# On profiniteness and Hausdorffness of topological residuated lattices

Jiang Yanga, Pengfei Heb,∗, Juntao Wangc School of Mathematics, Northwest
University, Xi’an, 710127, China  School of Mathematics and Information
Science, Shaanxi Normal University, Xi’an, 710119, China  School of Science,
Xi’an Shiyou University, Xi’an, 710065, China

###### Abstract

The aim of this paper is to study the profiniteness of compact topological
residuated lattices and the existence of Hausdorff topological residuated
lattices. Firstly, we study profinite residuated lattices and obtain
sufficient and necessary conditions for profiniteness in compact topological
residuated lattices. These conditions include topological and algebraic
characterizations. Moreover, it order to study the existence of Hausdorf
topological residuated lattices, we investigate finiteness conditions in
residuated lattices. Finally, we investigate linear topological residuated
lattices and give the class of residuated lattices that can be endowed with a
non-trivial Hausdorff topology.

###### keywords:

topological algebra; profiniteness; finiteness condition; Hausdorffness

##  1 Introduction

Recently, algebraic methods have been applied successfully to non-classical
logics via universal algebra and algebraic logic. While Boolean algebras are
algebraic semantics for two-valued logics, multiple-valued logic algebras
serve as algebraic semantics for non-classical logics. Among various multiple-
valued logic algebras, residuted lattices are very basic and important
algebraic structures. In fact, residuated lattices are structures that have
been studied by algebraists since the 1930s in [36], they include structures
such as lattices of ideals of a ring, lattice ordered groups, algebras of
relations. But the study has been revived recently as a study of mathematical
structures for substructural logics [17], that is, logics which lack some of
the three structural rules, namely contraction, weakening, and exchange. The
variety of residuated lattices is the equivalent algebraic semantics, in the
sense of Blok-Pigozzi [8], of the Full Lambek Calculus $FL$. The variety of
bounded commutative and integral residuated lattices, that is, the equivalent
algebraic semantics of $FL_{ew}$ , the calculus that results from $FL$ by
adding two structural rules: exchange and weakening. Important subvarieties of
bounded commutative and integral residuated lattices are Boolean algebras,
Heyting algebras, $MTL$-algebras, $BL$-algebras, $MV$-algebras.

Topology and algebra, two fundamental domains of mathematics, play
complementary roles. When an algebraic structure and a topology come naturally
together, the rules that describe the relationship between topology and
algebraic operation are almost always transparent, and naturally the operation
has to be continuous or semicontinuous. In recent decades, there has been a
lot of research [10, 11, 26, 27, 29, 37] on algebras endowed with a topology,
the resulting structures are called topological algebras. The most important
topologies on algebras to be considered are linear topologies. For example, it
has been noted by G. Birkhoff ([6] pp.52-54) that under certain conditions a
family of subgroups may be used to definite neighborhoods of the identity in a
group and thus introduce a topology. Such “subgroup topologies” called linear
topologies were introduced and their main properties follow from [14, 16, 22]
treatment of topological algebras as uniform structures. Moreover, Zahiri and
Borzooei [41] used a special family of filters on a $BL$-algebra $L$ to
construct a topology $\tau$ on $L$. This topology is not only linear but also
compatible with all operations of $L$. In [40], Yang et al. obtained that for
each filter $F$ of a residuated lattice $L$, the topology
$\mathcal{T}_{F}=\\{U\subseteq L:\forall x\in U,x/F\subseteq U\\}$ is not only
linear but also Alexandrov, furthermore, $(L,\mathcal{T}_{F})$ is a
topological residuated lattice. Also, a question of interest arises in [40],
that is, whether the sets $\mathcal{F}(L)$ and $LTRL(L)$ are equipotent, where
the set $\mathcal{F}(L)$ of all filters of $L$ and the set $LTRL(L)$ of all
linear topological residuated lattices with the underlying set $L$? In [20],
we prove that $|\mathcal{F}(L)|=|ZLTRL(L)|$ for a residuated lattice $L$,
where the set $ZLTRL(L)$ of all zero-dimensional linear topological residuated
lattices with the underlying set $L$. We will correct this result in this
paper. Moreover, a perfectly valid question is whether an algebra can in fact
be endowed with a Hausdorff, non-discrete (linear) topology. This question has
been settled in infinite Abelian groups, see Theorem 1.5 in [14] or Theorem
7.7 in [16]. In this paper, we shall discuss the same question for residuated
lattices, that is, the existence of Hausdorff topological residuated lattices.

It is well known that an algebra is profinite if it is representable as a
projective limit of finite algebras. Profinite algebras and profinite
completions have their origins in algebraic number theory and Galois theory.
Profiniteness is a property referring to the interplay between the topological
and the algebraic structure of a topological algebra, whereas for topological
groups, rings, semigroups, and distributive lattices(see for instance [24],
Sec.VI.2), profiniteness turns out to be a purely topological property as it
is equivalent to the underlying topological space being a Stone space (i.e.,
the topology is compact, Hausdorff, and zero-dimensional). N. Nikolov and D.
Segal [31, 32] have proved that the topological structure of a finitely
generated profinite group is completely determined by its algebraic structure;
more precisely, they proved that the subgroups of finite index of such a group
are precisely its open subgroups. This result is generalized to many algebras
of logic which are our main focus, including Heyting algebras [4],
$MV$-algebras [30], orthomodular lattices [13]. To the best of our knowledge,
up to now, profinite residuated lattices have received very little attention.
In this paper, we investigate profinite residuated lattices. In particular, we
give characterizations of the profiniteness of compact topological residuated
lattices.

##  2 Preliminaries

In this section, we summarize some basic definitions and results, which will
be used.

###### Definition 2.1

_[21] An algebraic structure
$\mathcal{L}=(L,\wedge,\vee,\odot,\rightarrow,0,1)$ of type $(2,2,2,\\\
2,0,0)$ is called a _residuated lattice_ if it satisfies the following
conditions:_

  * (1)

$(L,\wedge,\vee,0,1)$ is a bounded lattice;

  * (2)

$(L,\odot,1)$ is a commutative monoid;

  * (3)

$x\odot y\leq z$ if and only if $x\leq y\rightarrow z$, for all $x,y,z\in L$,
where $\leq$ is the partial order of the lattice $(L,\wedge,\vee,0,1)$.

Throughout this paper we will slightly abuse notation $L$ the universe of a
residuated lattice $\mathcal{L}=(L,\wedge,\vee,\odot,\rightarrow,0,1)$, when
there is no chance to confusion.

For convenience of readers, we provide some basic properties of residuated
lattices in the following proposition.

###### Proposition 2.2

[35] In any residuated lattice $(L,\wedge,\vee,\odot,\rightarrow,0,1)$, the
following properties hold: for any $x,y,z\in L$,

  1. $(R_{1})$

$1\rightarrow x=x$, $x\rightarrow 1=1$;

  2. $(R_{2})$

$x\leq y$ if and only if $x\rightarrow y=1$;

  3. $(R_{3})$

if $x\leq y$, then $y\rightarrow z\leq x\rightarrow z$, $z\rightarrow x\leq
z\rightarrow y$ and $x\odot z\leq y\odot z$;

  4. $(R_{4})$

$x\odot(x\rightarrow y)\leq y$;

  5. $(R_{5})$

$x\odot y\leq x\wedge y$, $x\leq y\rightarrow x$;

  6. $(R_{6})$

$x\rightarrow(y\rightarrow z)=(x\odot y)\rightarrow
z=y\rightarrow(x\rightarrow z)$;

  7. $(R_{7})$

$x\vee(y\odot z)\geq(x\vee y)\odot(x\vee z)$, hence $x^{m}\vee y^{n}\geq(x\vee
y)^{mn}$.

###### Definition 2.3

_[35] Let $(L,\wedge,\vee,\odot,\rightarrow,0,1)$ be a residuated lattice. A
_filter_ is a nonempty set $F\subseteq L$ such that for each $x,y\in L$,_

  1. (i)

$x,y\in F$ implies $x\odot y\in F$,

  2. (ii)

if $x\in F$ and $x\leq y$, then $y\in F$.

Note that in a residuated lattice $L$, a filter $F$ of $L$ is equivalent to a
_deductive system_ , that is, $F$ satisfies the following conditions: (i)
$1\in F$, and (ii) $x,~{}x\rightarrow y\in F$ implies $y\in F$.

Let $L$ be a residuated lattice. By $\mathcal{F}(L)$($Con(L)$), we mean the
set of all filters (congruences) of $L$. There is close correspondence between
congruences and filters of residuated lattices. For each congruence $\theta$
on the residuated lattice $L$, let $[1]_{\theta}=\\{x\in L:(1,x)\in\theta\\}$.
Then $[1]_{\theta}$ is a filter of $L$, called the filter _determined_ by a
congruence $\theta$. Conversely, for each filter $F$, $\theta_{F}=\\{(x,y)\in
L\times L:(x\rightarrow y)\odot(y\rightarrow x)\in F\\}$ is a congruence on
$L$, called the congruence _determined_ by a filter $F$. For any $x\in L$, let
$x/F$ be the equivalence class $x/\theta_{F}$. If we denote by $L/F$ the
quotient set $L/\theta_{F}$, then $L/F$ becomes a residuated lattice with the
operations induced from those of $L$. Moreover, the following result holds.

###### Theorem 2.4

[17] For every residuated lattice $L$, we have

$\mathcal{F}(L)\cong Con(L).$

We recall that the notions of _uniform structures_ and _uniform topologies_.
We shall use the following notation.

Let $X$ be a set. We denote by $\Delta_{X}$ the diagonal in $X\times X$,
namely $\Delta_{X}=\\{(x,x):x\in X\\}\subset X\times X$. Suppose that $R$ is a
subset of $X\times X$(in other words, $R$ is a binary relation on $X$). For
any $y\in X$, we define the set $R[y]=\\{x\in X:(x,y)\in R\\}$. The _inverse_
$R^{-1}\subset X\times X$ of $R$ is defined by $R^{-1}=\\{(x,y):(y,x)\in
R\\}$. One says that $R$ is symmetric if it satisfies $R^{-1}=R$. If $R$ and
$S$ are subsets of $X\times X$, we define their _composite_ $R\circ S\subset
X\times X$ by $R\circ S=\\{(x,y):\exists z\in Xs.t.~{}(x,z)\in
R~{}and~{}(z,y)\in S\\}$.

###### Definition 2.5

[23] Let $X$ be a set. A _uniform structure_ on $X$ is a non-empty set
$\mathcal{U}$ of subsets of $X\times X$ satisfying the following conditions:

  * (1)

if $V\in\mathcal{U}$, then $\Delta_{X}\subset V$;

  * (2)

if $V\in\mathcal{U}$ and $V\subset V^{\prime}\subset X\times X$, then
$V^{\prime}\in\mathcal{U}$;

  * (3)

if $V,W\in\mathcal{U}$, then $V\cap W\in\mathcal{U}$;

  * (4)

if $V\in\mathcal{U}$, then $V^{-1}\in\mathcal{U}$;

  * (5)

if $V\in\mathcal{U}$, then there exists $W\in\mathcal{U}$ such that $W\circ
W\subset V$.

_A set $X$ equipped with a uniform structure $\mathcal{U}$ is called a
_uniform space_ and the elements of $\mathcal{U}$ are called the _entourages_
of $X$. Let $X$ be a uniform space. One easily verifies that it is possible to
define a topology on $X$ by taking as open sets the subsets $\Omega\subset X$
which satisfy the following property: for each $x\in\Omega$, there exists an
entourage $V\subset X\times X$ such that $V[x]=\Omega$. One says that this
topology is the topology _associated_ with the uniform structure on $X$. A
subset $N\subset X$ is neighborhood of a point $x\in X$ for this topology if
and only if there exists an entourage $V$ such that $N=V[x]$. This topology is
Hausdorff if and only if the intersection of the entourages of $X$ coincides
with the diagonal $\Delta_{X}\subset X\times X$._

###### Proposition 2.6

[23] Let $X$ be a set and let $\mathcal{B}$ be a nonempty set of subsets of
$X\times X$. Then $\mathcal{B}$ is a base for some (necessarily unique)
uniform structure on $X$ if and only if it satisfies the following properties:

  * (a)

if $V\in\mathcal{B}$, then $\Delta_{X}\subset V$;

  * (b)

if $V\in\mathcal{B}$ and $W\in\mathcal{B}$, then there exists
$U\in\mathcal{B}$ such that $U\subset V\cap W$;

  * (c)

if $V\in\mathcal{U}$, then there exists $W\in\mathcal{B}$ such that $W\subset
V^{-1}$;

  * (d)

if $V\in\mathcal{U}$, then there exists $W\in\mathcal{B}$ such that $W\circ
W\subset V$.

Given two congruences $\theta_{1},\theta_{2}$ on an algebra $A$. If
$\theta_{1}\circ\theta_{2}=\theta_{2}\circ\theta_{1}$, then we say
$\theta_{1}$ and $\theta_{2}$ are _permutable_ , or $\theta_{1}$ and
$\theta_{2}$ _permute_. An algebra $A$ is _congruence permutable_ if every
pair of congruences on $A$ permutes. A class $\mathcal{K}$ of algebras is
_congruence permutable_ iff every algebra in $\mathcal{K}$ has congruence
permutable, see [9].

In what follows, we give a general result about topological algebras.

A _topological (universal) algebra_ is an object
$\mathbf{A}=(A,(f_{\alpha})_{\alpha\in I},\tau)$ where $A$ is a set,
$(f_{\alpha})_{\alpha\in I}$ a family of maps
$f_{\alpha}:X^{n_{\alpha}}\rightarrow X$, $n_{\alpha}$ the _arity_ of the
operation $f_{\alpha}$, and $\tau$ a topology on $A$ such that each
$f_{\alpha}$ maps the product space $(A,\tau)^{n_{\alpha}}$ continuously to
the space $(A,\tau)$. $(A,(f_{\alpha})_{\alpha\in I})$ is called the
_underlying algebra_ , and $(X,\tau)$ the _underlying space_.

By Proposition 2.6, we have the following result, which is of great
significance to study the topological algebras.

###### Theorem 2.7

For a congruence permutable algebra $\mathbf{A}=(A,(f_{\alpha})_{\alpha\in
I})$, if $(\mathscr{C},\subseteq)$ is a down-directed set, then
$(A,(f_{\alpha})_{\alpha\in I},\tau)$ is a topological algebra, where
$\mathscr{C}\subseteq ConA$ and $\tau$ is a topology induced by $\mathscr{C}$.

Clearly, $\mathscr{C}$ satisfies the conditions ($a$) and ($c$) of Proposition
2.6. The hypothesis that $(\mathscr{C},\subseteq)$ is a down-directed set
corresponds to the condition ($b$) of Proposition 2.6, and the congruence
permutability implies the condition ($d$) of Proposition 2.6. Thus, by
Proposition 2.6, $\mathscr{C}$ is a base for the uniform structure on $A$ and
$\tau$ is a topology associated with this uniform structure on $A$. Note that

$f_{\alpha}([a]_{\theta},[b]_{\theta})\subseteq[f_{\alpha}(a,b)]_{\theta},$

for any $a,b\in A$ and $\theta\in\mathscr{C}$. It follows that
$(A,(f_{\alpha})_{\alpha\in I},\tau_{\theta})$ is a topological algebra, where

$\tau_{\theta}=\\{U\subseteq A:\forall a\in U,[a]_{\theta}\subseteq U\\}.$

Since $\tau=\sup\\{\tau_{\theta}:\theta\in\mathscr{C}\\}$, we deduce that
$(A,(f_{\alpha})_{\alpha\in I},\tau)$ is a topological algebra.

The variety $\mathcal{RL}$ of residuated lattices is arithmetical. It has the
congruence extension property (CEP), and is congruence 1-regular, i.e., for
any congruence $\theta$, the coset of 1 uniquely determines $\theta$.
Furthermore, the congruences of a residuated lattice are completely determined
by filters (see Theorem 2.4). Thus, in [40], we studied topological residuated
lattices induced by a system of filters. Since the variety $\mathcal{RL}$ is
congruence permutable, it follows from Theorem 2.7 that a residuated lattice
endowed with the topology induced by a system of filters is a topological
residuated lattice, see Theorem 3.13 in [40].

A topological space $(X,\mathcal{T})$ is a _zero-dimensional space_ if
$\mathcal{T}$ has a clopen base. Let $\tau$ and $\tau^{\prime}$ be two
topologies on a given set $X$. If $\tau\subseteq\tau^{\prime}$, then we say
that $\tau^{\prime}$ _finer than_ $\tau$.

###### Definition 2.8

_[40] Let $\mathcal{F}$ be a family of filters of a residuated lattice $L$.
Then $\mathcal{F}$ called a _system of filters_ of $L$ if
$(\mathcal{F},\subseteq)$ is a down-directed set._

A topology $\tau$ of a residuated lattice $L$ is called a _linear topology_ on
$L$ if there exists a base $\beta$ for $\tau$ such that $B$ is a filter of
$L$, for any element $B$ of $\beta$ containing 1.

If $(\mathcal{F},\subseteq)$ is a system of filters of a residuated lattice
$L$, then, by Theorem 2.7 or Theorem 3.13 in [40],

$\mathcal{T}_{\mathcal{F}}=\\{U\subseteq L:\forall x\in U,\exists
F\in\mathcal{F}~{}s.t.,x/F\subseteq U\\}$

is a linear topology on $L$. We call it a _linear topology induced by
$\mathcal{F}$_. Also, $(L,\mathcal{T}_{\mathcal{F}})$ is a topological
residuated lattice.

##  3 Profinite residuated lattices

In this section, we study profinite residuated lattices and give several
characterizations of profiniteness in compact topological residuated lattices.

Let $I=(I,\leq)$ denote a _directed partially set_ or _directed poset_ , that
is, $I$ is a set with a binary relation $\leq$ such that $(I,\leq)$ is a poset
and if $i,j\in I$, there exists some $k\in I$ such that $i,j\leq k$.

###### Definition 3.1

_[19] (i) By an _inverse or projective system_ in a category $\mathcal{D}$ we
mean a family $\\{B_{i},\pi_{ij},I\\}$ of objects, indexed by a directed poset
$I$, with a family of morphisms $\pi_{ij}:B_{i}\rightarrow B_{j}$, for any
$j\leq i$, satisfying the following conditions:_

  * (1)

$\pi_{ik}=\pi_{jk}\circ\pi_{ij}$, for any $k\leq j\leq i$;

  * (2)

$\pi_{ii}=id_{B_{i}}$, for any $i\in I$.

_For brevity we say that $\\{B_{i},\pi_{ij},I\\}$ is an inverse system in
$\mathcal{D}$._

_(ii) The _inverse limit_ of an inverse system $\\{B_{i},\pi_{ij},I\\}$ in a
category $\mathcal{D}$ is an object $B$ of $\mathcal{D}$ together with a
family $\\{\phi_{i}:B\rightarrow B_{i}\\}_{i\in I}$ of morphisms (which is
often denoted by $\\{B,\phi_{i}\\}_{i\in I}$) satisfying the conditions:_

  * (1)

$\pi_{ij}\circ\phi_{i}=\phi_{j}$, for any $i,j\in I,j\leq i$(this condition
often is called _compatible condition_);

  * (2)

for any object $B^{\prime}$ of $\mathcal{D}$, together with a family of
morphisms $\lambda_{i}:B^{\prime}\rightarrow B,i\in I$, if
$\pi_{ij}\circ\lambda_{i}=\lambda_{j}$, for any $i,j\in I,j\leq i$ then there
exists a unique morphism $\lambda:B^{\prime}\rightarrow B$ such that
$\phi_{i}\circ\lambda=\lambda_{i}$, for any $i\in I$.

_The inverse limit of the above system is denoted by
$\lim\limits_{\longleftarrow}B_{i}$._

Recall from Grätzer [19] that the inverse limits of families of algebras are
constructed in the following way.

###### Theorem 3.2

[19] Let $\\{B_{i},\pi_{ij},I\\}$ be an inverse system of same algebras,
$\prod_{i\in I}B_{i}$ be its product and $\pi_{i}:\prod_{i\in
I}B_{i}\rightarrow B_{i}$ is defined by $\pi_{i}((x_{i})_{i\in I})=x_{i}$ for
any $i\in I$. Let

$B=\\{(b_{i})_{i\in I}\in\prod_{i\in I}B_{i}:\pi_{ij}(b_{i})=b_{j},j\leq
i\\}.$

Then B is a subalgebra of $\prod_{i\in I}B_{i}$ and $\\{B,\phi_{i}\\}_{i\in
I}$ is the inverse limit of $\\{B_{i},\pi_{ij},I\\}$, where
$\phi_{i}=\pi_{i}\upharpoonright_{B}$ for any $i\in I$.

Let $A$ be an algebra. Let $I$ denote the set of all congruences $\theta$ on
$A$ such that $A/\theta$ is finite. We denote the image of $a\in A$ in
$A/\theta$ by $[a]_{\theta}$. If $\theta\subseteq\theta^{\prime}$, then there
is a canonical projection
$\varphi_{\theta^{\prime}\theta}:A/\theta^{\prime}\rightarrow A/\theta$ given
by $\varphi_{\theta^{\prime}\theta}([a]_{\theta^{\prime}})=[a]_{\theta}$. Then
$(I,\supseteq)$ is a directed set, and
$(\\{A/\theta\\},\\{\varphi_{\theta^{\prime}\theta}\\},I)$ is an inverse
system of algebras. The _profinite completion $\widehat{A}$_ of $A$ is the
inverse limit of $(\\{A/\theta\\},\\{\varphi_{\theta^{\prime}\theta}\\},I)$.
By [19], we may identify $\widehat{A}$ with the subalgebra of
$\prod_{\theta\in I}A/\theta$ consisting of all $([a]_{\theta})_{\theta\in I}$
for which
$\varphi_{\theta^{\prime}\theta}([a]_{\theta^{\prime}})=[a]_{\theta}$ whenever
$\theta^{\prime}\subseteq\theta$. We define the canonical homomorphism
$e_{A}:A\rightarrow\widehat{A}$ by $e_{A}(a)=([a]_{\theta})_{\theta\in I}$.
Let $\pi_{\theta^{\prime}}:\prod_{\theta\in I}A/\theta\rightarrow
A/\theta^{\prime}$ denote the projection map. We also denote the restriction
of $\pi_{\theta^{\prime}}$ to $\widehat{A}$ by $\pi_{\theta^{\prime}}$. One
would expect a completion of an algebra $A$ to be an extension of $A$,
however, the canonical homomorphism $e_{A}:A\rightarrow\widehat{A}$ is not
always an embedding.

Let $(I,\leq)$ be a directed poset. Assume that $I^{\prime}$ is a subset of
$I$ in such a way that $(I^{\prime},\leq)$ becomes a directed poset. We say
that $I^{\prime}$ is _cofinal_ in $I$ if for every $i\in I$ there is some
$i^{\prime}\in I^{\prime}$ such that $i\leq i^{\prime}$. If
$\\{X_{i},\varphi_{ij},I\\}$ is an inverse system and $I^{\prime}$ is cofinal
in $I$, then $\\{X_{i},\varphi_{ij},I^{\prime}\\}$ becomes an inverse system
in an obvious way, and we say that $\\{X_{i},\varphi_{ij},I^{\prime}\\}$ is a
cofinal subsystem of $\\{X_{i},\varphi_{ij},I\\}$.

###### Proposition 3.3

Let $\\{A_{i},\varphi_{ij},I\\}$ be an inverse system of same algebras over a
directed poset $I$ and assume that $I^{\prime}$ is a cofinal subset of $I$.
Then

$\varprojlim_{i\in I}X_{i}\cong\varprojlim_{i^{\prime}\in
I^{\prime}}X_{i^{\prime}}.$

###### Proof 1

Assume that $\\{A_{i},\varphi_{ij},I^{\prime}\\}$ is a cofinal subsystem of
$\\{A_{i},\varphi_{ij},I\\}$ and denote by $(\varprojlim_{i^{\prime}\in
I^{\prime}},\varphi_{i^{\prime}}^{\prime})$ and $(\varprojlim_{i\in
I},\varphi_{i})$ their corresponding inverse limits, respectively. For any
$j\in I$, let $j^{\prime}\in I^{\prime}$ be such that $j\leq j^{\prime}$.
Define

$\overline{\varphi}_{j}:\varprojlim_{i^{\prime}\in
I^{\prime}}X_{i^{\prime}}\rightarrow X_{j}$

as the composition of canonical mappings
$\varphi_{j^{\prime}j},\varphi_{j^{\prime}}^{\prime}$. Observe that the maps
$\overline{\varphi}_{j}$ are well-defined and compatible. Hence they induce a
map

$\overline{\varphi}:\varprojlim_{i^{\prime}\in
I^{\prime}}X_{i^{\prime}}\rightarrow\varprojlim_{i\in I}X_{i}$

such that $\varphi_{j}\overline{\varphi}=\overline{\varphi}_{j}(j\in I)$. We
claim that the mapping $\overline{\varphi}$ is a bijection. Noth that if
$(x_{i^{\prime}})\in\varprojlim_{i^{\prime}\in I^{\prime}}X_{i^{\prime}}$ and
$\overline{\varphi}(x_{i^{\prime}})=(y_{i})$, then
$y_{i^{\prime}}=x_{i^{\prime}}$ for $i^{\prime}\in I^{\prime}$. Since
$I^{\prime}$ is cofinal in $I$, thus $\overline{\varphi}$ is an injection. To
see that $\overline{\varphi}$ is a surjection, let
$(y_{i})\in\varprojlim_{i\in I}X_{i}$ and consider the element
$(x_{i^{\prime}})$, where $x_{i^{\prime}}=y_{i^{\prime}}$ for every
$i^{\prime}\in I^{\prime}$. Then
$(x_{i^{\prime}})\in\varprojlim_{i^{\prime}\in I^{\prime}}X_{i^{\prime}}$ and
obviously, $\overline{\varphi}(x_{i^{\prime}})=(y_{i})$. This proves the
claim. The rest of the proof is obvious.

In what follows we shall be specially interested in the topological space $X$
that arise as inverse limits

$X=\lim_{\begin{subarray}{c}\longleftarrow\\\ i\in I\end{subarray}}X_{i}$

of finite spaces $X_{i}$ endowed with the discrete topology. We call such a
space _profinite space_ or a _Boolean space_. A topological space is _totally
disconnected_ if every point in the space is its own connected component.

###### Theorem 3.4

[33] Let $X$ be a topological space. Then the following conditions are
equivalent.

  1. (a)

$X$ is a profinite space;

  2. (b)

$X$ is compact Hausdorff and totally disconnected;

  3. (c)

$X$ is compact Hausdorff and admits a base of clopen sets for its topology.

###### Definition 3.5

_We call a residuated lattice $L$ _profinite_ if it is isomorphic to the
inverse limit of an inverse system of finite residuated lattices. The class of
this algebras denoted by $\mathbf{Pro}\mathcal{RL}$._

Thus, we see that if we are given a profinite algebra $A\cong\varprojlim_{i\in
I}A_{i}$, then $A$ is a topological algebra in its profinite topology, which
can definite in a natural way using the limiting cone $(\pi_{i}:A\rightarrow
A_{i})_{I}$. We will now record the fundamental fact that every profinite
topology is Boolean, i.e., every profinite topology is compact Hausdorff and
zero-dimensional.

###### Proposition 3.6

[2] If $A\cong\varprojlim_{i\in I}A_{i}$ is a profinite algebra, then $A$ is a
Boolean topological algebra in its profinite topology.

###### Remark 3.7

We have chosen to first describe profinite algebras using universal algebra,
via. as the limit of a diagram of finite algebras
$\\{A_{i},\varphi_{ij},I\\}$, and then to indicate that one can definite a
topology on $\varprojlim_{i\in I}A_{i}$, making $\varprojlim_{i\in I}A_{i}$ a
topological algebra. Alternatively, one can start by endowing each algebra
$A_{i}$ with the discrete topology, and then show that one can also take
limits of topological algebras, so that it follows immediately that
$\varprojlim_{i\in I}A_{i}$ is a topological algebra. One can then show that
$\varprojlim_{i\in I}A_{i}$ is a closed subalgebra of $\prod_{i\in I}A_{i}$,
so that it follows from general topology that the profinite topology on
$\varprojlim_{i\in I}A_{i}$ is a Boolean topology.

We begin this section by showing that an infinite profinite residuated lattice
cannot be countable. This is a general fact for locally compact topological
groups [16], but here we present a proof for profinite residuated lattices
only. The first part of the following proposition is similar to the classical
Baire category theorem, valid for locally compact spaces.

###### Proposition 3.8

Let $L$ be a profinite residuated lattice.

  1. (a)

Let $C_{1},C_{2},\ldots$ be countably infinite set of nonempty closets of $L$
having empty interior. Then

$L\neq\bigcup_{n=1}^{\infty}C_{i}.$

  2. (b)

The cardinality $|L|$ of $L$ is either finite or uncountable.

###### Proof 2

Part $(b)$ follows immediately from $(a)$. To prove $(a)$, assume that
$L=\bigcup_{n=1}^{\infty}C_{i}$, where each $C_{i}$ is a nonempty closed
subset of $L$ with empty interior. Then $D_{i}=L-C_{i}$ is a dense open subset
of $L$, for each $i\in\\{1,2,\ldots\\}$.

Next consider a nonempty open subset $U_{0}$ of $L$; then $U_{0}\cap D_{1}$ is
open and nonempty since $D_{1}$ is open and dense in $L$. By Theorem 3.4,
there is a nonempty clopen subset $U_{1}$ of $U_{0}\cap D_{1}$. Similarly,
$U_{1}\cap D_{2}$ is open and nonempty, therefore there is a nonempty clopen
subset $U_{2}$ of $U_{1}\cap D_{2}$. Proceeding in this manner we obtain a
nested sequence of clopen nonempty subsets

$U_{1}\supseteq U_{2}\supseteq\cdots\supseteq U_{i}\supseteq\cdots$

such that $U_{i}\subseteq D_{i}\cap U_{i-1}$ for each $i\in\\{1,2,\ldots\\}$.
Since $L$ is compact and the closed sets $U_{i}$ have the finite intersection
property, we have that

$\bigcap_{i=1}^{\infty}U_{i}\neq\emptyset.$

On the other hand,

$\bigcap_{i=1}^{\infty}U_{i}\subseteq\bigcap_{i=1}^{\infty}D_{i}=L-(\bigcup_{i=1}^{\infty}C_{i})=\emptyset,$

a contradiction.

###### Lemma 3.9

Let $\\{L_{i},\varphi_{ij},I\\}$ be an inverse system of topological
residuated lattices $L_{i}$. Then we have:

  1. (i)

the inverse limit $\underleftarrow{\lim}\\{L_{i}:i\in I\\}$ forms a subalgebra
of $\prod\\{L_{i}:i\in I\\}$;

  2. (ii)

if each topological residuated lattice in $\\{L_{i}:i\in I\\}$ is Hausdorff,
then the inverse limit $\underleftarrow{\lim}\\{L_{i}:i\in I\\}$ is
topologically closed in the product $\prod\\{L_{i}:i\in I\\}$;

  3. (iii)

if each topological residuated lattice in $\\{L_{i}:i\in I\\}$ is non-empty,
compact and Hausdorff, then the inverse limit
$\underleftarrow{\lim}\\{L_{i}:i\in I\\}$ is non-empty.

###### Proof 3

The proof of $(i)$ is easy. We sketch the proof of $(ii)$ and $(iii)$. First
assume that each $L_{i}$ is Hausdorff. For each $k\in I$, define the set

$A_{k}=\\{a\in\prod_{i\in I}:(\forall i\leq k)a_{i}=\varphi_{ki}(a_{k})\\}.$

Each $A_{k}$ is topological closed in $\Pi_{i\in I}L_{i}$, and
$\underleftarrow{\lim}\\{L_{i}:i\in I\\}=\bigcap_{k\in I}A_{k}$. So the
inverse limit is topological closed. Now assume further that each $L_{i}$ is
non-empty and compact. Since $I$ is direct and each $L_{i}$ is non-empty,
every finite intersection of members of $\\{A_{k}:k\in I\\}$ is non-empty. By
compactness, the set $\bigcap_{k\in I}A_{k}$ is non-empty.

###### Proposition 3.10

Let

$L=\varprojlim_{i\in I}L_{i},$

where $\\{L_{i},\varphi_{ij},I\\}$ is an inverse system of finite residuated
lattices $L_{i}$, and let

$\pi_{i}:L\rightarrow L_{i}(i\in I)$

be the projection homomorphisms. Then

$\\{S_{i}:S_{i}=Ker(\pi_{i})\\}$

is a fundamental system of open neighborhoods of 1 in L.

###### Proof 4

Consider the family of neighborhoods of 1 in $\prod_{i\in I}L_{i}$ of the form

$(\prod_{i\neq
i_{1},\ldots,i_{t}}L_{i})\times\\{1\\}_{i_{1}}\times\cdots\times\\{1\\}_{i_{t}},$

for any finite collection of indexes $i_{i},\ldots,i_{t}\in I$, where
$\\{1\\}_{i}$ denotes the subset of $L_{i}$ consisting of the identity
element. Since each $L_{i}$ is discrete, this family is fundamental system of
neighborhoods of the identity element of $\prod_{i\in I}L_{i}$. Let $i_{0}\in
I$ be such that $i_{1},\ldots,i_{t}\leq i_{0}$. Then

$L\cap[(\prod_{i\neq i_{0}})L_{i}\times\\{1\\}_{i_{0}}]=L\cap[(\prod_{i\neq
i_{1},\ldots,i_{t}}L_{i})\times\\{1\\}_{i_{1}}\times\cdots\times\\{1\\}_{i_{t}}].$

Therefore the family of neighborhoods of 1 in $L$, of the form

$L\cap[(\prod_{i\neq i_{0}})L_{i}\times\\{1\\}_{i_{0}}]$

is a fundamental system of open neighborhoods of 1. Finally, observe that

$L\cap[(\prod_{i\neq
i_{0}})L_{i}\times\\{1\\}_{i_{0}}]=Ker(\pi_{i_{0}})=S_{i_{0}}.$

###### Theorem 3.11

Let $(L,\tau)$ be a compact topological residuated lattice. Then the following
conditions are equivalent:

  1. (i)

$(L,\tau)$ is profinite;

  2. (ii)

there is a set $\mathcal{S}$ of clopen filters(congruences) on $L$ such that

  3. (a)

the set $(\mathcal{S},\supseteq)$ is directed, and

  4. (b)

for all $x,y\in L$ with $x\neq y$, there is a filter $F\in\mathcal{S}$ such
that $x/F\neq y/F$ (congruence $\theta\in\mathcal{S}$ such that
$[x]_{\theta}\neq[y]_{\theta}$).

###### Proof 5

First assume that $(L,\tau)$ is compact, and let $\mathcal{S}$ be a set of
clopen filters on $L$ such that $(a)$ and $(b)$ hold. We can set up an inverse
system on $\\{L/F:F\in\mathcal{S}\\}$, where $\mathcal{S}$ is ordered by
reverse inclusion: for all $F\subseteq G$ in $\mathcal{S}$, define the
homomorphism $\varphi_{FG}:L/F\rightarrow L/G$ by $\varphi_{FG}(x/F)=x/G$.
Since $(L,\tau)$ is compact, each quotient $L/F$ is finite. It is
straightforward to prove that we can definite an isomorphism
$\psi:L\rightarrow\underleftarrow{\lim}\\{L/F:F\in\mathcal{S}\\}$ by
$\psi(x)(F)=x/F$. (To prove that $\psi$ is surjective, use the compactness of
$(L,\tau)$) to find a member of $\bigcap\\{z(F):F\in\mathcal{S}\\}$, for each
$z\in\underleftarrow{\lim}\\{L/F:F\in\mathcal{S}\\}$.) This inverse limit is
clearly surjective. By condition $(b)$, it follows that $\psi$ is injective.

Now assume that $(L,\tau)$ is profinite. We can assume that it is equal to an
inverse limit $\underleftarrow{\lim}\\{L_{i}:i\in I\\}$ of finite residuated
lattices, that is, $L=\underleftarrow{\lim}\\{L_{i}:i\in I\\}$. Let
$\mathcal{S}=\\{Ker(\pi_{i}):i\in I\\}$. Then $Ker(\pi_{i})$ is a clopen
filters of $L$, for each $i\in I$. Thus the condition $(a)$ clearly holds. By
Proposition 3.10 and profinite topological is always Hausdorff, it follows
that $\bigcap\\{Ker(\pi_{i}):i\in I\\}=\\{1\\}$. Thus, it deduces that the
condition $(b)$ holds.

The following result is a particular case of a theorem of universal algebra.

###### Proposition 3.12

A residuated lattice $L$ is a subdirect product of a family $L_{i}$ of
residuated lattices if and only if there is a family $\\{F_{i}\\}_{i\in I}$ of
filters of $L$ such that

  * (i)

$L_{i}\cong L/F_{i}$ for each $i\in I$;

  * (ii)

$\bigcap_{i\in I}F_{i}=\\{1\\}$.

Recall that a topological algebra $(L,\tau)$ is _residually finite_ [34] if,
for any two distinct elements $x,y\in L$, there exists a finite discrete
topological algebra $F$ as well as continuous homomorphism
$\varphi:L\rightarrow F$ such that $\varphi(x)\neq\varphi(y)$.

###### Theorem 3.13

Let $(L,\tau)$ be a compact topological residuated lattice. Then the following
conditions are equivalent:

  1. (i)

$(L,\tau)$ is residually finite;

  2. (ii)

$(L,\tau)$ is profinite;

  3. (iii)

$(L,\tau)$ is a closed subdirect product of finite discrete residuated
lattices.

###### Proof 6

First, let us prove that $(i)$ implies $(iii)$. For $x,y\in L$ with $x\neq y$.
Then there exist a finite residuated lattice $F$ and a homomorphism
$\varphi_{x,y}:L\rightarrow F$ such that
$\varphi_{x,y}(x)\neq\varphi_{x,y}(y)$. It follows that $Ker(\varphi_{x,y})$
is a finite index clopen filter of $(L,\tau)$, for each $x,y\in L$ with $x\neq
y$. It is easy to check that $L$ is a subdiect product of
$\\{L/Ker(\varphi_{x,y}):x,y\in L~{}and~{}x\neq y\\}$. By Lemma 3.9 $(ii)$, it
deduces that $(iii)$ holds. Now assume that the condition $(iii)$ holds. Let
$L$ be a closed subdirect product of finite residuated lattices $\\{L_{i}:i\in
I\\}$. By Proposition 3.12, there exists a family $\\{F_{i}:i\in I\\}$ of
filters of $L$ such that $\bigcap_{i\in I}F_{i}=\\{1\\}$ and $L_{i}\cong
L/F_{i}$ for each $i\in I$. Thus $(L,\tau)$ is closed in $\prod_{i\in
I}L/F_{i}$, and hence $(L,\tau)$ is residually finite.

Since $(iii)$ implies $(ii)$ is obvious, it suffices to show the converse. Let
$L$ be a topological closed subalgebra of a product $\Pi_{i\in I}L_{i}$, where
$L_{i}$ is a finite residuated lattice for each $i\in I$. Let $\mathcal{S}$ be
the set of all finite non-empty subsets of $I$, ordered by inclusion. For each
$J\in\mathcal{S}$, let $L\upharpoonright_{J}:L\rightarrow\prod_{j\in J}L_{j}$
be the projection. For all $K\supseteq J$ in $\mathcal{S}$, let
$\varphi_{KJ}:L\upharpoonright_{K}\rightarrow L\upharpoonright_{J}$ be the
projection. Then $\\{L\upharpoonright_{J}:J\in\mathcal{S}\\}$ forms an inverse
system of topological residuated lattices. Define the inverse limit
$Y=\underleftarrow{\lim}\\{L\upharpoonright_{J}:J\in\mathcal{S}\\}$. We want
to define an isomorphism $\psi:L\rightarrow Y$ by
$\psi(x)(J)=x\upharpoonright_{J}$. For each $x\in L$, we have $\psi(x)\in Y$,
since
$\varphi_{KJ}((\psi(x))(K))=\varphi_{KJ}(x\upharpoonright_{K})=x\upharpoonright_{J}=\psi(x)(J)$,
for all $K\supset J$ in $\mathcal{S}$. The map $\psi$ is a homomorphism, since
the projections are homomorphisms. It is easy to check that $\psi$ is
injective. To see that $\psi$ is continuous, consider a subbasic open set
$U=\\{y\in Y:y(Y)(j)\in V\\}$ of $Y$, where $j\in J\in\mathcal{S}$ and $V$ is
open in $L_{j}$. We have $\psi^{-1}(U)=\\{x\in L:x(j)\in V\\}$, which is open
in $L$. It remains to verify that $\psi$ is surjective. Let $y\in Y$. For each
$J\in\mathcal{S}$, define the closed set $V_{J}=\\{x\in
L:x\upharpoonright_{J}=y\upharpoonright_{J}\\}$ in $L$. Since $y\in Y$, we
know that $y(J)\in L\upharpoonright_{J}$ and therefore $V_{J}\neq\emptyset$,
for each $J\in\mathcal{S}$. The set $\\{V_{J}:J\in\mathcal{S}\\}$ is closed
under the finite intersection, since $V_{J_{1}}\cap V_{J_{2}}\cap\cdots\cap
V_{J_{n}}=V_{J_{1}\cup\cdots\cup J_{n}}\neq\emptyset$. So, as $L$ is compact,
we can choose $x\in\bigcap\\{V_{J}:J\in\mathcal{S}\\}$. For each
$J\in\mathcal{S}$, we have $\psi(x)(J)=x\upharpoonright_{J}=y(J)$. Thus
$\psi(x)=y$.

##  4 Finiteness conditions in residuated lattices

The elementary theory of vector spaces concerns itself with spaces of finite
dimension, in such spaces there does not exist an infinite strictly ascending
or strictly descending chain of subspaces. Similarly, the elementary theory of
groups concerns itself with finite groups, or at any rate with groups which
are not “too infinite” in some sense. The purposes of this section is the
discussion of various finiteness conditions which can be imposed on a
residuated lattice.

So far we have considered a quite arbitrary residuated lattice. To go further,
however, and obtain deeper theorems we need to impose some conditions. The
most convenient way is in the form of “chain conditions”.

###### Definition 4.1

_Let $L$ be a residuated lattice. We call $L$ satisfying the _descending chain
condition_($DCC$ for short), if every strictly descending chain of filters_

$\cdots\subset F_{2}\subset F_{1}$

_is finite._

_An equivalent formulation is: if_

$\cdots\subset F_{2}\subset F_{1}$

_is an infinite descending sequence of filters, then there exists an integer
$n$ such that_

$F_{i}=F_{n},for~{}i=n+1,n+2\cdots.$

###### Definition 4.2

_A residuated lattice is said to satisfy the _minimal condition_ if every non-
empty collection $\mathcal{C}$ of filters has a minimal element, that is, if
there exists a filter in $\mathcal{C}$ which is not contained in any other
filter in the collection $\mathcal{C}$._

###### Proposition 4.3

Let $L$ be a residuated lattice. Then the following assertions are equivalent:

  * (1)

$L$ satisfies the descending chain condition;

  * (2)

$L$ satisfies the minimal condition;

  * (3)

each system of filters of $L$ has a minimum element.

###### Proof 7

The equivalence of conditions (1) and (2) is the standard method. Now, suppose
that the condition (2) is true. Let $(\mathcal{F},\subseteq)$ be a system of
filters. It must have a minimal filter $F$ in $\mathcal{F}$. For any another
filter $G$ in $\mathcal{F}$, since $(\mathcal{F},\subseteq)$ is down-directed,
we have $F\subseteq G$. Thus $F$ is a minimum filter in
$(\mathcal{F},\subseteq)$. It is valid that condition (3) implies condition
(1) , since every chain of filters is a system of filters.

###### Theorem 4.4

Let $L$ be a residuated lattice and $\mathcal{F}=\\{F_{i}:i\in I\\}$ be a
system of filters of $L$. Then the following statements are equivalent:

  * (i)

$(L,\mathcal{T}_{F})$ is a Hausdorff space;

  * (ii)

$(L,\mathcal{T}_{F})$ is a $T_{1}$-space;

  * (iii)

$(L,\mathcal{T}_{F})$ is a $T_{0}$-space;

  * (iv)

$\bigcap_{i\in I}F_{i}=\\{1\\}$;

  * (v)

$L$ is a subdirect product of a family $L/F_{i}$ of residuated lattices.

###### Proof 8

The equivalences of (i)-(iv) are similar to Proposition 3.20 in [38]. The
equivalence of (iv) and (v) follows from Proposition 3.12.

A filter $F$ of a residuated lattice $L$ is said to be _prime_ if $x\vee y\in
F$ implies $x\in F$ or $y\in F$ for all $x,y\in L$, see [25]. We denote the
set of all prime filters of a residuated lattice $L$ by $\mathcal{P}(L)$.

###### Theorem 4.5

[21] Let $L$ be a residuated lattice. For every element $a\in L$ with $a\neq
1$ there is a prime filter $P$ in $L$ with $a\notin P$.

###### Corollary 4.6

[21] Every proper filter of a residuated lattice is an intersection of prime
filters.

###### Corollary 4.7

[21] Let $L$ be a residuated lattice. Then $\bigcap\mathcal{P}(L)=\\{1\\}$.

Given a system $\mathcal{F}=\\{F_{i}:i\in I\\}$ of filters of a residuated
lattice $L$. It is easily checked that the following assertions hold:

  * (i)

$(L,\mathcal{T}_{F})$ is a discrete space if and only if
$\\{1\\}\in\mathcal{F}$;

  * (ii)

$(L,\mathcal{T}_{F})$ is an anti-discrete space if and only if
$\\{L\\}=\mathcal{F}$.

We call $\mathcal{F}(L)\backslash\\{1\\}$ is a _global system of filters_ if
it is a system of filters, i.e., $(\mathcal{F}(L)\backslash\\{1\\},\subseteq)$
is closed under finite intersections. Recall that a congruence $\theta$ on an
algebra is a _factor congruence_ if there is a congruence $\theta^{*}$ such
that $\theta\cap\theta^{*}=\triangle$, $\theta\cup\theta^{*}=\nabla$ and
$\theta$ permutes with $\theta^{*}$, see [9] for details. An algebra is
_directly indecomposable_ if it is not isomorphic to a direct product of two
nontrivial algebras.

###### Proposition 4.8

Let $L$ be a residuated lattice. Then the following conditions are equivalent:

  1. (1)

$\mathcal{F}(L)\backslash\\{1\\}$ is a global system of filters;

  2. (2)

any finite intersections of $\mathcal{F}(L)\backslash\\{1\\}$ is not equal to
$\\{1\\}$;

  3. (3)

there is no non-trivial factor congruence on $L$;

  4. (4)

$L$ is directly indecomposable.

###### Proof 9

It is easy to see that (1),(2) and (3) are equivalent. According to Corollary
7.7 in [9], the conditions (3) and (4) are equivalent.

An algebra $A$ is _subdirectly irreducible_ iff $A$ is trivial or there is a
minimum congruence in $Con(A)-\Delta$, see [9].

From Theorem 8.5 in [9], we know that a subdirectly irreducible residuated
lattice is directly indecomposable. But the converse, in general, is not true.
In what follows, we give some conditions the converse is also true.

###### Lemma 4.9

Let $L$ be a directly indecomposable residuated lattice. If $L$ is subdirectly
irreducible, then the corresponding zero-dimensional linear topological
residuated lattice of its global system of filters is non discrete.

###### Proof 10

Suppose that $L$ is subdirectly irreducible. By Theorem 8.4 in [9], the global
system of filters $\mathcal{F}(L)\backslash\\{1\\}$ has a minimum filter
$F_{0}$. Thus the zero-dimensional linear topology of its global system of
filters is equal to $\mathcal{T}_{F_{0}}$, and it obviously is non-trivial.

###### Theorem 4.10

Let $L$ be a directly indecomposable residuated lattice and $L$ satisfy $DCC$.
Then $L$ is subdirectly irreducible iff the corresponding zero-dimensional
linear topological residuated lattice of its global system of filters is non
discrete.

###### Proof 11

The “only if” part is Lemma 4.9. To prove the “if” part, by Proposition 4.3,
the global system of filters of $L$ has a minimum element $F_{0}$. Therefore,
by Theorem 8.4 in [9], we obatin that $L$ is subdirectly irreducible.

###### Proposition 4.11

If $L$ is a residuated lattice that satisfies the descending chain condition
then every element of $\mathcal{F}(L)\setminus\\{1\\}$ can be expressed
uniquely as an irredundant join of join-irreducible filters.

###### Proof 12

Note that the lattice $(\mathcal{F}(L),\subseteq)$ is a distributive lattice.
Thus, it directly follows from Theorem 5.2 in [7].

In the power set lattice $\mathcal{P}(X)$, the join-irreducible elements are
exactly the singleton subsets. Let $\mathcal{JF}(L)$ denote the set of join-
irreducible filters of a residuated lattice $L$. Thus generally the set
$\mathcal{JF}(L)$ of all join-irreducible filters of $L$ is not a down
directed set under the set inclusion. In the following we show that
$(\mathcal{JF}(L)\cup\\{1\\},\supseteq)$ is a directed set under some
finiteness conditions.

###### Lemma 4.12

If $L$ is a residuated lattice that satisfies the descending chain condition,
then $(\mathcal{JF}(L)\cup\\{1\\},\supseteq)$ is cofinal in
$(\mathcal{F}(L),\supseteq)$.

###### Proof 13

Suppose that $F,G\in\mathcal{JF}(L)$. If $F\cap G=\\{1\\}$, then there is
nothing to do, otherwise, without loss of generality we can assume that $F\cap
G$ is not a join-irreducible filter, thus, by Proposition 4.11, we have $F\cap
G=\bigvee_{i=1}^{n}F_{i}$, where $F_{i}\in\mathcal{JF}(L)$. It follows that
$F_{i}\subset F\cap G$, whence $(\mathcal{JF}(L),\supseteq)$ is a directed
poset. Let $F$ be a filter of $L$. Then, by Proposition 4.11, there are join-
irreducible filters $J_{1},\ldots,J_{n}$ such that $F=\bigvee_{i=1}^{n}J_{i}$.
Thus there some join-irreducible filter $J_{j}$ such that $J_{j}\subseteq F$.
Therefore $(\mathcal{JF}(L),\supseteq)$ is cofinal in
$(\mathcal{F}(L),\supseteq)$.

###### Theorem 4.13

Let $L$ be a finite residuated lattice. Then

$\widehat{L}=\varprojlim_{F\in\mathcal{JF}(L)\cup\\{1\\}}L/F.$

###### Proof 14

It follows from Proposition 3.3 and Lemma 4.12.

##  5 The existence of Hausdorff topological residuated lattices

In this section, we investigate linear topological residuated lattices and
discuss the existence of Hausdorff topological residuated lattices.

For a filter $F$ of a residuated lattice $L$. We can get a zero-dimensional
linear topological residuated lattice $(L,\mathcal{T}_{F})$, where
$\mathcal{T}_{F}=\\{U\subseteq L:\forall x\in U,x/F\subseteq U\\}$, see [20].
For simplicity, such a zero-dimensional linear topological residuated lattice
is called a _simple linear topological residuated lattice_ when is no danger
causing confusion.

Let $(L,\mathcal{U})$ be a zero-dimensional linear topological residuated
lattice. Considering $X_{\mathcal{U}}=\\{F:1\in F\in\mathcal{U}\\}$, by Lemma
3.4 and Theorem 3.7 in [20], we have the following result.

###### Theorem 5.1

Let $L$ be a residuated lattice. Then each zero-dimensional linear topology
$\mathcal{U}$ of $L$ is the supremum of simple linear topological residuated
lattices, i.e.,

$\mathcal{U}=\sup\\{\mathcal{T}_{F}:F\in X_{\mathcal{U}}\\}.$

Now, considering a special system of filters of a residuated lattice $L$, that
is, the set of all finite index filters, denoted it by $\mathcal{F}(L)_{fin}$.
It is easy to see that $(\mathcal{F}(L)_{fin},\subseteq)$ is a down-directed
set. Thus, $(L,\mathcal{T}_{\mathcal{F}(L)_{fin}})$ is a zero-dimensional
linear topological residuated lattice. Like the linear topology of abelian
group[16], this linear topology in a residuated lattice $L$ is called a
_finite index topology_ of $L$. Recall that an algebra $A$ is _finitely
approximable_ if $A$ is isomorphic to a subalgebra of a product of finite
algebras [28], the class of finitely approximable residuated lattices is
denoted by $\mathbf{Ap}\mathcal{RL}$. It follows that $A$ is finitely
approximable iff $A$ is a subdirect product of its finite homomorphic images.
Thus we have that a residuated lattice $L$ is finitely approximable if and
only if $\bigcap\mathcal{F}(L)_{fin}=\\{1\\}$.

As a consequence, we have the following characterization of finitely
approximable residuated lattices by using zero-dimensional linear topological
residuated lattices.

###### Theorem 5.2

A residuated lattice is finitely approximable if and only if its finite index
topology is Hausdorff.

###### Proof 15

It follows from the above statement and Theorem 4.4.

Let $(L,\mathcal{U})$ be a linear topological residuated lattice. Then there
exists a base $\beta$ for the topology $\mathcal{U}$ on $L$ such that any
element containing 1 of $\beta$ is a filter of $L$. Let $X$ be the set of all
elements containing 1 of $\beta$, $Y$ be the set of all finite intersections
of elements of $X$. Then $Y$ is a system of filters of $L$. The topology
induced by the system $Y$ of filters is denoted by $\mathcal{V}$.

In what follows, we give a simple proof of Theorem 3.7 in [20], which answered
an open problem in [41].

###### Theorem 5.3

If $(L,\mathcal{U})$ be a zero-dimensional linear topological residuated
lattice, then $\mathcal{V}$ is finer than $\mathcal{U}$.

###### Proof 16

Let $\beta^{\prime}=\\{x/F_{i}:i\in I,x\in L\\}$ be a base for topologies
$\mathcal{V}$ on $L$. Since $(L,\mathcal{U})$ is a zero-dimensional space,
there exists a clopen base $\gamma$ for $\mathcal{U}$. To complete the proof,
we need only show that for any $x\in U\in\gamma$ there exists
$x/F_{i}\in\beta^{\prime}$ such that $x/F_{i}\subseteq U$. Let $x\in
U\in\gamma$. Since $U$ is clopen and $\mathcal{U}$ is finer than
$\mathcal{V}$, one can obtain that $U$ is also clopen for $\mathcal{V}$. Thus
there is a clopen $V$ of $\mathcal{V}$ such that $x\in V\subseteq U$.
Therefore, there is some $F_{i}$ such that $x/F_{i}\subseteq V\subseteq U$,
since $\beta^{\prime}$ is a base for $\mathcal{V}$.

###### Corollary 5.4

The topology of a zero-dimensional linear topological residuated lattice is
completely determined by a fundament system of neighborhoods of 1.

###### Proof 17

Let $(L,\mathcal{U})$ be a zero-dimensional linear topological residuated
lattice. By Lemma 3.1 in [39], $X_{\mathcal{U}}=\\{F:1\in F\in\mathcal{U}\\}$
is a fundamental system of neighborhoods of 1. By Corollory 3.8 in [40], it
follows that the topology $\mathcal{U}$ is induced by the system
$X_{\mathcal{U}}$ of filters of $L$.

###### Corollary 5.5

A zero-dimensional linear topological residuated lattice is precisely the
linear topological residuated lattice induced by its the system of filters.

###### Proof 18

According to Theorem 3.14 in [20], every topological residuated lattice
induced by a system of filters is a zero-dimensional linear topological
residuated lattice. The rest of proof follows from Theorem 5.3 in the paper
and Corollory 3.8 in [40].

###### Definition 5.6

_Let $L$ be a residuated lattice and $(\mathcal{F}_{1},\subseteq)$ and
($\mathcal{F}_{2},\subseteq)$ be two systems of filters of $L$. We call that
$(\mathcal{F}_{1},\subseteq)$ is _equivalent_ to
$(\mathcal{F}_{2},\subseteq)$, if posets $(\mathcal{F}_{1},\subseteq)$ and
$(\mathcal{F}_{2},\subseteq)$ are coinitial each other, that is, for any
$F_{1}\in\mathcal{F}_{1}$ there exists $F_{2}\in\mathcal{F}_{2}$ such that
$F_{1}\subseteq F_{2}$ and vice versa._

###### Proposition 5.7

In a residuated lattice $L$, equivalent systems of filters have the same
topology.

###### Proof 19

The proof is straightforward.

###### Example 5.8

Let $\odot$ and $\rightarrow$ on the real unit interval $I=[0,1]$ be defined
as follows:

$x\odot y=min\\{x,y\\}$ and $x\rightarrow y=\begin{cases}1,&x\leq y,\\\
y,&otherwise.\\\ \end{cases}$

Then $\mathcal{I}=(I,min,max,\odot,\rightarrow,0,1)$ is a $BL$-algebra (called
the Gödel structure). It is easily checked that $\mathcal{F}=\\{(a,1]\mid
a\in[0,1)\\}$ is a system of filters. Since $\mathcal{F}$ satisfies the finite
intersection, we get that $\mathcal{F}=\mathcal{G}$, which is the set of all
finite intersections of elements of $\mathcal{F}$. One can check that $L$ does
not satisfy $DCC$.

Using the above example, we can show that the proof of Theorem 3.10 in [20] is
incorrect. In fact, in the proof of Theorem 3.10 in [20], suppose that
$(L,\mathcal{T})$ is a linear topological residuated lattice such that
$(L,\mathcal{T})$ is a zero-dimensional space. Then there exists a base
$\beta$ for the topology $\mathcal{T}$ on $L$ such that any element containing
1 of $\beta$ is a filter of $L$. Let $X$ be the set of all elements containing
1 of $\beta$ and $Y$ be the set of all intersections of elements of $X$. Then
$(X,\subseteq)$ and $(Y,\subseteq)$ are two systems of filters of a residuated
lattice $L$. In general, they are not equivalent. Then the topologies induced
by them are not same. For example, taking $X=\mathcal{F}$ in Example 5.8 and
using the construction of $Y$, we have $\\{1\\}\in Y$. Thus, $\mathcal{T}_{Y}$
is a discrete topology. But $\mathcal{T}_{X}$ is not a discrete topology.
Thus, $\mathcal{T}_{X}\neq\mathcal{T}_{Y}$, that is,
$\mathcal{T}\neq\mathcal{V}$ in the proof of Theorem 3.10 in [20]. It follows
that the mapping $\varphi$ from $\mathcal{F}(L)$ to $ZLTRL(L)$ defined by
$\varphi(F)=(L,\mathcal{T}_{F})$ for all $F$ in $\mathcal{F}(L)$ is not
surjective, where the set $ZLTRL(L)$ of all zero-dimensional linear
topological residuated lattices on a residuated lattice $L$, see [20].

Next, we will correct Theorem 3.10 in [20] in the following theorem.

###### Theorem 5.9

Let $L$ be a residuated lattice. Then $|\mathcal{F}(L)|=|ZLTRL(L)|$ if and
only if $L$ satisfies the descending chain condition.

###### Proof 20

Suppose that $L$ satisfies the descending chain condition. Let us define a
mapping $\varphi$ from $\mathcal{F}(L)$ to $ZLTRL(L)$ by
$\varphi(F)=(L,\mathcal{T}_{F})$ for all $F$ in $\mathcal{F}(L)$, where
$\mathcal{T}_{F}=\\{U\subseteq L:\forall x\in U,x/F\subseteq U\\}$. It follows
that $(L,\mathcal{T}_{F})$ is a zero-dimensional linear topological residuated
lattice. According to Remark 3.9 in [20], the mapping $\varphi$ is injective.
Thus, it suffices to show that $\varphi$ is a surjective mapping. Suppose that
$(L,\mathcal{U})$ is a zero-dimensional linear topological residuated lattice.
By Theorem 5.3 and its corollary, there is a system of filters
$X_{\mathcal{U}}$ such that $\mathcal{U}=\sup\\{\mathcal{T}_{F}:F\in
X_{\mathcal{U}}\\}$. By hypothesis and Proposition 4.3, there is a minimum
filter $F$ in $X_{\mathcal{U}}$. Therefore, we have
$\mathcal{U}=\mathcal{T}_{F}$.

Conversely, assume that $|\mathcal{F}(L)|=|ZLTRL(L)|$. By Proposition 4.3, it
suffices to show that each system of filters of $L$ has a minimum element.
Suppose that $(\mathcal{F},\subseteq)$ is a system of filters of $L$. Then the
system $(\mathcal{F},\subseteq)$ can induce a zero-dimensional linear
topological residuated lattice $(L,\mathcal{T}_{\mathcal{F}})$, and
$\mathcal{T}_{\mathcal{F}}=\sup\\{\mathcal{T}_{F}:F\in\mathcal{F}\\}$. By
hypothesis, it follows that the size of zero-dimensional linear topological
residuated lattices of $L$ is equal to the size of simple linear topological
residuated lattices of $L$. Thus there is a filter $G\in\mathcal{F}(L)$ such
that $\mathcal{T}_{G}=\sup\\{\mathcal{T}_{F}:F\in\mathcal{F}\\}$. It follows
that $\mathcal{T}_{F}\subseteq\mathcal{T}_{G}$, thus we obtain $G\subseteq F$
for any $F\in\mathcal{F}$. Particularly, $\mathcal{T}_{G}$ is an Alexandrov
topology and $1/G$ is a minimum open set of 1. According to Lemma 3.3 in [20],
it follows that there are filters $F_{1},F_{2},\ldots,F_{n}$ in $\mathcal{F}$
such that $1/G=1/\cap_{i=1}^{n}F_{i}$. Therefore, $G=\cap_{i=1}^{n}F_{i}$.
Since $(\mathcal{F},\subseteq)$ is a down-directed set, there is a filter
$F\in\mathcal{F}$ such that $F\subseteq\cap_{i=1}^{n}F_{i}=G$. Therefore, we
have $G=F\in\mathcal{F}$ and $G$ is a minimum filter of $\mathcal{F}$.

###### Lemma 5.10

Let $(L,\mathcal{U})$ be a topological residuated lattice. If $F$ is a filter
of $L$, then $F$ is open(closed) if and only if $a/F$ is open(closed) for any
$a\in L$.

###### Proof 21

There is no loss of generality in assuming that $F$ is an open filter of $L$.
Let $a\in L$. We define two maps $l_{a}:L\rightarrow L$ and
$g_{a}:L\rightarrow L$ by $l_{a}(x)=a\rightarrow x$ and $g_{a}(x)=x\rightarrow
a$ for any $x\in L$. Then clearly, $l_{a}$ and $g_{a}$ are continuous maps and
so $l_{a}^{-1}(F)$, $g_{a}^{-1}(F)$ are open in $(L,\mathcal{U})$. Since
$a/F=\\{x\in L:x\rightarrow a\in F~{}and~{}a\rightarrow x\in
F\\}=l_{a}^{-1}(F)\cap g_{a}^{-1}(F)$, then we have $a/F\in\mathcal{U}$.
Conversely, it follows the fact $F=1/F$.

We state next an easy consequence of compactness which is similar to
topological groups [33].

###### Theorem 5.11

In a compact zero-dimensional linear topological residuated lattice
$(L,\mathcal{U})$, a filter $G$ is open if and only if $G$ is closed of finite
index.

###### Proof 22

By hypothesis, let $X_{\mathcal{U}}=\\{F:1\in F\in\mathcal{U}\\}$. Suppose
that $G$ is an open filter of $L$, then there exists a filter $F$ in
$X_{\mathcal{U}}$ such that $F\subseteq G$. Let $L=\bigcup_{x\in L}x/F$. Since
$(L,\mathcal{U})$ is compact, there are $x_{1},\ldots,x_{n}\in L$ such that
$L=\bigcup_{i=1}^{n}x_{i}/F$. Due to $F\subseteq G$, it follows that
$x_{i}/F\subseteq x_{i}/G$. Thus we have $L=\bigcup_{i=1}^{n}x_{i}/G$. Without
loss of generality we can assume $x_{1}/G=1/G=G$, hence we have
$G=L\setminus\bigcup_{i=2}^{n}x_{i}/G$. By Lemma 5.10, each $x_{i}/G$ is open,
so $G$ is closed of finite index. Conversely, suppose $G$ is a closed filter,
then we can let $L=G\cup\bigcup_{i=1}^{n-1}x_{1}/G$, by Lemma 5.10,
$G=L\setminus\bigcup_{i=1}^{n-1}x_{1}/G$ is open. This proves the theorem.

###### Proposition 5.12

Let $\mathcal{F}(L)\setminus\\{1\\}$ be a global system of filters of a
residuated lattice $L$. If $(L,\mathcal{T}_{\mathcal{F}(L)\setminus\\{1\\}})$
is compact and Hausdorff, then $L$ is finitely approximable.

###### Proof 23

It follows from Theorem 4.4 and Theorem 5.11.

###### Theorem 5.13

Let $L$ be a non-trivial residuated lattice satisfying the descending chain
condition. Then $L$ is subdirectly irreducible if and only if it has a largest
non-trivial zero-dimensional linear topological residuated lattice.

###### Proof 24

It follows from Theorem 5.9 in the paper and Theorem 8.4 in [9].

By a chain of prime filters of a residuated lattice $L$ we mean a finite
strictly increasing sequence $P_{0}\subset P_{1}\subset\cdots\subset P_{n}$,
the _length_ of the chain is $n$.

###### Definition 5.14

_We define the _dimension_ of a residuated lattice $L$ to be the supremum of
the lengths of all chains of prime filters in $L$: it is an inter $n\geq 0$,
or $+\infty$(assuming $L$ is nontrivial). _

Noth that such a dimension also was called the _Krull dimension_ in
commutative rings (see for instance [1]).

###### Proposition 5.15

If a residuated lattice $L$ is subdirectly irreducible, then it does not have
a non-trivial Hausdorff zero-dimensional linear topological residuated
lattice.

###### Proof 25

Let $L$ be a subdirectly irreducible residuated lattice. Suppose that
$(L,\mathcal{U})$ is a Hausdorff zero-dimensional linear topological
residuated lattice. By Corollary 5.5, there is a system of filters
$(\mathcal{F},\subseteq)$ such that $\mathcal{U}=\sup\\{\mathcal{T}_{F}:F\in
X_{\mathcal{U}}\\}$. According to Theorem 4.4 and $\mathcal{U}$ is Hausdorff,
we have $\bigcap\mathcal{F}=\\{1\\}$. Since $\\{1\\}$ is completely meet-
irreducible in $(\mathcal{F}(L),\subseteq)$, it follows that
$\\{1\\}\in\mathcal{F}$. Therefore, $\mathcal{U}$ is a discrete topology.

###### Theorem 5.16

If a residuated lattice $L$ is not subdirectly irreducible and its dimension
is infinite, then there must be a non-trivial Hausdorff topological residuated
lattice on $L$.

###### Proof 26

Assume that $L$ is not subdirectly irreducible and its dimension is infinite.
According to Corollary 4.7, we have $\bigcap\mathcal{P}(L)=\\{1\\}$. Since $L$
is not subdirectly irreducible, we must have

$\bigcap(\mathcal{P}(L)\setminus\\{1\\})=\bigcap(\mathcal{F}(L)\setminus\\{1\\})=\\{1\\},$

otherwise it follows that $\\{1\\}\subset
F=\bigcap(\mathcal{F}(L)\setminus\\{1\\})$, by Proposition 4.6, $F$ is a
minimum filter in $\mathcal{F}(L)\setminus\\{1\\}$, which is a contradiction.
Since $L$ is an infinite dimension, there is a strictly increasing sequence of
$\mathcal{P}(L)\setminus\\{1\\}$, denote it by $\mathcal{P}$, such that

$\bigcap\mathcal{P}=\bigcap(\mathcal{P}(L)\setminus\\{1\\})=\\{1\\}.$

It is easily checked that $(\mathcal{P},\subseteq)$ is a system of filters,
thus $(L,\mathcal{T}_{\mathcal{P}})$ is a topological residuated lattice.
According to Proposition 4.4, $\mathcal{T}_{\mathcal{P}}$ is Hausdorff, which
is not-trivial, since the intersection of any finite of elements of
$\mathcal{P}$ is not equal to $\\{1\\}$.

###### Example 5.17

In Example 5.8, $(\mathcal{F},\subseteq)$ is a system of filters. Since the
Gödel structure $\mathcal{I}=(I,min,max,\odot,\rightarrow,0,1)$ is a linear
residuated lattice, it follows that each element of $\mathcal{F}$ is a prime
filter, and then $\mathcal{I}$ is an infinite dimension residuated lattice. It
is well known that $\mathcal{I}$ is not subdirectly irreducible(e.g.,see [15]
Theorem 2.4.11). Thus, by Theorem 5.16, we have $\mathcal{T}_{\mathcal{F}}$ is
a non-trivial Hausdorff zero-dimensional linear topological residuated
lattice. Routine calculation shows that for any element $(a,1]$ in
$\mathcal{F}$, we have $x/(a,1]=\\{x\\}$ if $x\leq a$, and $x/(a,1]=(a,x]$ if
$x>a$, that is, the topology is

$\mathcal{T}_{(a,1]}=\\{\\{x\\}:x\leq a\\}\cup(a,1].$

Since $\mathcal{T}_{\mathcal{F}}=\sup\\{\mathcal{T}_{(a,1]}:a\in[0,1)\\}$, it
follows that the Sorgenfry topology on $I$ is coarse than the topology
$\mathcal{T}_{\mathcal{F}}$.

##  6 Conclusions and future research topics

In this paper, we concluded that the equipotent of a profinite residuated
lattice is either finite or uncountable which used to judge whether the
topological residuated lattice is profinite with set-theoretical methods.
Giving a profinite algebra $A$, whether there is an algebra $B$ such that
$A\cong\widehat{B}$. In lattice theory, this problem is related to Grätzer’s
clebrated problem of representable posets, see [18] Problems 34-35, P 156. In
this paper, we concluded that
$\mathbf{Pro}\mathcal{RL}\subseteq\mathbf{Ap}\mathcal{RL}\cap\mathbf{C}\mathcal{RL}$,
where $\mathbf{C}\mathcal{RL}$ is the class of complete residuated lattices.
Likewise, the results for Heyting algebras [5], Grätzer’s problem of
representability for residuated lattices would be our further research topics.
It is well known that
$\mathbf{Pro}\mathcal{RL}\subseteq\mathbf{Sto}\mathcal{RL}$, consisting of
these Stone topological residuated lattices, in other words, the category
$\mathbf{Pro}\mathcal{RL}$ is equivalent to a full subcategory of
$\mathbf{Sto}\mathcal{RL}$. The question arises by P.T. Jonstone (see [24]
Sec. VI 2.6): given an algebra $A$, when can we say that every Stone
topological algebra is profinite? This question is also true for a wide class
of algebras, including groups, rings, bounded distributive lattices, Heyting
algebras and Boolean algebras. We would study this question for residuated
lattices.

Acknowledgments
This research is supported by a grant of National Natural Science Foundation
of China (12171294,11901371,12001423), Postdoctoral Science Foundation of
China (2019M660054XB).

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