{\;\;}3\hat{\;\;}t$ where $t$ is 0-1
sequence of the length $m-k-1$, we get that $p_{k}^{m}(u)=p_{k}^{m}(v)=x$ and
there is $2^{m-k-1}$ of such pairs. The same applies if we substitute $x$ with
$y$.

  2. 2.

We have $2^{m-k}$ pairs $(u,v)$ with $u<_{m}v$ and such that
$p_{k}^{m}(u)=x,p_{k}^{m}(v)=y$; they are of the form $u=x\hat{\;\;}t$ and
$v=y\hat{\;\;}t$ where $t$ is 0-1 sequence of the length $m-k$.

Additionally we will show that $(p_{k}^{m})^{-1}(\\{x,y\\})$ does not contain
isolated points. Let $v\in(p_{k}^{m})^{-1}(\\{x,y\\})$. Since $P_{m}$ does not
contain isolated points, there is $w$ with $w<_{m}v$ or $v<_{m}w$. Since
$p_{k}^{m}$ is order-preserving, $p_{k}^{m}(w)$ is comparable with
$p_{k}^{m}(v)\in\\{x,y\\}$. Thus $p_{k}^{m}(w)\in\\{x,y\\}$, and therefore $v$
is not isolated in $(p_{k}^{m})^{-1}(\\{x,y\\})$.

Now we fix $m$ such that $2^{m-k-1}$ is greater than the cardinality of
$\\{(x,y):x\leq_{H}y\\}$. For every $i\leq 2\cdot 4^{k-1}$ we will find a
quotient $g_{i}:(p_{k}^{m})^{-1}\left(D_{i}\right)\rightarrow
p^{-1}\left(D_{i}\right)$ such that diagram

${D_{i}}$${(p_{k}^{m})^{-1}\left(D_{i}\right)}$${p^{-1}\left(D_{i}\right)}$$\scriptstyle{p_{k}^{m}}$$\scriptstyle{g_{i}}$$\scriptstyle{p}$
(6)

commutes.

Suppose that we have already found such $g_{i}$’s. We define
$g:P_{m}\rightarrow H$ as follows: $g(x)=g_{i}(x)$ provided
$x\in(p_{k}^{m})^{-1}\left(D_{i}\right)$. Since $D_{i}\cap D_{j}=\emptyset$
for $i\neq j$, $g$ is well-defined. Let $y\in H$. Then $p(y)\in
P_{k}=\bigcup_{i}D_{i}$, thus there is $i$ with $p(y)\in D_{i}$, which implies
$y\in p^{-1}(D_{i})$. Since $g_{i}:(p_{k}^{m})^{-1}(D_{i})\to p^{-1}(D_{i})$
is a quotient map we can find $x\in(p_{k}^{m})^{-1}(D_{i})$ with $g_{i}(x)=y$.
Then $g(x)=g_{i}(x)=y$, thus $g$ is onto. Assume that $x\leq_{m}x^{\prime}$.
Then $p^{m}_{k}(x)\leq_{k}p^{m}_{k}(x^{\prime})$ and
$p^{m}_{k}(x),p^{m}_{k}(x^{\prime})$ belong to the same $D_{i}$. Since $g_{i}$
is homomorphism, then $g(x)=g_{i}(x)\leq_{H}g_{i}(x^{\prime})=g(x^{\prime})$
and $g$ is homomorphism as well. Let $u,u^{\prime}\in P_{k}$ with
$u<_{k}u^{\prime}$. There is $i$ with $D_{i}=\\{u,u^{\prime}\\}$. Since $p$ is
a quotient map, there are $y,y^{\prime}\in p^{-1}(D_{i})$ with
$y<_{H}y^{\prime}$ and $p(y)=u$, $p(y^{\prime})=u^{\prime}$. Since $g_{i}$ is
a quotient map, there are $x<_{m}x^{\prime}$ with $g_{i}(x)=u$ and
$g_{i}(x^{\prime})=u^{\prime}$, and $g$ is a quotient map as well. By (6) the
diagram

${P_{k}}$${P_{m}}$${H}$$\scriptstyle{p_{k}^{m}}$$\scriptstyle{g}$$\scriptstyle{p}$
(7)

commutes.

Now it is enough to construct $g_{i}$’s. Let $D_{i}=\\{L,U\\}$ with $L<U$. Now
we will categorize points and edges in $p^{-1}(D_{i})\subseteq H$. Isolated
points of $p^{-1}(D_{i})$ we divide into L-isolated (denoted by $LI$) and
U-isolated (denoted by $UI$) if $p$ maps them onto $L$ and $U$ respectively.
Further, pairs $(u,v)$ with $u<_{H}v$ in $p^{-1}(D_{i})$ are divided into $LP$
if $p(u)=p(v)=L$, $UP$ if $p(u)=p(v)=U$ and $LUP$ if $p(u)=L,p(v)=U$. Since
$p$ is a quotient map then $LUP\neq\emptyset$. Let us fix $r,b\in
p^{-1}(D_{i})$ such that $(r,b)\in LUP$.

Now we will do a similar thing for $(p_{k}^{m})^{-1}\left(D_{i}\right)$. We
have already noticed that it does not contain any isolated points. We divide
its pairs $(x,y)$ with $x<_{m}y$ into three categories: L-type $\mathbf{L}$ if
$p_{k}^{m}$ maps $x$ and $y$ into $L$; U-type $\mathbf{U}$ if $p_{k}^{m}$ maps
$x$ and $y$ into $U$; LU-type $\mathbf{LU}$ if $p_{k}^{m}$ maps $x$ into $L$
and $y$ into $U$.

Let $\ell_{1}=|LI|,\ell_{2}=|UI|,\ell_{3}=|LP|,\ell_{4}=|UP|,\ell_{5}=|LUP|$.
Then $\ell_{1},\ell_{2},\ell_{3},\ell_{4}\geq 0$ and $\ell_{5}\geq 1$. By
definition of $m$, $\ell_{1}$ and $\ell_{3}$ we obtain $|\mathbf{L}|\geq
2^{m-k-1}\geq|\\{(x,y):x\leq_{H}y\\}|$. Note that we can identify the set $LI$
with set of pairs $\\{(x,x):x\in LI\\}$ and this set is disjoint with $LUP$,
and therefore $|\\{(x,y):x\leq_{H}y\\}|\geq\ell_{1}+\ell_{3}$. So we can find
pairwise disjoint $\mathbf{L}_{1},\mathbf{L}_{2},\mathbf{L}_{3}$ such that
$\mathbf{L}_{1}\cup\mathbf{L}_{2}\cup\mathbf{L}_{3}=\mathbf{L}$ and
$|\mathbf{L}_{1}|=\ell_{1},|\mathbf{L}_{2}|=\ell_{3}$. Similarly we divide
$\mathbf{U}$ into pairwise disjoint sets
$\mathbf{U}_{1},\mathbf{U}_{2},\mathbf{U}_{3}$ such that
$\mathbf{U}_{1}\cup\mathbf{U}_{2}\cup\mathbf{U}_{3}=\mathbf{U}$ and
$|\mathbf{U}_{1}|=\ell_{2},|\mathbf{U}_{2}|=\ell_{4}$. Now we partition
$\mathbf{LU}$ into $\mathbf{LU}_{1},\mathbf{LU}_{2}$ with
$|\mathbf{LU}_{1}|=\ell_{5}$. Now we are ready to define $g_{i}$. Since
$P_{m}$ is isomorphic to $\dot{\bigcup}_{i<2\cdot 4^{m-1}}\mathbf{2}$, then we
will define $g_{i}$ for each pair $x,y\in P_{m}$ with $x<_{m}y$ as follows :

  * (1)

if $(x,y)$ is a $j$-th pair in $\mathbf{L}_{1}$, then $g_{i}(x)=g_{i}(y)$ is
the $j$-th element of $LI$;

  * (2)

if $(x,y)$ is a $j$-th pair in $\mathbf{L}_{2}$, then $(g_{i}(x),g_{i}(y))$ is
the $j$-th pair of $LP$;

  * (3)

if $(x,y)$ is in $\mathbf{L}_{3}$, then $g_{i}(x)=g_{i}(y)=r$ (recall that
$r,b\in p^{-1}(D_{i})$ have been fixed such that $(r,b)\in LUP$);

  * (4)

if $(x,y)$ is a $j$-th pair in $\mathbf{U}_{1}$, then $g_{i}(x)=g_{i}(y)$ is
the $j$-th element of $UI$;

  * (5)

if $(x,y)$ is a $j$-th pair in $\mathbf{U}_{2}$, then $(g_{i}(x),g_{i}(y))$ is
the $j$-th pair of $UP$;

  * (6)

if $(x,y)$ is a $j$-th pair in $\mathbf{U}_{3}$, then $g_{i}(x)=g_{i}(y)=b$;

  * (7)

if $(x,y)$ is a $j$-th pair in $\mathbf{LU}_{1}$, then $(g_{i}(x),g_{i}(y))$
is a $j$-th pair in $LUP$;

  * (8)

if $(x,y)$ is in $\mathbf{LU}_{2}$, then $(g_{i}(x),g_{i}(y))=(r,b)$.

By (1), (2), (4), (5) and (7) $g_{i}$ is onto $p^{-1}(D_{i})$. Fix
$(x,y)\in(p^{m}_{k})^{-1}(D_{i})\times(p^{m}_{k})^{-1}(D_{i})$ with $x<_{m}y$.
If $p^{m}_{k}(x)=p^{m}_{k}(y)=L$, then $p(g_{i}(x))=p(g_{i}(y))=L$ by (1)–(3).
If $p^{m}_{k}(x)=p^{m}_{k}(y)=U$, then $p(g_{i}(x))=p(g_{i}(y))=U$ by (4)–(6).
If $p^{m}_{k}(x)=L$ and $p^{m}_{k}(y)=U$, then $p(g_{i}(x))=L$ and
$p(g_{i}(y))=U$ by (7)–(8). This shows that diagram (6) commutes. The fact
that $g_{i}$ is a homomorphism follows easily from (1)-(8). Note that $g_{i}$
is quotient map by (2), (5) and (7). ∎

###  3.3 Properties of the inverse Fraïssé limit $\mathbb{P}$

Now we shall proof that in fact $\mathbb{P}$ is universal, in other words
every profinite poset is an image of some continuous quotient mapping from
$\mathbb{P}$. To achieve this we need to proof the following lemma.

######  Lemma 3.8.

Let $(G_{i})_{i\in\omega}$ be a sequence of non-empty, finite sets and
$q_{k}^{i}:G_{i}\rightarrow G_{k}$ be functions such that $q^{k}_{\ell}\circ
q^{i}_{k}=q^{i}_{\ell}$ for every $\ell<k<i$. We define

$T_{i}:=\\{(a_{0},a_{1},\ldots,a_{i})\in G_{0}\times G_{1}\times\ldots\times
G_{i}:q_{k}^{i}(a_{i})=a_{k}\text{ for every }k<i\\}$

for $i\in\omega$. If $T_{i}$ is non-empty for every $i\in\omega$ then there
exists a sequence $(x_{n})\in\prod_{i\in\omega}G_{i}$ such that
$q_{n}^{n+1}(x_{n+1})=x_{n}$ for all $n\in\omega$.

###### Proof.

Let us define $p_{k}^{i}(a_{0},a_{1},\ldots,a_{i})=(a_{0},a_{1},\ldots,a_{k})$
for $k\leq i$. Note that the condition $q^{i}_{k}(a_{i})=a_{k}$ for every
$k<i$ implies $q^{k}_{\ell}(a_{k})=a_{\ell}$ for every $\ell<k<i$. Therefore
$p_{k}^{i}\left[T_{i}\right]\subseteq T_{k}$ for all $i\in\omega$ and $k<i$.
Thus
$p_{0}^{i+1}\left[T_{i+1}\right]=p_{0}^{i}\left[p_{i}^{i+1}\left[T_{i+1}\right]\right]\subseteq
p_{0}^{i}\left[T_{i}\right]$ for every $i$. Furthermore as $T_{i}$ are finite
non-empty sets then $p_{0}^{i}\left[T_{i}\right]$ are finite and not-empty.
Because of that $\left(p_{0}^{i}\left[T_{i}\right]\right)_{i\in\mathbb{N}}$ is
a decreasing sequence of non-empty finite sets. Eventually this sequence must
stabilize so
$\bigcap_{i\in\mathbb{N}}p_{0}^{i}\left[T_{i}\right]=p_{0}^{\ell}\left[T_{\ell}\right]\neq\emptyset$
for some $\ell\in\omega$. Similarly
$\bigcap_{i\in\mathbb{N}}p_{k}^{i+k}\left[T_{i+k}\right]\neq\emptyset$ for all
$k\in\omega$. Let
$V_{k}:=\bigcap_{i\in\mathbb{N}}p_{k}^{i+k}\left[T_{i+k}\right]$ for
$k\in\omega$. Note that $p_{k}^{i}\left[V_{i}\right]=V_{k}$ for every $k\leq
i$. Now we are ready to define the desired sequence.

Let $x_{0}\in G_{0}$ such that $x_{0}\in V_{0}$. As
$p_{0}^{1}\left[V_{1}\right]=V_{0}$ we find $x_{1}\in G_{1}$ such that
$(x_{0},x_{1})\in V_{1}$, thus $q_{0}^{1}(x_{1})=x_{0}$. In general we can
assume that we have $(x_{0},x_{1}\ldots,x_{n})$ such that
$q_{k}^{k+1}(x_{k+1})=x_{k}$ for $k<n$. We can find $x_{n+1}\in G_{n+1}$ such
that $(x_{0},x_{1},\ldots,x_{n+1})\in V_{n+1}$ because
$p_{n}^{n+1}\left[V_{n+1}\right]=V_{n}$. Therefore
$q_{n}^{n+1}(x_{n+1})=x_{n}$. So we can find a sequence
$(x_{0},x_{1},\ldots)\in\prod_{i\in\omega}G_{i}$ such that
$q_{n}^{n+1}(x_{n+1})=x_{n}$ for every $n\in\omega$. ∎

In sequel we will use the following simple observation.

######  Observation 3.9.

Note that if the following diagrams

${A}$${B}$${B}$${E}$${C}$${D}$${D}$${F}$

commute, then diagram

${A}$${E}$${C}$${F}$

commutes as well.

The following Theorem follows from [9, Theorem 5.2]. However, the proof there
is very short and sketchy, with general model-theoretical assumptions. Anyone
who wants to really understand this theory needs to complete the reasoning.
Since we did not find detailed argument anywhere, we decided to present it
here. This can be adapted to general situation.

######  Theorem 3.10.

Let
$H=\varprojlim\langle\\{H_{n}\\}_{n\in\mathbb{N}},\\{h^{n}_{k}\\}_{k<n}\rangle$
and
$\mathbb{P}=\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$,
where $(P_{n})$ is a Fraïssé sequence. Then there is a continuous quotient map
$f:\mathbb{P}\rightarrow H$.

###### Proof.

We will define such quotient map using the fact that $\mathbb{P}$ is an
inverse limit of Fraïssé sequence
$\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$. Because of
(U) property we shall find quotient $f_{1}:P_{i_{1}}\rightarrow H_{1}$. Now
using (E) property we will find such $P_{i_{2}},i_{1}<i_{2}$ and
$f_{2}:P_{i_{2}}\to H_{2}$ that the diagram

${P_{i_{1}}}$${P_{i_{2}}}$${H_{1}}$${H_{2}}$$\scriptstyle{f_{1}}$$\scriptstyle{p_{i_{1}}^{i_{2}}}$$\scriptstyle{f_{2}}$$\scriptstyle{h_{1}^{2}}$
(8)

commutes. Now we can apply (E) property to $f_{2}$ and $h_{2}^{3}$ to get next
commuting diagram. Proceeding inductively we can apply this to $f_{n}$ and
$h_{n}^{n+1}$ to acquire respective commuting diagrams. Using Observation 3.9
we get that for every $n,k\in\mathbb{N},k<n$ the diagram

${P_{i_{k}}}$${P_{i_{n}}}$${H_{k}}$${H_{n}}$$\scriptstyle{f_{k}}$$\scriptstyle{p_{i_{k}}^{i_{n}}}$$\scriptstyle{f_{n}}$$\scriptstyle{h_{k}^{n}}$
(9)

commutes. Let $f:\mathbb{P}\to H$ be such that
$f(x_{1},x_{2},\ldots)=(f_{1}(x_{i_{1}}),f_{2}(x_{i_{2}}),\ldots)$. We shall
prove that it is a continuous quotient mapping onto $H$. It is easy to see
that for every $n\in\mathbb{N}$ we have
$h_{n}^{n+1}(f_{n+1}(x_{i_{n+1}}))=f_{n}(p_{i_{n}}^{i_{n+1}}(x_{i_{n+1}}))=f_{n}(x_{i_{n}})$.
So $f(x_{1},x_{2},\ldots)\in H$ for every $(x_{1},x_{2},\ldots)\in\mathbb{P}$.
We now show that $f$ is a homomorphism. Take
$(x_{1},\ldots)\leq_{\mathbb{P}}(y_{1},\ldots)$. Then for every
$n\in\mathbb{N}$ we have that $x_{i_{n}}\leq_{P_{i_{n}}}y_{i_{n}}$ so
$f_{n}(x_{i_{n}})\leq_{H_{n}}f_{n}(y_{i_{n}})$. Thus
$f(x)=(f_{1}(x_{i_{1}}),\ldots)\leq_{H}(f_{1}(y_{i_{1}}),\ldots)=f(y)$.

Now we will show that $f$ is onto $H$. Take $(z_{1},z_{2},\ldots)$ in $H$. It
is easy to see that $(f_{n}^{-1}\left[\\{z_{n}\\}\right])_{n\in\mathbb{N}}$ is
a sequence of non-empty finite sets as $P_{i_{n}}$ are finite and $f_{n}$ are
quotient maps. Let
$M:=\prod_{n\in\mathbb{N}}f_{n}^{-1}\left[\\{z_{n}\\}\right]$ and

$T_{l}:=\\{(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{l}})\in
f_{1}^{-1}\left[\\{z_{1}\\}\right]\times
f_{2}^{-1}\left[\\{z_{2}\\}\right]\times\ldots\times
f_{l}^{-1}\left[\\{z_{l}\\}\right]:p_{i_{k}}^{i_{l}}(x_{i_{l}})=x_{i_{k}}\text{
for }k<l\\}.$

We will show that $T_{l}$ is non-empty for every $l>1$. Take $x_{i_{l}}\in
f_{l}^{-1}\left[\\{z_{l}\\}\right]$. Then
$z_{l-1}=h_{l-1}^{l}(f_{l}(x_{i_{l}}))=f_{l-1}(p_{i_{l-1}}^{i_{l}}(x_{i_{l}}))$.
Thus $x_{i_{l-1}}:=p_{i_{l-1}}^{i_{l}}(x_{i_{l}})\in
f_{l-1}^{-1}\left[\\{z_{l-1}\\}\right]$. As diagram (9) commutes we can
continue this process to get a finite sequence in $T_{l}$ which makes it non-
empty. This also shows that $p_{i_{k}}^{i_{n}}$ maps
$f_{n}^{-1}\left[\\{z_{n}\\}\right]$ to $f_{k}^{-1}\left[\\{z_{k}\\}\right]$
for $k<n$. Furthermore $p_{i_{k}}^{i_{n}}\circ
p_{i_{n}}^{i_{l}}=p_{i_{k}}^{i_{l}}$ for $k<n<l$. Because of that we can use
Lemma 3.8 to get a sequence $(x_{i_{1}},x_{i_{2}},\ldots)\in M$ such that
$p_{i_{n}}^{i_{n+1}}(x_{i_{n+1}})=x_{i_{n}}$ for every $n\in\mathbb{N}$. Put
$x^{\prime}_{n}:=p_{n}^{i_{n}}(x_{i_{n}})$. Note that
$x^{\prime}_{i_{n}}=x_{i_{n}}$. For that reason we will write $x_{n}$ instead
of $x_{n}^{\prime}$, which will not lead to any confusion. Then
$f(x_{1},x_{2},\ldots)=(f_{1}(x_{i_{1}}),f_{2}(x_{i_{2}}),\ldots)=(z_{1},z_{2},\ldots)$
and
$p_{n}^{n+1}(x_{n+1})=p_{n}^{n+1}(p_{n+1}^{i_{n+1}}(x_{i_{n+1}}))=p_{n}^{i_{n}}(p_{i_{n}}^{i_{n+1}}(x_{i_{n+1}}))=p_{n}^{i_{n}}(x_{i_{n}})=x_{n}$
so $(x_{1},x_{2},\ldots)\in\mathbb{P}$.

Now we show that $f$ is a strict homomorphism. Let
$(u_{1},u_{2},\ldots),(v_{1},v_{2},\ldots)\in H$ such that
$(u_{1},u_{2},\ldots)\leq_{H}(v_{1},v_{2},\ldots)$. Then
$u_{n}\leq_{H_{n}}v_{n}$ for every $n\in\mathbb{N}$. Because $f_{n}$ is a
quotient map the sets $A_{n}:=\\{(x,y)\in P_{i_{n}}\times
P_{i_{n}}:x\leq_{P_{i_{n}}}y,f_{n}(x)=u_{n},f_{n}(y)=v_{n}\\}$ are non-empty.
Now for $n\in\mathbb{N}$, $(x_{i_{n}},y_{i_{n}})\in A_{n}$ and $k<n$ we define
$r_{k}^{n}(x_{i_{n}},y_{i_{n}}):=(p_{i_{k}}^{i_{n}}(x_{i_{n}}),p_{i_{k}}^{i_{n}}(y_{i_{n}}))$.
See that $(A_{i})_{i\in\mathbb{N}}$ is a sequence of non-empty, finite sets
and $r_{k}^{n}\circ r_{n}^{l}=r_{k}^{l}$ for $k<n<l$. We define

$E_{n}:=\\{((x_{i_{1}},y_{i_{1}}),(x_{i_{2}},y_{i_{2}}),\ldots,((x_{i_{n}},y_{i_{n}}))\in
A_{1}\times A_{2}\times\ldots\times
A_{n}:r_{k}^{n}(x_{i_{n}},y_{i_{n}})=(x_{i_{k}},y_{i_{k}})\\}.$

We will show that $\operatorname{rng}\left(r_{k}^{n}\right)\subset A_{k}$ for
any $n\in\mathbb{N}$ and $k<n$. Take any $n\in\mathbb{N},k<n$ and
$(x_{i_{n}},y_{i_{n}})\in A_{n}$. Because the diagrams (9) commute we get that
$f_{k}(p_{i_{k}}^{i_{n}}(x_{i_{n}}))=u_{k},f_{k}(p_{i_{k}}^{i_{n}}(y_{i_{n}}))=v_{k}$
and $p_{i_{k}}^{i_{n}}(x_{i_{n}})\leq_{P_{i_{k}}}p_{i_{k}}^{i_{n}}(y_{i_{n}})$
so $r_{k}^{n}(x_{i_{n}},y_{i_{n}})\in A_{k}$ for any $n\in\mathbb{N}$ and
$k<n$, thus $r_{k}^{n}$ maps $A_{n}$ to $A_{k}$ for $k<n,n\in\mathbb{N}$.
Furthermore this shows that $E_{n}$ is non-empty for every $n\in\mathbb{N}$.

Thus we can again apply Lemma 3.8 to get a sequence
$(x_{i_{n}},y_{i_{n}})_{n\in\mathbb{N}}\in\prod_{n\in\mathbb{N}}A_{n}$ such
that $r_{n}^{n+1}(x_{i_{n+1}},y_{i_{n+1}})=(x_{i_{n}},y_{i_{n}})$. So we have
got $(x_{i_{n}}),(y_{i_{n}})$ such that

$p_{i_{n}}^{i_{n+1}}(x_{i_{n+1}})=x_{i_{n}},p_{i_{n}}^{i_{n+1}}(y_{i_{n+1}})=y_{i_{n}},x_{i_{n}}\leq_{P_{i_{n}}}y_{i_{n}}\text{
and }f_{n}(x_{i_{n}})=u_{n},f_{n}(y_{i_{n}})=v_{n}.$

Now take
$x^{\prime}_{n}:=p_{n}^{i_{n}}(x_{i_{n}}),y^{\prime}_{n}:=p_{n}^{i_{n}}(y_{i_{n}})$.
Note that $x^{\prime}_{i_{n}}=x_{i_{n}}$ and $y^{\prime}_{i_{n}}=y_{i_{n}}$.
For that reason we will write $x_{n}$ and $y_{n}$ instead of $x^{\prime}_{n}$
and $y^{\prime}_{n}$, which will not lead to any confusion. Then

$f(x_{1},x_{2},\ldots)=(f_{1}(x_{i_{1}}),f_{2}(x_{i_{2}}),\ldots)=(u_{1},u_{2},\ldots),$
$f(y_{1},y_{2},\ldots)=(f_{1}(y_{i_{1}}),f_{2}(y_{i_{2}}),\ldots)=(v_{1},v_{2},\ldots),(x_{1},x_{2}\ldots)\leq_{\mathbb{P}}(y_{1},y_{2},\ldots).$

Furthermore for every $n\in\mathbb{N}$,
$p_{n}^{n+1}(x_{n+1})=p_{n}^{n+1}(p_{n+1}^{i_{n+1}}(x_{i_{n+1}})=p_{n}^{i_{n}}(p_{i_{n}}^{i_{n+1}}(x_{i_{n+1}})=p_{n}^{i_{n}}(x_{i_{n}})=x_{n}$.
The same applies to $y_{n}$. Thus $f$ is a strict homomorphism.

Now we will show that $f$ is continuous. Take any basic set
$U_{(z_{1},z_{2},\ldots,z_{n})}=(\\{z_{1}\\}\times\\{z_{2}\\}\times\ldots\times\\{z_{n}\\}\times
H_{n+1}\times\ldots)\cap H$. Let $(x_{1},x_{2},\dots)\in
f^{-1}\left[U_{(z_{1},z_{2},\ldots,z_{n})}\right]$. Then
$f_{1}(x_{i_{1}})=z_{1},\dots,f_{n}(x_{i_{n}})=z_{n}$. See that
$(x_{1},x_{2},\dots)\in(\\{x_{1}\\}\times\\{x_{2}\\}\times\dots\times\\{x_{i_{1}}\\}\times\dots\times\\{x_{i_{n}}\\}\times
P_{i_{n}+1}\times\dots)\cap\mathbb{P}\subset
f^{-1}\left[U_{(z_{1},z_{2},\ldots,z_{n})}\right]$ which means
$f^{-1}\left[U_{(z_{1},z_{2},\ldots,z_{n})}\right]$ is open, thus $f$ is
continuous. ∎

######  Remark 3.11.

The representation of a function $f$ in Theorem 3.10 is worth remembering, so
we remind it for reader’s convenience. Let $H$ a profinite poset generated by
a sequence $H_{n}$ and quotient mappings $h_{k}^{n}:H_{n}\to H_{k}$. There is
a sequence $i_{1}<i_{2}<\dots$ and quotient mappings $f_{n}:P_{i_{n}}\to
H_{n}$ such that the diagram

${P_{i_{n}}}$${P_{i_{k}}}$${H_{n}}$${H_{k}}$$\scriptstyle{f_{n}}$$\scriptstyle{p_{i_{n}}^{i_{k}}}$$\scriptstyle{f_{k}}$$\scriptstyle{h_{n}^{k}}$

commutes for every $n,k\in\mathbb{N},n<k$. Then we define a continuous
quotient mapping $f:\mathbb{P}\to H$ as follows
$f(x_{1},x_{2},\dots)=(f_{1}(x_{i_{1}}),f_{2}(x_{i_{2}}),\dots)$.

Now we will show that similarly all continuous quotient maps $f:\mathbb{P}\to
H$ where $H$ is a profinite poset are of this form.

######  Lemma 3.12.

Let $f:P\to L$ be a continuous quotient mapping where $L$ is a finite poset
and
$P=\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$
is a profinite poset. Let $i_{0}\in\mathbb{N}$. Then we can find $i>i_{0}$ and
quotient mapping $h:P_{i}\to L$ such that $f=h\circ p_{i}$ where $p_{i}:P\to
P_{i}$ is defined as follows: $p_{i}(x_{1},x_{2},\dots)=x_{i}$.

###### Proof.

We remind that the non-empty basic sets in $P$ have the following
representation:

$(\\{x_{1}\\}\times\\{x_{2}\\}\times\dots\times\\{x_{k}\\}\times P_{k+1}\times
P_{k+2}\dots)\cap P$

where $p_{j}^{n}(x_{n})=x_{j}$ for $j<n\leq k$.

We will say that this basic set has a rank $k$. As $P_{k}$ are all finite then
every basic set of rank $k$ is a finite disjoint union of some basic sets of
rank $k+1$.

Let $L=\\{l_{1},l_{2},\dots,l_{n}\\}$. As $L$ has a discrete topology, we get
that $f^{-1}[\\{l_{j}\\}]$ is a compact set as it is a closed subset of a
compact space. Take any open cover of $f^{-1}[\\{l_{j}\\}]$ consisting of
basic sets and take its finite sub-cover $\mathcal{V}_{j}$. Since
$f^{-1}[\\{l_{j}\\}]$ is open, we may assume that
$\bigcup\mathcal{V}_{j}=f^{-1}[\\{l_{j}\\}]$. Without the loss of generality
we can assume that all such sub-covers consist of sets with the same rank
greater than $i_{0}$, say $i$. From now on the rank $i$ is fixed.

Let

$(\\{x_{1}\\}\times\\{x_{2}\\}\times\dots\times\\{x_{i}\\}\times P_{i+1}\times
P_{i+2}\dots)\cap P$ (10)

be a basic set of rank $i$. Clearly for all non-empty sets
$(\\{x^{\prime}_{1}\\}\times\\{x^{\prime}_{2}\\}\times\dots\times\\{x^{\prime}_{i-1}\\}\times\\{x_{i}\\}\times
P_{i+1}\times P_{i+2}\dots)\cap P$ we have that $x^{\prime}_{j}=x_{j}$ for
$j<i$ as $p_{j}^{i}(x_{i})=x_{j}$. So each set of the form (10) is fully
described by $x_{i}$.

We define $h:P_{i}\to L$ as follows: $h(x_{i})=l$ if and only if
$(\\{p_{1}^{i}(x_{i})\\}\times\\{p_{2}^{i}(x_{i})\\}\times\dots\times\\{x_{i}\\}\times
P_{i+1}\times P_{i+2}\times\dots)\cap P\subset f^{-1}[\\{l\\}]$. By the
definition of $i$, each non-empty basic set of rank $i$ is contained in
exactly one fiber $f^{-1}[\\{l\\}]$, which shows that $h$ is well-defined.

Now we prove that $f=h\circ p_{i}$. Take any $(x_{1},x_{2},x_{3},\dots)\in P$.
Then $f(x_{1},\dots,x_{i},x_{i+1},\dots)=l_{j}$ for some $l_{j}\in L$. Clearly
$(\\{x_{1}\\}\times\\{x_{2}\\}\times\dots\times\\{x_{i}\\}\times
P_{i+1}\times\dots)\cap P\subset f^{-1}[\\{l_{j}\\}]$ then and we have that
$h(x_{i})=l_{j}$. So we have that
$h(p_{i}(x_{1},\dots,x_{i},x_{i+1},\dots))=f(x_{1},\dots,x_{i},x_{i+1},\dots)$

Now we prove that $h$ is onto $L$. Take any $l_{j}\in L$. We can find a basic
set
$W_{x_{i}}=(\\{x_{1}\\}\times\\{x_{2}\\}\times\\{x_{3}\\}\times\dots\times\\{x_{i}\\}\times
P_{i+1}\times\dots)\cap P$ in $\mathcal{V}_{j}$. Then $h(x_{i})=l_{j}$.

Now we prove that $h$ is a homomorphism. Take any $x_{i},y_{i}\in P_{i}$ such
that $x_{i}\leq_{P_{i}}y_{i}$. Define
$x_{j}:=p_{j}^{i}(x_{i}),y_{j}:=p_{j}^{i}(y_{i})$ for $j<i$. As $p_{j}^{i}$
are homomorphisms we get that $x_{j}\leq_{P_{j}}y_{j}$. As $p_{i}^{i+1}$ is a
quotient map we can find $x_{i+1},y_{i+1}\in P_{i+1}$ such that
$p_{i}^{i+1}(x_{i+1})=x_{i},p_{i}^{i+1}(y_{i+1})=y_{i}$ and
$x_{i+1}\leq_{P_{i+1}}y_{i+1}$. Proceeding inductively we get

$(x_{1},x_{2},\dots,x_{i},x_{i+1},\dots),(y_{1},y_{2},\dots,y_{i},y_{i+1},\dots)\in
P$

such that

$(x_{1},x_{2},\dots,x_{i},x_{i+1},\dots)\leq_{P}(y_{1},y_{2},\dots,y_{i},y_{i+1},\dots).$

So
$h(x_{i})=f(x_{1},x_{2},\dots,x_{i},x_{i+1},\dots)\leq_{L}f(y_{1},y_{2},\dots,y_{i},y_{i+1},\dots)=h(y_{i})$
thus $h$ is a homomorphism.

Now we show that $h$ is a strict homomorphism. Let $l,z\in L$ be such that
$l\leq_{L}z$. As $f$ is a quotient map we get that for some
$(x_{1},x_{2},\dots,x_{i},\dots),(y_{1},y_{2},\dots,y_{i},\dots)\in P$ such
that

$(x_{1},x_{2},\dots,x_{i},\dots)\leq_{P}(y_{1},y_{2},\dots,y_{i},\dots)$

and

$f(x_{1},x_{2},\dots,x_{i},\dots)=l,f(y_{1},y_{2},\dots,y_{i},\dots)=z.$

Then $x_{i}\leq_{P_{i}}y_{i}$ and $h(x_{i})=l,h(y_{i})=z$.

∎

Now we may prove the following:

######  Theorem 3.13.

Let
$P=\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$,
$H=\varprojlim\langle\\{H_{n}\\}_{n\in\mathbb{N}},\\{h^{n}_{k}\\}_{k<n}\rangle$
be profinite posets and let $f:P\to H$ be a continuous quotient mapping. Then
there is a strictly increasing sequence $(i_{n})_{n\in\mathbb{N}}$ and
quotient maps $g_{j}:P_{i_{j}}\to H_{j}$ such that
$f(x_{1},x_{2},\dots)=(g_{1}(x_{i_{1}}),g_{2}(x_{i_{2}}),\dots)$. Furthermore
the diagram

${P_{i_{k}}}$${P_{i_{n}}}$${H_{k}}$${H_{n}}$$\scriptstyle{g_{k}}$$\scriptstyle{p_{i_{k}}^{i_{n}}}$$\scriptstyle{g_{n}}$$\scriptstyle{h_{k}^{n}}$
(11)

commutes for every $n,k\in\mathbb{N}$ with $k<n$.

###### Proof.

Take any $(x_{1},x_{2},\dots)\in P$. Let $h_{k}:H\to H_{k}$ be the projection
on $k$-th coordinate, i.e. $h_{k}(y_{1},y_{2},\dots)=y_{k}$. Since $f$ and
$h_{k}$ are onto, then $\operatorname{rng}(h_{k}\circ f)=H_{k}$. So we can
apply Lemma 3.12 to continuous quotient map $h_{k}\circ f$. Using Lemma 3.12
inductively we find a strictly increasing sequence $(i_{k})$ of indices and
quotient mappings $g_{k}:P_{i_{k}}\to H_{k}$ such that $g_{k}\circ
p_{i_{k}}=h_{k}\circ f$ where $p_{i_{k}}(x_{1},x_{2},\dots)=x_{i_{k}}$. So we
get that $h_{k}(f(x_{1},x_{2},\dots))=g_{k}(x_{i_{k}})$. Thus the $k$-th
coordinate of $f(x_{1},x_{2},\dots)$ is $g_{k}(x_{i_{k}})$.

Now we are going to show that diagram (11) commutes. Take any
$n,k\in\mathbb{N}$ with $k<n$ and let $x_{i_{n}}\in P_{i_{n}}$. Let
$x_{s}:=p_{s}^{i_{n}}(x_{i_{n}})$ for $s<i_{n}$. Then
$(x_{1},x_{2},\dots,x_{i_{n}-1},x_{i_{n}},z_{i_{n}+1},\dots)\in P$ for some
$z_{i_{n}+\ell}\in P_{i_{n}+\ell}$ with $\ell\in\mathbb{N}$. Then

$f(x_{1},x_{2},\dots,x_{i_{n}-1},x_{i_{n}},z_{i_{n}+1},\dots)=(g_{1}(x_{i_{1}}),g_{2}(x_{i_{2}}),\dots,g_{k}(x_{i_{k}}),\dots,g_{n}(x_{i_{n}}),\dots)\in
H,$

so
$h_{k}^{n}(g_{n}(x_{i_{n}}))=g_{k}(x_{i_{k}})=g_{k}(p_{i_{k}}^{i_{n}}(x_{i_{n}})$,
which means $g_{k}\circ p_{i_{k}}^{i_{n}}=h_{k}^{n}\circ g_{n}$, thus diagram
(11) commutes. ∎

Now we are going to prove that if we have a continuous quotient map $f$ from
$\mathbb{P}$ onto a finite poset $A$ and a quotient mapping $g$ from finite
poset $B$ onto $A$, then we can find a continuous quotient map $h$ from
$\mathbb{P}$ onto $B$, such that $f=g\circ h$.

######  Theorem 3.14.

Let $f:\mathbb{P}\to A,g:B\to A$ be continuous quotient mappings, where $A,B$
are finite posets and
$\mathbb{P}=\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$
where $(P_{n})$ is a Fraïssé sequence. Then there exists a continuous quotient
mapping $h:\mathbb{P}\to B$, such that diagram

${A}$${\mathbb{P}{}}$${B}$$\scriptstyle{h}$$\scriptstyle{f}$$\scriptstyle{g}$

commutes.

###### Proof.

From Lemma 3.12 we get that there is there is a continuous quotient map
$t:P_{i_{1}}\to A$ such that diagram

${A}$${\mathbb{P}}$${P_{i_{1}}}$$\scriptstyle{p_{i_{1}}}$$\scriptstyle{f}$$\scriptstyle{t}$
(12)

commutes, where $p_{i_{1}}(x_{1},x_{2},\dots)=x_{i_{1}}$.

Using (E) property for $A,B,P_{i_{1}}$, we shall find $i_{2}>i_{1}$ and
quotient map $l:P_{i_{2}}\to P_{i_{1}}$, such that diagram

${P_{i_{1}}}$${P_{i_{2}}}$${A}$${B}$$\scriptstyle{t}$$\scriptstyle{p_{i_{1}}^{i_{2}}}$$\scriptstyle{l}$$\scriptstyle{g}$
(13)

commutes. Note that $l$ is continuous. Let
$p_{i_{2}}(x_{1},x_{2},\dots)=x_{i_{2}}$ Define $h:=l\circ p_{i_{2}}$. Clearly
$h$ is continuous and $g\circ h=g\circ l\circ p_{i_{2}}=t\circ
p_{i_{1}}^{i_{2}}\circ p_{i_{2}}=t\circ p_{i_{1}}=f$, thus we get the
assertion. ∎

Now we are going to show that $\mathbb{P}=\\{0,1,2,3\\}^{\omega}$ has a dense
$G_{\delta}$ set of isolated points with respect to its ordering.

######  Theorem 3.15.

Let $(X,d)$ be a compact metric space with closed relation $E$. Then the set
of $E$-isolated points is $G_{\delta}$ subset of $X$.

###### Proof.

Let $I$ be the set of all $E$-isolated points in $X$. Take any $x\in I$ and
$\epsilon>0$. Then the set $Y_{x,\epsilon}:=\\{y\in X:d(x,y)\geq\epsilon\\}$
is closed. Now, we will prove that for each $y\in Y_{x,\epsilon}$, there are
its open neighborhood $U_{y}$ and $\delta_{y}>0$ such that

$\text{ for every }x^{\prime},y^{\prime}\in X\text{ if
}d(x,x^{\prime})<\delta_{y}\text{ and }y^{\prime}\in U_{y}\text{, then
}(x^{\prime},y^{\prime})\notin E\text{ and }(y^{\prime},x^{\prime})\notin E$
($\spadesuit$)

Suppose to the contrary that there is $y\in Y_{x,\epsilon}$, such that for
every open neighborhood $U_{y}$ and $\delta_{y}>0$ there are
$x^{\prime},y^{\prime}\in X$, such that $d(x,x^{\prime})<\delta_{y}$ and
$y^{\prime}\in U_{y}$ and $(x^{\prime},y^{\prime})\in E$ or
$(y^{\prime},x^{\prime})\in E$. We can find sequences
$x_{n}^{\prime},y_{n}^{\prime}$, such that $d(x,x_{n}^{\prime})<\frac{1}{n+1}$
and $y_{n}^{\prime}\in B(y,\frac{1}{n+1})$. Then $x_{n}^{\prime}\to
x,y_{n}^{\prime}\to y$ and either $(x_{n}^{\prime},y_{n}^{\prime})\in E$ or
$(y_{n}^{\prime},x_{n}^{\prime})\in E$ for each $n\in\omega$. One of these
conditions, either $(x_{n}^{\prime},y_{n}^{\prime})\in E$ or
$(y_{n}^{\prime},x_{n}^{\prime})\in E$, is satisfied for infinitely many
$n$’s, so we can choose sub-sequences
$(x^{\prime}_{k_{n}}),(y^{\prime}_{k_{n}})$, which satisfy one of them for
every $n$. Assume that $(x^{\prime}_{k_{n}},y^{\prime}_{k_{n}})\in E$ for each
$n\in\omega$. Then as $E$ is closed we get that $(x,y)\in E$, which is a
contradiction. If $(y^{\prime}_{k_{n}},x^{\prime}_{k_{n}})\in E$ for each $n$
we reach a contradiction as well.

Since $Y_{x,\epsilon}$ is compact, there are open neighborhoods
$U_{y_{1}},\dots,U_{y_{n}}$ of $y_{1},\dots,y_{n}$, respectively, such that
$Y_{x,\epsilon}\subset\bigcup_{i=1}^{n}U_{y_{i}}$. Let
$\delta(x,\epsilon):=\min\\{\epsilon,\delta_{y_{1}},\delta_{y_{2}},\dots,\delta_{y_{n}}\\}$.
See that, if $x^{\prime}$ has a distance less that $\delta(x,\epsilon)$ from
$x$ and $y\in Y_{x,\epsilon}$, then $(x^{\prime},y)\notin E$ and
$(y,x^{\prime})\notin E$. For each $\epsilon>0$ define

$V_{\epsilon}:=\\{x^{\prime}\in X:d(x,x^{\prime})<\delta(x,\epsilon)\text{ for
some }x\in I\\}.$

Then $V_{\epsilon}$ is open. We will show that

$I=\bigcap_{n\geq 1}V_{\frac{1}{n}}.$

It is clear that $I\subset\bigcap_{n\geq 1}V_{\frac{1}{n}}$. On the other
hand, if $y\in X\setminus I$, then $(y,z)\in E$ or $(z,y)\in E$ for some $z\in
X$. Let $n\in\mathbb{N}$ be such that $\frac{2}{n}<d(z,y)$. We will show that
$y\notin V_{\frac{1}{n}}$. Suppose to the contrary that $y\in
V_{\frac{1}{n}}$. Let $x\in I$ be such that
$d(x,y)<\delta(x,\frac{1}{n})\leq\frac{1}{n}$. See that
$\frac{2}{n}<d(z,y)\leq d(z,x)+d(x,y)<d(z,x)+\frac{1}{n}$, so
$d(z,x)>\frac{1}{n}$, which means $z\in Y_{x,\frac{1}{n}}$. From condition
($\spadesuit$ ‣ 3.3) we get that $(y,z)\notin E$ and $(z,y)\notin E$, which
gives us a contradiction. This proves that
$I=\bigcap_{n\in\mathbb{N}}V_{\frac{1}{n}}$, which means $I$ is a $G_{\delta}$
set. ∎

######  Theorem 3.16.

$\mathbb{P}=\\{0,1,2,3\\}^{\omega}$ has a dense $G_{\delta}$ set of isolated
points.

###### Proof.

From Lemma 2.12, Theorem 3.6 and Theorem 3.15 we get that the set of isolated
points in $\mathbb{P}$ is $G_{\delta}$. So we only need to show that it is
dense.

Let $A:=\\{(x_{0},x_{1},\dots)\in\mathbb{P}:x_{n}=2$ for all but finitely many
$n$’s$\\}$. Immediately from the definition of order relation on $\mathbb{P}$
we obtain that $A\subset I$, where $I$ is the set of isolated points in
$\mathbb{P}$. We will show that $A$ is dense in $\mathbb{P}$. Take any basic
set
$\\{x_{0}\\}\times\\{x_{1}\\}\times\dots\times\\{x_{n}\\}\times\\{0,1,2,3\\}\times\dots$.
See that $(x_{0},x_{1},x_{2},\dots,x_{n},2,2,2,\dots)$ is an element of both
$A$ and
$\\{x_{0}\\}\times\\{x_{1}\\}\times\dots\times\\{x_{n}\\}\times\\{0,1,2,3\\}\times\dots$,
which means $A$ is dense, thus $I$ is also dense. ∎

##  4 The inverse system of order ideals

###  4.1 Induced quotient maps

Now we are going to study the following problem. Let
$\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$ be an
inductive sequence consisting of finite posets with quotient maps and
$P=\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$.
From Lemma 2.9 we know that each of $P_{n}$ embeds into its order ideal
$\mathcal{O}(P_{n})$ by mapping $x\mapsto\downarrow\\!\\!x$. The question is
whether we can find quotient mapping from $\mathcal{O}(P_{n+1})$ onto
$\mathcal{O}(P_{n})$ such that the following diagrams

${P_{1}}$${P_{2}}$${P_{3}}$${\ldots}$${P}$${\mathcal{O}(P_{1})}$${\mathcal{O}(P_{2})}$${\mathcal{O}(P_{3})}$${\ldots}$
? $\scriptstyle{\downarrow x}$$\scriptstyle{\downarrow
x}$$\scriptstyle{p_{1}^{2}}$$\scriptstyle{p_{2}^{3}}$$\scriptstyle{\downarrow
x}$$\scriptstyle{p_{3}^{4}}$$\scriptstyle{?}$$\scriptstyle{?}$$\scriptstyle{?}$$\scriptstyle{?}$
(14)

commute. Our aim is to show that question marks in the diagram can be
superseded by appropriate arrows and a limit object. We will study the
properties of that arrows and an object. Firstly, in the following Theorem we
define arrows between $\mathcal{O}(P_{n+1})$ and $\mathcal{O}(P_{n})$.

######  Theorem 4.1.

Let $P,Q$ be finite posets. Let $\varphi:P\to\mathcal{O}(P)$ and
$\psi:Q\to\mathcal{O}(Q)$ be the embeddings given by
$x\mapsto\downarrow\\!\\!x$. Let $p:Q\to P$ be a quotient map. Let
$\hat{p}:\mathcal{O}(Q)\to\mathcal{O}(P)$ be defined as follows:

  * •

$\hat{p}(\emptyset)=\emptyset$

  * •

$\hat{p}(\downarrow\\!\\!x)=\downarrow\\!\\!p(x)$ for every $x\in Q$

  * •

For every non-empty, non-principal down-set $A\in\mathcal{O}(Q)$ we take its
canonical decomposition $\\{\downarrow\\!\\!a_{i}:i<k\\}$ and we put
$\hat{p}(A)=\bigcup_{i=1}^{k}\hat{p}(\downarrow\\!\\!a_{i})$

Then $\hat{p}$ is a quotient map for posets, which is join-preserving and
preserves join-irreducible elements.

Furthermore the diagram

${P}$${Q}$${\mathcal{O}(P)}$${\mathcal{O}(Q)}$$\scriptstyle{\varphi}$$\scriptstyle{p}$$\scriptstyle{\psi}$$\scriptstyle{\hat{p}}$
(15)

commutes. We will say that quotient map $\hat{p}$ is induced by quotient map
$p$.

Let us stress the fact that one cannot obtain a quotient map for lattices in
such situation. We present an example that illustrates this. We say that a
mapping $p$ preserves meets, if $p(x\wedge y)=p(x)\wedge p(y)$ for all $x$ and
$y$.

######  Example 4.2.

Consider posets $Q:=\overline{\mathbf{1}}$ and $P:=\overline{\mathbf{2}}$ and
quotient map $p:P\rightarrow Q$ such that $p\equiv 1$. Let $\hat{p}$ be a
homomorphism induced by $p$. Then
$\hat{p}(\downarrow\\!\\!1\cap\downarrow\\!\\!2)=\hat{p}(\emptyset)=\emptyset$
and $\hat{p}(\downarrow\\!\\!1)=\downarrow\\!\\!1=\hat{p}(\downarrow\\!\\!2)$,
thus
$\hat{p}(\downarrow\\!\\!1)\cap\hat{p}(\downarrow\\!\\!2)=\downarrow\\!\\!1\neq\emptyset$.
Therefore $\hat{p}$ does not preserve meets.

Now we are ready to begin the proof of Theorem 4.1.

###### Proof.

By the definition of $\hat{p}$ the diagram (15) commutes and preserves join-
irreducible elements. Now we will show that $\hat{p}$ preserves set unions.
Firstly, we show that
$\hat{p}(\downarrow\\!\\!x\cup\downarrow\\!\\!y)=\hat{p}(\downarrow\\!\\!x)\cup\hat{p}(\downarrow\\!\\!y)$.
Let us consider two cases:

  1. (i)

If $\downarrow\\!\\!x\parallel\downarrow\\!\\!y$ then from Lemma 2.4 and
definition of $\hat{p}$ we get
$\hat{p}(\downarrow\\!\\!x\cup\downarrow\\!\\!y)=\hat{p}(\downarrow\\!\\!x)\cup\hat{p}(\downarrow\\!\\!y)$

  2. (ii)

If $\downarrow\\!\\!x\subset\downarrow\\!\\!y$ then $x\leq y$ which implies
$p(x)\leq p(y)$. Thus
$\hat{p}(\downarrow\\!\\!x)=\downarrow\\!\\!p(x)\subset\downarrow\\!\\!p(y)=\hat{p}(\downarrow\\!\\!y)$
so
$\hat{p}(\downarrow\\!\\!x)\cup\hat{p}(\downarrow\\!\\!y)=\hat{p}(\downarrow\\!\\!y)=\hat{p}(\downarrow\\!\\!x\cup\downarrow\\!\\!y)$.
(Case $\downarrow\\!\\!y\subset\downarrow\\!\\!x$ goes similarly).

By induction we obtain
$\hat{p}\left(\bigcup_{k=1}^{n}\downarrow\\!\\!x_{k}\right)=\bigcup_{k=1}^{n}\hat{p}(\downarrow\\!\\!x_{k})$
for any
$\downarrow\\!\\!x_{1},\dots,\downarrow\\!\\!x_{n}\in\mathcal{J}(\mathcal{O}(Q))$.
Now we deal with a general case. Let $X,Y\in\mathcal{O}(Q)$. Then
$X=\bigcup_{k=1}^{n}\downarrow\\!\\!x_{k}$ and
$Y=\bigcup_{k=1}^{m}\downarrow\\!\\!y_{k}$. So $X\cup
Y=\bigcup_{k=1}^{n}\downarrow\\!\\!x_{k}\cup\bigcup_{k=1}^{m}\downarrow\\!\\!y_{k}$,
thus $\hat{p}(X\cup
Y)=\bigcup_{k=1}^{n}\hat{p}(\downarrow\\!\\!x_{k})\cup\bigcup_{k=1}^{m}\hat{p}(\downarrow\\!\\!y_{k})=\hat{p}(X)\cup\hat{p}(Y)$.
Note that if $X\subset Y$, then $X\cup Y=Y$ and consequently
$\hat{p}(X)\subset\hat{p}(X\cup Y)=\hat{p}(X)\cup\hat{p}(Y)=\hat{p}(Y)$. This
shows that $\hat{p}$ is an order homomorphism as well.

Now we will show that $\hat{p}$ is surjective. Let $Z\in\mathcal{O}(P)$. Then
$Z=\bigcup_{k=1}^{n}\downarrow\\!\\!z_{k}$. For each $z_{k}$ we find $x_{k}$
such that $p(x_{k})=z_{k}$, in other words
$\widehat{p}(\downarrow\\!\\!x_{k})=\downarrow\\!\\!z_{k}$. Then

$\hat{p}\left(\bigcup_{k=1}^{n}\downarrow\\!\\!x_{k}\right)=\bigcup_{k=1}^{n}\hat{p}(\downarrow\\!\\!x_{k})=\bigcup_{k=1}^{n}\downarrow\\!\\!z_{k}=Z.$

Now we will show that $\hat{p}$ is a strict poset homomorphism. Let
$Y,Z\in\mathcal{O}(P)$ and $Y\subset Z$. Then
$Y=\bigcup_{k=1}^{n}\downarrow\\!\\!y_{k}\subset
Z=\bigcup_{k=1}^{m}\downarrow\\!\\!z_{k}$. Then for each $y_{k}$ we find
$z_{l_{k}}$ such that $\downarrow\\!\\!y_{k}\subset\downarrow\\!\\!z_{l_{k}}$
which means that $y_{k}\leq z_{l_{k}}$. As $p$ is strict, there are
$y_{k}^{\prime},z_{l_{k}}^{\prime}\in Q$ with $y_{k}^{\prime}\leq
z_{l_{k}}^{\prime}$, $p(y_{k}^{\prime})=y_{k}$ and
$p(z_{l_{k}}^{\prime})=z_{l_{k}}$.

Put $Y^{\prime}:=\bigcup_{k=1}^{n}\downarrow\\!\\!y_{k}^{{}^{\prime}}$. Then
$\hat{p}\left(Y^{\prime}\right)=Y$. Further for each $z_{p}$ such that $p\neq
l_{k}$ for all $k=1,\ldots,n$ we find $z_{p}^{{}^{\prime\prime}}\in Q$ such
that
$\hat{p}(\downarrow\\!\\!z_{p}^{{}^{\prime\prime}})=\downarrow\\!\\!z_{p}$.
Then $Y^{{}^{\prime}}\subset Z^{\prime}$ where
$Z^{\prime}:=\bigcup_{k=1}^{n}\downarrow\\!\\!z_{l_{k}}^{{}^{\prime}}\cup\bigcup\downarrow\\!\\!z_{p}^{{}^{\prime\prime}}$
and the latter union is over all $p$ not in $\\{l_{k}:k\leq n\\}$. Since
$\hat{p}(Z^{\prime})=Z$, $\hat{p}$ is a quotient map for posets. ∎

It turns out that the value of mapping $\hat{p}(A)$ does not depend on how one
represents $A$ as a union of principal down-sets.

######  Remark 4.3.

Let $P,Q$ be finite posets, $p:Q\to P$ be a quotient map and $\hat{p}$ be a
quotient map induced by $p$. Let $A\subset Q$ be a down-set with canonical
decomposition $\\{\downarrow\\!\\!x_{i}:i<n\\}$ and let
$\bigcup_{i=1}^{m}\downarrow\\!\\!y_{i}=A$. Then

$\bigcup_{i=1}^{m}\downarrow\\!\\!p(y_{i})=\bigcup_{i=1}^{m}\hat{p}(\downarrow\\!\\!y_{i})=\hat{p}\left(\bigcup_{i=1}^{m}\downarrow\\!\\!y_{i}\right)=\hat{p}(A)=\bigcup_{i=1}^{n}\downarrow\\!\\!p(x_{i}).$

It follows that if
$\bigcup_{i=1}^{m}\downarrow\\!\\!y_{i}=\bigcup_{i=1}^{l}\downarrow\\!\\!z_{i}\subset
Q$, then
$\bigcup_{i=1}^{m}\downarrow\\!\\!p(y_{i})=\bigcup_{i=1}^{l}\downarrow\\!\\!p(z_{i})$.

To properly define the inductive sequence of $\mathcal{O}(P_{n})$’s such that
diagram (14) commutes we need to show that
$\hat{p}^{n}_{k}=\hat{p}^{k+1}_{k}\circ\dots\circ\hat{p}^{n}_{n-1}$ for all
$n,k\in\mathbb{N}$ with $k<n$, which follows from the following.

######  Lemma 4.4.

Let $P,Q,H$ be finite posets and $p:Q\to P$, $q:P\to H$ be quotient mappings.
Then $\widehat{q\circ p}=\widehat{q}\circ\widehat{p}$.

###### Proof.

Let $A\in\mathcal{O}(Q)$. Consider the following cases:

  * •

if $A=\emptyset$, then $\widehat{q\circ
p}(A)=\emptyset=\hat{q}(\emptyset)=\hat{q}(\hat{p}(\emptyset))$.

  * •

if $A$ is a principal down-set $\downarrow\\!\\!x$, then $\widehat{q\circ
p}(\downarrow\\!\\!x)=\downarrow\\!\\!q(p(x))=\hat{q}(\downarrow\\!\\!p(x))=\hat{q}(\hat{p}(\downarrow\\!\\!x))=(\hat{q}\circ\hat{p})(\downarrow\\!\\!x)$.

  * •

if $A$ is non-empty, non-principal down-set, then let
$\\{\downarrow\\!\\!x_{i}:i<n\\}$ be its canonical decomposition. Then
$\widehat{q\circ p}(A)=\bigcup_{i=1}^{n}\widehat{q\circ
p}(\downarrow\\!\\!x_{i})=\bigcup_{i=1}^{n}\hat{q}(\hat{p}(\downarrow\\!\\!x_{i}))=\hat{q}(\bigcup_{i=1}^{n}\hat{p}(\downarrow\\!\\!x_{i}))=\hat{q}(\hat{p}(\bigcup_{i=1}^{n}\downarrow\\!\\!x_{i}))=(\hat{q}\circ\hat{p})(A)$.

Thus $\widehat{q\circ p}(A)=(\hat{q}\circ\hat{p})(A)$ for every
$A\in\mathcal{O}(Q)$, which gives the assertion. ∎

Now, we provide the equivalent conditions for the induced mapping to be
lattice homomorphism.

######  Theorem 4.5.

Let $P,Q$ be finite posets and $p:Q\to P$ be a quotient map. Then the
following are equivalent.

  1. (1)

$\hat{p}$ is a lattice-homomorphism.

  2. (2)

$\bigcup_{z\in\downarrow\\!\\!\>x\cap\downarrow\\!\\!\>y}\limits\downarrow\\!\\!p(z)\supset\downarrow\\!\\!p(x)\cap\downarrow\\!\\!p(y)$
for all $x,y\in Q$.

  3. (3)

If $t\leq p(x)$ and $t\leq p(y)$, then there is $z\in Q$, such that $z\leq
x,z\leq y$ and $t\leq p(z)$, for all $x,y\in Q$ and $t\in P$.

###### Proof.

The equivalence of (2) and (3) is pretty clear, so we will only show that (1)
and (2) are equivalent. Assume that (2) is true. Take any $x,y\in Q$. Observe
that

$\bigcup_{z\in\downarrow\\!\\!\>x\cap\downarrow\\!\\!\>y}\downarrow\\!\\!p(z)=\bigcup_{\downarrow\\!\\!\>z\subset\downarrow\\!\\!\>x\cap\downarrow\\!\\!\>y}\hat{p}(\downarrow\\!\\!z)\subset\hat{p}(\downarrow\\!\\!x\cap\downarrow\\!\\!y)$

It follows that
$\hat{p}(\downarrow\\!\\!x)\cap\hat{p}(\downarrow\\!\\!y)=\downarrow\\!\\!p(x)\cap\downarrow\\!\\!p(y)\subset\hat{p}(\downarrow\\!\\!x\cap\downarrow\\!\\!y)$.
Since $\hat{p}$ is an order-homomorphism, then
$\hat{p}(\downarrow\\!\\!x\cap\downarrow\\!\\!y)\subset\hat{p}(\downarrow\\!\\!x)\cap\hat{p}(\downarrow\\!\\!y)$.
Thus
$\hat{p}(\downarrow\\!\\!x)\cap\hat{p}(\downarrow\\!\\!y)=\hat{p}(\downarrow\\!\\!x\cap\downarrow\\!\\!y)$
Now take any two non-empty, down-sets $A,B$. Then
$A=\bigcup_{i=1}^{n}\downarrow\\!\\!a_{i},B=\bigcup_{i=1}^{m}\downarrow\\!\\!b_{i}$.
See that

$\hat{p}(A\cap
B)=\hat{p}(\bigcup_{i=1}^{n}\downarrow\\!\\!a_{i}\cap\bigcup_{i=1}^{m}\downarrow\\!\\!b_{i})=\hat{p}(\bigcup_{i=1}^{n}\bigcup_{j=1}^{m}\downarrow\\!\\!a_{i}\cap\downarrow\\!\\!b_{j})=\bigcup_{i=1}^{n}\bigcup_{j=1}^{m}\hat{p}(\downarrow\\!\\!a_{i}\cap\downarrow\\!\\!b_{j})=$
$=\bigcup_{i=1}^{n}\bigcup_{j=1}^{m}\hat{p}(\downarrow\\!\\!a_{i})\cap\hat{p}(\downarrow\\!\\!b_{j})=\bigcup_{i=1}^{n}\hat{p}(\downarrow\\!\\!a_{i})\cap\bigcup_{i=1}^{m}\hat{p}(\downarrow\\!\\!b_{i})=\hat{p}(A)\cap\hat{p}(B).$

Note that $\hat{p}(A\cap\emptyset)=\emptyset=\hat{p}(A)\cap\hat{p}(\emptyset)$
for any down-set $A$, thus $\hat{p}$ is a lattice-homomorphism.

Now, assume (1). Take any $x,y\in Q$. We need to consider two cases

  * •

$\downarrow\\!\\!x\cap\downarrow\\!\\!y=\emptyset$

  * •

$\downarrow\\!\\!x\cap\downarrow\\!\\!y\neq\emptyset$

If $\downarrow\\!\\!x\cap\downarrow\\!\\!y=\emptyset$, then
$\hat{p}(\downarrow\\!\\!x\cap\downarrow\\!\\!y)=\hat{p}(\downarrow\\!\\!x)\cap\hat{p}(\downarrow\\!\\!y)=\emptyset$.
Thus
$\downarrow\\!\\!p(x)\cap\downarrow\\!\\!p(y)\subset\bigcup_{z\in\downarrow\\!\\!\>x\cap\downarrow\\!\\!\>y}\downarrow\\!\\!p(z)$.

If $\downarrow\\!\\!x\cap\downarrow\\!\\!y\neq\emptyset$ and
$\downarrow\\!\\!x\cap\downarrow\\!\\!y=\bigcup_{i=1}^{m}\downarrow\\!\\!z_{i}=$
where $z_{i}$’s form its canonical decomposition, then
$\hat{p}(\downarrow\\!\\!x\cap\downarrow\\!\\!y)\neq\emptyset$. Let
$\hat{p}(\downarrow\\!\\!x\cap\downarrow\\!\\!y)=\bigcup_{i=1}^{n}\downarrow\\!\\!r_{i}$
be its canonical decomposition. Since
$\bigcup_{i=1}^{m}\downarrow\\!\\!p(z_{i})=\bigcup_{i=1}^{n}\downarrow\\!\\!r_{i}$,
then $\\{r_{1},\dots,r_{n}\\}\subset\\{p(z_{1}),\dots,p(z_{m})\\}$. Without
loss of generality we may assume that $r_{i}=p(z_{i})$ for $i=1,\dots,n$.
Finally

$\downarrow\\!\\!p(x)\cap\downarrow\\!\\!p(y)=\hat{p}(\downarrow\\!\\!x\cap\downarrow\\!\\!y)=\bigcup_{i=1}^{n}\downarrow\\!\\!p(z_{i})=\bigcup_{z\in\downarrow\\!\\!\>x\cap\downarrow\\!\\!\>y}\limits\downarrow\\!\\!p(z),$

which means that (2) holds. ∎

###  4.2 Properties of $\mathbb{O}(P)$

Let $\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$ be an
inductive sequence of finite posets with quotient mappings. Let
$\hat{p}_{k}^{n}:\mathcal{O}(P_{n})\to\mathcal{O}(P_{k})$ be a quotient
mapping induced by $p_{k}^{n}$ for $n\in\mathbb{N}$ and $k<n$, and let
$P=\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$.
Lemma 4.4 shows that
$\langle\\{\mathcal{O}(P_{n})\\}_{n\in\mathbb{N}},\\{\hat{p}^{n}_{k}\\}_{k<n}\rangle$
is an inductive sequence in the category of finite posets. We will now study
the propeties of
$\mathbb{O}(P):=\varprojlim\langle\\{\mathcal{O}(P_{n})\\}_{n\in\mathbb{N}},\\{\hat{p}^{n}_{k}\\}_{k<n}\rangle$.

######  Theorem 4.6.

Let
$\psi(x_{1},x_{2},\ldots)=(\downarrow\\!\\!x_{1},\downarrow\\!\\!x_{2},\ldots)$
for $(x_{1},x_{2},\dots)\in P$. Then $\psi:P\to\mathbb{O}(P)$ is an order-
embedding.

###### Proof.

We will show that $\psi$ is an order-embedding into $\mathbb{O}(P)$. Firstly
we show that $\psi$ is well defined, more precisely it actually maps $P$ into
$\mathbb{O}(P)$. Take $(x_{1},x_{2},\ldots)\in P$. We will show that
$\hat{p}_{n}^{n+1}(\downarrow\\!\\!x_{n+1})=\downarrow\\!\\!x_{n}$ for every
$n\in\mathbb{N}$. Due to Theorem 4.1 diagram

${P_{n}}$${P_{n+1}}$${\mathcal{O}(P_{n})}$${\mathcal{O}(P_{n+1})}$$\scriptstyle{\downarrow
x}$$\scriptstyle{p_{n}^{n+1}}$$\scriptstyle{\downarrow
x}$$\scriptstyle{\hat{p}_{n}^{n+1}}$ (16)

commutes for every $n\in\mathbb{N}$, so
$\hat{p}_{n}^{n+1}(\downarrow\\!\\!x_{n+1})=\downarrow\\!\\!p_{n}^{n+1}(x_{n+1})=\downarrow\\!\\!x_{n}$
for every $n\in\mathbb{N}$. Thus $\psi$ is properly defined.

Now we will show that it is an order embedding. Let
$(x_{1},x_{2},\ldots),(y_{1},y_{2},\ldots)\in P$. Then

$(x_{1},x_{2},\ldots)\leq(y_{1},y_{2},\ldots)\iff x_{n}\leq y_{n}\text{ for
every }n\in\mathbb{N}\iff$
$\downarrow\\!\\!x_{n}\subset\downarrow\\!\\!y_{n}\text{ for every
}n\in\mathbb{N}\iff\psi(x_{1},x_{2},\dots)=(\downarrow\\!\\!x_{1},\downarrow\\!\\!x_{2},\dots)\leq(\downarrow\\!\\!y_{1},\downarrow\\!\\!y_{2},\dots)=\psi(y_{1},y_{2},\dots),$

which makes $\psi$ an order embedding. ∎

######  Definition 4.7.

We say that $\mathbf{x}\in\mathbb{O}(P)$ is a _principal element_ if
$\mathbf{x}=(\downarrow\\!\\!a_{1},\downarrow\\!\\!a_{2},\dots)$ for some
$(a_{1},a_{2},\dots)\in P$.

We already know that $\mathbb{O}(P)$ is a poset (as an inverse limit of finite
posets with quotients). Surprisingly, despite the fact that induced quotient
mapping does not preserve meets, see Example 4.2, it turns out that
$\mathbb{O}(P)$ is actually a lattice.

######  Theorem 4.8.

$\mathbb{O}(P)$ is a lattice.

###### Proof.

We will show that for every $x,y\in\mathbb{O}(P)$ their supremum exists. Let
$(x_{1},x_{2},\ldots)$, $(y_{1},y_{2},\ldots)\in\mathbb{O}(P)$. It is easy to
see that $(z_{1},z_{2},\ldots):=(x_{1}\cup y_{1},x_{2}\cup y_{2},\ldots)$ is
an upper bound of $\\{x,y\\}$ because it is an element of $\mathbb{O}(P)$ as
$\hat{p}_{n}^{n+1}$ preserves set unions and $x_{n}\subset z_{n},y_{n}\subset
z_{n}$ for every $n\in\mathbb{N}$. Now we will show it is actually the
smallest upper bound. Let $(s_{1},s_{2},\ldots)$ be any upper bound of
$\\{x,y\\}$. Then for every $n\in\mathbb{N}$ we have $x_{n}\subset
s_{n},y_{n}\subset s_{n}$ thus $x_{n}\cup y_{n}\subset s_{n}$ so
$(z_{1},z_{2},\ldots)\leq(s_{1},s_{2},\ldots)$. Therefore
$(z_{1},z_{2},\ldots)$ is the supremum of $\\{x,y\\}$.

Now we will show that $\\{x,y\\}$ has an infimum. It is easy to see that
$(\emptyset,\emptyset,\ldots)$ is a lower bound of $\\{x,y\\}$. Consider the
set

$A_{1}:=\\{z_{1}\in\mathcal{O}(P_{1}):\text{ there is a lower bound of
}\\{x,y\\}\text{ such that }z_{1}\text{ is its first coordinate}\\}.$

Clearly it is a non-empty set. Because supremum of two lower bonds of
$\\{x,y\\}$ is still a lower bound of $\\{x,y\\}$ we get that $A_{1}$ has the
greatest element, say $r_{1}$. Now consider the set

$A_{2}:=\\{z_{2}\in\mathcal{O}(P_{2}):\text{ there is a lower bound of
}\\{x,y\\}\text{ such that }r_{1}\text{ is its first coordinate and }$
$z_{2}\text{ is the second}\\}.$

Similarly this set is non-empty and it has the greatest element $r_{2}$. There
is a lower bound of $\\{x,y\\}$ of the form
$(r_{1},r_{2},z_{3},z_{4},\dots)\in\mathbb{O}(P)$, which in particular means
that $\hat{p}^{2}_{1}(r_{2})=r_{1}$. Proceeding inductively we define
$A_{n}$’s and a sequence $(r_{1},r_{2},\ldots)$. By the construction
$\hat{p}^{n+1}_{n}(r_{n+1})=r_{n}$ for all $n\in\mathbb{N}$, which means
$(r_{1},r_{2},\ldots)\in\mathbb{O}(P)$. Furthermore it is a lower bound of
$\\{x,y\\}$.

Assume that $(s_{1},s_{2},\ldots)$ is a lower bound of $\\{x,y\\}$. Then by
the first part of the proof $(s_{1}\cup r_{1},s_{2}\cup r_{2},\ldots)$ is a
lower bound of $\\{x,y\\}$ as well and $r_{i}\leq s_{i}\cup r_{i}$. Thus by
the definition of $A_{i}$’s and simple induction we get $r_{i}=s_{i}\cup
r_{i}$. Thus $s_{i}\subset r_{i}$, and consequently
$(s_{1},s_{2},\ldots)\leq(r_{1},r_{2},\ldots)$. ∎

Let us present some properties of a limit lattice $\mathbb{O}(P)$. Firstly
note that $(\emptyset,\emptyset,\ldots)$ is zero (or the least) element of
$\mathbb{O}(P)$. Moreover, $(P_{1},P_{2},\ldots)$ is one (or the greatest)
element of $\mathbb{O}(P)$. As we have already proved
$\sup\\{(x_{1},x_{2},\ldots),(y_{1},y_{2},\ldots)\\}=(x_{1}\cup
y_{1},x_{2}\cup y_{2},\ldots)$. On the other hand there is not such easy
formula for infimum. We will focus on more involved properties of
$\mathbb{O}(P)$.

######  Theorem 4.9.

$\mathbb{O}(P)$ is atomic.

###### Proof.

Take any
$(x_{1},x_{2},\dots)\in\mathbb{O}(P)\setminus\\{(\emptyset,\emptyset,\dots)\\}$.
We will find an atom $(a_{1},a_{2},\dots)$ , such that
$(a_{1},a_{2},\dots)\leq(x_{1},x_{2},\dots)$. If $(x_{1},x_{2},\dots)$ is an
atom, then we can take $(a_{1},a_{2},\dots):=(x_{1},x_{2},\dots)$ and we get
the required condition. Assume that $(x_{1},x_{2},\dots)$ is not an atom. Then
there is $(y_{1},y_{2},\dots)\neq(\emptyset,\emptyset,\dots)$ with
$(y_{1},y_{2},\dots)<(x_{1},x_{2},\dots)$.

See that then set

$A_{1}:=\\{z_{1}\in\mathcal{O}(P_{1})\setminus\\{\emptyset\\}:\text{there is a
non-zero element of }\mathbb{O}(P)\text{ that is strictly less than }$
$(x_{1},x_{2},\dots)\text{ and }z_{1}\text{ is its first coordinate.}\\}$

is non-empty. As a non-empty subset of finite poset, it has a minimal element,
say $a_{1}$. Thus we can find $(a_{1},z_{2},z_{3},\dots)\in\mathbb{O}(P)$ with
$(\emptyset,\emptyset,\dots)<(a_{1},z_{2},z_{3},\dots)<(x_{1},x_{2},\dots)$.
Thus we get that set

$A_{2}:=\\{z_{2}\in\mathcal{O}(P_{2})\setminus\\{\emptyset\\}:\text{there is a
non-zero element of }\mathbb{O}(P)\text{ that is strictly less than }$
$(x_{1},x_{2},\dots)\text{ and }a_{1},z_{2}\text{ are its first and second
coordinate, respectively.}\\}$

is non-empty. Similarly, as a non-empty subset of a finite poset, it has a
minimal element, say $a_{2}$. Proceeding inductively, we define $A_{n}$’s and
sequence $(a_{1},a_{2},\dots)$. Clearly, $\hat{p}_{n}^{n+1}(a_{n+1})=a_{n}$
for each $n\in\mathbb{N}$. By the construction, we get that
$(a_{1},a_{2},\dots)\leq(x_{1},x_{2},\dots)$.

Now we will prove, that $(a_{1},a_{2},\dots)$ is an atom. Suppose that there
is $(s_{1},s_{2},\dots)\in\mathbb{O}(P)$ such that
$(\emptyset,\emptyset,\dots)\neq(s_{1},s_{2},\dots)<(a_{1},a_{2},\dots)$. Let
$i$ be the smallest index such that $s_{j}=a_{j}$ for $j<i$, and
$s_{i}<a_{i}$. Thus $s_{i}\in A_{i}$. But this contradicts the definition of
$a_{i}$. ∎

The following Lemmas show the structural properties of mappings induced by
quotients.

######  Lemma 4.10.

Let $Q,H$ be finite posets and $g:\mathcal{O}(Q)\rightarrow\mathcal{O}(H)$ be
a quotient map induced by quotient map from $Q$ onto $H$. Let
$q\in\mathcal{O}(Q)$. Then for every element $\downarrow\\!\\!a$ of the
canonical decomposition of $g(q)$ we can find an element $\downarrow\\!\\!b$
of the canonical decomposition of $q$ such that
$g(\downarrow\\!\\!b)=\downarrow\\!\\!a$.

###### Proof.

Let $p=g(q)$ and $\\{\downarrow\\!\\!p_{i}:i<m\\},\
\\{\downarrow\\!\\!q_{i}:i<n\\}$ be the canonical decompositions of $p$ and
$q$, respectively. Thus $p=\bigcup_{i<m}\downarrow\\!\\!p_{i}$ and
$q=\bigcup_{i<n}\downarrow\\!\\!q_{i}$. We know that $g$ preserves set unions,
so
$p=g(q)=g\left(\bigcup_{i<n}\downarrow\\!\\!q_{i}\right)=\bigcup_{i<n}g(\downarrow\\!\\!q_{i})$.
Thus
$\bigcup_{i<m}\downarrow\\!\\!p_{i}=\bigcup_{i<n}g(\downarrow\\!\\!q_{i})$.
Then $g(\downarrow\\!\\!q_{i})$ is a principal down-set for every $i<n$. Using
Lemma 2.5 we get that
$\\{\downarrow\\!\\!p_{i}:i<m\\}\subset\\{\downarrow\\!\\!g(q_{i}):i<n\\}$.
Thus for every $\ell<m$ we find $k<n$ such that
$\downarrow\\!\\!p_{\ell}=g(\downarrow\\!\\!q_{k})$ ∎

######  Lemma 4.11.

Let $\mathbf{p}=(p_{1},p_{2},\ldots)\in\mathbb{O(\text{P})}$. Let
$m\in\mathbb{N}$ and let $t$ be a principal down-set in the canonical
decomposition of $p_{m}$. Then there is a principal element
$(\downarrow\\!\\!x_{1},\downarrow\\!\\!x_{2},\ldots)$ such that

  1. (i)

$\downarrow\\!\\!x_{i}$ is in canonical decomposition of $p_{i}$ for every
$i>m$;

  2. (ii)

$\downarrow\\!\\!x_{m}=t$,

  3. (iii)

$\downarrow\\!\\!x_{i}\leq p_{i}$ for every $i<m$.

In particular
$(\downarrow\\!\\!x_{1},\downarrow\\!\\!x_{2},\ldots)\leq\mathbf{p}$.

###### Proof.

Fix a principal down-set $t$ in the canonical decomposition of $p_{m}$. Then
there is $x_{m}$ in $P_{m}$ with $t=\downarrow\\!\\!x_{m}$, which gives (ii).
Condition _(i)_ immediately follows from Lemma 4.10 by a simple induction. By
the definition of induced quotient map we obtain _(iii)_. ∎

Now we present a lemma that allows us to represent elements of $\mathbb{O}(P)$
as supremum of countably many principal elements
$(\downarrow\\!\\!x_{1},\downarrow\\!\\!x_{2},\ldots)$.

######  Lemma 4.12.

For any non-zero element $\mathbf{a}\in\mathbb{O}(P)$ there is a countable
family $\mathcal{A}$ consisting of principal elements, such that
$\sup(\mathcal{A})=\mathbf{a}$.

###### Proof.

Let $\mathbf{a}=(a_{1},a_{2},\ldots)\in\mathbb{O}(P)$ be non-zero. Each
coordinate of $\mathbf{a}$ can be expressed as a finite union of principal
down-sets from its canonical decomposition. Thus
$a_{n}=\bigcup_{i=0}^{m_{n}}\downarrow\\!\\!x_{n}^{(i)}$ for every
$n\in\mathbb{N}$ where $\\{\downarrow\\!\\!x_{n}^{(i)}:i<m_{n}\\}$ is the
canonical decomposition of $a_{n}$. For every $\downarrow\\!\\!x_{1}^{(i)}$,
using Lemma 4.11, we can fix a principal element $\mathbf{b}_{1}^{(i)}$ in
$\mathbb{O}(P)$, such that $\downarrow\\!\\!x_{1}^{(i)}$ is its first
coordinate and $\mathbf{b}_{1}^{(i)}\leq\mathbf{a}$. Let
$A_{1}=\\{\mathbf{b}_{1}^{(i)}:i\leq m_{1}\\}$. We already know that supremum
of two elements from $\mathbb{O}(P)$ is the sequence of suprema, and therefore
the first coordinate of $\sup(A_{1})$ is $a_{1}$. For every
$\downarrow\\!\\!x_{2}^{(i)}$ we can find a principal element
$\mathbf{b}_{2}^{(i)}$ in $\mathbb{O}(P)$ such that $x_{2}^{(i)}$ is its
second coordinate and $\mathbf{b}_{2}^{(i)}\leq\mathbf{a}$. Let
$A_{2}=\\{\mathbf{b}_{2}^{(i)}:i\leq m_{2}\\}$. Then the second coordinate of
$\sup(A_{2})$ is $a_{2}$ Proceeding inductively we obtain $A_{n}$’s. Now we
define $\mathcal{A}$ as $\bigcup_{n\in\mathbb{N}}A_{n}$. Note that as every
$A_{n}$ is finite, $\mathcal{A}$ is countable.

Now we show that $\sup(\mathcal{A})=\mathbf{a}$. By the definition of
$\mathcal{A}$, $\mathbf{x}\leq\mathbf{a}$ for all $\mathbf{x}\in\mathcal{A}$
so $\mathbf{a}$ is an upper bound of $\mathcal{A}$. Take any other upper bound
$s=(s_{1},s_{2},\ldots)$ of $\mathcal{A}$. Fix $n\in\mathbb{N}$. Then
$y_{n}\subset s_{n}$ for every $\mathbf{y}\in\mathcal{A}$. By the construction
$\bigcup\\{y_{n}:\mathbf{y}\in\mathcal{A}\\}$ equals $a_{n}$. Thus
$\mathbf{a}\leq s$ and $\sup(\mathcal{A})=\mathbf{a}$. ∎

Now we will show that $\mathbb{O}(P)$ is closed on taking supremum of
countably many principal elements.

######  Lemma 4.13.

Let
$\mathcal{A}:=\\{(\downarrow\\!\\!x_{1}^{(i)},\downarrow\\!\\!x_{2}^{(i)},\ldots)\in\mathbb{O}(P):i\in\omega\\}$.
Then

$\sup(\mathcal{A})=\left(\bigcup_{i\in\omega}\downarrow\\!\\!x_{1}^{(i)},\bigcup_{i\in\omega}\downarrow\\!\\!x_{2}^{(i)},\ldots\right).$

###### Proof.

We begin by proving that
$(\bigcup_{i\in\omega}\downarrow\\!\\!x_{1}^{(i)},\bigcup_{i\in\omega}\downarrow\\!\\!x_{2}^{(i)},\ldots)\in\mathbb{O}(P)$.
Since $\mathcal{O}(P_{n})$ is finite, there is $m_{n}$ with
$\bigcup_{i<m_{n}}\downarrow\\!\\!x_{n}^{(i)}=\bigcup_{i\in\omega}\downarrow\\!\\!x_{n}^{(i)}$.
We may assume $m_{n}\leq m_{n+1}$ for all $n\in\mathbb{N}$. Then

$\hat{p}_{n}^{n+1}\left(\bigcup_{i\in\omega}\downarrow\\!\\!x_{n+1}^{(i)}\right)=\hat{p}_{n}^{n+1}\left(\bigcup_{i<m_{n+1}}\downarrow\\!\\!x_{n+1}^{(i)}\right)=\bigcup_{i<m_{n+1}}\hat{p}_{n}^{n+1}\left(\downarrow\\!\\!x_{n+1}^{(i)}\right)=\bigcup_{i<m_{n+1}}\downarrow\\!\\!x_{n}^{(i)}=\bigcup_{i\in\omega}\downarrow\\!\\!x_{n}^{(i)}.$

So
$\left(\bigcup_{i\in\omega}\downarrow\\!\\!x_{1}^{(i)},\bigcup_{i\in\omega}\downarrow\\!\\!x_{2}^{(i)},\ldots\right)\in\mathbb{O}(P)$.

Clearly
$\left(\bigcup_{i\in\omega}\downarrow\\!\\!x_{1}^{(i)},\bigcup_{i\in\omega}\downarrow\\!\\!x_{2}^{(i)},\ldots\right)$
is an upper bound of $\mathcal{A}$ so we only need to show that it is the
least upper bound. Take any upper bound
$s=(s_{1},s_{2},\ldots)\in\mathbb{O}(P)$ of $\mathcal{A}$. Then
$\downarrow\\!\\!x_{n}^{(i)}\subset s_{n}$ for all
$n\in\mathbb{N},i\in\omega$. Thus
$\bigcup_{i\in\omega}\downarrow\\!\\!x_{n}^{(i)}\subset s_{n}$ for all
$n\in\mathbb{N}$. So
$\left(\bigcup_{i\in\omega}\downarrow\\!\\!x_{1}^{(i)},\bigcup_{i\in\omega}\downarrow\\!\\!x_{2}^{(i)},\ldots\right)\leq(s_{1},s_{2},\ldots)$,
and finally
$\sup(\mathcal{A})=\left(\bigcup_{i\in\omega}\downarrow\\!\\!x_{1}^{(i)},\bigcup_{i\in\omega}\downarrow\\!\\!x_{2}^{(i)},\ldots\right)$.
∎

The following is an immediate consequence of Lemma 4.12 and Lemma 4.13.

######  Corrolary 4.14.

For any non-zero element $\mathbf{a}\in\mathbb{O}(P)$ there are
$(x_{1}^{(i)},x_{2}^{(i)},\dots)$ in $P$, $i\in\omega$, such that
$\mathbf{a}=(\bigcup_{i\in\omega}\downarrow\\!\\!x_{1}^{(i)},\bigcup_{i\in\omega}\downarrow\\!\\!x_{2}^{(i)},\ldots)$.

We say that lattice $L$ is $\sigma$-complete, if for any countable set
$A\subset L$, both $\sup(A)$ and $\inf(A)$ exist. Now, we are ready to
formulate and prove the next property of $\mathbb{O}(P)$.

######  Theorem 4.15.

$\mathbb{O}(P)$ is $\sigma$-complete.

###### Proof.

Let $\mathcal{A}:=\\{(a_{1}^{(i)},a_{2}^{(i)},\dots),i\in\omega\\}$. From
Lemma 4.12 and Lemma 4.13 we get that
$\bigvee\mathcal{A}=\bigvee_{i\in\omega}\sup(\mathcal{A}_{i})$, where
$\sup(\mathcal{A}_{i})=(a_{1}^{(i)},a_{2}^{(i)},\dots)$ and $\mathcal{A}_{i}$
is a countable family consisting of principal elements. Now from Lemma 2.1 we
get that
$\bigvee_{i\in\omega}\sup(\mathcal{A}_{i})=\sup(\bigcup_{i\in\omega}\mathcal{A}_{i})$,
because $\sup(\bigcup_{i\in\omega}\mathcal{A}_{i})$ and
$\sup(\mathcal{A}_{i})$ exist as they are countable sets consisting of
principal elements.

Now we are going to show that $\bigwedge\mathcal{A}\in\mathbb{O}(P)$. Consider
the sequence
$x_{n}=(a_{1}^{(0)},a_{2}^{(0)},\dots)\wedge(a_{1}^{(1)},a_{2}^{(1)},\dots)\wedge\dots\wedge(a_{1}^{(n)},a_{2}^{(n)},\dots)$.
We can find $\ell_{0}$ such that for every $n\geq\ell_{0}$ the first
coordinate of $x_{n}$ is constant. We denote this constant coordinate by
$r_{1}$. Similarly we can find $\ell_{1}\geq\ell_{0}$ such that for every
$n\geq\ell_{1}$ the first coordinate of $x_{n}$ is $r_{1}$ and the second one
is constant. We denote this constant second coordinate by $r_{2}$. Proceeding
inductively we get a sequence $(r_{1},r_{2},\dots)$.

Now we will show that $\bigwedge\mathcal{A}=(r_{1},r_{2},\dots)$. See that for
every $n\in\mathbb{N}$, $\hat{p}_{n}^{n+1}(r_{n+1})=r_{n}$ as
$x_{\ell_{n}}\in\mathbb{O}(P)$. Thus $(r_{1},r_{2},\dots)\in\mathbb{O}(P)$.
Note that for every $n\in\omega$ we have $(r_{1},r_{2},\dots)\leq x_{n}$,
which implies $(r_{1},r_{2},\dots)\leq(a_{1}^{(n)},a_{2}^{(n)},\dots)$. In
other words, $(r_{1},r_{2},\dots)$ is a lower bound of $\mathcal{A}$. Let
$(s_{1},s_{2},\dots)$ be a lower bound of $\mathcal{A}$. Then
$(s_{1},s_{2},\dots)\leq x_{n}$ for every $n\in\omega$. By the definition of
$r_{i}$’s we obtain $s_{i}\subset r_{i}$, thus
$(s_{1},s_{2},\ldots)\leq(r_{1},r_{2},\ldots)$. Thus
$\bigwedge\mathcal{A}=(r_{1},r_{2},\dots)$. ∎

Finally, we are able to prove that $\mathbb{O(P)}$ is universal with respect
to induced quotient mappings, i.e. every inverse limit of order ideals
$\mathbb{O}(H)=\varprojlim\langle\\{\mathcal{O}(H_{n})\\}_{n\in\mathbb{N}},\\{\hat{h}^{n}_{k}\\}_{k<n}\rangle$
is a continuous quotient of $\mathbb{O(P)}$.

######  Theorem 4.16.

Let
$H=\varprojlim\langle\\{H_{n}\\}_{n\in\mathbb{N}},\\{h^{n}_{k}\\}_{k<n}\rangle$
and
$\mathbb{P}=\varprojlim\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$,
where $(P_{n})$ is a Fraïssé sequence and let $f:\mathbb{P}\to H$ be a
continuous quotient map. Then there is a continuous quotient map
$q:\mathbb{O(P)}\rightarrow\mathbb{O(\text{H})}$ which is join-preserving.
Furthermore the diagram

${H}$${\mathbb{P}}$${\mathbb{O}(H)}$${\mathbb{O(P)}}$$\scriptstyle{\psi}$$\scriptstyle{f}$$\scriptstyle{\phi}$$\scriptstyle{q}$
(17)

commutes.

###### Proof.

The quotient map $f$ has the following representation (see Theorem 3.13):
there are a sequence $i_{1}<i_{2}<\dots$ and quotient mappings
$f_{k}:P_{i_{k}}\to H_{k}$ such
$f(x_{1},x_{2},\dots)=(f_{1}(x_{i_{1}}),f_{2}(x_{i_{2}}),\dots)$ and diagram
(11) commutes for every $n,k\in\mathbb{N}$ with $k<n$.

Firstly we define $q$ for principal elements as follows:
$q(\downarrow\\!\\!x_{1},\downarrow\\!\\!x_{2},\ldots):=\psi(f(x_{1},x_{2},\ldots))=(\downarrow\\!\\!f_{1}(x_{i_{1}}),\downarrow\\!\\!f_{2}(x_{i_{2}}),\ldots)$.
Also we put $q(\emptyset,\emptyset,\ldots)=(\emptyset,\emptyset,\ldots)$.
Finally for arbitrary $\mathbf{a}=(a_{1},a_{2},\ldots)$ using Lemma 4.12 we
find a countable family
$\mathcal{A}=\\{(\downarrow\\!\\!x_{1}^{(t)},\downarrow\\!\\!x_{2}^{(t)},\ldots):t\in\omega\\}$
such that $\mathbf{a}=\sup\mathcal{A}$ and we put

$q(\mathbf{a}):=\sup\\{q(\downarrow\\!\\!x_{1}^{(t)},\downarrow\\!\\!x_{2}^{(t)},\ldots):t\in\omega\\}=\sup\\{(\downarrow\\!\\!f_{1}(x_{i_{1}}^{(t)}),\downarrow\\!\\!f_{2}(x_{i_{2}}^{(t)}),\ldots):t\in\omega\\}.$

Using Lemma 4.13 we get that in fact

$q(\mathbf{a})=\left(\bigcup_{t\in\omega}\downarrow\\!\\!f_{1}(x_{i_{1}}^{(t)}),\bigcup_{t\in\omega}\downarrow\\!\\!f_{2}(x_{i_{2}}^{(t)}),\dots\right)$

Clearly diagram (17) commutes. Now we will prove that the value of $q$ does
not depend on the choice of family $\mathcal{A}$ for $\mathbf{a}$. Assume that

$\mathbf{a}=(a_{1},a_{2},\dots)=\sup\\{(\downarrow\\!\\!x_{1}^{(t)},\downarrow\\!\\!x_{2}^{(t)},\ldots):t\in\omega\\}=\sup\\{(\downarrow\\!\\!y_{1}^{(t)},\downarrow\\!\\!y_{2}^{(t)},\ldots):t\in\omega\\}.$

We need to show that

$\sup\\{(\downarrow\\!\\!f_{1}(x_{i_{1}}^{(t)}),\downarrow\\!\\!f_{2}(x_{i_{2}}^{(t)}),\ldots):t\in\omega\\}=\sup\\{(\downarrow\\!\\!f_{1}(y_{i_{1}}^{(t)}),\downarrow\\!\\!f_{2}(y_{i_{2}}^{(t)}),\ldots):t\in\omega\\}$

which is, by Lemma 4.13 equivalent to

$\bigcup_{t\in\omega}\downarrow\\!\\!f_{l}(x_{i_{l}}^{(t)})=\bigcup_{t\in\omega}\downarrow\\!\\!f_{l}(y_{i_{l}}^{(t)})\text{
for all }l\in\mathbb{N}.$

Take any $l\in\mathbb{N}$. Again by Lemma we obtain
$\bigcup_{t\in\omega}\downarrow\\!\\!x_{i_{l}}^{(t)}=\bigcup_{t\in\omega}\downarrow\\!\\!y_{i_{l}}^{(t)}$.
Since $H_{i_{l}}$ is finite, we find $m_{l},n_{l}\in\mathbb{N}$ with

$\bigcup_{t\in\omega}\downarrow\\!\\!x_{i_{l}}^{(t)}=\bigcup_{t<m_{l}}\downarrow\\!\\!x_{i_{l}}^{(t)}\text{
and
}\bigcup_{t<n_{l}}\downarrow\\!\\!y_{i_{l}}^{(t)}=\bigcup_{t\in\omega}\downarrow\\!\\!y_{i_{l}}^{(t)}$

From Remark 4.3 we get that
$\bigcup_{t<m_{l}}\downarrow\\!\\!f_{l}(x_{i_{l}}^{(t)})=\bigcup_{t<n_{l}}\downarrow\\!\\!f_{l}(y_{i_{l}}^{(t)})$.
Note that

$\bigcup_{t\in\omega}\downarrow\\!\\!f_{l}(x_{i_{l}}^{(t)})=\bigcup_{t<m_{l}}\downarrow\\!\\!f_{l}(x_{i_{l}}^{(t)})=\bigcup_{t<n_{l}}\downarrow\\!\\!f_{l}(y_{i_{l}}^{(t)})=\bigcup_{t\in\omega}\downarrow\\!\\!f_{l}(y_{i_{l}}^{(t)}).$

Thus $q$ is well defined.

Now we will show that $q$ preserves suprema. Let
$\mathbf{a}=(a_{1},a_{2},\ldots),\mathbf{b}=(b_{1},b_{2},\ldots)$. We know
that both of these elements are countable supremum of principal elements
$(\downarrow\\!\\!x_{1}^{(t)},\downarrow\\!\\!x_{2}^{(t)},\dots)$ and
$(\downarrow\\!\\!y_{1}^{(t)},\downarrow\\!\\!y_{2}^{(t)},\dots)$,
respectively ($t\in\omega$). As value of $q$ does not depend on a chosen
decomposition of $\sup(\\{\mathbf{a},\mathbf{b}\\})$, we have that

$q(\sup(\\{\mathbf{a},\mathbf{b}\\}))=q\left(\bigcup_{t\in\omega}\downarrow\\!\\!x_{1}^{(t)}\cup\bigcup_{t\in\omega}\downarrow\\!\\!y_{1}^{(t)},\bigcup_{t\in\omega}\downarrow\\!\\!x_{2}^{(t)}\cup\bigcup_{t\in\omega}\downarrow\\!\\!y_{2}^{(t)},\ldots\right)=$
$\left(\bigcup_{t\in\omega}\downarrow\\!\\!f_{1}(x_{i_{1}}^{(t)})\cup\bigcup_{t\in\omega}\downarrow\\!\\!f_{1}(y_{i_{1}}^{(t)}),\bigcup_{t\in\omega}\downarrow\\!\\!f_{2}(x_{i_{2}}^{(t)})\cup\bigcup_{t\in\omega}\downarrow\\!\\!f_{2}(y_{i_{2}}^{(t)}),\ldots\right)=\sup(\\{q(\mathbf{a}),q(\mathbf{b})\\})$

Thus $q$ preserves suprema. And consequently, it preserves ordering as well.
Also $q$ is onto $\mathbb{O}(H)$. Indeed, take any
$\mathbf{g}\in\mathbb{O}(H)$. Then let
$\mathcal{G}:=\\{(\downarrow\\!\\!g_{1}^{(t)},\downarrow\\!\\!g_{2}^{(t)},\dots):t\in\omega\\}$
be a family of principal elements in $\mathbb{O}(H)$ such that
$\sup(\mathcal{G})=\mathbf{g}$. Observe that each element $\mathbf{x}$ of
family $\mathcal{G}$ can be expressed as a value
$\mathbf{x}=q(\mathbf{a})=\psi(f(\phi^{-1}(\mathbf{a})))$ of some principal
element $\mathbf{a}$ from $\mathbb{O(P)}$ as:

  1. (i)

$\psi$ is an isomorphism from $H$ to the set of all principal elements in
$\mathbb{O}(H)$.

  2. (ii)

$f$ is a quotient map from $\mathbb{P}$ to $H$.

  3. (iii)

$\phi^{-1}$ is an isomorphism from the set of all principal elements in
$\mathbb{O(P)}$ to $\mathbb{P}$.

Isomorphism is onto mapping and the composition of a onto mappings is an onto
mapping. Thus $q$ is a mapping from the set of all principal elements in
$\mathbb{O(P)}$ onto the set of all principal elements in $\mathbb{O}(H)$. For
each element $(\downarrow\\!\\!g_{1}^{(t)},\downarrow\\!\\!g_{2}^{(t)},\dots)$
in $\mathcal{G}$ we can find an element
$(\downarrow\\!\\!w_{1}^{(t)},\downarrow\\!\\!w_{2}^{(t)},\dots)$ such that
$q(\downarrow\\!\\!w_{1}^{(t)},\downarrow\\!\\!w_{2}^{(t)},\dots)=(\downarrow\\!\\!g_{1}^{(t)},\downarrow\\!\\!g_{2}^{(t)},\dots)$.
Let
$\mathcal{W}:=\\{(\downarrow\\!\\!w_{1}^{(t)},\downarrow\\!\\!w_{2}^{(t)},\dots)\in\mathbb{O(P)}:t\in\omega\\}$.
Clearly $q(\sup(\mathcal{W}))=\mathbf{g}$.

Now we must prove that $q$ is a quotient map for posets. Take
$\mathbf{a},\mathbf{b}\in\mathbb{O}(H)$ such that $\mathbf{a}\leq\mathbf{b}$.
Let $\mathcal{A},\mathcal{B}$ be families of principal elements such that
$\mathbf{a}=\sup(\mathcal{A})$ and $\mathbf{b}=\sup(\mathcal{B})$.

We will show that for any element
$(\downarrow\\!\\!a_{1}^{(t)},\downarrow\\!\\!a_{2}^{(t)},\ldots)$ of
$\mathcal{A}$ we will find
$(\downarrow\\!\\!b_{1}^{(m_{1})},\downarrow\\!\\!b_{2}^{(m_{2})},\ldots)$
such that
$(\downarrow\\!\\!a_{1}^{(t)},\downarrow\\!\\!a_{1}^{(t)},\ldots)\leq(\downarrow\\!\\!b_{1}^{(m_{1})},\downarrow\\!\\!b_{2}^{(m_{2})},\ldots)$
and $(\downarrow\\!\\!b_{n}^{(m_{i})})_{n\in\mathbb{N}}$ are elements of
$\mathcal{B}$ for every $i\in\mathbb{N}$.

Take any $t\in\omega$. Then
$(\downarrow\\!\\!a_{1}^{(t)},\downarrow\\!\\!a_{2}^{(t)},\ldots)\leq(\bigcup_{t\in\omega}\downarrow\\!\\!b_{1}^{(t)},\bigcup_{t\in\omega}\downarrow\\!\\!b_{2}^{(t)},\ldots)$.
Take any $n\in\mathbb{N}$. Then there is $m_{n}\in\omega$ (depending on $n$)
such that $\downarrow\\!\\!a_{n}^{(t)}\subset\downarrow\\!\\!b_{n}^{(m_{n})}$.

This proves that set (recall that $t$ is fixed!)

$G_{n}:=G_{n}^{(t)}=\\{(\downarrow\\!\\!a_{n}^{(t)},\downarrow\\!\\!b_{n}^{(m)})\in\mathcal{O}(H_{n})^{2}:\downarrow\\!\\!a_{n}^{(t)}\subset\downarrow\\!\\!b_{n}^{(m)}\text{
for some }m\\}$

is non-empty for all $n\in\mathbb{N}$. It is also finite as a subset of a
finite set.

By $T_{n}$ we denote that set of all elements
$((\downarrow\\!\\!a_{1}^{(t)},\downarrow\\!\\!b_{1}^{(m_{1})}),\ldots,(\downarrow\\!\\!a_{n}^{(t)},\downarrow\\!\\!b_{n}^{(m_{n})}))\in
G_{1}\times\ldots\times G_{n}$ such that

$(\hat{h}_{u}^{n}(\downarrow\\!\\!a_{n}^{(t)}),\hat{h}_{u}^{n}(\downarrow\\!\\!b_{n}^{(m_{n})}))=(\downarrow\\!\\!a_{u}^{(t)},\downarrow\\!\\!b_{u}^{(m_{u})})\text{
for any }u<n.$

Since $\hat{h}_{u}^{n}$ is a homomorphism,
$\downarrow\\!\\!a_{n}^{(t)}\subset\downarrow\\!\\!b_{n}^{(m_{n})}$ implies
$\downarrow\\!\\!a_{u}^{(t)}\subset\downarrow\\!\\!b_{u}^{(m_{u})}$ for every
$u<n$. Therefore $T_{n}$ is non-empty.

Using Lemma 3.8, for $q^{n}_{u}:G_{n}\to G_{u}$ given by
$q^{n}_{u}(\downarrow\\!\\!a_{n}^{(t)},\downarrow\\!\\!b_{n}^{(m_{n})})=(\hat{h}_{u}^{n}(\downarrow\\!\\!a_{n}^{(t)}),\hat{h}_{u}^{n}(\downarrow\\!\\!b_{n}^{(m_{n})}))$,
we get that there is a sequence
$((\downarrow\\!\\!a_{n}^{(t)},\downarrow\\!\\!b_{n}^{(m_{n})}))\in\prod_{n\in\mathbb{N}}G_{n}$
such that $\hat{h}_{n}^{n+1}(\downarrow\\!\\!a_{n+1}^{(t)})=a_{n}^{(t)}$ and
$\hat{h}_{n}^{n+1}(\downarrow\\!\\!b_{n+1}^{(m_{n+1})})=b_{n}^{(m_{n})}$ for
all $n\in\mathbb{N}$. Then
$(\downarrow\\!\\!b_{1}^{(m_{1})},\downarrow\\!\\!b_{2}^{(m_{2})},\dots)\leq\mathbf{b}$.
So, by the definition of $G_{n}^{(t)}$’s, we obtain that for every
$t\in\omega$ there is $(m_{n})\in\omega^{\mathbb{N}}$ such that
$(\downarrow\\!\\!a_{1}^{(t)},\downarrow\\!\\!a_{2}^{(t)},\ldots)\leq(\downarrow\\!\\!b_{1}^{(m_{1})},\downarrow\\!\\!b_{2}^{(m_{2})},\ldots)$.
(We should write $(m_{n}^{(t)})$, since the sequence $(m_{n})$ depends on $t$.
But we want our proof to be readable).

As $q$ is a quotient map for principal elements (this follows immediately from
the fact that $f$ is quotient), thus we will find
$(\downarrow\\!\\!c_{1}^{(t)},\downarrow\\!\\!c_{2}^{(t)},\ldots)$ and
$(\downarrow\\!\\!d_{1}^{(m_{1})},\downarrow\\!\\!d_{2}^{(m_{2})},\ldots)$ in
$\mathbb{O(P)}$ such that
$(\downarrow\\!\\!c_{1}^{(t)},\downarrow\\!\\!c_{2}^{(t)},\ldots)\leq(\downarrow\\!\\!d_{1}^{(m_{1})},\downarrow\\!\\!d_{2}^{(m_{2})},\ldots)$
and
$q(\downarrow\\!\\!c_{1}^{(t)},\downarrow\\!\\!c_{2}^{(t)},\ldots)=(\downarrow\\!\\!a_{1}^{(t)},\downarrow\\!\\!a_{2}^{(t)},\ldots),q(\downarrow\\!\\!d_{1}^{(m_{1})},\downarrow\\!\\!d_{2}^{(m_{2})},\ldots)=(\downarrow\\!\\!b_{1}^{(m_{1})},\downarrow\\!\\!b_{2}^{(m_{2})},\ldots)$.
Let $\mathcal{C}$ be the set of all
$(\downarrow\\!\\!c_{1}^{(t)},\downarrow\\!\\!c_{2}^{(t)},\ldots)$ and
$\mathcal{D}$ be the set of all corresponding
$(\downarrow\\!\\!d_{1}^{(m_{1})},\downarrow\\!\\!d_{2}^{(m_{2})},\ldots)$.

We know that for each
$(\downarrow\\!\\!b_{1}^{(m)},b_{2}^{(m)},\dots)\in\mathcal{B}$ we can find
$(\downarrow\\!\\!e_{1}^{(m)},\downarrow\\!\\!e_{2}^{(m)},\dots)\in\mathbb{O(P)}$
such that
$q(\downarrow\\!\\!e_{1}^{(m)},\downarrow\\!\\!e_{2}^{(m)},\dots)=(\downarrow\\!\\!b_{1}^{(m)},b_{2}^{(m)},\dots)$.
Let $\mathcal{E}$ be the set of all corresponding
$(\downarrow\\!\\!e_{1}^{(m)},\downarrow\\!\\!e_{2}^{(m)},\dots)$. Note that
$\sup(\mathcal{C})\leq\sup(\mathcal{D})\leq\sup(\mathcal{D})\vee\sup(\mathcal{E})$
and $q(\sup(\mathcal{C}))=\mathbf{a}$. Also
$q(\sup(\mathcal{D})\vee\sup(\mathcal{E}))=q(\sup(\mathcal{D}))\vee
q(\sup(\mathcal{E}))=\sup(q[\mathcal{D}])\vee\sup(q[\mathcal{E}])=\sup(\mathcal{B})=\mathbf{b}$.

At last we show that $q$ is continuous. Take any basic set
$U_{(z_{1},z_{2},\dots,z_{n})}\cap\mathbb{O}(H)$ and let
$(r_{1},r_{2},\dots)\in
q^{-1}[U_{(z_{1},z_{2},\dots,z_{n})}\cap\mathbb{O}(H)]=q^{-1}[U_{(z_{1},z_{2},\dots,z_{n})}]\cap\mathbb{O(P)}$.
Consider the following cases:

  * (i)

if $(r_{1},r_{2},\dots)$ is zero, then immediately from the definitions of $q$
and induced mapping we obtain

$q^{-1}[U_{(z_{1},z_{2},\dots,z_{n})}\cap\mathbb{O}(H)]=q^{-1}[\\{(\emptyset,\emptyset,\dots)\\}]=\\{(\emptyset,\emptyset,\dots)\\}.$

Note that

$\\{(\emptyset,\emptyset,\dots)\\}=(\\{\emptyset\\}\times
P_{2}\times\dots)\cap\mathbb{O(P)}$

so this set is open in $\mathbb{O(P)}$ and
$(r_{1},r_{2},\dots)\in\\{(\emptyset,\emptyset,\dots)\\}\subset
q^{-1}[U_{(z_{1},z_{2},\dots,z_{n})}]\cap\mathbb{O(P)}$.

  * (ii)

if $(r_{1},r_{2},\dots)$ is a non-zero element, then take a family of
principal elements
$\\{(\downarrow\\!\\!x_{1}^{(t)},\downarrow\\!\\!x_{2}^{(t)},\dots)\\}_{t\in\omega}$
such that
$(r_{1},r_{2},\dots)=\sup\\{(\downarrow\\!\\!x_{1}^{(t)},\downarrow\\!\\!x_{2}^{(t)},\dots):t\in\omega\\}$.
We will prove that
$(\\{r_{1}\\}\times\\{r_{2}\\}\times\dots\times\\{r_{i_{n}}\\}\times
P_{i_{n}+1}\times\dots)\cap\mathbb{O(P)}\subset
q^{-1}[U_{(z_{1},z_{2},\dots,z_{n})}\cap\mathbb{O}(H)]$. Let
$(l_{1},l_{2},\dots)\in(\\{r_{1}\\}\times\\{r_{2}\\}\times\dots\times\\{r_{i_{n}}\\}\times
P_{i_{n}+1}\times\dots)\cap\mathbb{O(P)}$ and take a family of principal
elements
$\\{(\downarrow\\!\\!y_{1}^{(t)},\downarrow\\!\\!y_{2}^{(t)},\dots):t\in\omega\\}$
with
$\sup\\{(\downarrow\\!\\!y_{1}^{(t)},\downarrow\\!\\!y_{2}^{(t)},\dots):t\in\omega\\}=(l_{1},l_{2},\dots)$.
Then

$\bigcup_{t\in\omega}\downarrow\\!\\!y_{i_{k}}^{(t)}=l_{i_{k}}=r_{i_{k}}=\bigcup_{t\in\omega}\downarrow\\!\\!x_{i_{k}}^{(t)}\text{
for }k\leq n,$

so

$z_{k}=\bigcup_{t\in\omega}\downarrow\\!\\!f_{k}(x_{i_{k}}^{(t)})=\bigcup_{t\in\omega}\downarrow\\!\\!f_{k}(y_{i_{k}}^{(t)})\text{
for }k\leq n.$

Since

$q(l_{1},l_{2},\dots)=(\bigcup_{t\in\omega}\downarrow\\!\\!f_{1}(y_{i_{1}}^{(t)}),\bigcup_{t\in\omega}\downarrow\\!\\!f_{2}(y_{i_{2}}^{(t)}),\dots,\bigcup_{t\in\omega}\downarrow\\!\\!f_{n}(y_{i_{n}}^{(t)}),\bigcup_{t\in\omega}\downarrow\\!\\!f_{n+1}(y_{i_{n+1}}^{(t)}),\dots)=$
$(z_{1},z_{2},\dots,z_{n},\bigcup_{t\in\omega}\downarrow\\!\\!f_{n+1}(y_{i_{n+1}}^{(t)}),\dots)$

we get $(l_{1},l_{2},\dots)\in
q^{-1}[U_{(z_{1},z_{2},\dots,z_{n})}\cap\mathbb{O}(H)]$, which means

$(\\{r_{1}\\}\times\\{r_{2}\\}\times\dots\times\\{r_{i_{n}}\\}\times
P_{i_{n}+1}\times\dots)\cap\mathbb{O(P)}\subset
q^{-1}[U_{(z_{1},z_{2},\dots,z_{n})}\cap\mathbb{O}(H)].$

This shows that $q^{-1}[U_{(z_{1},z_{2},\dots,z_{n})}\cap\mathbb{O}(H)]$ is a
neighborhood of all its points, which makes it open. Finally $q$ is
continuous.

∎

##  5 The category of finite lattices with order and join-irreducible
preserving quotient maps

In this section we are going to define a category in which the sequence
$\langle\\{\mathcal{O}(P_{n})\\}_{n\in\mathbb{N}},\\{\hat{p}^{n}_{k}\\}_{k<n}\rangle$
of order ideals related to
$\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{k}_{n}\\}_{k<n}\rangle$ (the
Fraïssé sequence (5)) is a Fraïssé sequence itself. To define that category we
need the notion of isomorphic arrows: we say that a pair $(g_{1},g_{2})$ is
the arrow from $f:A\to B$ to $f^{\prime}:A^{\prime}\to B^{\prime}$, if we the
following diagram

${A}$${B}$${A^{\prime}}$${B^{\prime}}$$\scriptstyle{g_{1}}$$\scriptstyle{f}$$\scriptstyle{g_{2}}$$\scriptstyle{f^{\prime}}$

commutes. The composition of these arrows is component-wise, i.e.
$(g_{1},g_{2})\circ(h_{1},h_{2})=(g_{1}\circ h_{1},g_{2}\circ h_{2})$. We say
that $(g_{1},g_{2})$ is an isomorphism, if there are $(h_{1},h_{2})$, such
that $(g_{1}\circ h_{1},g_{2}\circ
h_{2})=(\operatorname{id}_{A},\operatorname{id}_{B})$ and $(h_{1}\circ
g_{1},h_{2}\circ
g_{2})=(\operatorname{id}_{A^{\prime}},\operatorname{id}_{B^{\prime}})$. If
there is isomorphism between $f:A\to B$ and $f^{\prime}:A^{\prime}\to
B^{\prime}$, then we say $f$ and $f^{\prime}$ are isomorphic. Clearly, if
$g_{1}$ and $g_{2}$ are isomorphisms, then $(g_{1},g_{2})$ is also
isomorphism.

######  Theorem 5.1.

Let $L,T$ be finite distributive lattices, $p:L\to T$ be a mapping satisfying
the following conditions:

  * (i)

$p_{|\mathcal{J}(L)}$ is a quotient mapping;

  * (ii)

$p(\mathcal{J}(L))=\mathcal{J}(T)$;

  * (iii)

$p$ is a quotient mapping that is join-preserving.

Then $\hat{p}_{0}$ is isomorphic to $p$, where $p_{0}=p_{|\mathcal{J}(L)}$

###### Proof.

We will show that the following diagram

${T}$${L}$${\mathcal{O}(\mathcal{J}(T))}$${\mathcal{O}(\mathcal{J}(L))}$$\scriptstyle{\mathbf{B}_{T}}$$\scriptstyle{p}$$\scriptstyle{\mathbf{B}_{L}}$$\scriptstyle{\hat{p}_{0}}$

is commutative, where $\mathbf{B}_{T}$ and $\mathbf{B}_{L}$ are the
isomorphisms from Birkhoff’s Theorem (Theorem 2.10). Take any
$x\in\mathcal{J}(L)$. Then $\mathbf{B}_{T}(p(x))=\downarrow\\!\\!p(x)$ and
$\hat{p}_{0}(\mathbf{B}_{L}(x))=\hat{p}_{0}(\downarrow\\!\\!x)=\downarrow\\!\\!p(x)$,
so the diagram commutes for join-irreducible elements.

Now let $a\in L\setminus\mathcal{J}(L)$ be non-zero. Then
$\mathbf{B}_{L}(a)=\bigcup_{i=1}^{n}\downarrow\\!\\!a_{i}$, thus
$\hat{p}_{0}(\mathbf{B}_{L}(a))=\bigcup_{i=1}^{n}\downarrow\\!\\!p(a_{i})$.
See that
$\mathbf{B}_{T}(p(a))=\mathbf{B}_{T}(p(\mathbf{B}_{L}^{-1}(\bigcup_{i=1}^{n}\downarrow\\!\\!a_{i})))=\mathbf{B}_{T}(p(\bigvee_{i=1}^{n}a_{i}))=\bigcup_{i=1}^{n}\mathbf{B}_{T}(p(a_{i}))=\bigcup_{i=1}^{n}\downarrow\\!\\!p(a_{i})$.

Now see that any non-zero elements of $L$ are mapped to non-zero elements of
$T$ by $p$, and $p$ is onto, thus $p(\mathbf{0}_{L})=\mathbf{0}_{T}$. Hence
$\hat{p}_{0}(\mathbf{B}(\mathbf{0}_{L}))=\hat{p}_{0}(\emptyset)=\emptyset=\mathbf{B}(\mathbf{0}_{T})=\mathbf{B}(p(\mathbf{0}_{L}))$.
This proves that the diagram commutes. Since $\mathbf{B}$ is the isomorphism,
$p$ and $\hat{p}_{0}$ are isomorphic. ∎

Theorem 5.1 shows that we can identify mappings satisfying conditions
(i)–(iii) with induced quotient mappings. Conditions (i)–(iii) allows us to
have the inner definition of induced quotient mappings (up to isomorphism).
Using this we define a category of finite distributive lattices with mappings
isomorphic to induced quotient mappings. From now on we do not distinguish
mappings from that isomorphic to them.

Let $V$ be a countable class of all finite distributive lattices. Formally,
one can define $V$ as a set of all order ideals of finite posets on
$\mathbb{N}$. Let $A$ be the class of all induced quotient mappings between
finite distributive lattices; we say that $f$ is an arrow from $A$ to $B$,
provided $f$ is an induced quotient mapping from $B$ onto $A$. Then
$\operatorname{dom}f$ is the range of $f$ and $\operatorname{cod}f$ is the
domain of $f$. The composition $g\bullet f$ of two arrows, $f$ from $A$ to $B$
and $g$ from $B$ to $C$, is $f\circ g:C\to A$, where $\circ$ is the usual
mappings composition. Then $\mathfrak{L}:=\langle
V,A,\operatorname{dom},\operatorname{cod},\bullet\rangle$ is a category.

###  5.1 $\mathcal{O}(P_{n})$ is a Fraïssé sequence in $\mathfrak{L}$

Let $\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k<n}\rangle$ be the
inverse system defined in Section 3.2. We will prove that
$\langle\\{\mathcal{O}(P_{n})\\}_{n\in\mathbb{N}},\\{\hat{p}^{n}_{k}\\}_{k<n}\rangle$
is a Fraïssé sequence in $\mathfrak{L}$.

Let $L=\mathcal{O}(P)$ be a finite, distributive lattice. We know there exists
$n\in\mathbb{N}$ and quotient mapping $q:P\to P_{n}$. Then $\hat{q}$ is an
induced quotient mapping from $L$ onto $\mathcal{O}(P_{n})$. This shows that
the (U) condition holds.

Take any $n\in\mathbb{N}$ and an induced quotient mapping
$h:L\to\mathcal{O}(P_{n})$. There is a quotient $f:\mathcal{J}(L)\to P_{n}$
with $h=\hat{f}$. There is $m>n$ and $g$ such that the following diagram

${\mathcal{O}(P_{n})}$${L=\mathcal{O}(\mathcal{J}(L))}$${\mathcal{O}(P_{m})}$${P_{n}}$${\mathcal{J}(L)}$${P_{m}}$$\scriptstyle{h=\hat{f}}$$\scriptstyle{\hat{g}}$$\scriptstyle{f}$$\scriptstyle{g}$

commutes, where $g\circ f=p_{n}^{m}$ and up arrows are canonical poset
embeddings into order ideals. By Lemma 4.4,
$\hat{g}\circ\hat{f}=\hat{p}_{n}^{m}$, which shows property (A) and finishes
the proof.

###  5.2 A different approach to the inverse limit of
$\langle\\{\mathcal{O}(P_{n})\\}_{n\in\mathbb{N}},\\{\hat{p}^{n}_{k}\\}_{k<n}\rangle$.

Let $\langle\\{P_{n}\\}_{n\in\mathbb{N}},\\{p^{n}_{k}\\}_{k\leq n}\rangle$ be
the inverse system defined in Section 3.2, which inverse limit is
$\mathbb{P}$. We know that $P_{n}$ is isomorphic to $\dot{\bigcup}_{i<2\cdot
4^{n-1}}\mathbf{2}$. Moreover for any two $x,y\in P_{n}$ with $x<_{n}y$, there
are $x_{i},y_{i}\in P_{n+1}$, $i=1,2,3,4$, with $x_{i}<_{n+1}y_{i}$, and such
that

  * •

$p_{n}(x_{1})=p_{n}(x_{2})=x$, $p_{n}(y_{1})=p_{n}(y_{2})=y$, that is $p_{n}$
maps both $\mathbf{2}$-components $\\{x_{1},y_{1}\\}$ and $\\{x_{2},y_{2}\\}$
onto a $\mathbf{2}$-component $\\{x,y\\}$;

  * •

$p_{n}(x_{3})=p_{n}(y_{3})=x$, that is $p_{n}$ maps a $\mathbf{2}$-component
$\\{x_{3},y_{3}\\}$ onto $x$;

  * •

$p_{n}(x_{4})=p_{n}(y_{4})=y$, that is $p_{n}$ maps a $\mathbf{2}$-component
$\\{x_{4},y_{4}\\}$ onto $y$.

Note that exactly those properties of $\langle(P_{n}),(p^{n}_{k})_{k\leq
n}\rangle$ where needed to prove that it is a Fraisse sequence in category of
finite posets, see Theorem … and its proof.

Let $T_{n}=\\{x\in\\{0,1,2,3\\}^{n}:x(0)=0,1\\}$. Note that $|T_{n}|=2\cdot
4^{n-1}$ is the same as the number of $\mathbf{2}$-components of $P_{n}$. We
use $T_{n}$ to represent $\mathbf{2}$-components of $P_{n}$ in the following
way. $T_{1}=\\{0,1\\}$ and 0 and 1 represents two $\mathbf{2}$-components of
$P_{1}$. For $c\in T_{n}$ let $\\{x,y\\}$ with $x<y$ be a
$\mathbf{2}$-component of $P_{n}$. Then $c\hat{\;\;}0$ and $c\hat{\;\;}1$
represent those two $\mathbf{2}$-components of $P_{n+1}$ which are mapped onto
$\\{x,y\\}$, $c\hat{\;\;}2$ represents $\mathbf{2}$-component of $P_{n+1}$
which is mapped onto $x$, and $c\hat{\;\;}3$ represents $\mathbf{2}$-component
of $P_{n+1}$ which is mapped onto $y$.

Now, let us represent $\mathcal{O}(P_{n})$ as $\\{0,1,2\\}^{T_{n}}$. The idea
is the following. Any down-set can be represent as a union of principal down-
sets. Any principal down-set in $P_{n}$ is of the form $\\{x\\}$ or
$\\{x,y\\}$ where $c:=\\{x,y\\}$ with $x<y$ is a $\mathbf{2}$-component. Now,
a down-set $A\in\mathcal{O}(P_{n})$ can be represent as a function
$f:T_{n}\to\\{0,1,2\\}$ such that $f(c)=2$ is $A$ contains $\\{x,y\\}$,
$f(c)=1$ if $A$ contains $\\{x\\}$ but does not $\\{x,y\\}$, and $f(c)=0$ if
$A$ does not contain $\\{x\\}$.

For every $f,g\in\\{0,1,2\\}^{T_{n}}$ we define $(f\vee
g)(c)=\max\\{f(c),g(c)\\}$ and $(f\wedge g)(c)=\min\\{f(c),g(c)\\}$ for every
$c\in T_{n}$. Clearly $\\{0,1,2\\}^{T_{n}}$ is a lattice, with $\vee$ and
$\wedge$ as join and meet, respectively.

Note that the set of all join-irreducible elements
$\mathcal{J}(\\{0,1,2\\}^{T_{n}})$ in $\\{0,1,2\\}^{T_{n}}$ consists of those
$f\in\\{0,1,2\\}^{T_{n}}$ for which there is exactly one $c\in T_{n}$ with
$f(c)\neq 0$; consequently $f(c)\in\\{1,2\\}$. For given
$f\in\mathcal{J}(\\{0,1,2\\}^{T_{n}})$ we denote the unique $c\in T_{n}$ with
$f(c)\neq 0$ by $c_{f}$. We also say that $f$ is supported by $c$ if
$c=c_{f}$.

The following fact formalizes the connection between $\mathcal{O}(P_{n})$ and
$\\{0,1,2\\}^{T_{m}}$.

######  Lemma 5.2.

Let $\psi:\mathcal{O}(\dot{\bigcup}_{\mathbf{x}\in
T_{n}}\mathbf{2})\to\\{0,1,2\\}^{T_{n}}$ be defined step by step as follows

  * •

$\psi(\emptyset)=0$ (zero function)

  * •

For every $\mathbf{2}$-component $\\{x,y\\}$ with $x<y$ and $c_{0}\in T_{n}$
which represents that $\mathbf{2}$-component we put

$\psi(\downarrow\\!\\!x)(c)=\begin{cases}0,&c\neq c_{0}\\\
1,&c=c_{0},\end{cases}\;\;\text{ and
}\;\;\psi(\downarrow\\!\\!y)(c)=\begin{cases}0,&c\neq c_{0}\\\
2,&c=c_{0},\end{cases}$

  * •

For every non-empty join-irreducible element
$A\in\mathcal{O}(\dot{\bigcup}_{\mathbf{x}\in T_{n}}\mathbf{2})$ and its
canonical decomposition $\\{\downarrow\\!\\!a_{i}:i<k\\}$ we define
$\psi(A)=\bigvee_{i=1}^{k}\psi(\downarrow\\!\\!a_{i})$

Then $\psi$ is a lattice isomorphism.

We will call $\psi$ the canonical isomorphism between
$\mathcal{O}(\dot{\bigcup}_{\mathbf{x}\in T_{n}}\mathbf{2})$ and
$\\{0,1,2\\}^{T_{n}}$.

###### Proof.

By Lemma 2.3 we only need to show that $\psi$ is an order-isomorphism.

Firstly we show that it is join-preserving.

See that for every $x,y\in\dot{\bigcup}_{\mathbf{x}\in T_{n}}\mathbf{2}$ we
have that
$\psi(\downarrow\\!\\!x\cup\downarrow\\!\\!y)=\psi(\downarrow\\!\\!x)\vee\psi(\downarrow\\!\\!y)$.
To see it we need to consider the following two cases:

  * •

If $\downarrow\\!\\!x\subsetneq\downarrow\\!\\!y$ then $x<_{n}y$ by Lemma 2.2.
Then $\psi(\downarrow\\!\\!x\cup\downarrow\\!\\!y)=\psi(\downarrow\\!\\!y)=g$
and $g(c_{g})=2$. See that $\psi(\downarrow\\!\\!x)=f$ where $f$ is such that
$f(c_{g})=1$ and $f(c)=0$ for remaining $c$’s. Then $f\vee g=g$, and
consequently $\psi(\downarrow\\!\\!x\cup\downarrow\\!\\!y)=g=f\vee
g=\psi(\downarrow\\!\\!x)\vee\psi(\downarrow\\!\\!y)$.

  * •

If $\downarrow\\!\\!x\parallel\downarrow\\!\\!y$, then immediately from the
definition of $\psi$ we have
$\psi(\downarrow\\!\\!x\cup\downarrow\\!\\!y)=\psi(\downarrow\\!\\!x)\vee\psi(\downarrow\\!\\!y)$.

By induction it can be proven that
$\psi(\bigcup_{i=1}^{k}\downarrow\\!\\!x_{i})=\bigvee_{i=1}^{k}\psi(\downarrow\\!\\!x_{i})$.
This implies that $\psi(A\cup B)=\psi(A)\vee\psi(B)$ (cf. proof of Theorem
4.1). Thus $\psi$ is join-preserving. Consequently $A\subset B$ implies
$\psi(A)\leq\psi(B)$ for any $A,B\in\mathcal{O}(\dot{\bigcup}_{\mathbf{x}\in
T_{n}}\mathbf{2})$.

Now we will prove that $\psi(A)\leq\psi(B)$ implies $A\subset B$ for any
$A,B\in\mathcal{O}(\dot{\bigcup}_{\mathbf{x}\in T_{n}}\mathbf{2})$. We start
proving it for principal down-sets. Let $x,y\in P_{n}$ be such that
$\psi(\downarrow\\!\\!x)\leq\psi(\downarrow\\!\\!y)$. Let
$f=\psi(\downarrow\\!\\!x),g=\psi(\downarrow\\!\\!y)$. Then $f\leq g$ means
that $c_{f}=c_{g}$. Thus $x,y$ are in one $\mathbf{2}$-component of $P_{n}$.
Since $f(c_{f})\leq g(c_{g})$, then $x\leq_{P_{n}}y$. Consequently
$\downarrow\\!\\!x\subset\downarrow\\!\\!y$ by Lemma 2.2.

Now we consider the general case. Let
$A=\bigcup_{i=1}^{k}\downarrow\\!\\!a_{i},B=\bigcup_{i=1}^{l}\downarrow\\!\\!b_{i}\in\mathcal{O}(\dot{\bigcup}_{\mathbf{x}\in
T_{n}}\mathbf{2})$ and assume that
$\bigvee_{i=1}^{k}\psi(\downarrow\\!\\!a_{i})=\psi(A)\leq\psi(B)=\bigvee_{i=1}^{l}\psi(\downarrow\\!\\!b_{i})$.
By Lemma 2.7 for every $\downarrow\\!\\!a_{i}$ we can find
$\downarrow\\!\\!b_{j_{i}}$ such that
$\psi(\downarrow\\!\\!a_{i})\leq\psi(\downarrow\\!\\!b_{j_{i}})$. Then
$\downarrow\\!\\!a_{i}\subset\downarrow\\!\\!b_{j_{i}}$ for every
$i\in\\{1,\dots,k\\}$. Thus $A\subset B$.

The fact that $\psi$ is an injection follows from preserving order both ways.

Now we prove that $\psi$ is surjective. Take any $f\in\\{0,1,2\\}^{T_{n}}$.
Then $f=\bigvee_{i=1}^{k}f_{i}$ where
$f_{1},\dots,f_{k}\in\mathcal{J}(\\{0,1,2\\}^{T_{n}})$. For every $f_{i}$ we
can find $\downarrow\\!\\!x_{i}$ such that $x_{i}$ is in
$\mathbf{2}$-component represented by $c_{f_{i}}$. Depending on value of
$f(c_{f_{i}})$ we take $x_{i}$ to be lower or upper element of
$\mathbf{2}$-component $c_{f_{i}}$. Then $\psi(\downarrow\\!\\!x_{i})=f_{i}$.
Therefore
$\psi(\bigcup_{i=1}^{k}\downarrow\\!\\!x_{i})=\bigvee_{i=1}^{k}f_{i}=f$. ∎

Let $\psi$ be the canonical isomorphism between $\mathcal{O}(P_{n})$ and
$\\{0,1,2\\}^{T_{n}}$, and $\phi$ be the canonical isomorphism between
$\mathcal{O}(P_{n+1})$ and $\\{0,1,2\\}^{T_{n+1}}$.

We will define $q_{n}^{n+1}:\\{0,1,2\\}^{T_{n+1}}\to\\{0,1,2\\}^{T_{n}}$ such
that diagram

${\mathcal{O}(P_{n})}$${\mathcal{O}(P_{n+1})}$${\\{0,1,2\\}^{T_{n}}}$${\\{0,1,2\\}^{T_{n+1}}}$$\scriptstyle{\psi}$$\scriptstyle{\widehat{p}_{n}^{n+1}}$$\scriptstyle{\phi}$$\scriptstyle{q_{n}^{n+1}}$
(18)

commutes. As $\mathcal{O}(P_{n})$ is isomorphic to $\\{0,1,2\\}^{T_{n}}$, we
can express every non-zero, non join-irreducible element of
$\\{0,1,2\\}^{T_{n}}$ as a join of join-irreducible, non-comparable elements
of $\\{0,1,2\\}^{T_{n}}$. We only need to define $q_{n}^{n+1}$ for join-
irreducible elements of $T_{n+1}$ and expand it using the aforementioned
property. Let $f\in\mathcal{J}(\\{0,1,2\\}^{T_{n+1}})$. Then $f(c)\neq 0\iff
c=c_{f}$. We define $q_{n}^{n+1}(f)=g$ where

$g({c_{f}}\upharpoonright n)=\begin{cases}f(c_{f})&\text{ if
}c_{f}(n)\in\\{0,1\\}\\\ 1&\text{ if }c_{f}(n)=2\\\ 2&\text{ if
}c_{f}(n)=3\end{cases}\;\;\text{ and }\;\;g(c)=0\iff c\neq
c_{f}\upharpoonright n.$

Then $g\in\mathcal{J}(\\{0,1,2\\}^{T_{n}})$. Additionally, for $f\equiv 0$ we
put $q_{n}^{n+1}(f)\equiv 0\in\\{0,1,2\\}^{T_{n}}$.

In general, for $f\in\\{0,1,2\\}^{T_{n+1}}$ we have

$f=\bigvee_{c\in T_{n+1}}f_{c}=\max_{c\in T_{n+1}}f_{c}$

where $f_{c}$ is join-reducible supported by $c$. This implies that

$q_{n}^{n+1}(f)(c)=\max\big{\\{}f(c\hat{\;\;}0),f(c\hat{\;\;}1),\min\\{f(c\hat{\;\;}2),1\\},2\cdot\min\\{f(c\hat{\;\;}3),1\\}\big{\\}}$

for every $f\in\\{0,1,2\\}^{T_{n}}$ and $c\in T_{n}$.

We have proven the following.

######  Proposition 5.3.

The lattice $\mathbb{O(P)}$ is isomorphic to the inverse limit of inverse
system
$\langle\\{\\{0,1,2\\}^{T_{n}}\\}_{n\in\mathbb{N}},\\{q^{n}_{k}\\}_{k<n}\rangle$.

We hope that this representation of $\mathbb{O(P)}$ will allow to study its
properties in future.

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