Hostname: page-component-6898cfcb55-zkb6p Total loading time: 0 Render date: 2026-11-11T23:00:43.079Z Has data issue: false hasContentIssue false
  • English
  • Français

Not every complete lattice can support a unital quantale

Part of:LatticesOrdered structuresOrdered sets

Published online by Cambridge University Press:  29 October 2025

Shengwei Han and
Xin Wang
Shengwei Han*
Affiliation:
Department of Mathematics, Shaanxi Normal University, Xi'an, 710119, China e-mail: wangxin2020@snnu.edu.cn
Xin Wang
Affiliation:
Department of Mathematics, Shaanxi Normal University, Xi'an, 710119, China e-mail: wangxin2020@snnu.edu.cn
*
e-mail: hansw@snnu.edu.cn
Rights & Permissions [Opens in a new window]

Abstract

Quantales can be regarded as a combination of complete lattices and semigroups. Unital quantales constitute a significant subclass within quantale theory, which play a crucial role in the theoretical framework of quantale research. It is well known that every complete lattice can support a quantale. However, the question of whether every complete lattice can support a unital quantale has not been considered before. In this article, we first give some counter-examples to indicate that the answer to the above question is negative, and then investigate the complete lattices of supporting unital quantales.

Keywords

Complete latticesemigroupquantaleunital quantale

MSC classification

Primary:06F07: Quantales06A06: Partial order, general06B05: Structure theory

Information

Type
Article
Information
Canadian Mathematical Bulletin , Volume 69 , Issue 2 , June 2026, pp. 665 - 676
Check for updates
Copyright
© The Author(s), 2025. Published by Cambridge University Press on behalf of Canadian Mathematical Society

1 Introduction

Originally, quantales were invented to formalize the logic of observables in quantum physics, which became useful to describe the “spaces” of upper C Superscript asterisk $C^\ast $ -algebras (see [Reference Borceux, Rosický and Van Bossche3, Reference Kruml and Paseka11, Reference Kruml and Resende12]) and non-commutative geometry (see [Reference Connes4, Reference Mulvey and Pelletier13–Reference Mulvey and Resende15, Reference Protin and Resende17, Reference Resende19]). The literature on quantales often emphasizes that these structures are required to have units. Unital quantales, as a valued domain, have wide applications in theoretical computer science, logic, quantitative domain, enriched category, and many-valued topological space (see [Reference Abramsky and Vickers1, Reference Fan6, Reference Hofmann and Waszkiewicz10, Reference Resende18, Reference Yetter21, Reference Zhang22]). By definition, a quantale is a complete lattice Q with a semigroup multiplication ampersand $\&$ that distributes over arbitrary joins, that is,

(1.1) a ampersand left parenthesis backslash leftvee Underscript i element of upper I Endscripts b Subscript i Baseline right parenthesis equals backslash leftvee Underscript i element of upper I Endscripts left parenthesis a ampersand b Subscript i Baseline right parenthesis and left parenthesis backslash leftvee Underscript i element of upper I Endscripts b Subscript i Baseline right parenthesis ampersand a equals backslash leftvee Underscript i element of upper I Endscripts left parenthesis b Subscript i Baseline ampersand a right parenthesis $$ \begin{align}a\&(\bigvee_{i\in I}b_i)=\bigvee_{i\in I}(a\& b_i)\ \ \text{and}\ \ (\bigvee_{i\in I} b_i)\& a=\bigvee_{i\in I}(b_i\& a)\end{align} $$

for all a element of upper Q comma left brace b Subscript i Baseline right brace Subscript i element of upper I Baseline subset of or equal to upper Q $a\in Q, \{b_i\}_{i\in I}\subseteq Q$ . A quantale Q is said to be unital provided that there exists an element e element of upper Q $e\in Q$ such that q backslash ampersand e equals q equals e backslash ampersand q $q\&e=q=e\&q$ for all q element of upper Q $q\in Q$ . Q is said to be commutative provided that a backslash ampersand b equals b backslash ampersand a $a\&b=b\&a$ for all a comma b element of upper Q $a,b\in Q$ .

Definition 1.1 Let Q be a complete lattice. Then, we say that Q supports a (unital) quantale if there exists a semigroup multiplication ampersand $\&$ on Q such that left parenthesis upper Q comma ampersand right parenthesis $(Q, \&)$ becomes a (unital) quantale. An element e of Q can become a unit means that Q supports a unital quantale with unit e.

In [Reference Rosenthal20], one can see that every complete lattice Q can support a quantale left parenthesis upper Q comma ampersand Subscript a Baseline right parenthesis $(Q, \&_a)$ where a element of upper Q $a\in Q$ and ampersand Subscript a Baseline $\&_a$ is a semigroup multiplication determined by

(1.2) for all x comma y element of upper Q comma x ampersand Subscript a Baseline y equals StartLayout Enlarged left brace 1st Row 1st Column Blank 2nd Column 0 comma 3rd Column if x equals 0 or y equals 0 comma 2nd Row 1st Column Blank 2nd Column a comma 3rd Column otherwise period EndLayout $$ \begin{align}\forall x, y\in Q,\ x\&_a y=\left\{\begin{array}{@{}ll} 0, & \text{if}\ \ x=0\ \text{or}\ y=0,\\ a, & \text{otherwise.}\ \ \end{array}\right.\end{align} $$

In [Reference Paseka and Kruml16], Paseka and Kruml investigated the construction of unital quantales from a given quantale, and they proved that every quantale Q can be embedded into a unital quantale upper Q e $Q[e]$ . For some special quantales, by addition of an isolated unit, Gutiérrez García and Höhle considered the construction of unitally non-distributive quantales (see [Reference Gutiérrez García and Höhle8]). Based on the above work, Han and Lv further discussed the construction of unital quantales from a given quantale in [Reference Han and Lv9]. According to the construction of unital quantale upper Q e $Q[e]$ , by (1.2), one can easily see that if a complete lattice L is isomorphic to upper Q times bold 2 $Q\times \mathbf {2}$ for some complete lattice Q, then L can support a unital quantale, where bold 2 $\mathbf 2$ is the two-element lattice. So, we have the following two questions.

Question 1: Does every complete lattice support a unital quantale?

If the answer to Question 1 is negative, then we have Question 2 to ask.

Question 2: Which class of complete lattices can support unital quantales?

In order to answer the above questions, we need to review some concepts and notations. Let L be a complete lattice, a comma b element of upper L $a, b\in L$ . Then, L is said to be trivial if StartAbsoluteValue upper L EndAbsoluteValue equals 1 $\mid L\mid =1$ and a is said to be comparable if a less than or equals b $a\leq b$ or b less than or equals a $b\leq a$ for all b element of upper L $b\in L$ . Further, we let a comma b equals StartSet x element of upper L vertical bar a less than or equals x less than or equals b EndSet $[a, b]=\{x\in L \mid a\leq x\leq b\}$ and left parenthesis a comma b right bracket equals StartSet x element of upper L vertical bar a less than x less than or equals b EndSet $(a, b]=\{x\in L \mid a< x\leq b\}$ . a is called a lower-cover of b (or b is called an up-cover of a) if a not equals b $a\neq b$ and a comma b equals StartSet a comma b EndSet $[a, b]=\{a, b\}$ . A non-empty subset S of L is said to be a sublattice if S is closed under and $\wedge $ and or $\vee $ . We denote by upper L times upper M $L\times M$ the Cartesian product of complete lattices L and M. Then, upper L times upper M $L\times M$ is a complete lattice with pointwise order. In this article, we denote the bottom element and the top element in a complete lattice by 0 $0$ and 1 comma $1,$ respectively. For the notions and concepts, which are not explained, please refer to [Reference Birkhoff2, Reference Eklund, Gutiérrez García, Höhle and Kortelainen5, Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott7, Reference Rosenthal20].

2 Not every complete lattice can support a unital quantale

As we all know, every complete lattice can support a quantale. However, in this section, we will prove that not every complete lattice can support a unital quantale, that is, the answer to Question 1 is negative. To see this, we need to consider the following complete lattices.

Lemma 2.1 Let left parenthesis upper Q comma ampersand right parenthesis $(Q, \&)$ be a unital quantale with unit e. Then, a ampersand b less than or equals a and b $a\& b\leq a\wedge b$ for all a comma b element of 0 comma e $a, b\in [0, e]$ .

Example 2.2 Let upper N 5 equals StartSet 0 comma a comma b comma c comma 1 EndSet $N_5=\{0, a, b, c, 1\}$ be a complete lattice which is determined by Figure 1. Then, the top element 1 cannot become a unit.

Figure 1

The partial order on upper N 5 $N_5$ .

Hasse diagram with nodes labeled 0 at the bottom, then a, b, c in a pentagon, and 1 at the top, each connected by lines showing the partial order.

To see this, we assume that 1 $1$ is the unit in quantale left parenthesis upper N 5 comma ampersand right parenthesis $(N_5, \&)$ . Then, by Lemma 2.1, we have that b backslash ampersand a less than or equals b and a equals a $b\&a\leq b\wedge a=a$ and b backslash ampersand c less than or equals b and c equals 0 $b\&c\leq b\wedge c=0$ , which implies that b equals b ampersand 1 equals b ampersand left parenthesis a or c right parenthesis equals left parenthesis b backslash ampersand a right parenthesis or left parenthesis b backslash ampersand c right parenthesis equals b backslash ampersand a less than or equals a $b=b\&1=b\&(a\vee c)=(b\&a)\vee (b\&c)=b\&a\leq a$ , contradiction.

Example 2.3 Let upper M 3 equals StartSet 0 comma a comma b comma c comma 1 EndSet $M_3=\{0, a, b, c, 1\}$ be a complete lattice which is determined by Figure 2. Then, the top element 1 $1$ cannot become a unit.

Figure 2

The partial order on upper M 3 $M_3$ .

Hasse diagram with nodes labeled 0 at the bottom, a and c at the next level, b above them, and 1 at the top, showing the partial order structure.

To see this, we assume that 1 $1$ is the unit in quantale left parenthesis upper M 3 comma ampersand right parenthesis $(M_3, \&)$ . Then, by Lemma 2.1, we have that a backslash ampersand b less than or equals a and b equals 0 $a\&b\leq a\wedge b=0$ and a backslash ampersand c less than or equals a and c equals 0 $a\&c\leq a\wedge c = 0$ , which implies that a equals a ampersand 1 equals a ampersand left parenthesis b or c right parenthesis equals left parenthesis a backslash ampersand b right parenthesis or left parenthesis a backslash ampersand c right parenthesis less than or equals 0 $a=a\&1=a\&(b\vee c)=(a\&b)\vee (a\&c)\leq 0$ , contradiction.

Definition 2.4 Let L be a complete lattice and x element of upper L backslash StartSet 0 EndSet $x\in L\backslash \{0\}$ . Then, x is called an upper N 5 $N_5$ -element (an upper M 3 $M_3$ -element) of L if there exists some sublattice M of L such that x is the top element in M and M is isomorphic to upper N 5 $N_5$ ( upper M 3 $M_3$ ).

Lemma 2.5 Every upper N 5 $N_5$ -element or upper M 3 $M_3$ -element of a complete lattice L cannot become a unit of L.

Proof The proof follows from Examples 2.2 and 2.3.

Proposition 2.6 If every non-bottom element of a complete lattice L is either an upper N 5 $N_5$ -element or an upper M 3 $M_3$ -element, then L cannot support a unital quantale.

Remark 2.7 It is easy to obtain an equivalent statement of Proposition 2.6, that is, if a complete lattice supports a unital quantale, then it must have a non-bottom element which is neither an upper N 5 $N_5$ -element nor an upper M 3 $M_3$ -element. Based on this equivalent statement, we see that the converse of Proposition 2.6 is in general not true (see, Examples 2.9 and 2.10).

Example 2.8 Let upper L 1 equals left brace a Subscript i j Baseline vertical bar 1 less than or equals j less than or equals 2 Superscript i minus 1 Baseline comma i equals 1 comma 2 comma midline horizontal ellipsis comma n comma midline horizontal ellipsis right brace union StartSet 0 EndSet $L_1=\{a_{ij} \mid 1\leq j\leq 2^{i-1}, i=1, 2, \cdots ,n, \cdots \}\cup \{0\}$ be a complete lattice determined by Figure 3 in which every non-bottom element of upper L 1 $L_1$ has precisely two lower-covers, and every element of upper L backslash StartSet 0 comma a 11 EndSet $L\backslash \{0, a_{11}\}$ has precisely one up-cover.

Figure 3

The partial order on upper L 1 $L_1$ .

Hasse diagram with a subscripted a one one at the top, branching to a two one, a three two, a two two, then to a third row of a three one, a three two, a three three, a three four, all above zero.

It is easy to check that every non-bottom element of upper L 1 $L_1$ is an upper N 5 $N_5$ -element. By Proposition 2.6, we have that upper L 1 $L_1$ cannot support a unital quantale.

Example 2.9 Let upper L 2 equals upper L 1 union StartSet a 00 comma a 10 EndSet $L_2=L_1\cup \{a_{00}, a_{10}\}$ and the partial order on upper L 2 $L_2$ be determined by Figure 4. Then, upper L 2 $L_2$ is a complete lattice. Further, one can see that every element of upper L 2 backslash StartSet a 10 comma 0 EndSet $L_2\backslash \{a_{10}, 0\}$ is an upper N 5 $N_5$ -element, and a 10 $a_{10}$ is neither an upper N 5 $N_5$ -element or an upper M 3 $M_3$ -element. Surprisingly, upper L 2 $L_2$ can indeed support a unital quantale.

Figure 4

The partial order on upper L 2 $L_2$ .

Hasse diagram with a subscripted a zero zero at the top branching downward to a one zero and a one one, then further to a two one, two two, and so on, ending at zero at the bottom. See long description.
Figure 4 Long description

At the top is a zero zero. The next level has a one zero to the left and a one one to the right. From a one zero, a two one branches left and a two two branches right. From a one one, a two two branches left and a two two branches right. The third level has a three one, three two, three three, and three four, each connected to the previous level. Dashed lines connect each of these to the bottom node labeled zero. All connections are shown with straight or dashed lines, indicating the partial order structure.

To see this, we define now a binary operation ampersand $\&$ on upper L 2 $L_2$ as follows: for all x comma y element of upper L 2 $\forall x, y\in L_2$ ,

x ampersand y equals StartLayout Enlarged left brace 1st Row 1st Column Blank 2nd Column 0 comma 3rd Column if x equals 0 or y equals 0 comma 2nd Row 1st Column Blank 2nd Column 0 comma 3rd Column if x element of left parenthesis 0 comma a 21 right bracket comma y not an element of StartSet a 10 comma a 00 EndSet or x not an element of StartSet a 10 comma a 00 EndSet comma y element of left parenthesis 0 comma a 21 right bracket comma 3rd Row 1st Column Blank 2nd Column y comma 3rd Column if x equals a 10 comma 4th Row 1st Column Blank 2nd Column x comma 3rd Column if y equals a 10 comma 5th Row 1st Column Blank 2nd Column y comma 3rd Column if x equals a 00 comma y element of left parenthesis 0 comma a 21 right bracket comma 6th Row 1st Column Blank 2nd Column x comma 3rd Column if x element of left parenthesis 0 comma a 21 right bracket comma y equals a 00 comma 7th Row 1st Column Blank 2nd Column a 00 comma 3rd Column otherwise period EndLayout $$ \begin{align*}x\& y=\left\{\begin{array}{@{}ll} 0, & \text{if}\ \ x=0\ \text{or}\ y=0,\\ 0, & \text{if}\ \ x\in (0, a_{21}], y\not\in\{a_{10}, a_{00}\}\ \text{or}\ x\not\in\{a_{10}, a_{00}\}, y\in (0, a_{21}], \\ y, & \text{if}\ \ x=a_{10},\\ x, & \text{if}\ \ y=a_{10},\\ y, & \text{if}\ \ x=a_{00}, y\in (0, a_{21}],\\ x, & \text{if}\ \ x\in (0, a_{21}], y=a_{00},\\ a_{00}, & \text{otherwise.}\ \ \end{array}\right.\end{align*} $$

Then, left parenthesis upper L 2 comma ampersand right parenthesis $(L_2, \&)$ is a commutative unital quantale with unit a 10 $a_{10}$ .

Example 2.10 Let I denote the unit open interval left parenthesis 0 comma 1 right parenthesis $(0, 1)$ with usual order less than or equals $\leq $ , upper Sigma equals StartSet sigma Subscript r Baseline vertical bar r element of left parenthesis 0 comma 1 right parenthesis EndSet $\Sigma =\{\sigma _r \mid r\in (0, 1)\}$ and upper Gamma equals StartSet tau Subscript r Baseline vertical bar r element of left parenthesis 0 comma 1 right parenthesis EndSet $\Gamma =\{\tau _r \mid r\in (0, 1)\}$ . Further, we let r element of left parenthesis 0 comma 1 right parenthesis $r\in (0, 1)$ , upper X Subscript r Baseline equals StartSet x Subscript r comma i Baseline vertical bar i element of left parenthesis 0 comma 1 right parenthesis EndSet $X_r=\{x_{r,i} \mid i\in (0, 1)\}$ , upper Y Subscript r Baseline equals StartSet y Subscript r comma j Baseline vertical bar j element of left parenthesis 0 comma 1 right parenthesis EndSet comma $Y_r=\{ y_{r,j} \mid j\in (0, 1)\},$ and upper L 3 equals upper I union upper Sigma union upper Gamma union left parenthesis backslash leftcup Underscript r element of upper I Endscripts left parenthesis upper X Subscript r Baseline union upper Y Subscript r Baseline right parenthesis right parenthesis union StartSet down tack comma up tack EndSet $L_3=I\cup \Sigma \cup \Gamma \cup (\bigcup \limits _{r\in I}(X_r\cup Y_r))\cup \{\top , \bot \}$ . Next, we define a binary relation less than or equals Subscript upper L 3 Baseline $\leq _{L_3}$ on upper L 3 $L_3$ as follows: for all a comma r comma i comma j element of upper I $\forall a,r,i,j\in I$ , x element of upper L 3 $x\in L_3$ ,

i less than or equals Subscript upper L 3 Baseline j long left right double arrow i less than or equals j comma sigma Subscript i Baseline less than or equals Subscript upper L 3 Baseline sigma Subscript j Baseline long left right double arrow i less than or equals j long left right double arrow tau Subscript i Baseline less than or equals Subscript upper L 3 Baseline tau Subscript j Baseline comma $$ \begin{align*}i\leq_{L_3} j \Longleftrightarrow i\leq j,\ \sigma_i\leq_{L_3} \sigma_j \Longleftrightarrow i\leq j\Longleftrightarrow \tau_i\leq_{L_3}\tau_j,\end{align*} $$
x Subscript r comma i Baseline less than or equals Subscript upper L 3 Baseline x Subscript r comma j Baseline long left right double arrow i less than or equals j long left right double arrow y Subscript r comma i Baseline less than or equals Subscript upper L 3 Baseline y Subscript r comma j Baseline comma $$ \begin{align*}x_{r,i}\leq_{L_3} x_{r,j} \Longleftrightarrow i\leq j \Longleftrightarrow y_{r,i}\leq_{L_3} y_{r,j},\end{align*} $$
a less than or equals Subscript upper L 3 Baseline sigma Subscript r Baseline long left right double arrow a less than or equals r long left right double arrow a less than or equals Subscript upper L 3 Baseline tau Subscript r Baseline comma $$ \begin{align*}a\leq_{L_3} \sigma_r \Longleftrightarrow a\leq r \Longleftrightarrow a\leq_{L_3} \tau_r,\end{align*} $$
a less than or equals Subscript upper L 3 Baseline x Subscript r comma i Baseline long left right double arrow a less than or equals r long left right double arrow a less than or equals Subscript upper L 3 Baseline y Subscript r comma i Baseline comma $$ \begin{align*}a\leq_{L_3} x_{r,i} \Longleftrightarrow a\leq r \Longleftrightarrow a\leq_{L_3} y_{r,i},\end{align*} $$
x Subscript a comma i Baseline less than or equals Subscript upper L 3 Baseline sigma Subscript r Baseline long left right double arrow a less than or equals r long left right double arrow y Subscript a comma j Baseline less than or equals Subscript upper L 3 Baseline tau Subscript r Baseline comma $$ \begin{align*}x_{a,i}\leq_{L_3} \sigma_r \Longleftrightarrow a\leq r \Longleftrightarrow y_{a,j}\leq_{L_3} \tau_r,\end{align*} $$
up tack less than or equals Subscript upper L 3 Baseline x less than or equals Subscript upper L 3 Baseline down tack period $$ \begin{align*}\bot\leq_{L_3} x\leq_{L_3}\top.\end{align*} $$

It is easy to check that left parenthesis upper L 3 comma less than or equals Subscript upper L 3 Baseline right parenthesis $(L_3, \leq _{L_3})$ is a complete lattice (Figure 5). Further, we see that every element of upper Sigma union upper Gamma union StartSet down tack EndSet $\Sigma \cup \Gamma \cup \{\top \}$ is an upper N 5 $N_5$ -element, and every element of upper I union left parenthesis backslash leftcup Underscript r element of upper I Endscripts left parenthesis upper X Subscript r Baseline union upper Y Subscript r Baseline right parenthesis right parenthesis $I\cup (\bigcup \limits _{r\in I}(X_r\cup Y_r))$ is neither an upper N 5 $N_5$ -element nor an upper M 3 $M_3$ -element. However, upper L 3 $L_3$ indeed cannot support a unital quantale (see Proposition 2.12).

Figure 5

The partial order on upper L 3 $L_3$ .

A lattice diagram with three horizontal layers showing labeled nodes and edges representing a partial order structure. See long description.
Figure 5 Long description

At the bottom is a black dot labeled bottom. Above it is a white dot labeled 0. The next layer has three black dots labeled sigma sub a, a, and tau sub a, connected horizontally. The middle layer has three black dots labeled sigma sub r, r, and tau sub r, also connected horizontally. Vertical edges connect sigma sub a to sigma sub r, a to r, and tau sub a to tau sub r. Between sigma sub a and a is a black dot labeled x sub a comma i, and between a and tau sub a is y sub a comma j. Between sigma sub r and r is x sub r comma i, and between r and tau sub r is y sub r comma j. At the top is a white dot labeled 1, and above it a black dot labeled top. Dashed lines connect the outermost nodes to form a cube-like structure. All edges represent covering relations in the partial order.

For convenience, we use less than or equals $\leq $ instead of less than or equals Subscript upper L 3 Baseline $\leq _{L_3}$ . The following lemma lists some properties of complete lattice upper L 3 $L_3$ .

Lemma 2.11 Let upper L 3 $L_3$ be the complete lattice in Example 2.10 and a comma r comma i comma j element of upper I $a,r,i,j\in I$ . Then:

  1. (1) backslash leftvee Underscript b less than r Endscripts b equals r $\bigvee \limits _{b< r}b=r$ , backslash leftvee Underscript r element of upper I Endscripts r equals down tack $\bigvee \limits _{r\in I}r=\top $ and backslash leftvee Underscript r element of upper I Endscripts sigma Subscript r Baseline equals down tack equals backslash leftvee Underscript r element of upper I Endscripts tau Subscript r $\bigvee \limits _{r\in I}\sigma _r=\top =\bigvee \limits _{r\in I}\tau _r$ .

  2. (2) backslash leftwedge Underscript j element of upper I Endscripts y Subscript r comma j Baseline equals r equals backslash leftwedge Underscript i element of upper I Endscripts x Subscript r comma i $\bigwedge \limits _{j\in I}y_{r,j}=r=\bigwedge \limits _{i\in I}x_{r,i}$ , backslash leftvee Underscript j element of upper I Endscripts y Subscript r comma j Baseline equals tau Subscript r $\bigvee \limits _{j\in I}y_{r,j}=\tau _r$ and backslash leftvee Underscript i element of upper I Endscripts x Subscript r comma i Baseline equals sigma Subscript r $\bigvee \limits _{i\in I}x_{r,i}=\sigma _r$ .

  3. (3) backslash leftvee Underscript b less than r Endscripts x Subscript b comma j Baseline equals sigma Subscript r $\bigvee \limits _{b< r}x_{b,j}=\sigma _r$ and backslash leftvee Underscript b less than r Endscripts y Subscript b comma j Baseline equals tau Subscript r $\bigvee \limits _{b< r}y_{b,j}=\tau _r$ .

  4. (4) if a not equals r $a\neq r$ , then sigma Subscript r Baseline or a equals sigma Subscript r or a Baseline equals x Subscript r comma i Baseline or x Subscript a comma j $\sigma _r\vee a=\sigma _{r\vee a}=x_{r,i}\vee x_{a,j}$ and tau Subscript r Baseline or a equals tau Subscript r or a Baseline equals y Subscript r comma j Baseline or y Subscript a comma i $\tau _r\vee a=\tau _{r\vee a}=y_{r,j}\vee y_{a,i}$ .

  5. (5) sigma Subscript r Baseline and a equals r and a equals x Subscript r comma i Baseline and y Subscript a comma j $\sigma _r\wedge a=r\wedge a=x_{r,i}\wedge y_{a,j}$ and tau Subscript r Baseline and a equals r and a equals sigma Subscript r Baseline and tau Subscript a Baseline equals r and tau Subscript a $\tau _r\wedge a=r\wedge a=\sigma _r\wedge \tau _a=r\wedge \tau _a$ .

  6. (6) if a not equals r $a\neq r$ , then x Subscript r comma i Baseline and x Subscript a comma j Baseline equals a and r equals y Subscript r comma i Baseline and y Subscript a comma j $x_{r,i}\wedge x_{a,j}=a\wedge r = y_{r,i}\wedge y_{a,j}$ .

  7. (7) if a less than r $a< r$ , then sigma Subscript a Baseline and x Subscript r comma i Baseline equals a $\sigma _a\wedge x_{r,i}=a$ .

  8. (8) if x element of upper Sigma union left parenthesis backslash leftcup Underscript r element of upper I Endscripts upper X Subscript r Baseline right parenthesis comma y element of upper Gamma union left parenthesis backslash leftcup Underscript r element of upper I Endscripts upper Y Subscript r Baseline right parenthesis $x\in \Sigma \cup (\bigcup \limits _{r\in I}X_r), y\in \Gamma \cup (\bigcup \limits _{r\in I}Y_r)$ , then x or y equals down tack $x\vee y=\top $ .

Proposition 2.12 upper L 3 $L_3$ cannot support a unital quantale.

Proof Assume that upper L 3 $L_3$ supports a unital quantale left parenthesis upper L 3 comma ampersand right parenthesis $(L_3, \&)$ with unit upper Upsilon $\Upsilon $ . Then, we have the following different cases for unit upper Upsilon $\Upsilon $ .

Case 1: upper Upsilon not equals up tack $\Upsilon \neq \bot $ .

Assume that upper Upsilon equals up tack $\Upsilon =\bot $ . Then, by left parenthesis 1.1 right parenthesis comma $(1.1),$ one can see that up tack equals up tack ampersand down tack equals upper Upsilon ampersand down tack equals down tack $\bot =\bot \&\top =\Upsilon \&\top =\top $ , contradiction.

Case 2: upper Upsilon ̸ element of upper Sigma union upper Gamma union StartSet down tack EndSet $\Upsilon \not \in \Sigma \cup \Gamma \cup \{\top \}$ .

Assume that upper Upsilon element of upper Sigma union upper Gamma union StartSet down tack EndSet $\Upsilon \in \Sigma \cup \Gamma \cup \{\top \}$ . Since every element of upper Sigma union upper Gamma union StartSet down tack EndSet $\Sigma \cup \Gamma \cup \{\top \}$ is an upper N 5 $N_5$ -element, by Lemma 2.5, we have that upper Upsilon $\Upsilon $ cannot become a unit, contradiction.

Case 3: upper Upsilon ̸ element of upper I $\Upsilon \not \in I$ .

Assume that upper Upsilon equals r element of upper I $\Upsilon =r\in I$ . Then, we will prove that there exists d less than r $d< r$ such that sigma Subscript d Baseline less than or equals x Subscript d i $\sigma _d\leq x_{di}$ , contradiction.

To see this, we first give the following asserts.

(1) For all b element of upper I $b\in I$ with b less than r $b< r$ , we have that b less than or equals sigma Subscript r Baseline ampersand b less than or equals sigma Subscript b Baseline $b\leq \sigma _r\& b\leq \sigma _b$ .

There exists an element c element of upper I $c\in I$ with c less than b $c< b$ . By Lemma 2.11, we have b equals r ampersand b less than or equals sigma Subscript r Baseline backslash ampersand b equals left parenthesis sigma Subscript c Baseline or r right parenthesis backslash ampersand b equals left parenthesis sigma Subscript c Baseline backslash ampersand b right parenthesis or left parenthesis r backslash ampersand b right parenthesis less than or equals sigma Subscript c Baseline or b equals sigma Subscript b Baseline $b=r\& b\leq \sigma _r\&b=(\sigma _c\vee r)\&b=(\sigma _c\&b)\vee (r\&b)\leq \sigma _c\vee b=\sigma _b$ .

(2) There exists r greater than b 0 element of upper I $r>b_0\in I$ such that b 0 less than sigma Subscript r Baseline ampersand b 0 less than or equals sigma Subscript b 0 Baseline $b_0< \sigma _r\& b_0\leq \sigma _{b_0}$ .

Assume that for all b element of upper I $b\in I$ with b less than r $b< r$ , b equals sigma Subscript r Baseline ampersand b $b=\sigma _r\& b$ . Then, r equals backslash l e f t v e e Subscript b less than r Baseline b equals backslash l e f t v e e Subscript b less than r Baseline left parenthesis sigma Subscript r Baseline backslash ampersand b right parenthesis equals sigma Subscript r Baseline ampersand left parenthesis backslash l e f t v e e Subscript b less than r Baseline b right parenthesis equals sigma Subscript r Baseline backslash ampersand r equals sigma Subscript r Baseline $r=\bigvee \limits _{b< r}b = \bigvee \limits _{b< r}(\sigma _r\&b)=\sigma _r\&(\bigvee \limits _{b< r}b)=\sigma _r\&r=\sigma _r$ , which implies that r equals sigma Subscript r $r=\sigma _r$ , contradiction.

(3) For b 0 less than d less than r $b_0< d < r$ , we have that sigma Subscript r Baseline backslash ampersand d equals sigma Subscript d $\sigma _r\&d=\sigma _d$ .

Since b 0 less than sigma Subscript r Baseline backslash ampersand b 0 less than or equals sigma Subscript r Baseline backslash ampersand d $b_0< \sigma _r\&b_0\leq \sigma _r\&d$ , by (1) and Lemma 2.11 we have that sigma Subscript r Baseline backslash ampersand d equals sigma Subscript d $\sigma _r\&d=\sigma _d$ .

By (3) and Lemma 2.11, we have that sigma Subscript d Baseline equals sigma Subscript r Baseline backslash ampersand d equals left parenthesis x Subscript d comma i Baseline or r right parenthesis backslash ampersand d equals left parenthesis x Subscript d comma i Baseline backslash ampersand d right parenthesis or left parenthesis r backslash ampersand d right parenthesis equals left parenthesis x Subscript d comma i Baseline backslash ampersand d right parenthesis or d less than or equals left parenthesis x Subscript d comma i Baseline backslash ampersand r right parenthesis or d less than or equals x Subscript d comma i Baseline or d equals x Subscript d comma i $\sigma _d=\sigma _r\&d=(x_{d,i}\vee r)\&d=(x_{d,i}\&d)\vee (r\&d)=(x_{d,i}\&d)\vee d\leq (x_{d,i}\&r)\vee d\leq x_{d,i}\vee d=x_{d,i}$ , which implies that sigma Subscript d Baseline less than or equals x Subscript d comma i $\sigma _d\leq x_{d,i}$ , contradiction. Thus, upper Upsilon ̸ element of upper I $\Upsilon \not \in I$ .

Case 4: upper Upsilon ̸ element of backslash leftcup Underscript r element of upper I Endscripts left parenthesis upper X Subscript r Baseline union upper Y Subscript r Baseline right parenthesis $\Upsilon \not \in \bigcup \limits _{r\in I}(X_r\cup Y_r)$ .

Assume that upper Upsilon element of backslash leftcup Underscript r element of upper I Endscripts left parenthesis upper X Subscript r Baseline union upper Y Subscript r Baseline right parenthesis $\Upsilon \in \bigcup \limits _{r\in I}(X_r\cup Y_r)$ . Then, we will prove that there exist b comma r element of upper I $b, r\in I$ such that b less than r $b< r$ , left parenthesis upper Y Subscript r Baseline union StartSet tau Subscript r Baseline EndSet right parenthesis intersection StartSet m element of upper L 3 vertical bar m less than or equals tau Subscript b Baseline EndSet not equals backslash emptyset $(Y_r\cup \{\tau _r\})\cap \{m\in L_3 \mid m\leq \tau _b\}\neq \emptyset $ or left parenthesis upper X Subscript r Baseline union StartSet sigma Subscript r Baseline EndSet right parenthesis intersection StartSet m element of upper L 3 vertical bar m less than or equals sigma Subscript b Baseline EndSet not equals backslash emptyset $(X_r\cup \{\sigma _r\})\cap \{m\in L_3 \mid m\leq \sigma _b\}\neq \emptyset $ , contradiction.

Without loss of generality, we suppose that upper Upsilon equals x Subscript r comma i 0 Baseline element of upper X Subscript r $\Upsilon =x_{r,i_0}\in X_r$ for some r element of upper I $r\in I$ . Then, we have the following asserts.

(1) For all b element of upper I $b\in I$ with b less than r $b< r$ , we have sigma Subscript r Baseline backslash ampersand b equals b $\sigma _r\&b=b$ and b ampersand sigma Subscript r Baseline equals b $b\&\sigma _r=b$ .

Since b equals x Subscript r comma i 0 Baseline backslash ampersand b less than or equals sigma Subscript r Baseline backslash ampersand b equals left parenthesis sigma Subscript b Baseline or x Subscript r comma i 0 Baseline right parenthesis backslash ampersand b equals left parenthesis sigma Subscript b Baseline backslash ampersand b right parenthesis or left parenthesis x Subscript r comma i 0 Baseline backslash ampersand b right parenthesis less than or equals left parenthesis sigma Subscript b Baseline backslash ampersand x Subscript r comma i 0 Baseline right parenthesis or b equals sigma Subscript b $b=x_{r,i_0}\&b\leq \sigma _r\&b=(\sigma _b\vee x_{r,i_0})\&b=(\sigma _b\&b)\vee (x_{r,i_0}\&b)\leq (\sigma _b\&x_{r,i_0})\vee b= \sigma _b$ , we have that b less than or equals sigma Subscript r Baseline backslash ampersand b less than or equals sigma Subscript b $b\leq \sigma _r\&b\leq \sigma _b$ . For all x Subscript b comma j $x_{b,j}$ , we have that sigma Subscript r Baseline backslash ampersand b equals left parenthesis x Subscript b comma j Baseline or x Subscript r comma i 0 Baseline right parenthesis backslash ampersand b equals left parenthesis x Subscript b comma j Baseline backslash ampersand b right parenthesis or left parenthesis x Subscript r comma i 0 Baseline backslash ampersand b right parenthesis less than or equals left parenthesis x Subscript b comma j Baseline backslash ampersand x Subscript r comma i 0 Baseline right parenthesis or b equals x Subscript b comma j $\sigma _r\&b=(x_{b,j}\vee x_{r,i_0})\&b=(x_{b,j}\&b)\vee (x_{r,i_0}\&b)\leq (x_{b,j}\&x_{r,i_0})\vee b=x_{b,j}$ . By Lemma 2.11, we have b less than or equals sigma Subscript r Baseline backslash ampersand b less than or equals backslash l e f t w e d g e Subscript j element of upper I Baseline x Subscript b comma j Baseline equals b $b\leq \sigma _r\&b\leq \bigwedge \limits _{j\in I}x_{b,j}=b$ , which implies that sigma Subscript r Baseline backslash ampersand b equals b $\sigma _r\&b=b$ . Similarly, we can show that b ampersand sigma Subscript r Baseline equals b $b\&\sigma _r=b$ .

(2) sigma Subscript r Baseline backslash ampersand r equals r equals r ampersand sigma Subscript r Baseline $\sigma _r\&r=r=r\&\sigma _r$ and r backslash ampersand r equals r $r\&r=r$ .

By (1) and Lemma 2.11, we have r equals backslash l e f t v e e Subscript b less than r Baseline b equals backslash l e f t v e e Subscript b less than r Baseline left parenthesis sigma Subscript r Baseline backslash ampersand b right parenthesis equals sigma Subscript r Baseline ampersand left parenthesis backslash l e f t v e e Subscript b less than r Baseline b right parenthesis equals sigma Subscript r Baseline backslash ampersand r $r=\bigvee \limits _{b< r}b=\bigvee \limits _{b< r}(\sigma _r\&b)=\sigma _r\&(\bigvee \limits _{b< r}b)=\sigma _r\&r$ . Similarly, we have r equals r ampersand sigma Subscript r Baseline $r=r\&\sigma _r$ . For b element of upper I $b\in I$ with b less than r $b< r$ , we have sigma Subscript b Baseline backslash ampersand r less than or equals sigma Subscript b Baseline and r equals b $\sigma _b\&r\leq \sigma _b\wedge r=b$ and r equals sigma Subscript r Baseline backslash ampersand r equals left parenthesis sigma Subscript b Baseline or r right parenthesis backslash ampersand r equals left parenthesis sigma Subscript b Baseline backslash ampersand r right parenthesis or left parenthesis r backslash ampersand r right parenthesis less than or equals b or left parenthesis r backslash ampersand r right parenthesis less than or equals b or left parenthesis r ampersand sigma Subscript r Baseline right parenthesis less than or equals b or r equals r $r=\sigma _r\&r=(\sigma _b\vee r)\&r=(\sigma _b\&r)\vee (r\&r)\leq b\vee (r\&r)\leq b\vee (r\&\sigma _r)\leq b\vee r=r$ , which implies r backslash ampersand r equals r $r\&r=r$ .

(3) For all b element of upper I $b\in I$ with b less than r $b< r$ , we have r backslash ampersand b equals b equals b backslash ampersand r $r\&b=b=b\&r$ .

We let c element of upper I $c\in I$ and c less than b $c< b$ . Then, for all x Subscript c comma j $x_{c,j}$ , by (1) we have x Subscript c comma j Baseline backslash ampersand b less than or equals x Subscript c comma j Baseline ampersand x Subscript r comma i 0 Baseline equals x Subscript c comma j Baseline $x_{c,j}\&b\leq x_{c,j} \& x_{r,i_0}=x_{c,j}$ and x Subscript c comma j Baseline backslash ampersand b less than or equals sigma Subscript c Baseline backslash ampersand b less than or equals sigma Subscript r Baseline backslash ampersand b equals b $x_{c,j}\&b\leq \sigma _c\&b\leq \sigma _r\&b=b$ , which implies that x Subscript c comma j Baseline backslash ampersand b less than or equals x Subscript c comma j Baseline and b equals c $x_{c,j}\&b\leq x_{c,j}\wedge b=c$ . Thus, b equals sigma Subscript r Baseline backslash ampersand b equals left parenthesis x Subscript c comma j Baseline or r right parenthesis backslash ampersand b equals left parenthesis x Subscript c comma j Baseline backslash ampersand b right parenthesis or left parenthesis r backslash ampersand b right parenthesis $b=\sigma _r\&b=(x_{c,j}\vee r)\&b=(x_{c,j}\&b)\vee (r\&b)$ , which implies b equals r backslash ampersand b $b=r\&b$ . Similarly, we can prove that b equals b backslash ampersand r $b=b\&r$ .

(4) For all b element of upper I $b\in I$ with b less than r $b< r$ , we have r ampersand sigma Subscript b Baseline equals b equals sigma Subscript b Baseline backslash ampersand r $r\&\sigma _b=b=\sigma _b\&r$ .

By (2), we have that sigma Subscript b Baseline ampersand r less than or equals sigma Subscript b Baseline ampersand x Subscript r comma i 0 Baseline equals sigma Subscript b Baseline $\sigma _b\& r\leq \sigma _b\& x_{r,i_0}=\sigma _b$ and sigma Subscript b Baseline ampersand r less than or equals sigma Subscript r Baseline ampersand r equals r $\sigma _b\& r\leq \sigma _r\& r=r$ , which implies sigma Subscript b Baseline ampersand r less than or equals sigma Subscript b Baseline and r equals b $\sigma _b\& r\leq \sigma _b\wedge r=b$ . It follows from (3) that b equals b backslash ampersand r less than or equals sigma Subscript b Baseline backslash ampersand r less than or equals b $b=b\&r\leq \sigma _b\&r\leq b$ , that is, b equals sigma Subscript b Baseline backslash ampersand r $b=\sigma _b\&r$ . Similarly, we can show that r ampersand sigma Subscript b Baseline equals b $r\&\sigma _b=b$ .

(5) For all b element of upper I $b\in I$ with b less than r $b< r$ , we have r ampersand x Subscript b comma j Baseline equals b equals x Subscript b comma j Baseline backslash ampersand r $r\& x_{b,j}=b=x_{b,j}\&r$ .

By (3) and (4), we have that b equals r backslash ampersand b less than or equals r ampersand x Subscript b comma j Baseline less than or equals r ampersand sigma Subscript b Baseline equals b $b=r\&b\leq r\& x_{b,j}\leq r\& \sigma _b=b$ , that is, r ampersand x Subscript b comma j Baseline equals b $r\& x_{b,j}=b$ . Similarly, we have b equals x Subscript b comma j Baseline backslash ampersand r $b=x_{b,j}\&r$ .

(6) For all b element of upper I $b\in I$ with b less than r $b< r$ , we have tau Subscript r Baseline backslash ampersand b equals down tack ampersand b $\tau _r\&b=\top \& b$ and b ampersand tau Subscript r Baseline equals b ampersand down tack $b\&\tau _r=b\&\top $ .

By (1) and (3), we have that tau Subscript r Baseline ampersand b equals left parenthesis r or tau Subscript b Baseline right parenthesis backslash ampersand b equals left parenthesis r backslash ampersand b right parenthesis or left parenthesis tau Subscript b Baseline backslash ampersand b right parenthesis equals b or left parenthesis tau Subscript b Baseline backslash ampersand b right parenthesis equals left parenthesis sigma Subscript r Baseline backslash ampersand b right parenthesis or left parenthesis tau Subscript b Baseline backslash ampersand b right parenthesis equals left parenthesis sigma Subscript r Baseline or tau Subscript b Baseline right parenthesis backslash ampersand b equals down tack backslash ampersand b $\tau _r\& b=(r\vee \tau _b)\&b=(r\&b)\vee (\tau _b\&b)=b\vee (\tau _b\&b)=(\sigma _r\&b)\vee (\tau _b\&b)= (\sigma _r\vee \tau _b)\&b=\top \&b$ . Similarly, we have b ampersand tau Subscript r Baseline equals b ampersand down tack $b\&\tau _r=b\&\top $ .

(7) r less than tau Subscript r Baseline ampersand r equals down tack backslash ampersand r less than or equals tau Subscript r Baseline $r< \tau _r\& r=\top \&r\leq \tau _r$ .

By (6), we have that tau Subscript r Baseline ampersand r equals tau Subscript r Baseline ampersand left parenthesis backslash l e f t v e e Subscript b less than r Baseline b right parenthesis equals backslash l e f t v e e Subscript b less than r Baseline left parenthesis tau Subscript r Baseline backslash ampersand b right parenthesis equals backslash l e f t v e e Subscript b less than r Baseline left parenthesis down tack backslash ampersand b right parenthesis equals down tack ampersand left parenthesis backslash l e f t v e e Subscript b less than r Baseline b right parenthesis equals down tack backslash ampersand r $\tau _r\& r=\tau _r\&(\bigvee \limits _{b< r}b)=\bigvee \limits _{b< r}(\tau _r\&b)=\bigvee \limits _{b< r}(\top \&b) =\top \&(\bigvee \limits _{b< r}b)=\top \&r$ . Further, one can see that r equals r backslash ampersand r less than or equals x Subscript r comma i 0 Baseline backslash ampersand r less than or equals down tack backslash ampersand r equals tau Subscript r Baseline backslash ampersand r less than or equals tau Subscript r Baseline backslash ampersand x Subscript r comma i 0 Baseline equals tau Subscript r $r=r\&r\leq x_{r,i_0}\&r\leq \top \&r=\tau _r\&r\leq \tau _r\&x_{r,i_0}= \tau _r$ .

Assume that r equals tau Subscript r Baseline ampersand r $r=\tau _r\& r$ . It follows from (5) and Lemma 2.11 that for all b less than r $b< r$ , we have that r equals down tack ampersand r equals left parenthesis x Subscript b comma j Baseline or tau Subscript b Baseline right parenthesis backslash ampersand r equals left parenthesis x Subscript b comma j Baseline backslash ampersand r right parenthesis or left parenthesis tau Subscript b Baseline backslash ampersand r right parenthesis equals b or left parenthesis tau Subscript b Baseline backslash ampersand r right parenthesis less than or equals b or left parenthesis tau Subscript b Baseline backslash ampersand x Subscript r comma i 0 Baseline right parenthesis less than or equals b or tau Subscript b Baseline equals tau Subscript b Baseline $r=\top \& r=(x_{b,j}\vee \tau _b)\&r=(x_{b,j}\&r)\vee (\tau _b\&r)=b\vee (\tau _b\&r)\leq b\vee (\tau _b\&x_{r,i_0})\leq b\vee \tau _b=\tau _b$ , which implies r less than or equals tau Subscript b $r\leq \tau _b$ , contradiction. Thus, r less than tau Subscript r Baseline ampersand r equals down tack backslash ampersand r less than or equals tau Subscript r Baseline $r< \tau _r\& r=\top \&r\leq \tau _r$ .

By (1)–(7), for all b less than r $b< r$ , we see that r less than down tack backslash ampersand r equals left parenthesis sigma Subscript b Baseline or tau Subscript b Baseline right parenthesis backslash ampersand r equals left parenthesis sigma Subscript b Baseline backslash ampersand r right parenthesis or left parenthesis tau Subscript b Baseline backslash ampersand r right parenthesis equals b or left parenthesis tau Subscript b Baseline backslash ampersand r right parenthesis equals left parenthesis b ampersand r right parenthesis or left parenthesis tau Subscript b Baseline backslash ampersand r right parenthesis equals left parenthesis b or tau Subscript b Baseline right parenthesis ampersand r equals tau Subscript b Baseline ampersand r less than or equals tau Subscript r Baseline ampersand r less than or equals tau Subscript r Baseline $r<\top \&r=(\sigma _b\vee \tau _b)\&r=(\sigma _b\&r)\vee (\tau _b\&r)=b\vee (\tau _b\&r) =(b\& r)\vee (\tau _b\&r)=(b\vee \tau _b)\& r=\tau _b\& r\leq \tau _r\& r\leq \tau _r$ , which implies that r less than tau Subscript b Baseline backslash ampersand r less than or equals tau Subscript r $r<\tau _b\&r\leq \tau _r$ , that is, tau Subscript b Baseline backslash ampersand r element of upper Y Subscript r union StartSet tau Subscript r Baseline EndSet $\tau _b\&r\in Y_r\cup \{\tau _r\}$ . Obviously, tau Subscript b Baseline backslash ampersand r less than or equals tau Subscript b Baseline backslash ampersand x Subscript r comma i 0 Baseline less than or equals tau Subscript b $\tau _b\&r\leq \tau _b\&x_{r,i_0}\leq \tau _b$ . However, left parenthesis upper Y Subscript r Baseline union StartSet tau Subscript r Baseline EndSet right parenthesis intersection StartSet m element of upper L 3 vertical bar m less than or equals tau Subscript b Baseline EndSet equals backslash emptyset $(Y_r\cup \{\tau _r\})\cap \{m\in L_3 \mid m\leq \tau _b\}=\emptyset $ , contradiction. Thus, upper Upsilon ̸ element of backslash leftcup Underscript r element of upper I Endscripts upper X Subscript r $\Upsilon \not \in \bigcup \limits _{r\in I} X_r$ . Similarly, we can show that upper Upsilon ̸ element of backslash leftcup Underscript r element of upper I Endscripts upper Y Subscript r $\Upsilon \not \in \bigcup \limits _{r\in I} Y_r$ .

From Cases 1–4, it follows that upper Upsilon ̸ element of upper L 3 $\Upsilon \not \in L_3$ , contradiction. Therefore, upper L 3 $L_3$ cannot support a unital quantale.

By Example 2.8 and Proposition 2.12, we have the below conclusion.

Proposition 2.13 Not every complete lattice can support a unital quantale.

3 Complete lattices supporting unital quantales

In Section 2, we see that not all complete lattices can support unital quantales. In this section, we will investigate Question 2, that is, which class of complete lattices can support unital quantales.

An element x of quantale Q is called a zero-factor of Q provided that x not equals 0 $x\neq 0$ and there exists an element y element of upper Q backslash StartSet 0 EndSet $y\in Q\backslash \{0\}$ such that x ampersand y equals 0 $x\& y=0$ or y backslash ampersand x equals 0 $y\&x=0$ .

Proposition 3.1 Every complete chain supports a commutative unital quantale without zero-factors.

Proof Let I be a complete chain, e element of upper I backslash StartSet 0 EndSet $e\in I\backslash \{0\}$ . Then, we have two ways to define a binary operation on I such that I becomes a commutative unital quantale without zero-factors.

(1) The first way to define a binary operation ampersand $\&$ on I is given as follows:

for all x comma y element of upper I comma x ampersand y equals StartLayout Enlarged left brace 1st Row 1st Column Blank 2nd Column 0 comma 3rd Column if x equals 0 or y equals 0 comma 2nd Row 1st Column Blank 2nd Column x and y comma 3rd Column if x comma y element of left parenthesis 0 comma e right bracket comma 3rd Row 1st Column Blank 2nd Column x or y comma 3rd Column otherwise period EndLayout $$ \begin{align*}\forall x, y\in I,\ x\& y=\left\{\begin{array}{@{}ll} \ \ 0, & \text{if}\ \ x=0\ \text{or}\ y=0,\\ x\wedge y, & \text{if}\ \ x,y\in (0, e],\\ x\vee y, & \text{otherwise.}\ \ \end{array}\right.\end{align*} $$

It is easy to check that left parenthesis upper I comma ampersand right parenthesis $(I, \&)$ is a commutative unital quantale with unit e and L has no zero-factors.

(2) The second way to define a binary operation circled times $\otimes $ on I is given as follows:

for all x comma y element of upper I comma x circled times y equals StartLayout Enlarged left brace 1st Row 1st Column Blank 2nd Column 0 comma 3rd Column if x equals 0 or y equals 0 comma 2nd Row 1st Column Blank 2nd Column x and y comma 3rd Column if x comma y element of left parenthesis 0 comma e right bracket comma 3rd Row 1st Column Blank 2nd Column y comma 3rd Column if x element of left parenthesis 0 comma e right bracket comma y element of left parenthesis e comma 1 right bracket comma 4th Row 1st Column Blank 2nd Column x comma 3rd Column if y element of left parenthesis 0 comma e right bracket comma x element of left parenthesis e comma 1 right bracket comma 5th Row 1st Column Blank 2nd Column 1 comma 3rd Column otherwise period EndLayout $$ \begin{align*}\forall x, y\in I,\ x\otimes y=\left\{\begin{array}{@{}ll} \ 0, & \text{if}\ \ x=0\ \text{or}\ y=0,\\ x\wedge y, & \text{if}\ \ x,y\in (0, e],\\ \ y, & \text{if}\ \ x\in (0, e],\ y\in (e, 1],\\ \ x, & \text{if}\ \ y\in (0, e],\ x\in (e, 1],\\ \ 1, & \text{otherwise.}\ \ \end{array}\right.\end{align*} $$

It is easy to check that left parenthesis upper I comma circled times right parenthesis $(I, \otimes )$ is a commutative unital quantale with unit e and L has no zero-factors.

Proposition 3.2 Let Q be a complete lattice and e element of upper Q backslash StartSet 0 EndSet $e\in Q\backslash \{0\}$ . If 0 comma e $[0, e]$ supports a left parenthesis $($ commutative right parenthesis $)$ unital quantale with unit k and Q satisfies the condition left parenthesis asterisk right parenthesis $(\ast )$

(*) for all x element of left parenthesis 0 comma e right bracket comma y element of upper Q backslash 0 comma e comma x less than or equals y long left right double arrow e less than or equals y comma $$ \begin{align}\forall x\in (0, e], y\in Q\backslash [0, e],\ x\leq y\Longleftrightarrow e\leq y, \end{align} $$

then Q supports a left parenthesis $($ commutative right parenthesis $)$ unital quantale with unit k.

Proof Let left parenthesis 0 comma e comma ampersand right parenthesis $([0, e], \&)$ be a (commutative) unital quantale with unit k. Then, ampersand $\&$ can be extended to a binary operation circled times $\otimes $ on Q as follows:

for all x comma y element of upper Q comma x circled times y equals backslash lbra c e StartLayout 1st Row 1st Column at l l 2nd Column 0 3rd Column comma 2nd Row 1st Column backslash t e x t 2nd Column i f 3rd Column x equals 0 backslash t e x t o r y equals 0 comma y comma 3rd Row 1st Column backslash t e x t 2nd Column i f 3rd Column x element of left parenthesis 0 comma e right bracket comma y element of upper Q backslash comma x comma 4th Row 1st Column backslash t e x t 2nd Column i f 3rd Column x element of upper Q backslash comma y element of left parenthesis 0 comma e right bracket comma x backslash ampersand y comma 5th Row 1st Column backslash t e x t 2nd Column i f 3rd Column x comma y element of left parenthesis 0 comma e right bracket comma 1 comma 6th Row 1st Column backslash t e x t 2nd Column i f 3rd Column x comma y element of upper Q backslash period EndLayout $$ \begin{align*} \forall x, y\in Q,\ x\otimes y=\left\{\begin{array}{@{}ll} \ 0, & \text{if}\ \ x=0\ \text{or}\ y=0,\\ \ y, & \text{if}\ \ x\in (0, e],\ y\in Q\backslash [0, e],\\ \ x, & \text{if}\ \ x\in Q\backslash [0, e],\ y\in (0, e],\\ x\&y, & \text{if}\ \ x,y\in (0, e],\\ \ 1, & \text{if}\ \ x,y\in Q\backslash [0, e]. \end{array}\right. \end{align*} $$

It can be verified that left parenthesis upper Q comma circled times right parenthesis $(Q, \otimes )$ is a (commutative) unital quantale with unit k. Further, if left parenthesis 0 comma e comma ampersand right parenthesis $([0, e], \&)$ has no zero-factors, then left parenthesis upper Q comma circled times right parenthesis $(Q, \otimes )$ has no zero-factors.

Corollary 3.3 Let Q be a complete lattice and e element of upper Q backslash StartSet 0 EndSet $e\in Q\backslash \{0\}$ . If 0 comma e $[0, e]$ is a chain and Q satisfies the condition left parenthesis asterisk right parenthesis $(\ast )$ , then Q supports a commutative unital quantale without zero-factors.

Proof The proof follows immediately from Propositions 3.1 and 3.2.

Corollary 3.4 Every complete lattice with non-bottom minimal element supports a commutative unital quantale without zero-factors.

Proof Let Q be a complete lattice with non-bottom minimal element e. Then, 0 comma e equals StartSet 0 comma e EndSet $[0, e]=\{0, e\}$ and Q satisfies the condition left parenthesis asterisk right parenthesis $(\ast )$ . By Corollary 3.3, we have that Q supports a commutative unital quantale without zero-factors.

Corollary 3.5 Every finite complete lattice supports a commutative unital quantale without zero-factors.

Proposition 3.6 Let Q be a complete lattice and a element of upper Q backslash StartSet 0 comma 1 EndSet $a\in Q\backslash \{0, 1\}$ be a comparable element. Then, we have:

  1. (1) if a comma 1 $[a, 1]$ is a chain, then Q supports a commutative unital quantale.

  2. (2) if 0 comma a $[0, a]$ is a chain, then Q supports a commutative unital quantale without zero-factors.

Proof (1) Assume that a comma 1 $[a, 1]$ is a chain. Then, we can define a binary operation circled times $\otimes $ on Q as follows:

for all x comma y element of upper Q comma x circled times y equals backslash lbra c e StartLayout 1st Row 1st Column at l l 2nd Column 0 3rd Column comma 2nd Row 1st Column backslash text 2nd Column i f 3rd Column x comma y element of comma x and y comma 3rd Row 1st Column backslash text 2nd Column o t h e r w i s e 3rd Column period EndLayout $$ \begin{align*}\forall x, y\in Q,\ x\otimes y=\left\{\begin{array}{@{}ll} \ 0, & \text{if}\ \ x, y\in [0, a],\\ x\wedge y, & \text{otherwise}. \end{array}\right.\end{align*} $$

It is easy to verify that left parenthesis upper Q comma circled times right parenthesis $(Q, \otimes )$ is a commutative unital quantale with unit 1 $1$ .

(2) The proof follows directly from Propositions 3.1 and 3.2.

Let Q be a complete lattice and StartSet up tack comma e comma down tack EndSet intersection upper Q equals backslash emptyset $\{\bot , e, \top \}\cap Q=\emptyset $ . Next, we will consider three types of extensions upper Q Subscript up tack Baseline comma upper Q Subscript e Baseline comma upper Q Subscript down tack Baseline $Q_\bot , Q_e, Q_\top $ of Q, where upper Q Subscript up tack Baseline equals upper Q union StartSet up tack EndSet comma upper Q Subscript e Baseline equals upper Q union StartSet e EndSet comma upper Q Subscript down tack Baseline equals upper Q union StartSet down tack EndSet $Q_\bot =Q\cup \{\bot \}, Q_e=Q\cup \{e\}, Q_\top =Q\cup \{\top \}$ . We, respectively, define the relations less than or equals Subscript upper Q Sub Subscript up tack Subscript Baseline comma less than or equals Subscript upper Q Sub Subscript e Subscript Baseline comma less than or equals Subscript upper Q Sub Subscript down tack Subscript Baseline $\leq _{Q_\bot }, \leq _{Q_e}, \leq _{Q_\top }$ as follows:

less than or equals Subscript upper Q Sub Subscript up tack Subscript Baseline equals less than or equals Subscript upper Q Baseline union StartSet left parenthesis up tack comma x right parenthesis vertical bar x element of upper Q Subscript up tack Baseline EndSet comma $$ \begin{align*}\leq_{Q_\bot}=\ \leq_Q\cup\{(\bot, x) \mid x\in Q_\bot\},\end{align*} $$
less than or equals Subscript upper Q Sub Subscript e Subscript Baseline equals less than or equals Subscript upper Q Baseline union StartSet left parenthesis 0 comma e right parenthesis comma left parenthesis e comma e right parenthesis comma left parenthesis e comma 1 right parenthesis EndSet comma $$ \begin{align*}\leq_{Q_e}=\ \leq_Q\cup\{(0, e), (e, e), (e, 1)\},\end{align*} $$
less than or equals Subscript upper Q Sub Subscript down tack Subscript Baseline equals less than or equals Subscript upper Q Baseline union StartSet left parenthesis x comma down tack right parenthesis vertical bar x element of upper Q Subscript down tack Baseline EndSet period $$ \begin{align*}\leq_{Q_\top}=\ \leq_Q\cup\{(x, \top) \mid x\in Q_\top\}.\end{align*} $$

It is easily checked that left parenthesis upper Q Subscript up tack Baseline comma less than or equals Subscript upper Q Sub Subscript up tack Subscript Baseline right parenthesis comma left parenthesis upper Q Subscript e Baseline comma less than or equals Subscript upper Q Sub Subscript e Subscript Baseline right parenthesis comma $(Q_\bot , \leq _{Q_\bot }), (Q_e, \leq _{Q_e}),$ and left parenthesis upper Q Subscript down tack Baseline comma less than or equals Subscript upper Q Sub Subscript down tack Subscript Baseline right parenthesis $(Q_\top , \leq _{Q_\top })$ are complete lattices, called the extensions of Q. By Corollary 3.3 and Proposition 3.6, we have the following result.

Corollary 3.7 For any complete lattice Q, the extensions upper Q Subscript up tack Baseline comma upper Q Subscript e Baseline comma upper Q Subscript down tack Baseline $Q_\bot , Q_e, Q_\top $ can all support unital quantales.

The proof of Theorem 3.8 can be found in Theorem 2.4 in [Reference Han and Lv9]. For the convenience of readers, we will give the main details.

Theorem 3.8 Let Q be a complete lattice with upper Q equals upper M times upper K $Q=M\times K$ . If M or K supports a commutative unital quantale without zero-factors, then Q supports a unital quantale.

Proof Without loss of generality, we assume that left parenthesis upper K comma ampersand right parenthesis $(K, \&)$ is a commutative unital quantale without zero-factors and e is the unit. By (1.2), we have that left parenthesis upper M comma ampersand Subscript a Baseline right parenthesis $(M, \&_a)$ is a quantale with a element of upper M $a\in M$ . Next, we define a binary operation ampersand Subscript asterisk Baseline $\&_\ast $ on Q as follows: for all left parenthesis m comma k right parenthesis comma left parenthesis n comma k Superscript prime Baseline right parenthesis element of upper M times upper K $\forall (m, k), (n, k^\prime )\in M\times K$ ,

left parenthesis m comma k right parenthesis ampersand Subscript asterisk Baseline left parenthesis n comma k Superscript prime Baseline right parenthesis equals left parenthesis left parenthesis m ampersand Subscript a Baseline n right parenthesis or left parenthesis m asterisk k Superscript prime Baseline right parenthesis or left parenthesis n asterisk k right parenthesis comma k ampersand k Superscript prime Baseline right parenthesis comma $$ \begin{align*} (m, k)\&_{\ast}(n, k^\prime)=((m\&_a n)\vee(m\ast k^\prime)\vee(n\ast k), {k}{\&}{k}^{\prime}), \end{align*} $$

where asterisk $\ast $ is a module operator from upper M times upper K $M\times K$ into M defined by

for all m element of upper M comma k element of upper K comma m asterisk k equals StartLayout Enlarged left brace 1st Row 1st Column Blank 2nd Column 0 comma 3rd Column if k equals 0 comma 2nd Row 1st Column Blank 2nd Column m comma 3rd Column otherwise period EndLayout $$ \begin{align*} \forall m\in M, k\in K, m\ast k=\left\{\begin{array}{@{}ll}0, & \text{if } k=0,\\m, & \text{otherwise}. \end{array}\right. \end{align*} $$

It can be verified that left parenthesis upper Q comma ampersand Subscript asterisk Baseline right parenthesis $(Q, {\&}_{\ast })$ is a unital quantale with unit left parenthesis 0 comma e right parenthesis $(0, e)$ .

Corollary 3.9 Let Q be a complete lattice with upper Q equals upper M times upper K $Q=M\times K$ satisfying one of the following conditions:

  1. (1) M or K is a non-trivial complete chain;

  2. (2) M or K is a non-trivial finite complete lattice;

  3. (3) M or K has a comparable element e such that 0 comma e $[0, e]$ is a chain.

Then, Q supports a unital quantale.

Proof The proof follows from Proposition 3.1, Corollary 3.5, Proposition 3.6(2), and Theorem 3.8.

In the following, we will give a complete lattice which does not belong to any of the aforementioned types of complete lattices supporting unital quantales.

Example 3.10 Let I denote the unit open interval left parenthesis 0 comma 1 right parenthesis $(0, 1)$ with usual order less than or equals $\leq $ , upper Sigma equals StartSet sigma Subscript r Baseline vertical bar r element of left parenthesis 0 comma 1 right parenthesis EndSet $\Sigma =\{\sigma _r \mid r\in (0, 1)\}$ . Further, we let upper X equals StartSet x Subscript r Baseline vertical bar r element of left parenthesis 0 comma 1 right parenthesis EndSet $X=\{x_{r} \mid r\in (0, 1)\}$ and upper L 4 equals upper I union upper Sigma union upper X union StartSet down tack comma up tack EndSet $L_4=I\cup \Sigma \cup X\cup \{\top , \bot \}$ . We define a binary relation less than or equals Subscript upper L 4 Baseline $\leq _{L_4}$ on upper L 4 $L_4$ as follows: for all r comma i comma j element of upper I $\forall r,i,j\in I$ , x element of Subscript upper L 4 Baseline $x\in _{L_4}$ ,

i less than or equals Subscript upper L 4 Baseline j long left right double arrow i less than or equals j long left right double arrow sigma Subscript i Baseline less than or equals Subscript upper L 4 Baseline sigma Subscript j Baseline comma $$ \begin{align*}i\leq_{L_4} j \Longleftrightarrow i\leq j \Longleftrightarrow\sigma_i\leq_{L_4} \sigma_j,\end{align*} $$
i less than or equals Subscript upper L 4 Baseline sigma Subscript j Baseline long left right double arrow i less than or equals j long left right double arrow i less than or equals Subscript upper L 4 Baseline x Subscript j Baseline comma $$ \begin{align*}i\leq_{L_4} \sigma_j \Longleftrightarrow i\leq j\Longleftrightarrow i\leq_{L_4} x_j,\end{align*} $$
x Subscript r Baseline less than or equals Subscript upper L 4 Baseline sigma Subscript j Baseline long left right double arrow r less than or equals j comma $$ \begin{align*}x_{r}\leq_{L_4} \sigma_j \Longleftrightarrow r\leq j,\end{align*} $$
up tack less than or equals Subscript upper L 4 Baseline x less than or equals Subscript upper L 4 Baseline down tack period $$ \begin{align*}\bot\leq_{L_4} x\leq_{L_4}\top.\end{align*} $$

It is easy to check that left parenthesis upper L 4 comma less than or equals Subscript upper L 4 Baseline right parenthesis $(L_4, \leq _{L_4})$ is a complete lattice (Figure 6).

Figure 6

The partial order on upper L 4 $L_4$ .

Line diagram with two vertical chains labeled bottom to top as bottom, 0, r, 1, top and bottom, sigma sub r, x sub r, 1, top, joined by dashed lines.

We let upper I Subscript r Baseline equals StartSet a element of upper I vertical bar a less than or equals r EndSet comma upper X Subscript r Baseline equals StartSet x Subscript a Baseline element of upper X vertical bar a less than or equals r EndSet comma $I_r=\{a\in I \mid a\leq r\}, X_r=\{x_a\in X \mid a\leq r\},$ and upper Sigma Subscript r Baseline equals StartSet sigma Subscript a Baseline element of upper Sigma vertical bar a less than or equals r EndSet $\Sigma _r=\{\sigma _a\in \Sigma \mid a\leq r\}$ . Next, we define a binary operation circled times $\otimes $ on upper L 4 $L_4$ as follows: for all x comma y element of upper L 4 $\forall x, y\in L_4$ ,

x circled times y equals StartLayout Enlarged left brace 1st Row 1st Column Blank 2nd Column normal up tack comma 3rd Column if x equals normal up tack or y equals normal up tack comma 2nd Row 1st Column Blank 2nd Column x and y comma 3rd Column if x comma y element of upper I Subscript r Baseline comma 3rd Row 1st Column Blank 2nd Column x Subscript a and c Baseline comma 3rd Column if x equals x Subscript a Baseline element of upper X Subscript r Baseline comma y equals x Subscript c Baseline element of upper X Subscript r Baseline comma 4th Row 1st Column Blank 2nd Column sigma Subscript a and c Baseline comma 3rd Column if x equals sigma Subscript a Baseline element of normal upper Sigma Subscript r Baseline comma y equals sigma Subscript c Baseline element of normal upper Sigma Subscript r Baseline comma 5th Row 1st Column Blank 2nd Column a and c comma 3rd Column if x equals a element of upper I Subscript r Baseline comma y equals x Subscript c Baseline element of upper X Subscript r Baseline or x equals x Subscript a Baseline element of upper X Subscript r Baseline comma y equals c element of upper I Subscript r Baseline comma 6th Row 1st Column Blank 2nd Column a and c comma 3rd Column if x equals a element of upper I Subscript r Baseline comma y equals sigma Subscript c Baseline element of normal upper Sigma Subscript r Baseline or x equals sigma Subscript a Baseline element of normal upper Sigma Subscript r Baseline comma y equals c element of upper I Subscript r Baseline comma 7th Row 1st Column Blank 2nd Column sigma Subscript a and c Baseline comma 3rd Column if x equals x Subscript a Baseline element of upper X Subscript r Baseline comma y equals sigma Subscript c Baseline element of normal upper Sigma Subscript r Baseline or x equals sigma Subscript c Baseline element of normal upper Sigma Subscript r Baseline comma y equals x Subscript a Baseline element of upper X Subscript r Baseline comma 8th Row 1st Column Blank 2nd Column y comma 3rd Column if x element of upper I Subscript r Baseline union normal upper Sigma Subscript r Baseline union upper X Subscript r Baseline comma y element of upper L 4 backslash left parenthesis upper I Subscript r Baseline union normal upper Sigma Subscript r Baseline union upper X Subscript r Baseline right parenthesis comma 9th Row 1st Column Blank 2nd Column x comma 3rd Column if y element of upper I Subscript r Baseline union normal upper Sigma Subscript r Baseline union upper X Subscript r Baseline comma x element of upper L 4 backslash left parenthesis upper I Subscript r Baseline union normal upper Sigma Subscript r Baseline union upper X Subscript r Baseline right parenthesis comma 10th Row 1st Column Blank 2nd Column normal down tack comma 3rd Column otherwise period EndLayout $$ \begin{align*}x\otimes y=\left\{\begin{array}{@{}ll} \ \ \bot, & \ \text{if} \ \ x=\bot\ \text{or}\ y=\bot,\\ x\wedge y, &\ \text{if}\ \ x, y\in I_r,\\ x_{a\wedge c}, & \ \text{if}\ \ x=x_a\in X_r, y=x_c\in X_r,\\ \sigma_{a\wedge c}, & \ \text{if}\ \ x=\sigma_a\in \Sigma_r, y=\sigma_c\in \Sigma_r,\\ a\wedge c, & \ \text{if}\ \ x=a\in I_r,\ y=x_c\in X_r\ \text{or}\ x=x_a\in X_r, y=c\in I_r,\\ a\wedge c, & \ \text{if}\ \ x=a\in I_r,\ y=\sigma_c\in \Sigma_r\ \text{or}\ x=\sigma_a\in \Sigma_r, y=c\in I_r,\\ \sigma_{a\wedge c}, & \ \text{if}\ \ x=x_a\in X_r, y=\sigma_c\in\Sigma_r\ \text{or}\ x=\sigma_c\in\Sigma_r, y=x_a\in X_r,\\ \ \ y, & \ \text{if}\ \ x\in I_r\cup\Sigma_r\cup X_r,\ y\in L_4\backslash(I_r\cup\Sigma_r\cup X_r),\\ \ \ x, & \ \text{if}\ \ y\in I_r\cup\Sigma_r\cup X_r,\ x\in L_4\backslash(I_r\cup\Sigma_r\cup X_r),\\ \ \ \top, & \ \text{otherwise}.\ \ \end{array}\right.\end{align*} $$

It is easy to verify that left parenthesis upper L 4 comma circled times right parenthesis $(L_4, \otimes )$ is a commutative unital quantale with unit x Subscript r $x_r$ and upper L 4 $L_4$ has no zero-factors.

Further, we will obtain more interesting observations for complete lattice upper L 4 $L_4$ .

(1) Since every element of upper Sigma $\Sigma $ is an upper N 5 $N_5$ -element, it follows from Lemma 2.5 that sigma Subscript r Baseline element of upper Sigma $\sigma _r\in \Sigma $ cannot become a unit.

(2) Every element of upper I union upper X $I\cup X$ is neither an upper N 5 $N_5$ -element nor an upper M 3 $M_3$ -element. Based on the above description, we see that every element x Subscript r $x_r$ of X can become a unit. However, every element of I cannot become a unit (see the proof of Case 3 in Proposition 2.12).

(3) The top element down tack $\top $ is neither an upper N 5 $N_5$ -element nor an upper M 3 $M_3$ -element. Also, down tack $\top $ cannot become a unit.

Assume that down tack $\top $ can become a unit in quantale left parenthesis upper L 4 comma ampersand right parenthesis $(L_4, \&)$ . For d comma r element of upper I $d,r\in I$ , we have that d ampersand sigma Subscript r Baseline less than or equals d and sigma Subscript r Baseline less than or equals r $d\&\sigma _r\leq d\wedge \sigma _r\leq r$ , which implies that sigma Subscript r Baseline equals down tack ampersand sigma Subscript r Baseline equals left parenthesis backslash leftvee Underscript r less than d Endscripts d right parenthesis ampersand sigma Subscript r Baseline equals backslash leftvee Underscript r less than d Endscripts left parenthesis d ampersand sigma Subscript r Baseline right parenthesis less than or equals r $\sigma _r=\top \&\sigma _r=(\bigvee \limits _{r< d}d)\&\sigma _r=\bigvee \limits _{r< d}(d\&\sigma _r)\leq r$ , contradiction.

(4) upper L 4 $L_4$ does not belong to any of the aforementioned types of complete lattices supporting unital quantales in Propositions 3.1, 3.2, and 3.6, and Theorem 3.8.

To see this, upper L 4 $L_4$ is obviously not a chain and also upper L 4 $L_4$ has no non-trivial comparable elements. Further, we see that there are no non-bottom elements e in upper L 4 $L_4$ such that upper L 4 $L_4$ satisfies the condition left parenthesis asterisk right parenthesis $(\ast )$ , and upper L 4 $L_4$ is not isomorphic to any product lattice upper M times upper K $M\times K$ for all non-trivial complete lattices M and K.

4 Conclusion

From Remark 2.7, we see that if a complete lattice supports a unital quantale, then it must have a non-bottom element which is neither an upper N 5 $N_5$ -element nor an upper M 3 $M_3$ -element. However, the converse is in general not true. From Examples 2.10 and 3.10, one can see that for the non-bottom elements of a complete lattice which are neither an upper N 5 $N_5$ -element nor an upper M 3 $M_3$ -element, some can become a unit while others cannot. So, it is not easy to determine whether a complete lattice can support a unital quantale. In other words, it is very difficult to give a necessary and sufficient condition for a complete lattice to support a unital quantale.

Acknowledgments

The authors are very grateful to the referees for the very careful reading of the manuscript and very useful comments.

Footnotes

This work was supported by the National Natural Science Foundation of China (Grant Nos. 12471436 and 12331016).

References

Abramsky, S. and Vickers, S., Quantale, observational logic and process semantics . Math. Struct. Comput. Sci. 3(1993), no. 2, 161–227.CrossRefGoogle Scholar
Birkhoff, G., Lattice theory. American Mathematical Society, Providence, RI, 1940.CrossRefGoogle Scholar
Borceux, F., Rosický, J., and Van Bossche, G., Quantales and upper C Superscript asterisk ${C}^{\ast }$ -algebras . J. Lond. Math. Soc. 40(1989), no. 3, 398–404.CrossRefGoogle Scholar
Connes, A., Noncommutative geometry. Academic Press, San Diego, CA, 1994.Google Scholar
Eklund, P., Gutiérrez García, J., Höhle, U., and Kortelainen, J., Semigroups in complete lattices: Quantales, modules and related topics, Developments in Mathematics, 54, Springer, Berlin, 2018.CrossRefGoogle Scholar
Fan, L., A new approach to quantitative domain theory . Electron. Notes Theor. Comput. Sci. 45(2001), 77–87.CrossRefGoogle Scholar
Gierz, G., Hofmann, K. H., Keimel, K., Lawson, J. D., Mislove, M., and Scott, D. S., A compendium of continuous lattices. Springer-Verlag, New York, NY, 1980.CrossRefGoogle Scholar
Gutiérrez García, J. and Höhle, U., Unitally nondistributive quantales . Port. Math. 82(2025), nos. 3–4, 259–280.CrossRefGoogle Scholar
Han, S. and Lv, J., Construction of unital quantales . Canad. Math. Bull. 68(2025), no. 3, 761–776.CrossRefGoogle Scholar
Hofmann, D. and Waszkiewicz, P., Approximation in quantale-enriched categories . Topol. Appl. 158(2011), no. 8, 963–977.CrossRefGoogle Scholar
Kruml, D. and Paseka, J., Algebraic and categorical aspects of quantales . Handbook Algebra 5(2008), 323–326.Google Scholar
Kruml, D. and Resende, P., On quantales that classify upper C Superscript asterisk ${C}^{\ast }$ -algebras . Cah. Topol. Géom. Différ. Catég. 45(2004), no. 4, 287–296.Google Scholar
Mulvey, C. J. and Pelletier, J. W., On the quantisation of points . J. Pure Appl. Algebra 159(2001), nos. 2–3, 231–295.CrossRefGoogle Scholar
Mulvey, C. J. and Pelletier, J. W., On the quantisation of spaces . J. Pure Appl. Algebra 175(2002), nos. 1–3, 289–325.CrossRefGoogle Scholar
Mulvey, C. J. and Resende, P., A noncommutative theory of Penrose tilings . Internat. J. Theoret. Phys. 44(2005), no. 6, 655–689.CrossRefGoogle Scholar
Paseka, J. and Kruml, D., Embeddings of quantales into simple quantales . J. Pure Appl. Algebra 148(2000), no. 2, 209–216.CrossRefGoogle Scholar
Protin, M. C. and Resende, P., Quantales of open groupoids . J. Noncommut. Geom. 6(2012), no. 2, 199–247.CrossRefGoogle Scholar
Resende, P., Quantales, finite observation and strong bismulation . Theoret. Comput. Sci. 254(2001), nos. 1–2, 95–149.CrossRefGoogle Scholar
Resende, P., Étale groupoids and their quantales . Adv. Math. 208(2007), no. 1, 147–209.CrossRefGoogle Scholar
Rosenthal, K. I., Quantales and their applications. Longman Scientific & Technical, New York, NY, 1990.Google Scholar
Yetter, D. N., Quantales and (noncommutative) linear logic . J. Symbolic Logic 55(1990), no. 1, 41–64.CrossRefGoogle Scholar
Zhang, D., An enriched category approach to many valued topology . Fuzzy Sets and Systems 158(2007), 349–366.CrossRefGoogle Scholar
Figure 0

Figure 1The partial order on $N_5$N5.

Figure 1

Figure 2The partial order on $M_3$M3.

Figure 2

Figure 3The partial order on $L_1$L1.

Figure 3

Figure 4The partial order on $L_2$L2.Figure 4 Long description.

Figure 4

Figure 5The partial order on $L_3$L3.Figure 5 Long description.

Figure 5

Figure 6The partial order on $L_4$L4.