数理解析研究所講究録
1525 巻 2006 年 51-61

51

COMPLETIONS OF GBL-ALGEBRAS AND ACYCLIC MODALALGEBRAS: NEGATIVE RESULTSTADEUSZ LITAK (JOINT WORK WITH TOMASZ KOWALSKI)

ABSTRACT. We prove that several well-known varieties of modal and GBLalgebras are not closed under completions.

1. INTRODUCTION
In recent years, a lot of attention has been devoted to completions in algebraic
logic –in particular, to varieties of algebras closed under canonical extensions and
Dedekind-McNeille completions. The most general notion of closure under completions is clearly the following: we say that a variety is closed under completions.
if for every
there a lattice complete
and an embedding :Observe that we do not require to be a complete homomorphisms, that is, topreserve arbitrary joins. Now, if we can prove that a variety is not closed undercompletions, it follows that it cannot be can nical. Notice also, that strictly speaking we do not need to assume that algebras in our variety have lattice reducts:
being partially ordered would be sufficient. In the paper however, we will deal
exclusively with varieties that do have lattice reducts. We show that a number
nding to well-knownof varieties
dal and substructural logics fail to beclosed under completions.
Closure under completions should not be confused with being completely generated, that is, with the existence of a class $X$ of lattice-complete algebras such that
$V=HSP(X)$ . Being completely generated is a much weaker property then closure
under completions. In particular, every variety generated by its finite members iscompletely generated; most varieties discussed in the present paper have the finite
model property and hence are completely generated.
Failure of closure under completions seems harder to show than failure of canonicity, at least in the sense that there are non-canonical varieties closed under completions. Good examples can be found in the modal realm, for instance, Wolter [17]shows that the variety corresponding to tense logics of the reals is not canonical, butevery algebra in this variety can be embedded in a dual algebra of a Kripke frame
satisfying the axioms of this logics. This property is known as complenity and isproperly contained between canonicity and closure under completions. As follows
$V$

$f$

$\mathfrak{B}\in V$$\mathfrak{U}\in V$$\mathfrak{U}\mapsto \mathfrak{B}$$f$$\mathit{0}$$V$$\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}\dot{\mathrm{o}}$$\mathrm{m}o$is perhaps not the most fortunate name, it is however justified by the fact that the dual
algebra of a Kripke frame is a particular case of a complex algebra.
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TADEUSZ LITAK (JOINT WORK WITH TOMASZ KOWALSKI)

from results of Surendonk [15] and Wang [16], the variety corresponding to McKinsey axiom
is not complex, let alone canonical (failure of canonicity
was proven earlier by Goldblatt) but is closed under completions.
Nevertheless, as our results show, a proof that a given variety is not closed undercompletions does not necessarily have to be harder than a proof of non-canonicity.
The present paper uses fairly straightforward algebraic calculations. In particular,
the precise way operations on the original algebra induce operations on its canonicalextension is irrelevant here. Instead, we focus only on (possibly infinitely) equations
that have to hold in the whole variety. This is yet another instance of a wellknown mathematical paradox: a stronger theorem can be sometimes easier to prove.
-canonical varieties closed under completions –requireIntermediate
more involved techniques.
The first group of examples in this paper is based on a generalization of a result
from author’s
Thesis [11]. We provide a generic theorem which allows to showthat some well-known varieties of modal algebras are not closed under completions.
These varieties include the variety of L\"ob (diagonalizable) algebras and Grzegorczyk algebras; thus, we generalize the known observations that these logics are notcanoncial. Essentially, the varieties covered by our method correspond to logics
determined by those classes of frames which contain frames of arbitrary finite size
but no frames with either infinite ascending chains or two-point clusters. Thus, wecall our base variety acyclic or weakly Grzegorczyk.
The second group of examples is takena recently submitted paper withTomasz Kowalski, which is the basis of this report. Again, we provide a generictheorem allowing to show that several well-known varieties of -algebras are notclosed under completions. More specifically, the background for our results is provided by the variety of GBL-algebras, defined first (as far as we know) in [10].
GBL-algebras form a rather wide class including, in particular,-algebras, BLalgebras, Brouwerian algebras and -groups. Our results generalize very broadly
several already existing ones, for example Priestley and Gehrke [7] observation that-algebras are not canonical.
$\square 0parrow 0\square p$$\mathrm{c}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{s}-\mathrm{n}\mathrm{o}\mathrm{n}$$\mathrm{P}\mathrm{h}\mathrm{D}$$\mathrm{h}\mathrm{o}\mathrm{m}$$\mathrm{F}\mathrm{L}$$\mathrm{M}\mathrm{V}$$\ell$$\mathrm{M}\mathrm{V}$

2. ACYCLIC ALGEBRAS AND THEIR SUBVARIETIES, whereA modal algebra is a structure
is a boolean. Other operations (algebra and preserves V
andis its dual. A modal algebra is called acyclic
the standard way; $0^{+}x:=x\vee\theta x,$
or weakly Grzegorczyk (cf. [12]) if it satisfies the following inequality:$\mathfrak{U}:=\langle A, \wedge, \neg, \theta, \perp\rangle$$0$$\mathrm{a}\mathrm{n}\mathrm{d}\perp$$\langle A, \vee, \neg, \perp\rangle$$\wedge,$$arrow,$$\mathrm{T}$$\coprod^{+}$$x\leq\neg$ ( $xarrow C$ ( $\neg x$ A $0x$ )).

Recall that a Grzegorzyk algebra satisfies)).
(
A
Thus, every Grzegorczyk algebra is weakly Grzegorczyk. The reader may be($x\leq\neg$ $xarrow 0$$\neg x$$\phi x$more familiar with the definition of Grzegorczyk algebras as those satisfying$(\square (yarrow$ .

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COMPLETIONS OF GBL-ALGEBRAS AND ACYCLIC MODAL ALGEBRAS: NEGATIVE RESULTS

To see that these definitions are equivalent, negate both sides of the lower inequality, replace withand recall that in boolean algebras $aarrow b=\neg barrow\neg a$ .
Another important class of acyclic algebras is the class of L\"ob algebras or diagonalizable
defined by the inequality$\neg x$$y$$al_{\mathit{9}}ebras$$\phi x\leq 0(x\wedge\neg\theta x)$,or, dually,$\square (\square yarrow y)\leq\square y$

Theorem 2.1. Diagonalizable algebras are acyclic.

Proof. It is perhaps easier here to work with dual forms. First, observe that

diagonalizable algebras satisfy transitinity, i.e., $\square y\leq$ as$\square y\leq\square$($\square ^{2}y$A $yarrow y$ ) $=$ ($\coprod^{2}y$A$\square yarrow\square y\wedge y$)$\leq$(the first and second inequality holding by boolean laws, the last by the L\"ob inequality). Also, we have that$\square (\square (yarrow$(the first inequality by boolean Iaws, the second by transitivity, the third by the
L\"ob inequality). Finally,$\square (yarrow$(the first inequality by boolean laws, the second by transitivity). In this way weget that in diagonalizable algebras$\square ^{+}(\square (yarrow\square y)arrow y)\leq\square y\wedge(\square yarrow y)\leq y$,

as desired.
Recall that nontrivial L\"ob algebras are not Grzegorczyk.

3.

$\mathrm{G}\mathrm{B}\mathrm{L}$-ALGEBRAS AND THEIR SUBVARIETIBS

is a
A residuated lattice is an algebrasuch thatlattice, $A,$
is a monoid and the following residuation law is satisfied:iff
iff $x\leq z/y$ .
The class of residuated lattices is equationally definable, cf. e.g. [10] for an
equational basis. In connection to substructural logics (see, e.g. [14] for that connection), it is convenient to add another constant, called , to the type. This$1,0)$
defines a class of algebras known as $FL$ -algebras, i.e., algebras
asdefinedsuch thatlatticesaResiduatedis residuated lattices.
-algebras defined
above are in this setting term-equivalent to the subvariety ofby the equation $0=1$ . Alternatively, they are also -free reducts of FL-algebras.
These are essentially just two different ways of removing from the signature, so
we will not be particular about which way is preferred. We refer the reader to theforthcoming [4] for more of -algebras, residuated lattices and algebraic semantics
for substructural logics.
$\mathfrak{U}:=\langle A, \wedge, \vee, \cdot, \backslash , /, 1\rangle$$\langle$$\cdot,$$\langle A, \wedge, \vee\rangle$$1)$$\mathrm{r}\mathrm{e}\mathrm{s}:x\cdot y\leq z$$y\leq x\backslash z$$0$$\langle A,$$\langle A, \wedge, \vee, \cdot, \backslash , /, 1\rangle$$\mathrm{F}\mathrm{L}$

$0$

$0$

$\mathrm{F}\mathrm{L}$$\wedge,$$\vee,$$\backslash ,$$\cdot,$$/,$54
TADEUSZ LITAK (JOINT WORK WITH TOMASZ KOWALSKI)

Little of the theory of residuated lattices is needed for our results. One fact that, then for
we will use is that in any residuated lattice if $X$ exists for some$y\inA$
the following holdan arbitrary$X\subseteq A$(1) $y\cdot X=\{yx:x\in X\}$ ,(2) $X\backslash y=\wedge\{x\backslash y:x\in X\}$ .
Of importance, as we mentioned already, will be the variety of Generalized BLalgebras (see [5] for more on GBL). These are residuated lattices satisfying thefollowing law:
A ] $=x\wedge y=[(x\wedge y)/x]x$ .
GBL: [
To get even more leeway with the extra constant , we will take as our base,the variety of pointed $GBL$ -algebras defined as residuated lattices satisfying GBL
and endowed with an additional constant (proper residuated lattice setting), or
as -algebras satisfying GBL (FL setting).
This is a rather comprehensive class of algebras. It contains, on the one hand,the variety of lattice-ordered groups ( -groups) as a subvariety. Recall thatgroups are (term-equivalent to) residuated lattices satisfying $x(x\backslash 1)=1$ (by setting
and $1/x$ are, respectively, the left and right. Then, the elements$x\backslash
1=1/x$ holds.
On the other hand, it also
inverse of , and they coincide, i.e.,
A residuatedpointed
GBL-algebras.
integralcommutative
varietyofthecontains$1\geqx$, i.e., if the unit of the monoid is the greatest
lattice is integral if it satisfies. A residuated latticeeasilythat
follows
element. From integrality it
redundant, as
is commutative if it satisfies $x\cdot y=y\cdot x$ . In such a$x\backslash y=y/x$ . Thus, for commutative residuated lattices, weof the signa2. Brouwerian algebras are residuated lattices satisfying
ture and$x\cdot y=x\wedge y$ . The variety of Brouwerian algebras is precisely the class of O-freesubreducts of Heyting algebras, and thus Brouwerian algebras provide standard
examples of integra.1 commutative GBL-algebras.
Recall that a lattice is bounded if it has both the largest and the smallest element. Every bounded GBL-algebra is integral, in fact, every GBL-algebra is a direct
product of an -group and an integral GBL-algebra ([1, 5]). An integral FL-algebrain which is the lower bound is called zero-bounded. Zero-bounded integral pointed
GBL-algebras that satisfy in addition
Lin: $(xarrow y)\vee(yarrow x)=1$
are precisely Hajek’s $BL$ -algebras (cf. [9]).
$x$$x\backslash (x$

$y)$

$0$

$0$

$\mathrm{F}\mathrm{L}$$\ell-$$\ell$$x^{-1}=1\backslash x)$$x\backslash 1$$x$$x\cdot y\leq x\wedge y$$\mathrm{c}\mathrm{a}\mathrm{s}\mathrm{e}/\mathrm{b}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{s}$$\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{v}\mathrm{e}/\mathrm{o}\mathrm{u}\mathrm{t}$$\mathrm{r}\mathrm{e}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{c}\mathrm{e}\backslash \mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}arrow$$\ell$$0$

Two subvarieties of -algebras are of particular interests for us. The first is thesubvariety of $MV$-algebras. These are BL–algebras satisfying$\mathrm{B}\mathrm{L}$$\mathrm{M}\mathrm{V}:(xarrow 0)arrow 0=x$.

. The second is the subvariety of productis usually written as
where
algebras [3] defined by
II: $\neg\neg x\leq(xarrow xy)arrow y(\neg\neg y)$ .
$xarrow \mathrm{O}$$\neg x$, an unfortunate clash of terminology. In some papers, Brouwerian algebras are defined
to be the duals of Heyting algebras, not reducts without the bottom element. Given Brouwer’s
views on algebraic logic, it is rather ironic that his name is employed so often in the field.
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COMPLETIONS OF GBL-ALGEBRAS AND ACYCLIC MODAL ALGEBRAS: NEGATIVE RESULTS

The definition of a complete lattice is standard. Recall that every complete
lattice is bounded. It is then very
to conclude that some classes of residuatedlattices do not contain any complete algebra.
$\mathrm{e}\mathrm{a}s\mathrm{y}$Fact 3.1. No non-trivial -group is complete.
$\ell$

Proof. No non-trivial -group is bounded, as every such algebra contains $y>1$$\ell$and multiplication is strictly order-preserving by cancellativity.

$\square$So, lattice-incompleteness of -groups arises somehow trivially. In this case, amore interesting notion seems to be local completeness. A lattice is locally complete if every interval of is complete. Reals or integers (both examples of -groups)
with standard order are locally complete but not complete. For bounded lattices,-algebras or modal algebras, local completeness is equivalent to completelike
ness. Notice also that by well-known Makinson’s Theorem in modal logic, such asituation cannot arise for modal algebras. In order to obtain examples of varieties
with no complete algebras, one needs boolean algebras with at least two diamonds;
see [12] for details. It is curious that also in this case, one of operators is supposed
to satisfy the weak Grzegorczyk axiom.
A class of algebras $K$ is closed under (local) completions if every algebra from
$K$ can beembedded into a (locally) complete algebra from $K$ . Let us stress once
again: we do not require this embedding to preserve infinite meets and joins, only
finitary operations. Canonical extensions, for example, do not preserve infinitary
operations. Thus, when proving that closure under completions fails, we cannot
use too much information about suprema and infima in an algebra witnessing thisfailure. As we said in the introduction, we only use laws that hold in the wholevariety.
$\ell$$L$$L$$\ell$$\mathrm{B}\mathrm{L}$4. ACYCLIC V.ARIETIES WHICH ARE NOT CLOSED UNDER COMPLETIONSFirst, let us provide a criterion of failure of closure under completions for weakly

Grzegorczyk algebras:

Theorem 4.1. No modal algebra
following property for every :$L$containing a sequence$\{a_{n}\}_{n\in\omega}\in L$with the$n$$\perp\neq a_{n}\leq 0$($\neg a0\wedge\cdots$

A$\neg a_{n}$A $a_{n+1}$ )can be embedded in a complete acyclic algebra.

Proof. Define$b_{n}:=\neg a_{0}\wedge\cdots$

A $\neg a_{2n-1}$ A$a_{2n}$,$c_{n}:=\neg a0\wedge\cdots$

A$\neg a_{2n}$A $a2n+1$.

and assume and are respective suprema of both sequences. It is straightforward
to see that for every(1) $\perp\neq b_{m}\leq B$ ,(2)
Ahence $B\leq\neg C$ ,(3) $b_{m}\leq 0c_{m},$ $c_{m}\leq 0b_{m+1}$ , and hence
$B$

$C$

$m\in\omega$$b_{m}$$C=\perp \mathrm{a}\mathrm{n}\mathrm{d}$(4) $B\leq 0C\leq 0(\neg B\wedge\theta B)$ .

56
TADEUSZ LITAK (JOINT WORK WITH TOMASZ KOWALSKI)

But then if we replace
hand side becomes equal
not hold.
For a subset$A\subseteq\omega$$x$in the weak Gregorczyk inequality, the rightwith, so the inequality does
, while left is greater$B$$\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{n}\perp$$\mathrm{t}\mathrm{o}\perp$$\square$, let$\underline{\triangleright}A:=\{n\in\omega|\exists a\in A.n\leq a\}$,$\triangleright A:=\{n\in\omega|\exists a\in A.n<a\}$.

Let be the boolean algebra of finite-cofinite subsets of natural numbers. It isleft to the reader to verify that:(1)
is a Grzegorczyk algebra;(2)
is a diagonalizable algebra;(3) in both cases, the sequence $\{\{n\}|n\in\omega\}$ satisfies the assumption ofTheorem 4.1.
Thus, we get the following$\mathfrak{B}$$\langle \mathfrak{B},\underline{\triangleright}\rangle$$\langle \mathfrak{B}, \triangleright\rangle$

is closed(1) No variety of acyclic algebras containing
Corollary 4.2.
under completions. Hence, Grzegorczyk algebras are not closed underpletions.
is closed under comple(2) No variety of acyclic algebras containing
closednotunder completions.
algebrasareHence,
diagonalizabletions.
$\langle \mathfrak{B},\underline{\triangleright}\rangle$$comarrow$$\langle \mathfrak{B}, \triangleright\rangle$

This result can be generalized in several directions. For example. the weakGrzegorczyk inequality can be replaced with inequality expressing the fact that
corresponding frames have no three-point cycles or no four-point cycles –or stillin the inequality can be replaced
weaker for arbitrary finite . Alternatively,or any possibility form in the sense of Goldblatt [8].
with$\coprod^{+}$$n$$\coprod^{\leq n}$

5. GENERIC THEOREM FOR$\mathrm{G}\mathrm{B}\mathrm{L}$-ALGEBRAS

In order to introduce similiar criterion for GBL-algebras, we need a following

Definition 5.1. Let be a $GBL$ -algebra. Suppose there are subsets and ofwith the folloutng properties:
1
is a nonprincipal ideal urithout supremum in .
2 $B<A,$ $i.e.$ , for each $a\in A$ and $b\in B$ we have $b<a$ .
3 there are elements $w,$ $u\in B$ with $w<u$ such thatfor all $b\in B$ ,and,a for each $a\in A$ there is $b\in B$ urith
for each $b\in B$ there is $a\in A$ utth $ba\leq w$ .
$(B, A)$ is then called
residuation discontinuity and itself–a discontinuous
GBL-algebra.
$A$

$L$

$B$

$L$

$L$

$B$

$b\backslash u\in A$$b\backslash u\leq a$$\mathrm{b}$

$L$

$a$

The choice of the name will become clear from the proof of the next theorem,
is not continuous at a certain
where indeed it turns out that the function
to look at picturesencouraged).
readeris(namelyThe
presumed
to existpoint
and proofs of corollaries in Section 6 to see concrete examples of such a discontinuity.
Pictures are also meant to make the proof below appear less abstract.
$x(x\backslash u)$$b_{\infty}$

57
COMPLETIONS OF GBL-ALGEBRAS AND ACYCLIC MODAL ALGEBRAS: NEGATIVE RESULTJ$a_{2}a_{1}[^{1}a_{0}$$a_{1}|_{1}a_{2}\neg b_{1}a_{0}|\neg b_{2}1$1$\mathfrak{l}$

1

1

1
1

1
1

.
$\neg a_{i\cdot 0}$$\mathrm{i}$

1
1
1
1

$\mathfrak{P}$$b_{0}b_{1}b_{2}$$\not\subset$

FIGURB 1. Chang’s chain$\not\subset$

and product algebra$\mathfrak{P}$Theorem 5.2 (Completion Discontinuity). No variety of $GBL$ -algebras containing a discontinuous $GBL$ -algebra is closed under sectional completions.

is a discontinuous GBL-algebra. Let $(B, A)$ be a residuation discontinuity. Define $b_{\infty}=B$ and
A . Notice that we make noassumptions whatsoever about. It may well belong to , or even to .
. On the other hand,
Consider. As $b_{\infty}>u$ we obtain$B\cdot(B\backslash u)=B\cdot(\wedge\{b\backslash u:b\in B\})$ . Since every $a\in A$ is minorised by
for
$b\in
B$
Therefore,.
somebelongs to , weand each$B\cdot(\wedge\{b\backslash u:b\in B\})=B\cdot a_{\infty}=\{b\cdot a_{\infty} : b\in B\}$ . Now, for each $b\in B$ choose.
. Sincean $a_{b}\in A$ withfor all $a\in A$ , we getThus,, a contradiction.

Proof. Suppose$\dot{\mathfrak{M}}$$A$$a_{\infty}=$

$A$

$L$

$a_{\infty}$$b_{\infty}(b_{\infty}\backslash u)=u$$b_{\infty}(b_{\infty}\backslash u)$$b\backslash u$$A$$b\backslash u$$b\cdot a_{b}\leq w$$\mathrm{h}\mathrm{a}\mathrm{v}\mathrm{e}\wedge\{b\backslash u:b\in B\}=a_{\infty}$$b\cdot a_{\infty}\leq b\cdot a_{b}\leq w<u$$a_{\infty}\leq a$$b_{\infty}(b_{\infty}\backslash u)<u$

6. APPLICATIONS OF THE GENERIC THEOREM-algebra is Chang’s chain $C[2]$-algebras. A particular example of an$a_{0}:=1$
and $b_{0}:=0$ . The lattice
whose universe iswithif
if $n\leq m,$order is defined as follows: every is below all$n\geq m$ . Multiplication is defined by$\mathrm{M}\mathrm{V}$$\mathrm{M}\mathrm{V}$$C:=\{a_{n}\}_{n\in\omega}\cup\{b_{n}\}_{n\in v}$($a_{n}’ \mathrm{s},$$b_{n}$$b_{n}\cdot b_{m}:=0$$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}-$,$a_{n}\cdot b_{m}:=b_{m-n}$,$a_{n}\leq a_{m}$$b_{n}\leq b_{m}$$a_{n}\cdot a_{m}:=a_{n+m}$

stands for truncated subtraction. See Figure 2.

,

58
TADEUSZ LITAK (JOINT WORK WITH TOMASZ KOWALSKI)

Lemma 6.1. Chang’s chain$\mathrm{C}$is a discontinuous GBL-al9ebra.

-algebra having a subalgebra
isomorphic
is a complete,and . Define
to ; we can identify elements of$w=b_{0},$ $u=b_{1}$ . Conditions 1 and 2 of Definition 5.1 are immediate. For 3, observe
and $b_{i+1}arrow b_{1}=a_{i}$ .
that for every

Proof. Assume$\mathbb{C}’$$\mathrm{M}\mathrm{V}$$\mathfrak{U}$$\mathrm{C}’$$\mathrm{C}$$A=\{a_{n}\}_{n\in\omega},$$\not\subset$$B=\{b_{n}\}_{n\in\omega}$$\neg b_{i}=a_{i}$$i,$$\square$Corollary 6.2. No variety of zero-bounded $GBL$ -algebras containing Chang’s
chain (in particular, $MV$-algebras and $BL$ -algebras) is closed under completions.
-algebras, closure under completions
Another curious aside: for varieties ofis actually equivalent to canonicity (we observed in the introduction that it is notalways the case) and both in turn are equivalent to failure of finite generation. Let
us prove it in more detail. The lemma below is folklore, but to make the paper
self-contained we provide a proof.
$\mathrm{M}\mathrm{V}$Lemma 6.3. A $va7\dot{?}etyV$ of $MV$-algebras does not containfinitely generated.

$C$

if and only if it is-algebra satisfies the identity $x^{n+1}=x^{n}$ for some$n\in\omega$ , we get that finitely generated
does not contain . For the converse,
suppose is not finitely generated. Then for every
there is a subdirectly$x^{k+1}=x^{k}$. Choose a suitable $a_{k}\in A_{k}$ witnessing
falsifying
irreducible
for some nonprincipal . Then,that fact. Let be the ultraproduct$a^{n+1}<a^{n}$for every. Also, is linearly
the element $a=(a_{k} : k\in\omega)/U$ hasordered, since allare. It is now easy to verify that the subalgebra of generated
by is isomorphic to C.

Proof. Since each finite$\mathrm{M}\mathrm{V}$

$V$

$C$

$k\in\omega$$V$$\mathfrak{U}_{k}\in V$$\mathfrak{U}$$U$$\prod_{k\in\omega}\mathfrak{U}_{k}/U$$n\in\omega$$\mathfrak{U}$$\mathfrak{U}$$\mathfrak{U}_{k}$$\square$$a$

The second lemma is due to Gehrke and holds essentially because canonicalextensions commute with homomorphisms and Boolean products. Since every direct
product can be rendered as a Boolean product of ultraproducts, and ultrapowers offinite algebras are these algebras themselves, we have:

Lemma 6.4.

$A$

finitely generated variety is canonical.

, we give [6] as the reference as it is also
Although it was proven in the latean excellent general introduction to canonical extensions. Combining the lemmas
together we obtain:$1980\mathrm{s}$Theorem 6.5. For a variety $V$ of $MV$-algebras the following are equivalent:(1) $V$ is canonical,(2) $V$ is closed under completions,(3) $V$ is finitely generated,.
(4) $V$ is -potent for some$n$$n\in\omega$

Lattice-ordered groups. As we have already mentioned, -groups can be viewed. Mundici showed in [13] thatas a subvariety of GBL algebras satisfying-algebras and -groups. The -group
there is a categorical equivalence between
obtained Rom Chang’s chain via this equivalence is, as should be expected, a good
example that -groups are not closed under local mpletions. This algebra is isomorphic to the lexicographic product of integers with themselves. To be precise, we$\ell$$x(x\backslash 1)=1$$\mathrm{M}\mathrm{V}$$\ell$$\ell$$\mathrm{c}o$$\ell$

59
COMPLETIONS OF GBL-ALGEBRAS AND ACYCLIC MODAL ALGEBRAS: NEGATIVE RESULTS
1
1
1
$1$

1
$1$

$1$

$(1, 0)$ $\}_{Z}$
$1$

$1$

$1$

1
$1$

$!$

$(0,0)|_{Z}$

$111$

1
$1$

$1$

$1$

$\downarrow$$(-1,0)\}_{Z}$$|_{\mathrm{I}}^{(1,1)=u}\mathrm{l}(=_{1,0)=w}(-1,-1)$

$1$

1
$1$

$1$

1
$111$

3
FIGURE 2. Algebra 7 (left) and its magnified detail (right), where and are respectively thedefine the algebra
order, and multiplication
lexicographic
respect
minimum and maximum withto theand inverse are defined pointwise, as addition and unary minus.
$2=\langle Z\cross Z, \wedge, \vee, \cdot, -1, (0,0)\rangle$$\wedge$$\vee$

Theorem 6.6. The algebra 7 is a discontinuous $GBL$ -algebra. Thus, no nontrivial variety of -groups is closed under (local) completions.
$\ell$

Proof. Define $A=\{(\ell, m)\in I:\ell\geq 0\}$ and $B=I\backslash A=\{(k, n)\in I:k\leq-1\}$ .

satisfy conditions 1 and 2. Now putting $u=(-1,1)$ and
So definedandhold as well. As all theseand$w=(-1,0)$ we can verify that conditions
verifications are routine, we will only show . Take any $a=(\ell, m)\in A$ . There are
two cases to consider.
$A$

$B$

$3\mathrm{a}$$3\mathrm{a}$$3\mathrm{b}$60
TADEUSZ LITAK (JOINT WORK WITH TOMASZ KOWALSKI)

If $\ell>0$ , we take the element $b=(x, y)$ with $x=-\ell$ and arbitrary. Observethat belongs to $B$ . We have $b\backslash u=-(-\ell, y)+(-1,1)=(P-1,1-y)<(\ell, m)=a$since $\ell-1<\ell$ .
If $\ell=0$ , we take the element $b=(x, y)$ with $x=-1$ and $y=m-1$ . This again
belongs to $B$ . Then, we have $b\backslash u=-(-1,1-m)+(-1,1)=(0, m)=a$ .
$y$

$b$

$\square$Product algebras. The variety of product algebras is generated by the realinterval $[0,1]$ as a residuated lattice under natural order and natural multiplication.
In particular, the algebra best described as the upper halfof Chang’s
chain with an additional zero element at the bottom, is a product algebra. See
we have $\neg a_{i}=0$ . The algebra
Figure 2 for a picture of ; notice that for any
being its monolith congruence filter. In fact,
is subdirectly irreducible withalso generates the whole variety of product algebras. Unlike , the algebra
is complete itself. It will be our example of a complete algebra whose ultrapower
is not embeddable in any complete algebra from the variety it generates. Take anultrapower St
for a nonprincipal .
$\{a_{n} :$\mathfrak{P}$asn\in\omega\}$$i\in\omega$$\mathfrak{P}$$A$$\mathrm{C}$$\mathfrak{P}$$\mathfrak{P}$$U$$=\mathfrak{P}^{I}/U$

Theorem 6.7. The algebra is a discontinuous $GBL$ -algebra. Thus, the variety
of product dgebras is not closed under completions. The same holds for every larger
variety of GBL-algebras.
$\Re$

Proof. The algebra is subdirectly irreducible, with the monolith congruence
filter isomorphic to . Pick elementswith $0<w\prec u$ . This is always
$\Re$

$A$

$w,$ $u\in R\backslash A$

possible by properties of ultraproducts, moreover, every element $e\in R\backslash A$ has an
immediate successor (i.e., a cover with respect to the natural ordering). Define
inductively $u_{0}=u$ and $u_{n+1}=u_{n}’$ . Let $B$ be the downward closure of.
Clearly, $B$ is an ideal without supremum in $R$ . Notice thathas the following$e^{l}$$\{u_{k} :k\in\omega\}$$\mathfrak{P}$property:
for every and , if $x>0$ and is the k-th successor of , then $yarrow x$
is the k-th predecessor of 1, i.e., equals .
The property above is first-order expressible, and thus carries over to . Therefore, $u_{k}arrow u=a_{k}\in A$ . Since every $b\in B$ is either smaller than or equal towefor some
have $barrow u\in A$ for every $b\in B$ . Further, for eachget
. This takes care of all the required properties up to 3(a). To show,
that3(b)ho1ds as we11, note first that ifb
so it remains to show that for each
there is an $a\in A$ with $au_{j}\leq w$ . To thisend, observe thathas the following, first-order expressible, property.
for every and , if $x>0$ and is the k-th predecessor of , then the$k+1$ -th successor of 1, i.e.,
has $a_{k+1}\cdot y\prec x$ .
We conclude that $a_{k+1}\cdot u_{k}=w\prec u$ holds in . This completes the list ofrequirements. The conclusion now follows by Theorem 5.2.
$x$$\bullet$

$x$

$y$

$y$

$a_{k}$

$\Re$

$u$

$k\in\omega,$$u_{k}$$j\in\omega$$\mathrm{w}\dot{\mathrm{e}}$$a_{j}=u_{j}arrow \mathrm{u}$$\leq w,$$\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}xb\leq w\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{y}x\mathrm{w}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{s}\mathrm{o}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{r}$$j\in\omega$$\mathfrak{P}$$\bullet$

$x$

$x$

$y$

$y$

$a_{k+1}$$\Re$$\square$

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