Beyond left-continuity: t-norms on the fuzzy unit interval induced by pairs of t-norms

Introduction

Triangular norms, or t-norms, are associative, commutative, non-decreasing binary operations on the real unit interval $[0,1]$ with neutral element $1$. They were introduced in 1963 by Schweizer and Sklar [17, 18] to formulate the triangle inequality in probabilistic metric spaces. In fuzzy set theory, t-norms have been extensively used to model fuzzy intersections [1, 16] and conjunction operators in fuzzy logic [6, 11].

In 1967, Goguen introduced $L$-fuzzy sets [8], where the codomain of a fuzzy set is a partially ordered set (poset) $L$. Noting that the definition of a t-norm on $[0,1]$ can be naturally extended to any bounded poset, De Cooman and Kerre generalized the concept of t-norms to bounded posets [5], thereby providing fuzzy intersection operators for Goguen’s $L$-fuzzy sets.

Following Goguen’s general framework, many different types of posets $L$ have been employed as codomains of fuzzy sets. Among these, an important class consists of posets into which the unit interval $[0,1]$ can be embedded as a subset, because such posets allow a more precise description of concepts with subtle fuzzy features. A prominent example is Zadeh’s type-2 fuzzy sets [20, 21], where the codomain is taken to be the set $\mathbb{M} = [0,1]^{[0,1]}$ of all fuzzy subsets of $[0,1]$. To extend the usual order on $[0,1]$ to $\mathbb{M}$, two compatible partial orders $\sqsubseteq_{\sqcap}$ and $\sqsubseteq_{\sqcup}$ were introduced on $\mathbb{M}$, yielding the posets $(\mathbb{M},\sqsubseteq_{\sqcap})$ and $(\mathbb{M},\sqsubseteq_{\sqcup})$ [12]. Also in [12], the authors lifted t-norms on $[0,1]$ to binary operations on these two posets by using the meet operation $\wedge$ on the codomain $[0,1]$.

In fuzzy topology, Hutton introduced the fuzzy unit interval $\mathbb{I}(L)$ in 1975 as a lattice-theoretic extension of the real unit interval, where $L$ is a complete lattice [13]. In the special case $L = [0,1]$, the fuzzy unit interval can be characterized equivalently as the collection of all right-continuous non-increasing functions $f : [0,1] \longrightarrow[0,1]$ satisfying $f(1) = 0$, which forms a subset of $\mathbb{M}$. We denote this particular fuzzy unit interval by $\mathbb{I}$. Endowed with the pointwise order, $\mathbb{I}$ is a subposet of $\mathbb{M}$.

In fact, $\mathbb{I}$ enjoys even stronger structural properties. The unit interval $[0,1]$ embeds into $\mathbb{I}$ as a complete sublattice, and $\mathbb{I}$ is precisely the tensor product of $[0,1]$ with itself in the category SUP, whose objects are complete lattices and whose morphisms are supremum-preserving maps. Consequently, t-norms on $\mathbb{I}$ can be constructed by forming the tensor product of two quantales with the same underlying set $[0,1]$ [7, 9, 10]. Recall that $([0,1],T)$ is a quantale if and only if $T$ is left-continuous. Thus, the quantale tensor product construction of a t-norm on $\mathbb{I}$ requires both input t-norms to be left-continuous. This observation leads to a natural question: given two arbitrary t-norms $T_1$ and $T_2$ on $[0,1]$ that are not necessarily left-continuous, can one still construct a t-norm on $\mathbb{I}$ from them?

In this paper, we give sufficient conditions under which this construction yields a t-norm. We present a general method for constructing a binary operation $T_1 \star T_2$ on $\mathbb{I}$ starting from two arbitrary t-norms $T_1$ and $T_2$ on $[0,1]$. When both t-norms are left-continuous, this operation coincides with the quantale tensor product construction already known in the literature; the existence of the resulting unital quantale follows from [9 Theorem 4.2]. Our proof in this case is a direct verification for the unit interval, rather than a new tensor-product construction. We then prove that $T_1 \star T_2$ is a t-norm on $\mathbb{I}$ whenever at least one of $T_1$, $T_2$ is continuous. We also provide examples illustrating two complementary phenomena: on the one hand, if neither component t-norm is continuous, $T_1 \star T_2$ need not be associative and may therefore fail to be a t-norm on $\mathbb{I}$; on the other hand, although the drastic product t-norm $T_D$ is neither continuous nor left-continuous, $T_D \star T_D$ is still a t-norm on $\mathbb{I}$.

The main associativity theorem goes beyond the case of two left-continuous components because the second component may fail left-continuity when the first is continuous. The pair $(T_D,T_D)$ shows that both components may fail left-continuity, whereas $(T_D,T_{\mathrm{nM}})$ shows that left-continuity of only one component does not suffice. These results provide sufficient conditions and contrasting examples, not a classification of all admissible pairs.

The paper is organized as follows. Section 2 recalls basic notions concerning fuzzy unit intervals and t-norms on $[0,1]$. In Section 3 we introduce the operation $T_1 \star T_2$, establish its fundamental properties, and discuss its Archimedean-type properties. Section 4 is devoted to proving our main result: if either $T_1$ or $T_2$ is continuous, then $T_1 \star T_2$ is a t-norm on $\mathbb{I}$; two important examples are also discussed there. Section 5 concludes the paper and outlines directions for future work.

Preliminaries

Unless explicitly stated otherwise, $(P,\leq)$ denotes a bounded partially ordered set. Its bottom and top elements are denoted by $\bot$ and $\top$, respectively. We always assume that $\bot\ne\top$.

Equipped with the usual order of real numbers, the unit interval $[0,1]$ is a bounded partially ordered set with $\bot=0$ and $\top=1$. It is also a complete lattice. For a subset $A \subseteq[0,1]$, we write $\sup A$ and $\inf A$ for the least upper bound and greatest lower bound of $A$, respectively. In particular, $\sup\varnothing=0$ and $\inf\varnothing=1$.

Let $f:[0,1]\longrightarrow[0,1]$ be a non-increasing function. Define

for all $t\in[0,1]$. In particular,

Hutton’s fuzzy unit interval [13] is defined by taking a general complete lattice $L$ as the codomain of the fuzzy sets. In the present paper we restrict to the case $L=[0,1]$.

The quotient set $\mathbb{M}^{R}/\sim$ is called the fuzzy unit interval.

We denote by $\mathbb{I}$ the collection of all right-continuous non-increasing maps $f:[0,1]\longrightarrow[0,1]$ satisfying $f(1)=0$. Then a map $f:[0,1]\longrightarrow[0,1]$ belongs to $\mathbb{I}$ if and only if $f(\inf A)=\sup f(A)$ for all $A\subseteq[0,1]$.

Proof. Given an element $\lambda\in\mathbb{M}^{R}$, let $\bar{\lambda}(t)=\lambda(t^{+})$ for all $t \in[0,1]$. Then $\bar{\lambda} \in\mathbb{I}$. We first show that the map $[\lambda]\mapsto\bar{\lambda}$ is well-defined from $\mathbb{M}^{R}/\sim$ to $\mathbb{I}$. By the definition of the equivalence relation $\sim$ on $\mathbb{M}^{R}$, if $\lambda\sim\mu$, then clearly $\bar{\lambda}=\bar{\mu}$. Thus, the assignment is independent of the choice of representatives.

To prove injectivity, assume $[\lambda]\ne[\mu]$. By the definition of $\sim$, either $\lambda(t_{0}^{+})\ne\mu(t_{0}^{+})$ or $\lambda(t_{0}^{-})\ne\mu(t_{0}^{-})$ for some $t_{0}\in[0,1]$. We consider two cases:

1. If $\lambda(t_{0}^{+})\ne\mu(t_{0}^{+})$ for some $t_{0}$, then clearly $\bar{\lambda}\ne\bar{\mu}$.

2. If $\lambda(t_{0}^{-})\ne\mu(t_{0}^{-})$ for some $t_{0}$, without loss of generality assume $\lambda(t_{0}^{-})<\mu(t_{0}^{-})$. Choose $\delta>0$ such that for all $t\in(t_{0}-\delta,t_{0})$,

For all $t\in(t_{0}-\delta,t_{0})$, since $\mu(t)$ is greater than the constant $\frac{\lambda(t_{0}^{-})+\mu(t_{0}^{-})}{2}$, we have $\frac{\lambda(t_{0}^{-})+\mu(t_{0}^{-})}{2}\le\mu(t^{+})$. Consequently, for all $t\in(t_{0}-\delta,t_{0})$,

which implies $\bar{\lambda}\ne\bar{\mu}$.

Thus, the map is injective.

For surjectivity, let $\lambda\in\mathbb{I}$ be arbitrary. Define $\widetilde{\lambda}\in\mathbb{M}^{R}$ by

Then $\bar{\widetilde{\lambda}}=\lambda$, showing that the map is also surjective.

Therefore, the assignment $[\lambda]\mapsto\bar{\lambda}$ is a bijection as desired.

Proof. Fix $y$ and put $A_y=\{x:f(x)\le y\}$. Since $f$ takes arbitrary infima to suprema, $f(\inf A_y)=\sup f(A_y)\le y$. Thus $A_y=[f^{*}(y),1]$, proving the equivalence and uniqueness. For every $B\subseteq[0,1]$ and $x\in[0,1]$, this equivalence gives

Consequently $f^{*}(\inf B)=\sup_{b\in B}f^{*}(b)$, including $B=\varnothing$, and therefore $f^{*}\in\mathbb{I}$. The symmetry of the adjunction equivalence gives $(f^{*})^{*}=f$. If $f\le g$, then $\{x:g(x)\le y\}\subseteq\{x:f(x)\le y\}$, so $f^{*}\le g^{*}$. Applying this implication to $f^{*},g^{*}$ proves the converse.

Thus $f\mapsto f^{*}$ is an order-preserving involution, hence a complete lattice automorphism.

Step functions play a fundamental role in the structure of $\mathbb{I}$. For $r,a\in[0,1]$, define the step function $\langle r,a\rangle:[0,1]\longrightarrow[0,1]$ by

Equipped with the pointwise order, $\mathbb{I}$ is a complete lattice whose top and bottom elements are $\langle1,1\rangle$ and $\langle0,0\rangle$, respectively. Suprema are calculated pointwise: for every family $(f_i)_{i\in J}\subseteq\mathbb{I}$,

Indeed, the right-hand side is non-increasing, vanishes at $1$, and satisfies, for $x<1$,

It therefore belongs to $\mathbb{I}$ and is the least upper bound. This also covers the empty family. Let

The following lemma shows that $\mathcal{E}$ is sup-dense in $\mathbb{I}$.

Proof. (1) We have

(2) First, for every $x\in[0,1]$, we have

Hence,

Second, notice that $f(r)\leq f(r^-)$ since $f$ is non-increasing. By (1), it holds that

Hence,

so $f=\sup_{r\in(0,1]}\langle r,f(r^-)\rangle$. By the definition of step function,

Thus, it holds that

There are some important basic t-norms on the interval $[0,1]$; see, e.g., [2, 14]. We list them below:

• minimum: $T_M(x,y)=\min\{x,y\}$,

• product: $T_P(x,y)=x\cdot y$,

• Łukasiewicz t-norm: $T_L(x,y)=\max\{x+y-1,0\}$,

• drastic product: $T_D(x,y)=\begin{cases}\min\{x,y\}, & \max\{x,y\}=1,\\ 0, & \max\{x,y\}<1,\end{cases}$

• the nilpotent minimum t-norm: $T^{\mathrm{nM}}(x,y)=\begin{cases}\min\{x,y\}, & x+y>1,\\ 0, & x+y\le1.\end{cases}$

The t-norms $T_M$, $T_P$ and $T_L$ are continuous. The nilpotent minimum t-norm $T^{\mathrm{nM}}$ is left-continuous but not continuous, whereas the drastic product $T_D$ is not left-continuous.

Operations on the fuzzy unit interval induced by pairs of t-norms

Given t-norms $T_1$ and $T_2$ on $[0,1]$, we obtain a binary operation

This prescription is independent of the representation of a step function. Indeed, every nonzero step has unique parameters $r,a>0$, whereas $\langle r,a\rangle=\bot$ precisely when $r=0$ or $a=0$. Since every t-norm satisfies $T(0,t)=0$, a zero input always gives the zero step, independently of its parameters. For nonzero steps, $\langle r,a\rangle\le\langle s,b\rangle$ is equivalent to $r\le s$ and $a\le b$. Consequently the operation on $\mathcal{E}$ is monotone. Its commutativity, associativity, and neutral element $\langle1,1\rangle$ follow from the corresponding properties of $T_1,T_2$. Since $\mathcal{E}$ is a sup-dense subset of $\mathcal{I}$, the operation $T_1\star T_2$ can be naturally extended to $\mathcal{I}$ as follows:

By monotonicity, when both inputs are steps, every term in the supremum is bounded by their product in (3), and that product itself is an admissible term. Thus the extension agrees with (3) on $\mathcal{E}$.

Proof. (1) Since both $T_1$ and $T_2$ are commutative, the commutativity of $T_1\star T_2$ follows directly from its definition.

(2) Assume that $f_1\le f_2$. For each step function $\langle r,a\rangle\in\mathcal{E}$, if $\langle r,a\rangle\le f_1$ then $\langle r,a\rangle\le f_2$. Therefore, it follows that

by the definition of $T_1\star T_2$ straightforwardly.

(3) Since $\langle1,1\rangle$ is the top in $\mathcal{I}$, we have

The second equality follows from $t,b\le1$, the monotonicity of $T_1,T_2$, and the fact that $t=b=1$ is admissible. The last equality is Lemma 2.5(2).

If both $T_1$ and $T_2$ are left-continuous t-norms, then both $([0,1],T_1)$ and $([0,1],T_2)$ are quantales. Recall from [7, 9, 10] that the monoidal operation $T_1\otimes T_2$ of the tensor product $([0,1],T_1)\otimes([0,1],T_2)$ is given by

Particularly, for each point $x\in[0,1]$,

Proof. On the one hand, by Proposition 3.3, we obtain

On the other hand, for any $p\in[0,1]$, if $p<T_1\star T_2(f,g)(x)$, then there exists a pair $(r,s)$ with $T_1(r,s)>x$ such that $p<T_2(f(r^-),g(s^-))$. Since $T_1$ is left-continuous, there is a pair $(r',s')$ satisfying $r'<r$, $s'<s$ and $x<T_1(r',s')\leq T_1(r,s)$. Thus,

Since $p$ is arbitrary, we conclude that

Therefore, we have shown that $T_1\star T_2=T_1\otimes T_2$, as desired.

:::

(2) Recall from [12] that the convolution of a t-norm $T$ is the operation $\mathcal{T}:\mathbb{M}\times\mathbb{M}\longrightarrow\mathbb{M}$ defined by

Let $T$ be continuous and satisfy

Then $T\star T_M(f,g)=\mathcal{T}(f,g)$ for all $f,g\in\mathbb{I}$. Indeed, Theorem 3.4 gives

For $0\leq x<1$, continuity of $T$ permits lowering a coordinate until $T(r,s)=x$, proving that this supremum is at most $\mathcal{T}(f,g)(x)$. Conversely, any pair with $T(r,s)=x$ and $r,s<1$ can be approached from above by pairs with $T(r_n,s_n)>x$; right-continuity of $f,g$ gives the reverse inequality. Pairs with $r=1$ or $s=1$ contribute zero, and at $x=1$ both expressions vanish. Consequently, the convolution leaves $\mathbb{I}$ invariant and coincides there with $T\star T_M$.

In what follows, we discuss the Archimedean property of $T_1\star T_2$.

Let $T$ be a t-norm on $[0,1]$. For each $x\in[0,1]$, define $x_1=x$ and $x_{n+1}=T(x_n,x)$ for all $n\geq1$. Recall that $T$ is Archimedean if, for all $x,y\in(0,1)$, there exists $n\geq1$ such that $x_n<y$. Equivalently, $\inf_{n\geq1}x_n=0$ for every $x\in(0,1)$; see [14]. The definition of Archimedean property of t-norms on a bounded poset were proposed by De Baets and Mesiar as follows.

We first establish a property of $T_1\star T_2$ concerning finite suprema.

Remark 3.10. When $T_1\star T_2$ is associative, the sequence $\{f_n\}$ in Proposition 3.9 consists of the powers of $f$ with respect to $T_1\star T_2$. It should be pointed out that (4) does not imply the stronger condition

In fact, take $T_1=T_2=T_P$. Then $T_1\star T_2$ is a t-norm by Theorem 3.4, and both $T_1$ and $T_2$ are Archimedean. Let

By the formula for $T_1\star T_2$ on step functions, we have

For every $n\ge1$, taking $x=2^{-n-1}$ gives

Hence $f_n\nleq g$ for all $n\ge1$, even though $\bot<g<f<\top$.

Associativity of the induced operation

This section is devoted to the associativity of the induced operation $T_1\star T_2$. We first prove a general sufficient condition and then examine two complementary examples which show that the condition is meaningful but not necessary.

The proof of the main theorem relies on the following approximation lemma, which allows us to pass from the pointwise formula for $T_1\star T_2$ to the associativity identity.

It follows that

and

Taking the supremum over all admissible triples $(u,v,w)$ yields

On the other hand, we prove the inequality in the opposite direction, i.e.

It is sufficient to show that for any $r,w\in[0,1]$ satisfying $x<T_1(r,w)$, there exist $u_r,v_r$ with $x<T_1(T_1(u_r,v_r),w)$ such that

By the continuity of $T_1$, there are real numbers $s_0,s_1\in[0,1]$ such that

Since $T_1\star T_2(f,g)$ is non-increasing, we have

By Lemma 4.1, there exist $u_1,v_1\in[0,1]$ such that

Now take $u_r=u_1$ and $v_r=v_1$. Then we obtain

and

as desired. Therefore, we have shown that

Second, the equality

can be checked similarly.

Because both $T_1$ and $T_2$ are associative, we obtain the equality

Thus, $T_1\star T_2$ is associative as desired.

Proof. It suffices to verify associativity. If any of $f,g,h \in\mathbb{I}$ equals the neutral element $\langle1,1\rangle$, the identity

follows directly from Proposition 3.1.

Assume now that none of $f,g,h$ is the neutral element. We first claim that

Recall that

and note that $\langle T_D(u,v),T_D(a,b)\rangle\neq\langle0,0\rangle$ iff both $T_D(u,v)\neq0$ and $T_D(a,b)\neq0$.

(i) If $u=1$, then $\langle1,a\rangle\leq f$ implies $a\leq f(1^-)<1$. From $T_D(a,b)\neq0$ we obtain $b=1$. The condition $\langle v,b\rangle\leq g$ implies $\langle b,v\rangle\leq g^*$, and hence $v\leq g^*(b)=g^*(1^-)$. Consequently,

(ii) If $v=1$, a symmetric argument yields

Therefore,

Now, we prove the reverse inequality. By Lemma 2.5(1), $\langle1,f^*(1^-)\rangle\leq f^*$ and $\langle1,g(1^-)\rangle\leq g$. Besides, $\langle1,f^*(1^-)\rangle\leq f^*$ implies $\langle f^*(1^-),1\rangle\leq f$. Hence,

Similarly,

which establishes the claim.

Since $f\neq\langle1,1\rangle$ and $g\neq\langle1,1\rangle$,

Similarly, $f^*\neq\langle1,1\rangle$ and $g^*\neq\langle1,1\rangle$ imply

Then by the above claim, we have

For any $\langle s,t\rangle\leq h<\langle1,1\rangle$, either $s<1$ or $t<1$, thus

Therefore,

Analogously, $T_D\star T_D(f,T_D\star T_D(g,h))=\langle0,0\rangle$. Consequently,

which completes the proof of associativity.

The following example shows that $T_1 \star T_2$ need not be associative when neither component t-norm is continuous.

Conclusion

We have shown that for any two t-norms $T_1,T_2$ on $[0,1]$, the operation $T_1 \star T_2$ defined on the fuzzy unit interval $\mathbb{I}$ is always commutative, monotone and has $\langle1,1\rangle$ as its neutral element. Associativity—and hence the property of being a t-norm—holds in the following cases:

1. At least one of $T_1,T_2$ is continuous.

2. Both $T_1$ and $T_2$ are left-continuous; in this case $T_1 \star T_2$ is precisely the tensor product of the quantales $([0,1],T_1)$ and $([0,1],T_2)$.

The example $T_D \star T_D$ shows that neither continuity nor left-continuity is necessary. On the other hand, the case of $T_D \star T^{\mathrm{nM}}$ shows failure of associativity, showing that the interaction between a left-continuous t-norm and a non-left-continuous t-norm can destroy associativity. For the reader’s convenience, we summarize the known cases established or recalled in this paper in the following table.

$T_1$	$T_2$	Is $T_1 \star T_2$ a t-norm?
left-continuous	left-continuous	yes
continuous	arbitrary	yes
arbitrary	continuous	yes
$T_D$	$T_D$	yes
$T_D$	$T^{\mathrm{nM}}$	no

The case of two left-continuous components belongs to the existing quantale tensor-product theory. The new associativity result permits one component to be arbitrary when the other is continuous; the drastic pair shows that even two components failing left-continuity may work. The contrasting example shows why these results do not extend to arbitrary pairs.

The Archimedean discussion in § 3 answers a separate question: two Archimedean components imply that the fixed recursive powers of each $f < \top$ have infimum $\bot$ in $\mathbb{I}$. If the induced operation is associative, this gives an Archimedean t-norm.

However, a complete characterization of those pairs $(T_1,T_2)$ for which $T_1 \star T_2$ is a t-norm remains an open problem.

Acknowledgements

Rongjun Wu would like to express gratitude to the Southwest Minzu University Research Startup Funds for Grant (#RQD2021100), which supported the writing of this article. Jie Sun would like to express gratitude to the National Natural Science Foundation of China (No. 12601905) and the Southwest Minzu University Research Startup Funds for Grant (#RQD2024011). Hongliang Lai would like to express gratitude to the National Natural Science Foundation of China (No. 12671089).