2.1 Constructing unital quantales by Q-algebras

Let Q be a commutative (unital) quantale and $(M, \cdot , \otimes )$ be a (unital) Q-algebra. We define a semigroup multiplication $\&_\cdot $ on $M\times Q$ and a map $\star \colon (M\times Q)\times Q\rightarrow M\times Q$ as follows

(2.1) $$ \begin{align}(m, q)\&_\cdot(n, p)=((m\otimes n)\vee(m\cdot p)\vee(n\cdot q), q\&p)\end{align} $$

and

$$ \begin{align*}(m, q)\star p=(m\cdot p, q\&p).\end{align*} $$

Note that if $(M, \cdot , \otimes )$ be a Q-algebra with trivial module operator, then $(M\times Q, \otimes )=(M\times Q, \&_\cdot )$ .

Lemma 2.3 Let Q be a commutative quantale without zero-factors. Then every complete lattice M can give rise to a Q-module $(M, \ast )$ with unital module operator $\ast $ , that is,

$$ \begin{align*}\forall m\in M, q\in Q,\ \ m\ast q=\begin{cases} 0, & \ \mbox{if}\ \ q=0,\\ m, & \ \mbox{otherwise.}\ \ \end{cases}\end{align*} $$

Thus, every quantale $(M, \otimes )$ can give rise to a Q-algebra $(M, \ast , \otimes )$ .

Proof By Lemma 2.3, we have that $(M, \ast , \otimes )$ is a unital K-algebra with unital module operator. It follows from Corollary 2.2 that $(M\times K, \&_\ast )$ is a unital quantale. Clearly, we see that $\forall (m, k), (n, k^\prime )\in M\times K$ ,

$$ \begin{align*}(m, k)\&_\ast(n, k^\prime)=((m\otimes n)\vee(m\ast k^\prime)\vee(n\ast k), k\&k^\prime) \end{align*} $$

and $(0, e)$ is the unit of $(M\times K, \&_\ast )$ , where e is the unit of K.

Figure 1 The partial order on $\mathbf {K_6}$ .

Figure 2 The partial order on $\mathbf {K_8}$ .

Let L be a complete lattice and let $\triangleleft $ be the totally below relation on L, i.e., given $\alpha , \beta \in L$ , we write $\beta \triangleleft \alpha $ if for any subset $A\subseteq L$ with $\alpha \leq \bigvee A$ there is an element $\gamma \in A$ such that $\beta \leq \gamma $ . An element $\alpha $ of L is called $\triangleleft $ -approximable if $\alpha =\bigvee \{\beta \in L\ \colon \beta \triangleleft \alpha \}$ . A lattice L is called strictly non-distributive if there exist $\alpha , \beta , x\in L$ such that $x\wedge (\alpha \vee \beta )\not \leq (x\wedge \alpha )\vee (x\wedge \beta )$ and $x\not \leq \alpha \vee \beta $ (see [Reference Gutiérrez García and Höhle8]). Let $L, M$ be complete lattices such that L or M is strictly non-distributive. Then $L\times M$ is also strictly non-distributive.

The results in Corollary 2.6 generalize the work of Paseka and Kruml for unital quantales in [Reference Paseka and Kruml19].

In [Reference Gutiérrez García and Höhle8], Guriérrez García and Höhle showed that under mild conditions, quantales on non-distributive lattices can be extended to unitally non-distributive quantales by addition of an isolated unit. The following theorem indicates that every quantale can be extended to a unitally non-distributive quantale. First, we recall the concept of unitally non-distributive quantales.

A unital quantale Q is called unitally non-distributive if it satisfies the following properties:

1. (1) the unit e is $\triangleleft $ -approximable.

2. (2) the unit e is non-distributive, that is, there exists a subset A of Q such that $e\wedge (\bigvee A)\not \leq \bigvee _{a\in A}(e\wedge a)$ .

Note that $\mathbf {K_8}$ is a unitally non-distributive quantale and the unit f is strictly non-distributive, that is, there exist $\alpha , \beta \in Q$ such that $f\wedge (\alpha \vee \beta )\not \leq (f\wedge \alpha )\vee (f\wedge \beta )$ and $f\not \leq \alpha \vee \beta $ . In $\mathbf {K_6}$ , the unit e is $\triangleleft $ -approximable, but it is not non-distributive.

Proof By Theorem 2.4, we have that $(Q\times K, \&_\ast )$ is a commutative unital quantale. Let $(M, \cdot , \otimes )$ be a Q-algebra. Then it follows from Lemma 2.3 that $(M, \ast , \otimes )$ is a unital K-algebra with unital module operator. We define a map $\odot \colon M\times (Q\times K)\rightarrow M$ as follows

$$ \begin{align*}\forall m\in M,\ (q, k)\in Q\times K,\ m\odot(q, k):=(m\cdot q)\vee (m\ast k).\end{align*} $$

Let $m, n, m_i\in M,\ (q, k), (p, k^\prime ), (q_i, k_i)\in Q\times K$ . Then by Lemma 2.3, we have

$$ \begin{align*}\begin{array}{l@{\;}l@{\;}l}(\bigvee_i m_i)\odot (q, k)&=& ((\bigvee_i m_i)\cdot q)\vee((\bigvee_i m_i)\ast k)\\ &=& (\bigvee_i m_i\cdot q)\vee(\bigvee_i m_i\ast k)\\ &=& \bigvee_i (m_i\cdot q)\vee(m_i\ast k)\\ &=& \bigvee_i m_i\odot (q, k), \end{array}\end{align*} $$

$$ \begin{align*}\begin{array}{l@{\;}l@{\;}l}m\odot (\bigvee_i(q_i, k_i))&=& m\odot (\bigvee_i q_i, \bigvee_i k_i)\\ &=& (m\cdot (\bigvee_i q_i))\vee(m\ast(\bigvee_i k_i))\\ &=& (\bigvee_i (m\cdot q_i)\vee(m\ast k_i))\\ &=& \bigvee_i m\odot( q_i, k_i), \end{array}\end{align*} $$

$$ \begin{align*}\begin{array}{l@{\;}l@{\;}l}(m\otimes n)\odot(q, k)&=& ((m\otimes n)\cdot q)\vee((m\otimes n)\ast k)\\ &=& (m\otimes(n\cdot q))\vee (m\otimes(n\ast k))\\ &=& m\otimes((n\cdot q)\vee(n\ast k))\\ &=& m\otimes(n\odot(q, k))\\ &=& (m\otimes(n\cdot q))\vee (m\otimes(n\ast k))\\ &=& ((m\cdot q)\otimes n)\vee((m\ast k)\otimes n)\\ &=& (m\odot(q, k))\otimes n, \end{array}\end{align*} $$

$$ \begin{align*}\!\!\!\!\begin{array}{l@{\;}l@{\;}l}m\odot((q, k)\&_\ast (p, k^\prime)&=& m\odot((q\& p)\vee(p\ast k)\vee(q\ast k^\prime), k\&k^\prime)\\ &=& (m\cdot((q\& p)\vee(p\ast k)\vee(q\ast k^\prime)))\vee(m\ast(k\&k^\prime))\\ &=& (m\cdot(q\& p))\vee(m\cdot(p\ast k))\vee(m\cdot(q\ast k^\prime))\vee (m\ast(k\&k^\prime))\\ &=& ((m\cdot q)\cdot p)\vee((m\ast k)\cdot p)\vee((m\cdot q)\ast k^\prime)\vee((m\ast k)\ast k^\prime)\\ &=& (((m\cdot q)\vee(m\ast k))\cdot p)\vee(((m\cdot q)\vee(m\ast k))\ast k^\prime)\\ &=& ((m\cdot q)\vee(m\ast k))\odot(p, k^\prime)\\ &=& (m\odot(q, k))\odot(p, k^\prime), \end{array}\end{align*} $$

and $m\odot (0, e)=m$ where e is the unit of K.

By the above description, we have that $(M, \odot , \otimes )$ is a unital $(Q\times K)$ -algebra.

Let Q be a commutative quantale and K be a commutative unital quantale without zero-factors. Then for a given Q-algebra $(M, \cdot , \otimes )$ , using the following methods we obtain two unital quantales $((M\times Q)\times K, \&_\ast )$ and $(M\times (Q\times K), \&_\odot )$ .

Proof We define a map $f\colon (M\times Q)\times K\rightarrow M\times (Q\times K)$ as follows

$$ \begin{align*}\forall ((m, q), k)\in (M\times Q)\times K,\ \ f((m, q), k)=(m, (q, k)).\end{align*} $$

It is easy to show that f is an isomorphism of quantales, that is, $((M\times Q)\times K, \&_\ast )\cong (M\times (Q\times K), \&_\odot )$ .

2.2 Constructing unital quantales by upper sets

In order to provide a unified semantics for a wide class of substructural logics, Rump and Yang introduced the concept of quantum B-algebras (see [Reference Rump23, Reference Rump and Yang24]). Quantum B-algebras can be regarded as implicational subreducts of quantales. A quantum B-algebra is a poset X with two binary operations $\rightarrow $ and $\rightsquigarrow $ satisfying the following three conditions

$$ \begin{align*}y\leq z\Longrightarrow x\rightarrow y\leq x\rightarrow z\end{align*} $$

$$ \begin{align*}x\leq y\rightarrow z\Longleftrightarrow y\leq x\rightsquigarrow z\end{align*} $$

$$ \begin{align*}x\rightsquigarrow(y\rightarrow z) = y\rightarrow (x\rightsquigarrow z)\end{align*} $$

for all $x,y,z\in X$ . A quantum B-algebra X is called unital if there exists an element $u\in X$ such that $u\rightarrow x =x= u \rightsquigarrow x$ for all $x \in X$ .

Note that by (1.1) we see that a quantale is a quantum B-algebra. Rump and Yang used upper sets to build a relation between quantum B-algebras and quantales. Let X be a quantum B-algebra, $U(X)$ denote the set of all upper sets of X. Then $U(X)$ is a complete lattice with respect to set-theoretic union. For $A,B\in U(X)$ , we define a semigroup multiplication $\odot $ on $U(X)$ as follows

(2.2) $$ \begin{align}A\odot B:=\{x\in X : \exists b\in B, b\rightarrow x\in A\}.\end{align} $$

Note that $A\odot B =\{x\in X : \exists a\in A, a\rightsquigarrow x\in B\}=\{x\in X : \exists a\in A, b\in B, s.t.\ a\leq b\rightarrow x\}= \{x\in X : \exists a\in A, b\in B, s.t.\ b\leq a\rightsquigarrow x\}$ .

By upper-set quantales, Rump and Yang proved that the category of quantum B-algebras and the category of logical quantales are dually equivalent (see [Reference Rump and Yang24]). In [Reference Han, Xu and Qin10], Han, Xu, and Qin gave a sufficient and necessary condition for the upper-set quantale $U(X)$ to be a unital quantale. A subset E of a quantum B-algebra X is called positive provided that E is an upper set of X, and $x\rightsquigarrow y\in E\Longleftrightarrow x\leq y \Longleftrightarrow x\rightarrow y\in E$ for all $x,y\in X$ . Note that if a quantum B-algebra has a positive subset, then it is unique (see [Reference Han and Bao9, Proposition 3.6]). A quantum B-algebra is called positive if it has a positive subset. Clearly, a unital quantum B-algebra is positive, but a positive quantum B-algebra is in general not unital (see [Reference Han, Xu and Qin10]).

Example 2.15 Let $Q=\{0, a, b, 1\}$ be a complete lattice determined by Figure 3 and $\&$ be a semigroup multiplication on Q determined by the table below.

Figure 3 The partial order on Q.

It is easy to verify that $(Q, \&)$ is a non-unital quantale, and $\{a, b, 1\}$ is a positive subset of Q. Thus, by Corollary 2.13 we have that $(U(Q), \odot )$ is a unital quantale.

When quantales are considered as quantum B-algebras, we can use upper sets to construct upper-set quantales. Further, if quantales are positive, then the corresponding upper-set quantales are unital. When quantales are considered as partially ordered semigroups (posemigroups for short), we can use lower sets to construct lower-set quantales (see [Reference Han and Zhao13, Reference Kruml and Paseka15]). A natural question is under what condition the lower-set quantale $L(Q)$ of a quantale Q is unital.

Let $(S, \cdot , \leq )$ be a posemigroup, $L(S)$ denote the set of all lower sets of S. Then $L(S)$ is a complete lattice with respect to set-theoretic union. We now define a semigroup multiplication $\otimes $ on $L(S)$ as follows

(2.3) $$ \begin{align}A\otimes B =\{ x\in S\ \colon \exists a\in A, b\in B,\ x\leq a\cdot b\}.\end{align} $$

A subset K of a posemigroup $(S, \cdot , \leq )$ is called normal provided that K satisfies the following three conditions:

1. (1) K is a lower set of S.

2. (2) $k\cdot q\leq q$ and $q\cdot k\leq q$ for all $q\in S, k\in K$ .

3. (3) For any $q\in S$ , there exist $k, k^\prime \in K$ such that $q=k\cdot q$ and $q=q\cdot k^\prime $ .

Note that if a posemigroup has a normal subset, then it is unique. To see this, we let K and $K^\prime $ be two normal subsets. Then for $k\in K$ there exists $k^\prime \in K^\prime $ such that $k=k^\prime \cdot k\leq k^\prime \in K^\prime $ , which implies that $k\in K^\prime $ , that is, $K\subseteq K^\prime $ . Similarly, we have $K^\prime \subseteq K$ . Thus, $K=K^\prime $ . A posemigroup is called normal if it has a normal subset. A unital posemigroup is normal, but the converse is in general not true (see Example 2.18).

Example 2.18 Let $S=\{0, a, b, c, d\}$ be a poset determined by Figure 4 and $\&$ be a semigroup multiplication on S determined by the table below.

Figure 4 The partial order on S.

It is easy to verify that $(S, \cdot , \leq )$ is a non-unital posemigroup, and $\{0, a, b, c\}$ is a normal subset of S. Thus, by Proposition 2.17 we have that $L(S), \otimes )$ is a unital quantale.

If quantales are considered as posemigroups, then we have the following interesting result.

2.3 Constructing unital quantales by adding elements

For some special quantales, Guriérrez García and Höhle investigated the constructions of unital quantales by addition of an isolated element (see [Reference Gutiérrez García and Höhle8]). Next, based on the work of Guriérrez García and Höhle, we shall continue the research on the construction of unital quantales.

2.3.1 Adding two elements to construct unital quantales

Let L be a complete lattice, $\alpha \in L\backslash \{0, 1\}$ . Then $\alpha $ is called isolated (see [Reference Gutiérrez García and Höhle8]) in L if there exist two elements $\alpha ^-\leq \alpha $ and $\alpha ^+\geq \alpha $ of L such that the following properties hold:

$$ \begin{align*}(\downarrow\! \alpha)\backslash\{\alpha\}=\downarrow\!\alpha^-\ \mbox{and}\ (\uparrow\!\alpha)\backslash\{\alpha\}=\uparrow\!\alpha^+.\end{align*} $$

Lemma 2.20 $($ Guriérrez García and Höhle [Reference Gutiérrez García and Höhle8] $)$ Let L be a complete lattice, and $\top $ and e be two different elements satisfying the condition $L\cap \{\top , e\}=\emptyset $ . Then for every $\gamma \in L$ there exists a unique complete lattice-structure on $L^\gamma =L\cup \{\top , e\}$ satisfying the following conditions:

1. (1) $L^\gamma $ is an extension of L.

2. (2) The element $\top $ is the universal upper bound of $L^\gamma $ .

3. (3) The element e is isolated and satisfies the conditions $e^-=\gamma $ and $e^+=\top $ .

Theorem 2.21 $($ Guriérrez García and Höhle [Reference Gutiérrez García and Höhle8] $)$ Let $(Q, \&)$ be a quantale, $\gamma \in Q\backslash \{1\}$ and let $Q^\gamma =Q\cup \{\top , e\}$ be the extension of the underlying lattice by an isolated element e in the sense of Lemma 2.20. Then there exists a unique quantale structure on $Q^\gamma $ with unit e and subquantale Q if and only if Q satisfies the following conditions for all $\alpha \in Q$ and $\beta \in Q$ with $\beta \not \leq \gamma $ :

(2.4) $$ \begin{align}(\gamma\&\alpha)\vee(\alpha\&\gamma)\leq \alpha,\end{align} $$

(2.5) $$ \begin{align}1\&\alpha\leq(\beta\&\alpha)\vee\alpha\ \mbox{and}\ \alpha\&1\leq(\alpha\&\beta)\vee\alpha.\end{align} $$

Remark 2.22

1. (1) If a quantale $(Q, \&)$ has an element $\gamma $ satisfying (2.4) and (2.5), then Q can be extended to a unital quantale $Q^\gamma $ with unit e where the semigroup multiplication $\&^\gamma $ on $Q^\gamma $ is determined by

   $$ \begin{align*}\forall x,y\in Q^\gamma,\ x\&^\gamma y =\begin{cases} x\& y, & \mbox{if}\ \ x,y\in Q,\\ y, & \mbox{if}\ \ x=e,\\ x, & \mbox{if}\ \ y=e,\\ (1\&y)\vee y, & \mbox{if}\ \ x=\top, y\in Q,\\ x\vee(x\&1), & \mbox{if}\ \ y=\top, x\in Q,\\ \top, & \mbox{if}\ \ x=y=\top.\ \ \end{cases} \end{align*} $$

2. (2) Since every non-top element $\gamma $ of a two-sided quantale satisfies the properties (2.4) and (2.5), every two-sided quantale Q can be extended to a unital quantale  $Q^\gamma $ .

Using a method similar to adding isolated elements, we shall give a new method to construct unital quantales.

Lemma 2.23 Let L be a complete lattice, and $\bot $ and e be two different elements satisfying the condition $L\cap \{\bot , e\}=\emptyset $ . Then for every $\gamma \in L$ there exists a unique complete lattice-structure on $L_\gamma =L\cup \{\bot , e\}$ satisfying the following conditions:

1. (1) $L_\gamma $ is an extension of L.

2. (2) The element $\bot $ is the universal lower bound of $L_\gamma $ .

3. (3) The element e is isolated and satisfies the conditions $e^+=\gamma $ and $e^-=\bot $ .

Let $(Q, \&)$ be a quantale and $\gamma $ be a non-zero element of Q. Define a semigroup multiplication $\&_\gamma $ as follows

$$ \begin{align*}\forall x,y\in Q_\gamma,\ x\&_\gamma y =\begin{cases} x\& y, & \mbox{if}\ \ x,y\in Q,\\ y, & \mbox{if}\ \ x=e,\\ x, & \mbox{if}\ \ y=e,\\ \bot, & \mbox{if}\ \ x=\bot\ \mbox{or}\ y=\bot.\ \ \end{cases} \end{align*} $$

Theorem 2.24 Let $\gamma $ be a non-zero element of a quantale Q and let $Q_\gamma =Q\cup \{\bot , e\}$ be the extension of the underlying lattice by an isolated element e in the sense of Lemma 2.23. Then $(Q_\gamma , \&_\gamma )$ is a unital quantale with unit e if and only if $\gamma $ is the unit of Q.

Proof The proof is similar to that of Theorem 2.21.

2.3.2 Adding one element to construct unital quantales

Inspired by the work of Guriérrez García and Höhle, we will consider the method of adding one element to construct unital quantales. Let r be a non-zero element of quantale Q and $P_r(Q)\stackrel {\triangle }=\{y\in Q \mid (r\&y)\vee (x\&y)= y\vee (x\&y)\ \mbox {and} \ (y\&r)\vee (y\&x)=y\vee (y\&x)\ \mbox {for all}\ x\in Q \backslash \{0\}\}$ .

Proposition 2.25 Let Q be a quantale, $r\in Q\backslash \{0\}$ . Then we have

1. (1) $P_r(Q)$ is a subquantale of Q.

2. (2) $1\in P_r(Q) \Longleftrightarrow r\&1=1=1\&r$ .

3. (3) if Q is a unital quantale with unit u, then $u\in P_r(Q) \Longleftrightarrow u=r$ .

4. (4) if Q is a unital quantale with unit u, then $P_u(Q)=Q$ .

Proof Proof is straightforward.

Definition 2.26 A non-zero element r of quantale Q is called a weak unit if $P_r(Q)=Q$ .

Remark 2.27

1. (1) If Q is a quantale with $\&=\&_1$ defined by (1.2), then every non-zero element of Q is a weak unit.

2. (2) If Q is a quantale with $\&=\&_a$ and $a\neq 1$ defined by (1.2), then no element of Q is a weak unit.

3. (3) If Q is a unital quantale, then by Proposition 2.25(3) we see that the unit is the unique weak unit in Q. Conversely, if a quantale Q has a unique weak unit, then Q is not necessarily a unital quantale (see Example 2.28).

Example 2.28 Let $Q=\{0, a, b, 1\}$ be a complete lattice with $0< a< b< 1$ and $\&$ be a semigroup multiplication on Q determined by the table below.

It is easy to verify that Q is a non-unital quantale and there is a unique weak unit 1 in Q.

Let r be a non-zero element of quantale Q and $e\not \in Q$ . We denote by $\overline {Q_r}$ the set $Q\cup \{e\}$ and define a partial order $\leq _r$ on $\overline {Q_r}$ and a semigroup multiplication $\&_e$ on $\overline {Q_r}$ as follows

$$ \begin{align*}\leq_r=\leq \cup\{(e, x) \mid r\leq x, x\in Q\}\cup\{(e, e), (0,e)\}\end{align*} $$

and

(2.6) $$ \begin{align}\forall x,y\in \overline{Q_r},\ \ x\&_e y=\begin{cases} x\&y, & \ \mbox{if}\ \ x,y\in Q,\\ y, & \ \mbox{if}\ \ x=e,\\ x, & \ \mbox{if}\ \ y=e.\end{cases}\end{align} $$

Proposition 2.29 Let Q be a quantale and r be a non-zero element of Q. Then we have

1. (1) $(\overline {Q_r}, \leq _r)$ is a complete lattice and $e\vee x=r\vee x$ for all $x\in Q\backslash \{0\}$ .

2. (2) e is an isolated element in $\overline {Q_r}$ and $e^+=r, e^-=0$ .

3. (3) $(\overline {Q_r}, \leq _r, \&_e)$ is a unital posemigroup if and only if $x\leq (x\&r)\wedge (r\&x)$ for all $x\in Q$ .

Proof Proof is straightforward.

Proposition 2.30 Let Q be a quantale and r be a non-zero element of Q. Then $(\overline {Q_r}, \leq _r, \&_e)$ is a unital quantale if and only if r is a weak unit of Q.

Proof Let $(\overline {Q_r}, \leq _r, \&_e)$ be a unital quantale with unit e. Then by Proposition 2.29 we have that $(r\&y)\vee (x\&y)=(r\vee x)\& y=(e\vee x)\&_e y=(e\&_e y)\vee (x\&_e y)=y\vee (x\&y)$ for all $y\in Q, x\in Q\backslash \{0\}$ . Similarly, we have $(y\& r)\vee (y\&x)=y\vee (y\&x)$ . Thus, r is a weak unit of Q.

Conversely, we let r be a weak unit of Q. By Proposition 2.29, it is easy to check that the semigroup multiplication $\&_e$ distributes over arbitrary joins, that is, $(\overline {Q_r}, \leq _r, \&_e)$ is a unital quantale.

Corollary 2.31 If quantale Q has a weak unit r, then Q can be embedded into a unital quantale $\overline {Q_r}$ .

Example 2.32 $($ Guriérrez García and Höhle [Reference Gutiérrez García and Höhle8] $)$ Let $(G, \cdot , e)$ be a group with ${\mid G\mid \geq 2}$ . We provide G with the discrete order $"=\text{"}$ and subsequently we apply the MacNeille completion. This construction leads to a complete lattice $G_\infty =G\cup \{\bot , \top \}$ by adding the universal bounds to the discretely ordered set G. The binary operation $\ast $ on $G_\infty $ is determined by

$$ \begin{align*}\forall x,y\in Q_\infty,\ \ x\ast y=\begin{cases} x\cdot y, & \ \mbox{if}\ \ x,y\in G,\\ \bot, & \ \mbox{if}\ \ x=\bot\ \mbox{or}\ y=\bot,\\ \top, & \ \mbox{otherwise}.\end{cases}\end{align*} $$

It is easy to check that $(G_\infty , \ast )$ is a unital quantale and the unit e is an isolated element of $G_\infty $ with $e^+=\top , e^-=\bot $ .

Proposition 2.33 Let Q be a unital quantale such that the unit e is an isolated element and $e^-=0$ . Then $Q\cong \overline {K_r}$ for some quantale K with weak unit r if and only if $Q\backslash \{e\}$ is a subquantale of Q.

Proof Let K be a quantale with weak unit r. Then by Proposition 2.30 we have that $\overline {K_r}$ is a unital quantale with unit e and K is a subquantale of $\overline {K_r}$ . Since $Q\cong \overline {K_r}$ , we have that $Q\backslash \{e\}\cong K$ , which implies that $Q\backslash \{e\}$ is a subquantale of Q.

Conversely, we let $r=e^+$ and $K=Q\backslash \{e\}$ . Then by Proposition 2.29 we have that r is a weak unit of K. Further, we see that $Q=\overline {K_r}$ .

Corollary 2.34 Let Q be a unital quantale such that the unit e is an isolated element and $e^-=0$ . Then $Q\cong \overline {K_r}$ for some quantale K with weak unit r if and only if $x\&y=e$ implies that $x=e=y$ for all $x,y\in Q$ .

Remark 2.35

1. (1) Let $(G_\infty , \ast )$ be the unital quantale in Example 2.32. Then $(G_\infty , \ast )$ is not of form $\overline {K_r}$ , that is, $G_\infty \not \cong \overline {K_r}$ .

2. (2) Let $Q=\{0, a, 1\}$ be a quantale with $\&=\&_1$ . Then $1$ is a weak unit of Q. By Proposition 2.30, we see that $\overline {Q_1}$ is a unital quantale determined by Figure 5 and the following table.

   

   Figure 5 The partial order on $\overline {Q_1}$ .

Similarly, we can give another method of adding one element to construct unital quantales. Let Q be a quantale with $e\not \in Q$ . For a non-top element r of Q, we denote by $\overline {Q^r}$ the set $Q\cup \{e\}$ and define a partial order $\leq _r$ on $\overline {Q^r}$ and a binary operation $\&_e$ on $\overline {Q^r}$ as (2.6)

$$ \begin{align*}\leq_r=\leq \cup\{(x, e) \mid x\leq r, x\in Q\}\cup\{(e, e), (e,1)\}.\end{align*} $$

Lemma 2.36 Let Q be a quantale and r be a non-top element of Q. Then we have

1. (1) $(\overline {Q^r}, \leq _r)$ is a complete lattice and $e\vee x=1$ for all $x\in Q$ with $x\not \leq r$ .

2. (2) e is an isolated element in $\overline {Q^r}$ and $e^+=1, e^-=r$ .

3. (3) $(\overline {Q^r}, \leq _r, \&_e)$ is a posemigroup with unit e if and only if $(x\&r)\vee (r\&x)\leq x\leq (x\&1)\wedge (1\&x)$ for all $x\in Q$ .

Theorem 2.37 Let $(Q, \&)$ be a quantale with non-top element r and let $\overline {Q^r}$ be the extension of the underlying lattice in the sense of Lemma 2.36. Then there exists a unique quantale structure on $\overline {Q^r}$ with unit e and subquantale Q if and only if Q satisfies the following conditions for all $x\in Q$ and $y\in Q$ with $y\not \leq r$ :

$$ \begin{align*}(r\&x)\vee(x\&r)\leq x,\end{align*} $$

$$ \begin{align*}1\&x=(y\&x)\vee x\ \mbox{and}\ x\&1=(x\&y)\vee x.\end{align*} $$

Proof The proof is similar to that of Theorem 2.21.

Let Q be a quantale with $\top \not \in Q$ and $\widehat {Q}_{\top }$ denote the set $Q\cup \{\top \}$ . We define a partial order $\leq _{\top }$ on $\widehat {Q}_{\top }$ and a semigroup multiplication $\&_{\top }$ on $\widehat {Q}_{\top }$ as follows

$$ \begin{align*}\leq_{\top}=\leq \cup\{(x, \top)\ \colon x\in \widehat{Q}_\top\}\end{align*} $$

and

$$ \begin{align*}\forall x,y\in \widehat{Q}_\top,\ \ x\&_\top y=\begin{cases} x\&y, & \ \mbox{if}\ \ x,y\in Q,\\ y, & \ \mbox{if}\ \ x=\top,\\ x, & \ \mbox{if}\ \ y=\top.\end{cases}\end{align*} $$

Proposition 2.38 Let $(Q, \&)$ be a quantale. Then $(\widehat {Q}_\top , \&_\top )$ is a unital quantale if and only if Q is a two-sided quantale.