0
t
t
0 0
c
d
0 c
d
0 c
d
0 c
d
. . .
0 c
d
Figure 2.2: Illustration: induction steps for Claim 3 in Theorem 2.3
2 we have that the graph of thed-level function restricted to[d, e]is a subset ofE too. By Corollary 1.23, we conclude that∗◦is uniquely determined on[c, e]×[d, e], hencec∗◦dis uniquely determined.
Due to Claim 3 it remains to prove that the values of∗◦onD¬uniquely determinex→∗◦0. Assume that two continuous Archimedean t-norms,∗◦1and∗◦2are given such that they coincide onD¬, and assume, on the contrary, that there existsc∈ [0,1]such thatc→∗◦1 0 < c→∗◦2 0. Then we have c∗◦1(c→∗◦2 0)>0andc∗◦2(c→∗◦2 0) = 0by residuation. Due to continuity, forε=c∗◦1(c→∗◦2 0) there exists δ > 0such that for any x ∈ ]c→∗◦2 0,(c→∗◦2 0) +δ] we have0 < c∗◦2x < ε. We have0< c∗◦2x < ε < c∗◦1xandc∗◦1x=c∗◦2xby Claim 3; a contradiction.
2.4 Applying rotation-invariance
As a by-product of Theorem 1.33 about the structure of involutive elements, which was obtained using rotation-invariance, we obtain results concerning left-continuous t-norms (Theorem 2.6, Theorem 2.10) and continuous Archimedean t-norms (Theorem 2.12). We shall need the follow-ing statement:
Proposition 2.4 x≤ ¬cy ⇐⇒ y ≤ ¬cx
Proof. We have x ≤ ¬cy ⇐⇒x∗◦y ≤ c⇐⇒ y∗◦x ≤ c⇐⇒ y ≤ ¬cxby the adjointness property, by the commutativity of∗◦, and by the adjointness property, respectively.
Theorem 2.5 Let A = {an |n ∈ N} ⊂ [0,1[ such that limn→∞an = 1. Any continuous Archimedean t-norm is determined by its values in{an×[0, an]|an ∈A}.
Proof. By Theorem 2.1, any continuous Archimedean t-norm is theϕ-transform of either the Łukasiewicz t-norm, or the product t-norm. Since{ϕ(an)|n ∈ N}is a sequence converging to 1, and the set{an×[0, an]|an ∈A}is mapped to the set{ϕ(an)×[0, ϕ(an)]|an ∈A}it suffices to prove the that the Łukasiewicz t-norm and the product t-norm are determined by their values in {bn×[0, bn]|bn∈A}wherebnis any sequence from[0,1[converging to1. The following claim is in [79]: Let T be any t-norm, x ∈ (0,1)arbitrary, and suppose that the value of T(x, y) is determined fory ∈[0, x]. Then the values ofT(T(x, . . . , x
| {z }
i
), y)are determined fori = 2,3, . . ., y∈[0,1].
Assume that the values of a continuous Archimedean t-normT coincide with the Łukasiewicz (resp. product) t-norm on{bn×[0, bn]|n∈N}, wherebnis any sequence from[0,1[converging to1. We shall showT =TL(resp. T =TP). By the continuity ofT it is sufficient to prove it on a dense subset of[0,1]2. Let[a, b]⊂[0,1],a < bbe arbitrary. Fixn ∈Nsuch thatbn >1−b−a2 . Lettingx=bnin the above statement we conclude thatT coincides with the Łukasiewicz (resp.
product) t-norm onmax(0,1−i·(1−bn))×[0,1](resp. onbin×[0,1]),i∈N,i >1. Since we have1−bn < b−a2 , in both cases there existsi ∈ N, i >1such that1−i·bn ∈ [a, b], and this ends the proof.
a1 a a2 3...an
Figure 2.3: A subdomain of uniqueness, see Theorem 2.5
Theorem 2.6 Let ∗◦ be a left-continuous t-norm with the following property: There exists {cn|n∈N} ⊂]0,1]such thatlimn→∞cn= 0and
x→∗◦cn = min (1,1 +cn−x) (2.5) holds for allx∈[0,1]. Then∗◦is the Łukasiewicz t-norm.
Proof. Observe thatcn’s are involutive. First, we shall show that the set of involutive elements is dense in[0,1]. Let[a, b] ⊂ [0,1]arbitrary. Taken such thatcn < b−a, and choosem ∈ N such thatcm < cn. Thencn−cm < b−a, and there exists ani∈Nsuch thati·(cn−cm)∈[a, b].
Thus it is sufficient to prove that for any i ∈ N, min (1, i·(cn−cm))is involutive. In fact we shall prove more: Letd0 =cm, d1 = cn, and fori = 2,3, . . . letdi =di−1→∗◦di−2 →∗◦di−1. A
2.4. APPLYING ROTATION-INVARIANCE 53 straightforward induction using Theorem 1.33 shows that for i ∈ N, di is involutive. We state that
di = min (1, cm+i·(cn−cm))
x→∗◦di = min (1,1 +di−x) forx∈[0,1]. (2.6) Indeed, these hold fori= 0,1. By induction assume that the statements holds for0,1, . . . , i−1.
We shall prove the first statement: Ifdi−1 = 1 then by using (T4) we havedi = 1→∗◦di−2 →∗◦
1 = di−2 →∗◦ 1 = 1. Therefore, we may suppose di−1 < 1 which together with the induction hypothesis yields d0 < . . . < di−2 < di−1 < 1. By using the induction hypothesis we obtain di = di−1→∗◦di−2 →∗◦ di−1 = 1 +di−2−di−1 →∗◦ di−1 = 1−(cn−cm) →∗◦ di−1. In case di−1 ≥1−(cn−cm)we getdi = 1by (2.6). Ifdi−1 <1−(cn−cm)then1−(cn−cm)→∗◦di−1 = 1 +di−1 −(1−(cn−cm)) =cm+i·(cn−cm). The proof of the first statement is concluded.
Now we verify the second statement, as follows: The casedi = 1being straightforward we may assumedi <1. Thenx→∗◦di = x→∗◦di−1→∗◦di−2→∗◦di−1 =x→∗◦di−1 →∗◦di−2→∗◦ di−1 by (1.12). Ifx < di then both sides of (2.6) are equal to 1. If x ≥ di = cm +i·(cn−cm)then the induction hypothesis yieldsx→∗◦di−1→∗◦di−2→∗◦ di−1 = 1 +di−1−x→∗◦ di−2→∗◦di−1 = 1 +di−2−(1 +di−1−x) →∗◦ di−1 = di−2−di−1+x →∗◦ di−1 = x−(cn−cm)→∗◦ di−1 = 1+di−1−(x−(cn−cm)) = 1+cm+(i−1)·(cn−cm)−x+(cn−cm) = 1+cm+i·(cn−cm)−x= 1+di−x, which concludes the proof of the second statement. Thus, the set of involutive elements is dense in[0,1], as stated.
Next we prove
x→∗◦0 = 1−x (2.7)
for x ∈ [0,1]. Observe that it suffices to prove it for x ∈]0,1]since 0→∗◦ 0 = 1by (T4). Let x ∈]0,1] arbitrary. Choosen0 ∈ Nsuch that for any n > n0 we havecn < x. Then by (2.2), respectively, we havex→∗◦0 = limcn↓0x→∗◦cn= limcn↓01 +cn−x= 1−x.
Letn, m∈Nsuch thatcm < cn, and construct, for anyi∈N,dias above. As shown, the set of suchdi’s is dense in[0,1]. By using the exchange property and (2.7) we have, forx ∈ [0,1], x∗◦di→∗◦0→∗◦0 = x→∗◦ di→∗◦0→∗◦0 = x→∗◦ di. By this and (2.6) we obtainx∗◦1−di = 1−min(1,1 +di−x). But1−min(1,1 +di−x) = max(0,1−x−(1−di)) = TL(x,1−di).
Since thedi’s are dense in[0,1], the set of1−di’s is dense in[0,1]too. That is,∗◦coincides with the Łukasiewicz t-norm on a dense subset of[0,1]2. Finally, left-continuity of∗◦entails∗◦=TL.
Remark 2.7 The set{cn|n ∈N} ⊂]0,1]in Theorem 2.6 is minimal in the following sense:
Of course, dropping out any subset from a convergent sequence such that the cardinality of the remaining sequence is still infinite results in a convergent sequence with the same limit, thus such a subset can always be left out from {cn | n∈N}. However, as shown by Example 2.8, antecedents of Theorem 2.6 can not be relaxed such that the set{cn}becomes finite. Moreover, not even a convergent sequence is sufficient if its limit differs from0.
Example 2.8 Increasing bijections from [0,1] to [0,1] are called automorphisms of [0,1].
With any automorphism ϕand with any t-norm T one can define Tϕ, which is a t-norm and is called theϕ-transformationofT, as follows:
Tϕ(x, y) =ϕ−1(T(ϕ(x), ϕ(y)))
Letf andg be two automorphisms of[0,1], as depicted respectively in Figure 2.4. It is easy to verify that thef-, and g-transformations of the Łukasiewicz t-norm have elements, such that the corresponding level sets satisfy (2.5). For a visualization see Figure 2.5.
0 0.2 0.4 0.6 0.8 1
0.2 0.4 0.6 0.8 1 0
0.2 0.4 0.6 0.8 1
0.2 0.4 0.6 0.8 1
Figure 2.4: Two automorphisms of[0,1]
Figure 2.5: Tf andTg, see Example 2.8
Example 2.9 The condition in (2.5) can not be relaxed by simply saying that x →∗◦ cn is involutive. A counterexample is the rotation ([73]) of the product t-norm given as follows: LetT be the linear transformation of the product t-norm into[12,1], that is,T(x, y) = (2x−1)(2y−1)+1
2 and
let
(TP)rot(x, y) =
T(x, y) ifx, y ∈]12,1]
1−max{t∈[12,1]|T(x, t)≤1−y} ifx∈]12,1]andy∈[0,12] 1−max{t∈[12,1]|T(y, t)≤1−x} ifx∈[0,12]andy∈]12,1]
0 ifx, y ∈[0,12]
.
Then each element in [0,12[ in involutive. The rotation of the product t-norm (depicted in Fig-ure 1.14 on page 42) has exactly one point of discontinuity; hence it is not isomorphic to the Łukasiewicz t-norm.
2.4. APPLYING ROTATION-INVARIANCE 55 Theorem 2.10 Let ∗◦ be a left-continuous t-norm with the following property: There exists {cn|n∈N} ⊂]0,1]such thatlimn→∞cn= 0,limn→∞ cn
Proof. Observe thatcn’s are involutive. First, we shall show that the set of involutive elements is dense in [0,1]. Let [a, b] ⊂ [0,1], a < b arbitrary. Take n such that cn < b−a, and 0 <
cn
cn+1 −1< b−a. Suchnexists. Indeed, if there weren0 ∈Nsuch that for alln > n0we would have ccn
n+1 ≤1then this would yield thatcnis nondecreasing, which contradicts tolimn→∞cn= 0, cn>0. Letd0 =cn+1,d1 =cn, and fori= 2,3, . . .letdi =di−1→∗◦di−2→∗◦di−1. We state that
. The proof of (2.9) is concluded. Now we verify the second statement, as follows: The case di = 1 being straightforward we may assume di < 1. Then x→∗◦ di = x→∗◦ di−1→∗◦di−2→∗◦di−1 = x→∗◦di−1→∗◦di−2 →∗◦ di−1 by (1.12). Ifx ≥ di =
Therefore, there existsi ∈ Nsuch thatdi ∈ [a, b]. Thus, the set of involutive elements is dense in[0,1], as stated.
To conclude the proof, we shall show that ∗◦ coincides with the product t-norm on a dense subset of[0,1]2. Indeed, then left-continuity of∗◦ensures∗◦=TP. To this end, it suffices to prove that for anyε > 0, ∗◦ coincides with the product t-norm on a dense subset of]0,1]×[ε,1]. For all n ∈ N construct, for any i ∈ N, di as above, and denote their set by D. As shown, D is dense in[0,1]. Letx ∈]0,1] arbitrary. Fixn ∈ N such thatcn < x, and cxn < ε. SinceD is dense in [0,1] we have that H = {cdn | d ∈ D, cn ≤ d < x} is dense in [cxn,1], whence it is dense in [ε,1]. Let d ∈ H arbitrary. By using the involutive nature of x→∗◦ cn on [cn,1], the exchange property, the involutive nature of x→∗◦ cn on [cn,1], and (2.8), respectively, we have x∗◦ cdn →∗◦cn =x∗◦d→∗◦cn→∗◦cn =x→∗◦d→∗◦cn→∗◦cn =x→∗◦d = dx. Taking into account that the right-hand side is smaller that 1it implies that the left-hand side is smaller than 1, and hence it is equal to x∗◦cncn
d
. Thus we obtainedx∗◦ cdn = x· cdn = TP x,cdn
forx ∈]0,1], d ∈ H.
The proof is concluded.
Proposition 2.11 Let ∗◦ be a continuous t-norm, c be an involutive element. If x ≥ cthen fc(x) =yis equivalent tox∗◦y=c.
Theorem 2.12 LetA={cn|n∈N} ⊂]0,1]such thatlimn→∞cn = 0. Any nilpotent t-norm is determined by itscn-level sets. If in addition we havelimn→∞ cn
cn+1 = 1then any strict t-norm is determined by itscn-level sets.
Proof. Denote∗◦the nilpotent (resp. strict) t-norm, andTL(resp. TP) the Łukasiewicz (resp.
product) t-norm. By Theorem 2.1 the Łukasiewicz (resp. product) t-norm is aϕ-transform of any nilpotent (resp. strict) t-norm. Forn ∈N, letbn=ϕ−1(cn). Then we have{bn|n ∈N} ⊂]0,1]
andlimn→∞bn = 0. As shown by the isomorphismϕ, it is sufficient to prove that the Łukasie-wicz (resp. product) t-norm is determined by itsbn-level sets, where{bn|n∈N}is an arbitrary subset of]0,1]withlimn→∞bn = 0(resp. withlimn→∞bn= 0,limn→∞ bbn
n+1 = 1). To conclude assume thatT is a nilpotent (resp. strict) t-norm,{bn|n ∈N}is a set as described above, and the bn-level sets of∗◦coincide with thebn-level sets of the Łukasiewicz (resp. product) t-norm. Since any nilpotent (resp. strict) t-norm is continuous, thebn-level sets determine the fbn functions of
∗
◦, whence the hypothesis of Theorem 2.6 (resp. Theorem 2.10) are fulfilled. Thus,∗◦=TL(resp.
∗
◦=TP).
Remark 2.13 Since the counterexamples of Examples 2.8, 2.9 are based on continuous Arc-himedean t-norms we obtain the minimality of the set{cn|n∈N}in Theorem 2.12.