What makes a Stone topological algebra Profinite

Abstract

This paper is a contribution to understanding what properties should a topological algebra on a Stone space satisfy to be profinite. We reformulate and simplify proofs for some known properties using syntactic congruences. We also clarify the role of various alternative ways of describing syntactic congruences, namely by finite sets of terms and by compact sets of continuous self mappings of the algebra.

1 Introduction

Profinite algebras, that is, inverse limits of inverse systems of finite algebras, appear naturally in several contexts. There is an extensive theory of profinite groups, which appear as Galois groups but are also studied as a generalization of finite groups [19, 27]. One may also view p-adic number theory as an early study of special profinite algebraic structures [28]. In the context of general algebras, profinite topologies seem to have first appeared in [12]. A fruitful line of development came about with the discovery that formal equalities between elements of free profinite algebras may be used to describe pseudovarieties [11, 25], which are classes of finite algebras of a fixed type closed under taking homomorphic images, subalgebras, and finite direct products. Since pseudovarieties play an important role in algebraic theories developed for computer science, profinite algebras are also a useful tool in that context. In particular, profinite semigroups have been extensively used (see, for instance, [2, 4,5,6, 9, 26, 30]).

As the underlying topological structure of a profinite algebra is compact and 0-dimensional, that is, a Stone space, profinite algebras may also be viewed as dual spaces of Boolean algebras augmented by continuous operations. For finitely generated relatively free profinite algebras, the corresponding Boolean algebras have special significance [2, Theorem 3.6.1] and are particularly relevant in the applications of finite semigroup theory to the theory of formal languages, where they appear as Boolean algebras of regular languages. The dual role of the algebraic operations in relatively free profinite algebras has also been investigated [20,21,22].

For certain classes of (topological) algebras, it turns out that being a Stone space is sufficient to guarantee profiniteness. Special cases were considered in [24] but the essential ingredient lies in the fact that syntactic congruences are determined by finitely many terms [1, 16] using an idea of Hunter [23] that may be traced back to Numakura [24]. For such classes of algebras, profiniteness is thus a purely topological property, although this is not true in general.

Recently, several characterizations of profiniteness in a Stone topological algebra A have been obtained in [29]. They are formulated in terms of topological properties of the translation monoid of the algebra A, which is a submonoid of the monoid of continuous transformations of A, which is itself a topological monoid under the compact-open topology.

We explore further the role of syntactic congruences in the characterization of profiniteness. This leads us to an extension to topological algebras over topological signatures of Gehrke’s sufficient condition for a quotient of a profinite algebra to be profinite, with a simplified proof. The proofs of Schneider and Zumbrägel’s characterizations of profiniteness in this language are also somewhat simplified. The key ingredient is quite simple: a Stone topological algebra is profinite if and only if the syntactic congruence of every clopen subset is clopen (Theorem 4.1).

As already mentioned above, the existence of descriptions of the syntactic congruences of clopen subsets by a finite number of terms is a sufficient condition for profiniteness. In such terms, all but one variable are evaluated to arbitrary values in the algebra, which means that potentially infinitely many polynomials in one variable are used. In a profinite algebra, each such syntactic congruence may in fact be described by finitely many linear polynomials in one variable, which may depend on the congruence. We explore more generally the property of a syntactic congruence being determined by a compact set of continuous self mappings of the algebra, a property that, for a clopen subset of a locally compact algebra is equivalent to the congruence being clopen. This leads to several further characterizations of profiniteness for a Stone topological algebra in terms of how their syntactic congruences may be described. This is summarized at the very end of the paper, in Theorem 6.1.

2 Topological algebras

We say that an equivalence relation on a topological space X is closed (respectively open or clopen) if it is a closed (respectively open or clopen) subset of the product space $X\times X$. It is easy to verify that an equivalence relation is open if and only if it is clopen, if and only if its classes are open, if and only if its classes are clopen (see, for instance, [6, Exercise 3.40]). For a closed equivalence relation, the classes are closed, but the converse fails in general [6, Exercise 3.39]. Given an equivalence relation $\theta$ on a topological space X, the quotient set $X/\theta$ is endowed with the largest topology that renders continuous the natural mapping $X\rightarrow X/\theta$. In particular, a set of $\theta$-classes is closed (respectively, open) if and only if so is its union in X.

Unlike some literature on topology, we require the Hausdorff separation property for a space to be locally compact or compact.

The following observation may be considered as an exercise in topology, based on [13, Chapter I, §10.4, Proposition 8].

Proposition 2.1

Let X be a compact space and $\theta$ an equivalence relation on X. Then the quotient space $X/\theta$ is compact if and only if $\theta$ is closed.

Following [29], by a signature we mean a sequence $\Omega =(\Omega _n)_{n\in {\mathbb {N}}}$ of sets of operation symbols arranged by arity. We say that it is a topological signature if each set $\Omega _n$ is endowed with a topology.

An $\Omega$-algebra is a pair (A, E), where A is a set and $E=(E_n^A)_{n\in {\mathbb {N}}}$ is a sequence of evaluation mappings $E_n^A:\Omega _n\times A^n\rightarrow A$. In case A is a topological space and $\Omega$ is a topological signature, we say that the $\Omega$-algebra is a topological $\Omega$-algebra if each mapping $E_n^A$ is continuous. For an $\Omega$-algebra (A, E) and $w\in \Omega _n$, we let $w_A:A^n\rightarrow A$ be the operation defined by

$$\begin{aligned} w_A(a_1,\ldots ,a_n)=E_n^A(w,a_1,\ldots ,a_n). \end{aligned}$$

Reference to the sequences E and $\Omega$, which should be understood from the context, will usually be omitted and so we simply say that A is an algebra.

We view finite algebras as discrete topological algebras. Note that the requirement that the evaluation mappings be continuous may still be nontrivial. For instance, consider the signature $\Omega$ with one binary operation symbol and $\Omega _1$ the one-point ($\infty$) compactification of ${\mathbb {N}}$ (in which the open sets are all subsets of ${\mathbb {N}}$ together with the subsets of the form $\{\infty \}\cup ({\mathbb {N}}\setminus F)$, where F is a finite subset of ${\mathbb {N}}$). We may then consider finite semigroups as $\Omega$-algebras by interpreting the binary operation symbol as the semigroup multiplication, each unary operation symbol $n\in {\mathbb {N}}$ as the n! power and $\infty$ as the unique idempotent power. The evaluation mappings are continuous, and so we obtain a topological $\Omega$-algebra. However, it is sometimes useful to consider an (“unnatural”) interpretation of the operation symbol $\infty$ that makes the evaluation mapping $E_1$ discontinuous. This idea underlies the recent paper [7].

For a class $\mathcal {K}$ of topological algebras, we say that a topological algebra A is residually $\mathcal {K}$ if, for every pair $a,a'$ of distinct elements of A, there is a continuous homomorphism $\varphi :A\rightarrow B$ into a member B of $\mathcal {K}$ such that $\varphi (a)\ne \varphi (a')$.

A topological algebra A is said to be profinite if it is an inverse limit of finite discrete algebras. Equivalently, A is residually finite and compact (see, for instance, [5]). On the other hand, we say that A is a Stone topological algebra if its underlying topological space is a Stone space (note that “Stone algebra” has a different meaning in the literature). While it is not hard to see that every profinite algebra is a Stone topological algebra, the converse fails, because Stone topological algebras need not be residually finite. We present two examples of such algebras. The first example is borrowed from the paper [8, Example 5.2], which deals with unary algebras. Variations of this example are also found in [29, Examples 6.2 and 6.3] and [16, Example 7.2].

Fig. 1

A Stone topological unary algebra which is not profinite

Example 2.2

Consider the one-point compactification $U={\mathbb {N}}\cup \{\infty \}$ of ${\mathbb {N}}$. Equip U with the unary operation a defined by $a(n) = {\max \{0,n-1\}}$ and $a(\infty )=\infty$, as in Figure 1. This gives a Stone topological algebra over a signature consisting of a single unary operation symbol. Moreover, all continuous homomorphisms defined on this topological algebra and taking values in finite unary algebras are constant. Indeed, all continuous mappings from U to a finite discrete space must be eventually constant, and eventually constant homomorphisms must in fact be constant.

Within the context of this paper, the previous example can serve as a basic negative example, on which one can test various conditions equivalent to profiniteness. We further include the next example which shows that the argument from Example 2.2 can occur also in a richer algebra. The example is interesting in itself, as it is a Stone topological (modular) lattice which is not profinite. It already appeared in [16, Example 7.4] and can be traced back to an earlier paper of Clinkenbeard [17].

Fig. 2

A Stone topological modular lattice which is not profinite

Example 2.3

Let $U={\mathbb {N}}\cup \{\infty \}$ be the one-point compactification of ${\mathbb {N}}$, equipped with the partial ordering depicted in Figure 2. One can check that U is indeed a Stone topological modular lattice. Let us briefly suggest how to prove that it is not residually finite. Like in the previous example, continuous mappings from U onto finite discrete spaces are eventually constant. Moreover, eventually constant homomorphisms defined on U are in fact constant: the key is to show that a homomorphism $\varphi$ which identifies $(n,n+3)$ for $n\in 3{\mathbb {N}}$ also identifies $(n-3,n)$. In particular, eventually constant homomorphisms must be constant on $3{\mathbb {N}}\cup \{\infty \}$, and thus also constant on all of U.

A continuous mapping from a topological space X to a topological algebra A is said to be a generating mapping if its image generates (algebraically) a dense subalgebra of A. In case X is a subset of A, we say that the topological algebra A is generated by X if the inclusion $X\hookrightarrow A$ is a generating mapping. We also say that A is X-generated if there is a generating mapping $X\rightarrow A$ and that it is finitely generated if it is X-generated for some finite set X.

By a congruence on an $\Omega$-algebra A we mean an equivalence relation $\theta$ that is compatible with the $\Omega$ operations. The set $A/\theta$ of all congruence classes $a/\theta$ inherits a natural structure of $\Omega$-algebra: for $w\in \Omega _n$, $w_{A/\theta }(a_1/\theta ,\ldots ,a_n/\theta )=w_A(a_1,\ldots ,a_n)/\theta$. In case A is a compact $\Omega$-algebra and $\theta$ is a closed congruence, the quotient algebra $A/\theta$ is a compact $\Omega$-algebra for the quotient topology [29, Lemma 4.3].

For a set X, let $T_\Omega (X)$ be the $\Omega$-term algebra, that is, the absolutely free $\Omega$-algebra. The elements of $T_\Omega (X)$ are usually viewed in computer science as trees whose leaves are labeled by elements of X or of $\Omega _0$ and each non-leaf node is labeled by an element of some $\Omega _n$, in which case the node has exactly n children.

For instance, for $u\in \Omega _2$, $v\in \Omega _3$, and $w\in \Omega _0$, the term

$$\begin{aligned} u(v(x_1,u(w,x_1),x_3),u(x_3,x_2)) \end{aligned}$$

is represented by the labeled tree pictured in Figure 3.

Fig. 3

A labeled tree representing an element of an $\Omega$-term algebra

If the tree representing the term t has exactly one occurrence of $x\in X$ as a leaf label, then we say that t is linear in x. For instance, the term above is linear only in $x_2$.

Let $\varphi :X\rightarrow A$ be any mapping. Since the term algebra $T_\Omega (X)$ is the absolutely free algebra over the set X, there is a unique homomorphism ${\hat{\varphi }}:T_\Omega (X)\rightarrow A$ such that ${\hat{\varphi }}\circ \iota =\varphi$, where $\iota$ is the inclusion mapping of X in $T_\Omega (X)$. In case $X=\{x_1,\ldots ,x_m\}$, we also denote ${\hat{\varphi }}(t)$ by $t_A\bigl (\varphi (x_1),\ldots ,\varphi (x_m)\bigr )$ for $t\in T_\Omega (X)$. In this case, given elements $a_j\in A$ (${j\ne i}$), a term $t\in T_\Omega (X)$ determines a polynomial transformation of A given by

$$\begin{aligned} a\mapsto t_A(a_1,\ldots ,a_{i-1},a,a_{i+1},\ldots ,a_m). \end{aligned}$$

The translation monoid M(A) consists of all such polynomial transformations given by terms that are linear in the distinguished variable $x_i$.

For two topological spaces X and Y, let ${\mathcal {C}}(X,Y)$ be the set of continuous mappings from X to Y; it is endowed with the compact-open topology, for which a subbase consists of all sets of the form

$$\begin{aligned}{}[K,U]=\bigl \{f\in \mathcal {C}(X,Y) : f(K)\subseteq U\bigr \}, \end{aligned}$$

where $K\subseteq X$ and $U\subseteq Y$ are respectively compact and open. Subsets of ${\mathcal {C}}(X,Y)$ are endowed with the induced topology. We also write ${\mathcal {C}}(X)$ for ${\mathcal {C}}(X,X)$. Note that, in case A is a topological algebra, M(A) is a submonoid of the monoid $\mathcal {C}(A)$.

Note that, if X is compact 0-dimensional (that is, a Stone space), then we may restrict the choice of both K and U to be clopen subsets of X in the subbasic open sets [K, U] of the compact-open topology of ${\mathcal {C}}(X)$ (cf. [14, Chapter X, §3.4, Remark 2]).

3 Syntactic congruences

We say that an equivalence relation $\theta$ on a set A saturates a subset L of A if L is a union of $\theta$-classes. The following purely algebraic result is well known. We provide a proof for the sake of completeness.

Lemma 3.1

For an algebra A and a subset L of A, the set of all pairs ${(a,a')\in A\times A}$ such that

$$\begin{aligned} \forall f\in M(A)\ \bigl (f(a)\in L \iff f(a')\in L\bigr ) \end{aligned}$$

is the largest congruence on A saturating L.

Proof

Let $\theta$ denote the set of pairs in the statement of the lemma. Note that it is an equivalence relation on A. Let $w\in \Omega _n$ and $(a_i,a_i')\in \theta$ for $i=1,\ldots ,n$ and suppose that $f\in M(A)$ is such that $f\bigl (w_A(a_1,\ldots ,a_n)\bigr )\in L$. Considering the polynomial transformation g given by $g(x)=f\bigl (w_A(x,a_2,\ldots ,a_n)\bigr )$, we deduce from $(a_1,a_1')\in \theta$ that $f\bigl (w_A(a_1',a_2,\ldots ,a_n)\bigr )\in L$. Proceeding similarly on each of the remaining components, we see that $f\bigl (w_A(a_1',\ldots ,a_n')\bigr )\in L$. Hence, $\theta$ is a congruence on A. Taking for f the identity transformation of A we conclude that $\theta$ saturates L.

Now, suppose that $\rho$ is a congruence on A saturating L and let ${(a,a')\in \rho }$ and $f\in M(A)$. Because polynomial transformations of A preserve $\rho$-equivalence, we have $\bigl (f(a),f(a')\bigr )\in \rho$. Since $\rho$ saturates L, it follows that $f(a)\in L$ if and only if $f(a')\in L$, which shows that $(a,a')\in \theta$. Hence, $\rho$ is contained in $\theta$, thereby completing the proof of the lemma. $\square$

The congruence of the lemma is called the syntactic congruence of L on A and it is denoted $\sigma _L^A$.

Remark 3.2

Let L be a subset of an algebra A and let $\alpha _L$ be the equivalence relation whose classes are L and $A\setminus L$. Then we may reformulate the definition of the syntactic congruence by saying that $\sigma _L^A$ is the largest congruence contained in $\alpha _L$. This equivalent definition can also be expressed by the following useful formula:

$$\begin{aligned} \sigma _L^A =\bigcap _{f\in M(A)}(f\times f)^{-1}(\alpha _L) =\bigcap _{f\in M(A)}\alpha _{f^{-1}(L)}. \end{aligned}$$

(3.1)

Corollary 3.3

Let L be a clopen subset of a topological algebra A. Then the syntactic congruence $\sigma _L^A$ is closed.

Proof

Consider the equivalence relation $\alpha _L$ of Remark 3.2, so that formula (3.1) holds. Since $\alpha _L$ is a clopen subset of $A\times A$ and each mapping $f\in M(A)$ is continuous, so that $f\times f\in {\mathcal {C}}(A\times A)$, it follows that $\sigma _L^A$ is an intersection of clopen sets of the form $(f\times f)^{-1}(\alpha _L)$ and, therefore, it is closed. $\square$

The following result is another simple application of Lemma 3.1.

Proposition 3.4

Let $\varphi :A\rightarrow B$ be an onto homomorphism between two $\Omega$-algebras and let L be a subset of B. Then we have

$$\begin{aligned} (\varphi \times \varphi )^{-1}\sigma _L^B=\sigma _{\varphi ^{-1}(L)}^A, \end{aligned}$$

so that $\varphi$ induces an isomorphism $A/\sigma _{\varphi ^{-1}(L)}^A\rightarrow B/\sigma _L^B$.

Proof

By the general correspondence theorem of universal algebra [15, Theorem 6.20], $(\varphi \times \varphi )^{-1}\sigma _L^B$ is a congruence on A. It saturates $\varphi ^{-1}(L)$ since, if $(a,a')\in (\varphi \times \varphi )^{-1}\sigma _L^B$ and $a\in \varphi ^{-1}(L)$, then $(\varphi (a),\varphi (a'))\in \sigma _L^B$ and $\varphi (a)$ belongs to L and, therefore, so does $\varphi (a')$. This shows that $(\varphi \times \varphi )^{-1}\sigma _L^B\subseteq \sigma _{\varphi ^{-1}(L)}^A$.

Conversely, suppose that $(a,a')\in \sigma _{\varphi ^{-1}(L)}^A$ and $t(x_1,\ldots ,x_m)$ is a term linear in $x_1$ such that $t_B(\varphi (a),b_2,\ldots ,b_m)\in L$ with the $b_j$ in B. For each $j\in \{2,\ldots ,m\}$, let $a_j\in A$ be such that $b_j=\varphi (a_j)$. Then, we must have $t_A(a,a_2,\ldots ,a_m)\in \varphi ^{-1}(L)$, which yields $t_A(a',a_2,\ldots ,a_m)\in \varphi ^{-1}(L)$, which in turn entails $t_B(\varphi (a'),b_2,\ldots ,b_m)\in L$. This establishes the reverse inclusion $\sigma _{\varphi ^{-1}(L)}^A\subseteq (\varphi \times \varphi )^{-1}\sigma _L^B$. $\square$

4 Profiniteness

To give an application of Proposition 3.4, we first need some connections of syntactic congruences with profiniteness. Recall, as stated in Corollary 3.3, that the syntactic congruence of a clopen subset of a topological algebra is always closed.

Theorem 4.1

A Stone topological algebra A is profinite if and only if, for every clopen subset L of A, the syntactic congruence $\sigma _L^A$ is clopen.

Proof

($\Rightarrow$) It follows from residual finiteness and compactness that there is a continuous homomorphism $\varphi :A\rightarrow B$ onto a finite algebra B such that $L=\varphi ^{-1}(\varphi (L))$ [3, Lemma 4.1]. The kernel of $\varphi$ (by which we mean the equivalence relation on A given by $\{(s_1,s_2)\in A^2 : \varphi (s_1)=\varphi (s_2)\}$) is thus a clopen congruence on A that saturates L. By Lemma 3.1, $\sigma _L^A$ contains the kernel of $\varphi$. Hence, the classes of $\sigma _L^A$ are also clopen, as they are unions of classes of the kernel of $\varphi$.

($\Leftarrow$) We need to show that A is residually finite. Given distinct points $a,a'\in A$, since A is a Stone space, there is a clopen subset $L\subseteq A$ such that $a\in L$ and $a'\notin L$. Since $\sigma _L^A$ is clopen, the canonical homomorphism $A\rightarrow A/\sigma _L^A$ is a continuous homomorphism onto a finite (discrete) algebra that separates the points a and $a'$, as $\sigma _L^A$ saturates L by Lemma 3.1. $\square$

We are now ready to prove the following result. The special case where the signature is discrete was first proved in [20, Theorem 4.3] using duality theory.

Theorem 4.2

Let A be a profinite algebra and let $\theta$ be a closed congruence on A such that $A/\theta$ is 0-dimensional. Then the quotient $A/\theta$ is profinite.

Proof

Let $B=A/\theta$ and let $\varphi :A\rightarrow B$ be the canonical homomorphism. By assumption and Proposition 2.1, B is a Stone topological algebra. We apply the criterion for B to be profinite of Theorem 4.1. So, let L be a clopen subset of B. By Proposition 3.4, we obtain the equality

$$\begin{aligned} (\varphi \times \varphi )^{-1}\sigma _L^B=\sigma _{\varphi ^{-1}(L)}^A. \end{aligned}$$

Since $\varphi$ is continuous, the set $\varphi ^{-1}(L)$ is clopen. Hence, by Theorem 4.1, $\sigma _{\varphi ^{-1}(L)}^A$ is a clopen congruence, and so $A/\sigma _{\varphi ^{-1}(L)}^A$ is finite. By Proposition 3.4, the algebra $B/\sigma _L^B$ is finite. Since $\sigma _L^B$ is a closed congruence by Corollary 3.3, it follows that $\sigma _L^B$ is clopen. $\square$

A compact space X has a natural uniform structure in the sense of [13, Chapter II] with base consisting of the open neighborhoods of the diagonal $\{(x,x) : x\in X\}$, that is, its entourages are the subsets of $X\times X$ that contain such neighborhoods. This is the only uniform structure that determines the topology of X. In case X is a Stone space, one may take as basic members the sets of the form $\bigcup _{i=1}^nL_i\times L_i$ where the $L_i$ constitute a clopen partition of X; we call the union $\bigcup _{i=1}^nL_i\times L_i$ the entourage determined by the partition $L_1,\ldots ,L_n$. Suppose A is a profinite algebra and $L_1,\ldots ,L_n$ is a clopen partition of A. By Theorem 4.1, the set $\theta =\bigcap _{i=1}^n\sigma _{L_i}^A$ is a clopen congruence on A that saturates each set $L_i$. Hence, to obtain a subbase of the uniform structure of A one may consider only the clopen partitions defined by clopen congruences.

Let X be a uniform space and let $\mathcal {F}\subseteq \mathcal {C}(X)$. We say that $\mathcal {F}$ is equicontinuous if for every entourage $\alpha$ of X and every $x\in X$, there is an open subset $U\subseteq X$ such that $x\in U$ and, for every $f\in \mathcal {F}$, $f(U)\times f(U)\subseteq \alpha$. We say that $\mathcal {F}$ is uniformly equicontinuous if for every entourage $\alpha$ of X, there is an entourage $\beta$ of X such that, for every $f\in \mathcal {F}$, $(f\times f)(\beta )\subseteq \alpha$. In case X is compact, the two properties are equivalent [14, Chapter X, §2.1, Corollary 2].

Here is another characterization of profiniteness for Stone topological algebras. It is taken from [29, Theorem 4.4]. The proof presented here is basically the same as that in [29] although it is slightly simplified thanks to the usage of syntactic congruences.

Theorem 4.3

A Stone topological algebra A is profinite if and only if M(A) is equicontinuous.

Proof

($\Rightarrow$) Let $L_1,\ldots ,L_n$ be a clopen partition of A and let $\theta =\bigcap _{i=1}^n\sigma _{L_i}^A$. Then, by Lemma 3.1, for every $f\in M(A)$, we have

$$\begin{aligned} (f\times f)(\theta )\subseteq \bigcup _{i=1}^nL_i\times L_i. \end{aligned}$$

($\Leftarrow$) We apply the criterion of Theorem 4.1. So, let L be a clopen subset of A and consider the entourage $\alpha _L$ determined by the clopen partition $L,A\setminus L$. By (uniform) equicontinuity, there exists a clopen partition $L_1,\ldots ,L_n$ of A, determining an entourage $\beta$ of X, such that, for every $f\in M(A)$, the inclusion $(f\times f)(\beta )\subseteq \alpha _L$ holds or, equivalently, $\beta \subseteq \bigcap _{f\in M(A)}(f\times f)^{-1}(\alpha _L)$. By Remark 3.2, we have

$$\begin{aligned} \bigcap _{f\in M(A)}(f\times f)^{-1}(\alpha _L)=\sigma _L^A. \end{aligned}$$

Hence, we have $\beta \subseteq \sigma _L^A$. It follows that each class of $\sigma _L^A$ is a union of some of the $L_i$ and, therefore it is clopen. Hence, $\sigma _L^A$ is a clopen congruence and A is profinite by Theorem 4.1. $\square$

A subset of a topological space X is said to be relatively compact if it is contained in a compact subset of X. The following is a reformulation of Theorem 4.3. As observed in [29], the equivalence between the criteria of Theorems 4.3 and 4.4 is an immediate application of the Arzelà-Ascoli theorem of functional analysis [14, Chapter X, §2.5, Corollary 3].

Theorem 4.4

([29]). A Stone topological algebra A is profinite if and only if M(A) is relatively compact in $\mathcal {C}(A)$.

5 Determination of syntactic congruences

In this section, we examine several ways of describing syntactic congruences.

5.1 Determination by terms and functions

Moving forward, we drop the superscripts when denoting syntactic congruences, as it will always be clear from the context to which algebra they pertain. We say that a syntactic congruence $\sigma _L$ of an algebra A is finitely determined by terms if there exists a finite subset F of $T_\Omega (\{x_1,\ldots ,x_m\})$ such that $\sigma _L$ consists of all pairs $(a,a')\in A^2$ for which

$$\begin{aligned} \forall t\in F\ \forall b_2,\ldots ,b_m\in A\ \bigl ( t_A(a,b_2,\ldots ,b_m)\in L \iff t_A(a',b_2,\ldots ,b_m)\in L \bigr ). \end{aligned}$$

We then also say that the set of terms F determines $\sigma _L$. In Lemma 5.13 below, it is observed that if $\sigma _L$ is determined by a finite set of terms, then it is also determined by such a set in which every term is linear in the first variable. Finite determination occurs in many important examples: in semigroups, monoids, rings and distributive lattices, all syntactic congruences are determined by finite sets of terms, even in a uniform way in the sense that the same finite set of terms works for all syntactic congruences on all algebras of the chosen type. This holds because in each case, there are only finitely many types of linear polynomials (in rings for instance, the linear polynomials in the variable x are all of the form $x+c$, $ax+c$, $xb+c$, or $axb+c$).

Generalizing a result of Numakura [24] (for the cases of semigroups and distributive lattices), the first author [1] has shown that a sufficient condition for a Stone topological algebra to be profinite is that its syntactic congruences of clopen subsets are finitely determined (see also [10, 16]).

There is a stronger form of determination of a syntactic congruence, which is suggested by Lemma 3.1. We say that the syntactic congruence $\sigma _L$ of a subset L of an algebra A is S-determined if S is a set of functions $A\rightarrow A$ such that, for all $a,a'\in A$, $(a,a')\in \sigma _L$ holds if and only if

$$\begin{aligned} \forall f\in S\ \bigl (f(a)\in L \iff f(a')\in L\bigr ). \end{aligned}$$

Note that $\sigma _L$ is M(A)-determined by Lemma 3.1. As in Remark 3.2, note that $\sigma _L$ is S-determined if and only if

$$\begin{aligned} \sigma _L = \bigcap _{f\in S}(f\times f)^{-1}(\alpha _L) = \bigcap _{f\in S}\alpha _{f^{-1}(L)}. \end{aligned}$$

For example, in the unary algebra U from Example 2.2, the syntactic congruence $\sigma _{\{0\}}$ (which is in fact the diagonal relation) is S-determined for $S\subseteq M(U)$ if and only if $S=M(U)$.

Proposition 5.1

Let A be an algebra and L a subset of A. If the syntactic congruence $\sigma _L$ has finite index then it is F-determined for some finite subset F of M(A).

Proof

Consider the syntactic homomorphism $\eta :A\rightarrow A/\sigma _L$, which sends each $a\in A$ to its syntactic class $a/\sigma _L$. By assumption, the algebra $A/\sigma _L$ is finite. Hence, $M(A/\sigma _L)$ is a finite monoid. For each $f\in M(A/\sigma _L)$, we may choose a term $t\in T_\Omega (\{x_1,\ldots ,x_{k+1}\})$ which is linear in $x_1$ and elements $a_2,\ldots ,a_{k+1}\in A$ such that $f(x/\sigma _L)=t_A(x,a_2,\ldots ,a_{k+1})/\sigma _L$ for every $x\in A$. We let ${\hat{f}}(x)=t_A(x,a_2,\ldots ,a_{k+1})$ for each $x\in A$, which defines an element ${\hat{f}}$ of M(A) such that $\eta \circ {\hat{f}}=f\circ \eta$. Let $F=\{{\hat{f}} : f\in M(A/\sigma _L)\}$. We claim that $\sigma _L$ is F-determined.

Since $F\subseteq M(A)$, we have $\sigma _L\subseteq \bigcap _{g\in F}(g\times g)^{-1}(\alpha _L)$. Conversely, suppose $a,a'\in A$ are such that, for every $f\in M(A/\sigma _L)$, we have ${\hat{f}}(a)\in L$ if and only if ${\hat{f}}(a')\in L$. Since $\sigma _L$ saturates L, $x\in L$ is equivalent to ${\eta (x)\in \eta (L)}$. Thus, we get $f(\eta (a))\in \eta (L)$ if and only if $f(\eta (a'))\in \eta (L)$ for every $f\in M(A/\sigma _L)$. By definition of the syntactic congruence, we deduce that ${(\eta (a),\eta (a'))\in \sigma _{\eta (L)}}$, that is, $(a,a')\in (\eta \times \eta )^{-1}(\sigma _{\eta (L)})$. By Proposition 3.4, it follows that $(a,a')\in \sigma _L$, which establishes the claim. $\square$

Note that Proposition 5.1 applies in particular to clopen syntactic congruences $\sigma _L$ on a compact algebra A, which are necessarily determined by a clopen subset L of A. One may ask whether the same finite subset F of M(A) may be used to determine all syntactic congruences of clopen subsets L of A. This is trivially the case for finite algebras but here is an infinite example for which such a finite set F also exists.

Example 5.2

Let A be a Stone space and consider a constant binary operation on A. This turns A into a profinite semigroup. Note that, for every $L\subseteq A$, we have $\sigma _L=\alpha _L$. Thus, all syntactic congruences are F-determined for every subset F of M(A). The same conclusion would be reached if we took the empty signature instead.

The next result shows that such an example cannot be simultaneously finitely generated and infinite.

Proposition 5.3

If A is an infinite finitely generated profinite algebra then, for every finite subset F of M(A), there exists a clopen subset L of A such that $\sigma _L$ is not F-determined.

Proof

Suppose that F is a finite subset of M(A) that determines the syntactic congruence of every clopen subset L of A, that is, its syntactic congruence $\sigma _L$ is given by the formula $\sigma _L=\bigcap _{f\in F}\alpha _{f^{-1}(L)}$. Note that each equivalence relation $\alpha _{f^{-1}(L)}$ has at most two classes. Hence, $\sigma _L$ has at most $2^{|F|}$ classes, that is, the syntactic algebra $A/\sigma _L$ has at most $2^{|F|}$ elements. Now, given a finite set S and a finite generating set X of A, there are only finitely many functions $X\rightarrow S$. Thus, on subsets of S there are only finitely many algebraic structures which are homomorphic images of A. There are only finitely many congruences that are kernels of these homomorphisms from A to S. In particular, there are only finitely many syntactic congruences $\sigma _L$ of clopen subsets L of A. Since the corresponding syntactic homomorphisms $A\rightarrow A/\sigma _L$ suffice to separate the points of A, we conclude that A is finite, in contradiction with the hypothesis.

$\square$

5.2 Compact determination

Suppose that A is a topological algebra and $L\subseteq A$. We say that the syntactic congruence $\sigma _L$ is compactly determined if $\sigma _L$ is C-determined for some compact subset C of ${\mathcal {C}}(A)$. Similarly, $\sigma _L$ is finitely determined if $\sigma _L$ is F-determined for some finite subset F of ${\mathcal {C}}(A)$. This is not to be confused with $\sigma _L$ being determined by a finite set of terms, which is, a priori, a weaker property in case $F\subseteq M(A)$. Theorem 5.16 at the end of Subsection 5.3 shows that all these properties are equivalent in case A is compact and $L\subseteq A$ is clopen.

In this subsection, among other results, we establish that, if all syntactic congruences of clopen subsets of a Stone topological algebra A are compactly determined, then A is profinite. Several of our results are stated for locally compact algebras because they hold in that more general setting.

Let us briefly introduce a convenient notation for partial evaluation. Given $f:Y\times Z\rightarrow X$, we define $f^\sharp :Z\rightarrow X^Y$ by:

$$\begin{aligned} f^\sharp (z)(y) = f(y,z). \end{aligned}$$

The following proposition regroups a few useful properties of the compact-open topology. For proofs, we refer the reader to [14, Chapter X, §3.4, Theorem 3 and Proposition 9].

Proposition 5.4

Let X, Y and Z be topological spaces.

• (1) If $f\in {\mathcal {C}}(Y\times Z, X)$, then $f^\sharp \in {\mathcal {C}}(Z, {\mathcal {C}}(Y,X))$.

Assume, additionally, that Y is locally compact.

• (2) Composition is a continuous mapping ${\mathcal {C}}(Y,X)\times {\mathcal {C}}(Z,Y)\rightarrow {\mathcal {C}}(Z,X)$.

• (3) Evaluation is a continuous mapping $Y\times {\mathcal {C}}(Y,X)\rightarrow X$.

In particular, from Proposition 5.4(2) it follows that, for every locally compact topological space X, ${\mathcal {C}}(X)$ is a topological monoid. In case A is a topological algebra, M(A) is a submonoid of ${\mathcal {C}}(A)$, whence so is its closure $\overline{M(A)}$ if A is locally compact.

The following statement is an elementary observation in topology whose proof is presented for the sake of completeness.

Proposition 5.5

Let X and Y be two topological spaces, with X locally compact, and let K be a compact subspace of ${\mathcal {C}}(X,Y)$.

• (1) The following mapping is closed:

  $$\begin{aligned} \Phi _K:2^Y&\rightarrow 2^X\\ V&\mapsto \bigcup _{f\in K}f^{-1}(V). \end{aligned}$$

• (2) The following mapping is open:

  $$\begin{aligned} \Psi _K:2^Y&\rightarrow 2^X\\ U&\mapsto \bigcap _{f\in K}f^{-1}(U). \end{aligned}$$

Proof

Clearly, (1) and (2) imply each other by taking the complement: $\Phi _K(2^Y\setminus U)=2^X\setminus \Psi _K(U)$. To prove (1), consider a closed subset V of Y. Let $\{ x_i \}_{i\in I}$ be a net in $\Phi _K(V)$ converging to $x\in X$. By definition of $\Phi _K$, there exists a net $\{f_i\}_{i\in I}$ in K such that, for all $i\in I$, $f_i(x_i)\in V$. Since K is compact, we may assume by taking a subnet that the first component of $\{ (f_i,x_i) \}_{i\in I}$ converges, to say $f\in K$. Since V is closed and evaluation is continuous by Proposition 5.4(3), we obtain

$$\begin{aligned} f(x) = \lim _{i\in I}f_i(x_i) \in V. \end{aligned}$$

It follows that $x\in \Phi _K(V)$, thus showing that $\Phi _K(V)$ is closed. $\square$

A further result in topology that we require for later considerations is the following.

Lemma 5.6

Let X be a Hausdorff topological space. The diagonal mapping $\Delta :{\mathcal {C}}(X)\rightarrow {\mathcal {C}}(X\times X)$ defined by $\Delta (f) = f\times f$ is continuous with respect to the compact-open topologies.

Proof

By [18, Lemma 3.4.6], the sets of the form $[K,U_1\times U_2]$, where K ranges over all compact subsets of $X\times X$ and $U_1$ and $U_2$ range over all open subsets of X, constitute a subbase for the compact-open topology of ${\mathcal {C}}(X\times X)$. Therefore, it suffices to show that for each compact subset $K\subseteq X\times X$ and open subsets $U_1, U_2\subseteq X$, the set $\Delta ^{-1}[K,U_1\times U_2]$ is open in the compact-open topology of ${\mathcal {C}}(X)$.

Let $p_1, p_2$ be the natural projections of $X\times X$ on its first and second components, respectively. Note that, for $i=1,2$, we have $p_i\circ \Delta (f) = f\circ p_i$. It follows that:

$$\begin{aligned} \Delta (f)(K)\subseteq U_1\times U_2&\iff p_i(\Delta (f)(K))\subseteq U_i&\text {for i=1,2} \\&\iff f(p_i(K))\subseteq U_i&\text {for i=1,2}. \end{aligned}$$

This shows that $\Delta ^{-1}[K,U_1\times U_2] = [p_1(K),U_1]\cap [p_2(K),U_2]$. But for $i=1,2$, the projection $p_i$ is continuous, so $p_i(K)$ is a compact subset of X and $[p_i(K),U_i]$ is open in the compact-open topology of ${\mathcal {C}}(X)$. Hence, so is $\Delta ^{-1}[K,U_1\times U_2]$. $\square$

The next proposition shows that, in order to be compactly determined, the syntactic congruence of a clopen subset merely needs to be determined by a relatively compact subset of ${\mathcal {C}}(A)$.

Proposition 5.7

Let A be a locally compact algebra and L be a clopen subset of A. If $F\subseteq {\mathcal {C}}(A)$ is such that $\sigma _L$ is F-determined, then $\sigma _L$ is also ${\overline{F}}$-determined.

Proof

Clearly, we have:

$$\begin{aligned} \bigcap _{f\in {\overline{F}}}(f\times f)^{-1}(\alpha _L) \subseteq \bigcap _{f\in F}(f\times f)^{-1}(\alpha _L) = \sigma _L. \end{aligned}$$

To prove the reverse inclusion, let us fix $x\in \sigma _L$ and take an arbitrary $f\in {\overline{F}}$. Let $\{ f_i \}_{i\in I}$ be a net in F converging to f. By assumption, $(f_i\times f_i)(x)$ belongs to $\alpha _L$ for every $i\in I$. Since $\alpha _L$ is closed and the evaluation and diagonal mappings are continuous, respectively by Proposition 5.4(3) and Lemma 5.6, we deduce that

$$\begin{aligned} (f\times f)(x) = \lim _{i\in I}(f_i\times f_i)(x) \in \alpha _L. \end{aligned}$$

This shows that $x\in (f\times f)^{-1}(\alpha _L)$, as required. $\square$

Combining Proposition 5.7 with Theorem 4.4, we get the following result.

Corollary 5.8

Let A be a profinite algebra and L be a clopen subset of A. Then $\sigma _L$ is compactly determined.

The following result shows the significance of the notion of compactly determined syntactic congruence.

Theorem 5.9

Let A be a locally compact algebra and L a subset of A. Then L is clopen and $\sigma _L$ is compactly determined if and only if $\sigma _L$ is clopen.

Proof

Suppose first that L is clopen and $\sigma _L$ is K-determined, where K is a compact subset of ${\mathcal {C}}(A)$. Since $\sigma _L$ is K-determined, we may write $\sigma _L = \Psi _{\Delta K}(\alpha _L)$ in the notation of Proposition 5.5 and Lemma 5.6. By Lemma 5.6, $\Delta$ is continuous with respect to the compact-open topologies, so $\Delta K$ is a compact subset of ${\mathcal {C}}(A\times A)$. By Proposition 5.5 applied to $X=Y=A\times A$, it follows that $\Psi _{\Delta K}$ maps open relations on A to open relations on A. But since L is clopen, so is $\alpha _L$, whence $\sigma _L$ is clopen.

For the converse, assume that $\sigma _L$ is clopen. If $\sigma _L$ is the universal relation, then L is either $\emptyset$ or A and $\sigma _L$ is K-determined for every subset K of ${\mathcal {C}}(A)$. Otherwise, we may choose elements $a,b\in A$ that are not $\sigma _L$-equivalent. Let K be the set of all mappings $A\rightarrow \{a,b\}$ that are constant on each $\sigma _L$-class. Since $\sigma _L$ is clopen, K is contained in ${\mathcal {C}}(A)$. Consider the mapping

$$\begin{aligned} \varphi :{\mathcal {C}}(A/\sigma _L,\{a,b\})&\rightarrow K\\ f&\mapsto f\circ \eta , \end{aligned}$$

where $\eta :A\rightarrow A/\sigma _L$ is the natural quotient mapping. Note that $A/\sigma _L$ is a discrete space under the quotient topology, whence it is locally compact. By Proposition 5.4(2), $\varphi$ is a continuous mapping. As both spaces $A/\sigma _L$ and $\{a,b\}$ are discrete, the space ${\mathcal {C}}(A/\sigma _L,\{a,b\})$ is in fact the product space $\{a,b\}^{A/\sigma _L}$. Since $\varphi$ is onto and continuous, we deduce that K is compact. The proof is achieved by observing that $\sigma _L$ is K-determined. $\square$

Note that the second part of the above proof does not use the hypothesis that the algebra A is locally compact. In the locally compact case, the proof could also be given by invoking the Arzelà-Ascoli theorem.

Theorems 5.9 and 4.1 yield the following result.

Corollary 5.10

Let A be a Stone topological algebra, and suppose that for every clopen subset L of A, $\sigma _L$ is compactly determined. Then A is profinite.

Given a profinite algebra A and a clopen subset L of A, one may wonder what are the compact subsets of ${\mathcal {C}}(A)$ determining $\sigma _L$. The following shows that, at the very least, there is always one that is minimal.

Proposition 5.11

Let A be a locally compact algebra, L be a clopen subset of A. Then, every compact subset of ${\mathcal {C}}(A)$ determining $\sigma _L$ contains a minimal compact subset determining $\sigma _L$.

Proof

We apply Zorn’s lemma. Let F be a compact subset of ${\mathcal {C}}(A)$ determining $\sigma _L$. Fix a descending chain $\{F_i\}_{i\in I}$ of closed subsets of F that determines $\sigma _L$ and let:

$$\begin{aligned} F' = \bigcap _{i\in I}F_i. \end{aligned}$$

We want to show that $F'$ determines $\sigma _L$. Fixing an arbitrary $i\in I$, the inclusion $F'\subseteq F_i$ gives:

$$\begin{aligned} \bigcap _{f\in F'}(f\times f)^{-1}(\alpha _L) \supseteq \bigcap _{f\in F_i}(f\times f)^{-1}(\alpha _L) = \sigma _L. \end{aligned}$$

(5.1)

It remains to show the reverse inclusion. Let us suppose that $x\in A^2\setminus \sigma _L$. Then, we have

$$\begin{aligned} \forall i\in I\ \exists f_i\in F_i,\ (f_i\times f_i)(x)\notin \alpha _L. \end{aligned}$$

This defines a net $\{ f_i \}_{i\in I}$ in F, which by compactness has a converging subnet $\{ f_{i_j} \}_{j\in J}$; let $f'\in F$ denote its limit.

For each $i\in I$, there is $k\in J$ such that $i_k\geqslant i$, and so the net $\{f_{i_j}\}_{j\in J}$ is eventually in $F_i$. Since $F_i$ is closed, it follows that $f'\in F_i$, and this holds for each $i\in I$. Thus, $f'$ belongs to $F'$. Furthermore, as the evaluation and diagonal mappings are continuous, again respectively by Proposition 5.4(3) and Lemma 5.6, and $\alpha _L$ is clopen, we get

$$\begin{aligned} (f'\times f')(x) = \lim _{j\in J}(f_{i_j}\times f_{i_j})(x) \notin \alpha _L. \end{aligned}$$

This shows that:

$$\begin{aligned} x\in \bigcup _{f\in F'}(f\times f)^{-1}(A^2\setminus \alpha _L) = A^2\setminus \bigcap _{f\in F'}(f\times f)^{-1}(\alpha _L). \end{aligned}$$

Thus, the reverse inclusion in (5.1) is proved and $\sigma _L$ is $F'$-determined. $\square$

In case A is a compact algebra and L is a clopen subset of A, one may use the ideas in the proof of Proposition 5.1 to show that every minimal compact subset of ${\mathcal {C}}(A)$ determining $\sigma _L$ is finite. Combining with Proposition 5.11, it follows that every compact subset of ${\mathcal {C}}(A)$ determining $\sigma _L$ contains a finite such set. The following example shows that Proposition 5.11 cannot be improved in the same direction for arbitrary locally compact algebras.

Example 5.12

Consider the additive monoid ${\mathbb {N}}$ of natural numbers under the discrete topology. Let L be an infinite subset of ${\mathbb {N}}$ containing no infinite arithmetic progression, for instance the set of all powers of 2 or the set of all primes. We claim that:

• (1) $\sigma _L$ is the equality relation;

• (2) $\sigma _L$ is not finitely determined.

As ${\mathbb {N}}$ is locally compact, being discrete, and $\sigma _L$ is clopen, for the same reason, Theorem 5.9 yields that $\sigma _L$ is compactly determined. Hence, by (2), $\sigma _L$ is an example of a compactly determined syntactic congruence of a clopen subset of a locally compact algebra that is not finitely determined.

To prove Claim (1), since ${\mathbb {N}}$ is a commutative monoid, a pair (m, n) of natural numbers belongs to $\sigma _L$ if and only if

$$\begin{aligned} \forall x\in {\mathbb {N}}\ (m+x\in L \iff n+x\in L). \end{aligned}$$

(5.2)

Suppose that Property (5.2) holds with $m<n$. Since L is infinite, there exists k such that $m+k$ belongs to L and whence so does $n+k$. Using (5.2), we deduce that all the elements in the arithmetic progression starting with $m+k$ with period $n-m$ lie in L, which contradicts the assumption on the set L. Hence no two distinct elements of ${\mathbb {N}}$ can be $\sigma _L$-equivalent, thereby proving (1).

To establish Claim (2), it suffices to observe that the argument at the beginning of the proof of Proposition 5.3 shows that a finitely determined syntactic congruence has finite index, which is not the case of $\sigma _L$ by (1).

5.3 Compact determination versus finite determination by terms

We next show that, for compact algebras, a syntactic congruence is compactly determined if and only if it is determined by a finite set of terms, in the sense introduced at the beginning of Subsection 5.1.

We start with some notation. Let A be a topological algebra, F be a subset of ${\mathcal {C}}(A)$, and S be a subset of ${\mathcal {C}}(A^{k+1},A)$. Given $s\in S$, in view of Proposition 5.4(1), we obtain a function $s^\sharp \in {\mathcal {C}}(A^k,{\mathcal {C}}(A))$. We define:

$$\begin{aligned} F_S = \{ f\in F : \exists s\in S\ \exists v\in A^k, f = s^\sharp (v) \}. \end{aligned}$$

We also abbreviate $F_{\{s\}}$ by $F_{s}$. Note that

$$\begin{aligned} F_S =\bigcup _{s\in S} F_s =F\cap \left( \bigcup _{s\in S}s^\sharp (A^k)\right) \subseteq {\mathcal {C}}(A). \end{aligned}$$

In case T is subset of $T_\Omega (\{x_1,\dots ,x_{k+1}\})$, we also write $F_T$ for $F_S$, where $S=\{t_A : t\in T\}$.

Lemma 5.13

Let L be a subset of an algebra A. Then $\sigma _L$ is determined by a finite set of terms if and only if it is determined by some set of the form $F_T$, where T is a finite subset of $T_\Omega (\{x_1,\dots ,x_{k+1}\})$ for some $k\geqslant 0$, consisting of terms linear in $x_1$, and $F\subseteq M(A)$.

Proof

The if part of the statement of the lemma is trivial. For the converse, we first associate with each term $t\in T_\Omega (\{x_1,\ldots x_{k+1}\})$ a term $s\in T_\Omega (X)$ with $X=\{y_1,\ldots ,y_r,x_2,\ldots ,x_{k+1}\}$ by replacing each occurrence of $x_1$ by a distinct $y_i$. Let

$$\begin{aligned} s_i=s_{T_\Omega (X)}(y^{(i-1)},x,z^{(r-i)},x_2\ldots ,x_{k+1}) \in T_\Omega (x,y,z,x_2,\ldots ,x_{k+1}), \end{aligned}$$

where $u^{(\ell )}$ stands for $\ell$ components equal to u. Then each term $s_i$ is linear in x and the following formulas hold for $a,a'\in A$ and $v\in A^k$:

$$\begin{aligned} (s_1)_A^\sharp (a,a',v)(a')&=t_A^\sharp (v)(a')\\ (s_r)_A^\sharp (a,a',v)(a)&=t_A^\sharp (v)(a)\\ (s_i)_A^\sharp (a,a',v)(a)&=(s_{i+1})_A^\sharp (a,a',v)(a') \quad (i=1,\ldots ,r-1). \end{aligned}$$

It follows that

$$\begin{aligned} \sigma _L&\subseteq \bigcap _{i=1,\ldots ,r;\ b,b'\in A} \bigl ((s_i)_A^\sharp (b,b',v)\times (s_i)_A^\sharp (b,b',v)\bigr )^{-1}(\alpha _L) \\&\subseteq \bigl (t_A^\sharp (v)\times t_A^\sharp (v)\bigr )^{-1}(\alpha _L), \end{aligned}$$

which shows that $\sigma _L$ is also determined by a finite set of terms that are linear in $x_1$ by simply taking, for a finite set of terms determining $\sigma _L$, the union of the sets of terms constructed above for each term in the given set. $\square$

The next lemma examines how the operation $F\mapsto F_T$ behaves with respect to topological closure when T is a finite set of terms.

Lemma 5.14

Let A be a compact algebra, F be a subset of ${\mathcal {C}}(A)$ and T be a finite set of terms in $T_\Omega (\{x_1,\dots ,x_{k+1}\})$. Then, $\overline{F_T}$ is contained in ${\overline{F}}_T$.

Proof

Let $f\in \overline{F_T}$. Then, we can write f as a limit

$$\begin{aligned} f = \lim _{i\in I}f_i, \end{aligned}$$

where $\{f_i\}_{i\in I}$ is a net in $F_T$. For each $i\in I$, choose $t_i\in T$ and $v_i\in A^k$ such that $f_i = (t_i)_A^\sharp (v_i)$. Since T is finite and $A^k$ is compact, we may extract a subnet $\{f_{i_j}\}_{j\in J}$ such that $\{t_{i_j}\}_{j\in J}$ takes a constant value $t\in T$ and $\{v_{i_j}\}_{j\in J}$ converges to v in $A^k$. By Proposition 5.4(1), $t_A^\sharp$ is continuous and it follows that

$$\begin{aligned} f = \lim _{j\in J}f_{i_j} = \lim _{j\in J} t_A^\sharp (v_{i_j}) = t_A^\sharp (v). \end{aligned}$$

Hence, f lies in ${\overline{F}}_T$. $\square$

We are now ready to achieve the goal announced at the beginning of this subsection.

Proposition 5.15

Let A be a compact algebra, and L be a clopen subset of A. If $\sigma _L$ is determined by a finite set of terms, then it is compactly determined.

Proof

By Lemma 5.13, there is a finite subset T of $T_\Omega (\{x_1,\dots ,x_{k+1}\})$ consisting of terms linear in $x_1$ and a subset F of M(A) such that $F_T$ determines $\sigma _L$. Then, by Proposition 5.7, $\sigma _L$ is also determined by $\overline{F_T}$. By Lemma 5.14, we have

$$\begin{aligned} \overline{F_T} \subseteq {\overline{F}}_T \subseteq M(A). \end{aligned}$$

Since both $\overline{F_T}$ and M(A) determine $\sigma _L$, so does ${\overline{F}}_T$. We claim that ${\overline{F}}_T$ is compact. Indeed, for each $t\in T$, we have

$$\begin{aligned} {\overline{F}}_t = {\overline{F}}\cap t_A^\sharp (A^k). \end{aligned}$$

But note that $t_A^\sharp (A^k)$ is compact, because $A^k$ is compact and $t_A^\sharp$ is continuous by Proposition 5.4(1). Therefore, ${\overline{F}}_t$, being an intersection of a closed with a compact subset, is itself compact. It follows that ${\overline{F}}_T =\bigcup _{t\in T}{\overline{F}}_t$ is compact, as claimed. $\square$

Combining Theorem 5.9 and Propositions 5.1 and 5.15, we obtain the following main result of this section.

Theorem 5.16

The following conditions are equivalent for a compact algebra A and a subset $L\subseteq A$:

• (1) $\sigma _L$ is clopen;

• (2) L is clopen and $\sigma _L$ is compactly determined;

• (3) L is clopen and $\sigma _L$ is finitely determined;

• (4) L is clopen and $\sigma _L$ is finitely determined by a set of terms;

• (5) the quotient algebra $A/\sigma _L$ is finite and discrete.

The following example shows that it is not possible to extend Proposition 5.15 for the case of locally compact algebras. In fact, we exhibit a locally compact semigroup and a clopen subset whose syntactic congruence is not clopen. This syntactic congruence is not compactly determined by Theorem 5.9, while it is determined by a finite set of terms as in fact every syntactic congruence of a semigroup has this property.

Example 5.17

We consider the topological semigroup obtained as the direct product of the following locally compact semigroups A and B, so that it is locally compact. Let $A={\mathbb {N}}$ be the discrete semigroup of natural numbers with maximum as operation, and let B be the one point compactification of the usual additive semigroup of natural numbers ${\mathbb {N}}$. While the description of the locally compact semigroup A is clear, we describe the compact semigroup B in more detail: we have $B={\mathbb {N}}\cup \{\infty \}$, where $\infty$ is the new point for which we put $\infty + n=n + \infty =\infty +\infty =\infty$, for $n\in {\mathbb {N}}$.

Now we take $L=\{(n,n) \mid n\in {\mathbb {N}}\}$, which is clopen in $A\times B$ because

$$\begin{aligned} L=\bigcup _{n\in {\mathbb {N}}} \{(n,n)\} \quad \text {and} \quad (A\times B ) \setminus L =\bigcup _{n\in {\mathbb {N}}} \{n\}\times (B\setminus \{n\}). \end{aligned}$$

We claim that one class of the syntactic congruence $\sigma _L$ is $A\times \{\infty \}$. All elements in $A\times \{\infty \}$ are $\sigma _L$-related as it is not possible to multiply them by any element and obtain a result in L. To show that elements from $A\times \{\infty \}$ are not $\sigma _L$-related with other elements, consider pairs (i, j) and $(k,\infty )$ with $i,j,k\in {\mathbb {N}}$. Then $(i+j,i) \cdot (i,j)=(i+j,i+j)\in L$ but $(i+j,i) \cdot (k,\infty )=(\max \{i+j,k\},\infty )\not \in L$. Hence, (i, j) and $(k,\infty )$ are not $\sigma _L$-equivalent. Finally, we claim that $A\times \{\infty \}$ is not open, which establishes that $\sigma _L$ is not clopen. To show that $A\times \{\infty \}$ is not open, recall first that in the product space $A\times B$, a base of the topology consists of open subsets $O \times O'$ with O and $O'$ open subsets respectively of A and B. However, if we consider $(k,\infty )$ in such $O \times O'$, then there is also some element $(k,\ell )$ in $O \times O'$ with the same first coordinate and $\ell \in {\mathbb {N}}$.

This example has another feature that it is worth noting. The algebra $A\times B$ is residually finite. Hence, while the condition that the syntactic congruence of a clopen subset of a locally compact 0-dimensional algebra is always clopen implies that the algebra is residually discrete (as indeed, the quotient of a topological algebra by a clopen congruence is a discrete algebra under the quotient topology), the converse fails even under the stronger assumption of residual finiteness. This is in contrast with the case of compact algebras, for which the two conditions are equivalent by Theorem 4.1.

6 Summary of results and conclusion

In conclusion, we have the following result building on the various characterizations of profiniteness in Stone topological algebras presented in this paper.

Theorem 6.1

The following conditions are equivalent for a Stone topological algebra A:

• (1) A is profinite;

• (2) for every clopen subset $L\subseteq A$, $\sigma _L$ is a clopen congruence;

• (3) M(A) is equicontinuous;

• (4) M(A) is relatively compact in $\mathcal {C}(A)$;

• (5) the closure of M(A) in $\mathcal {C}(A)$ is a profinite submonoid;

• (6) for every clopen subset $L\subseteq A$, there exists a continuous homomorphism $\varphi :A\rightarrow B$ onto a finite algebra B such that $L=\varphi ^{-1}(\varphi (L))$;

• (7) for every clopen subset $L\subseteq A$, the congruence $\sigma _L$ is determined by some finite set of terms;

• (8) for every clopen subset $L\subseteq A$, the congruence $\sigma _L$ is F-determined by some finite subset F of M(A);

• (9) for every clopen subset $L\subseteq A$, the congruence $\sigma _L$ is F-determined by some finite subset F of $\mathcal {C}(A)$;

• (10) for every clopen subset $L\subseteq A$, the congruence $\sigma _L$ is C-determined by some compact subset C of $\mathcal {C}(A)$.

Proof

By Theorems 4.1, 4.3 and 4.4, we have the equivalences (2)$\,\Leftrightarrow \,$(1) $\,\Leftrightarrow \,$(3) $\,\Leftrightarrow \,$(4). In fact, in the proof of Theorem 4.1, we showed that (1)$\,\Rightarrow \,$(6)$\,\Rightarrow \,$(2), whence we also have (1)$\,\Leftrightarrow \,$(6)

The implication (5)$\,\Rightarrow \,$(4) is obvious. To establish the reverse implication (4)$\,\Rightarrow \,$(5) it is enough to prove that $\overline{M(A)}$ is a 0-dimensional topological monoid, which holds if so is ${\mathcal {C}}(A)$. Now, given a clopen subset $L\subseteq A$, we have

$$\begin{aligned} \mathcal {C}(A)\setminus [K,L]=\bigcup _{a\in K}[a,A\setminus L], \end{aligned}$$

which shows that [K, L] is closed for every subset K of A. Thus, if $K,L\subseteq A$ are clopen, then so is [K, L]. Hence, $\mathcal {C}(A)$ is 0-dimensional and it was already observed after Proposition 5.4 that ${\mathcal {C}}(A)$ is a topological monoid.

The equivalence of (1), (7), (8), and (10) follows from Theorems 5.16 and 4.1. To conclude the proof, it remains to observe that the implications (8)$\,\Rightarrow \,$(9) and (9)$\,\Rightarrow \,$(10) are trivial. $\square$

Some of the results of Section 5 suggest looking at locally compact residually discrete algebras as a generalization of profinite algebras and, more generally at locally compact 0-dimensional algebras as a generalization of Stone topological algebras. However, it is not clear where such a study might lead, perhaps for lack of interesting examples. Examples 5.12 and 5.17 show that much of the good behavior observed in the compact case breaks down for locally compact algebras.