Fuzzy Sets
1965 · paper · Mathematics and statistics
Introduces fuzzy sets, classes whose members have grades of membership between zero and one. Extends notions such as inclusion, union, intersection, complement and convexity to them, and proves a separation theorem for convex fuzzy sets without requiring disjointness.
Checked: PDF from dcsc.tudelft.nl, fetched 5 Oct 2026. Page numbers are those of that PDF.
2 findings
Substance Shadow assertion "S_H(A) = S_H(B) for all H ⇒ A = B" (p.350) rests on the step "f_B(x) ≦ β for all x on F" (p.351). That step does not follow from F merely supporting Γ_β, and a checked counterexample shows the implication fails for bounded convex fuzzy sets as defined.
- p. 13 Shadow of a fuzzy set, statement and proof
S_H(A) = S_H(B) for all H ⇒ A = B
- p. 13 Shadow proof
the set Γ_β = {x | f_B(x) > β} is convex, and hence there exists a hyperplane F supporting Γ_β and passing through x_0
- p. 14 Shadow proof, first line
f_B(x) ≦ β for all x on F, we have f_{S_H(B)}(x_0*) ≦ β
Only the authors can say what is right here.
Judge's note: Checked pp.350–351 (PDF p13–14). A supporting hyperplane F through x_0 puts Γ_β={f_B>β} in a closed half-space, but points of Γ_β may lie on F, so 'f_B ≤ β on F' is unjustified. My counterexample in E^2: Γ_B = {x2>0, x1²+x2²<1} ∪ {(t,0): 0<t<1}, f_B = indicator of Γ_B. Γ_A = Γ_B ∪ {(0,0)}, f_A = indicator of Γ_A. Both are convex: segments between upper-half-disk points and points (t,0) or the origin stay in the set. So every level set Γ_α, α∈(0,1], is convex, as definition (24) requires, and both sets are bounded. f_A and f_B differ only at the origin. Projection lines not through the origin give equal sups. A non-horizontal line through the origin meets the open upper half-disk, so both sups are 1. The horizontal line contains (1/2,0) with f_B=1. So S_H(A)=S_H(B) for every H, yet A≠B. In the proof (x_0 = origin, α=1, β=0), the only supporting line through x_0 is x2=0, and f_B(1/2,0)=1>β on it. The stated result is false as printed without an extra regularity condition (e.g. closed level sets). The 'more general' forms on p.351 (A = conv B) fail for the same pair, since B is convex. A stated result is wrong, so S1.
Second check: another model agreed
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Substance Separation theorem "Then D = 1 − M" (p.352): Case 2 claims "for all points on the minus side of H, f_A(x) ≦ M". With the paper's closed sides ("h(x) ≧ 0 … h(x) ≦ 0", footnote 8: sides "have H in common"), this does not follow, and the theorem fails for bounded convex fuzzy sets with non-closed level sets.
- p. 14 Definition of degree of separation and footnote 8
all points for which h(x) ≧ 0 being on one side of H and all points for which h(x) ≦ 0 being on the other side
- p. 16 Proof, Case 2
for all points on the minus side of H, f_A(x) ≦ M, and for all points on the plus side of H, f_B(x) ≦ M
- p. 15 Theorem statement
Then D = 1 − M.
Only the authors can say what is right here.
Judge's note: Checked pp.351–353 (PDF p14–16). The definition on p.351 uses closed sides {h≥0}, {h≤0}, and footnote 8 confirms that they share H. Ordinary separation of Γ_A={f_A>M} and Γ_B={f_B>M} only places them in closed half-spaces, so points of Γ_A may lie on H, which belongs to the side where f_A ≤ M is claimed. Counterexample I checked in E^1: f_A = indicator of [−1,0], f_B = indicator of (0,1]. Both are convex and bounded. M = sup min = 0 < Min(M_A, M_B) = 1, so this is Case 2, and the theorem gives D=1. For any point H={p}, both ways of assigning the sides include a point with membership 1 for the relevant set: p≤0 gives f_A(0)=1 on [p,∞), and p>0 gives f_B=1 on (0,p] ⊂ (−∞,p]; similarly f_A(−1)=1 or f_B(1/2)=1 for the other pairing. So M_H=1 for all H, M̄=1, D=0≠1. With open sides the theorem would hold here (H={0} gives M_H=0). Under the definition as printed, the stated theorem is false, so S1. The authors must say which convention is intended.
Second check: another model agreed
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8 minor (typos and the like; most checked by the judge only)
Wording Eq. (21) "f_Λ(x) = (f_C(x) − f_B(x))/(f_A(x) − f_B(x)), x ∈ X" is stated for all x ∈ X but is 0/0 wherever f_A(x) = f_B(x); the existence claim for Λ still holds.
- p. 8 Eq. (21)
f_Λ(x) = (f_C(x) − f_B(x)) / (f_A(x) − f_B(x)), x ∈ X.
Only the authors can say what is right here.
Judge's note: Checked p.345 (PDF p8). If f_A(x)=f_B(x)=a, the hypothesis A∩B ⊂ C ⊂ A∪B gives Min ≤ f_C ≤ Max, so f_C(x)=a, and (21) becomes (a−a)/(a−a)=0/0. Where f_A≠f_B: f_B + Λ(f_A−f_B) = f_C, and Λ∈[0,1] because f_C lies between f_A and f_B. So the formula holds wherever it is defined. At the degenerate points any value of f_Λ in [0,1] works, so the claim 'one can always find a fuzzy set Λ' stands. The formula only needs to be restricted to those x or completed there. The step is faulty but the result stands, so S3.
Second check: another model agreed
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Wording Γ with a level subscript is defined with ≧ in Eq. (24) ("Γα = {x | fA(x) ≧ α}", PDF p10) and under Boundedness (PDF p11), but the shadow proof (PDF p13) writes "the set Γβ = {x | fB(x) > β} is convex" with a strict inequality, citing only that B is a convex fuzzy set.
- p. 10 Eq. (24)
Γα = {x | fA(x) ≧ α}
- p. 11 Boundedness
{x | fA(x) ≧ α} are bounded for all α > 0
- p. 13 Shadow proof, last paragraph
the set Γβ = {x | fB(x) > β} is convex
Judge's note: Read Eq. (24) on p10, the Boundedness definition on p11 and the shadow proof on p13 from the page images; the inequality signs are as quoted (≧, ≧, >). The same symbol is used for two different sets. The strict set is restated locally and is itself convex under definition (25): if fB(x1) > β and fB(x2) > β then fB on the segment is ≥ Min[fB(x1), fB(x2)] > β. So the argument is not affected; this is a notational inconsistency only, and either form could be intended locally.
Second check: another model agreed
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Wording "In particular, if A is strongly convex and x0 is attained, then x0 is unique.": the thing attained is the maximal grade M (at the point x0), not x0; the sentence should read "M is attained".
- p. 12 Section V, paragraph beginning 'Let A be a convex fuzzy set and let M = Sup'
convex and x0 is attained, then x0 is unique
- p. 12 Section V, paragraph after 'Strict and strong convexity', sentence beginning 'In particular'
x0 is attained, then x0 is unique
- p. 12 Same paragraph, preceding sentence
either M is attained for some x, say x0
Judge's note: Read the paragraph on the crop and the full PDF p12 image. The preceding sentence says "either M is attained for some x, say x0, or there is at least one point x0 at which M is essentially attained", and the justification that follows starts "For, if M = fA(x0) and M = fA(x1)". The paper uses "attained" for M throughout (Lemma on p11, core definition and theorem on p12, corollary on p13). The intended meaning is recoverable, so this is a wording slip, not a mathematical error.
Second check: another model agreed
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Wording Lemma proof (p.349): "x_1, x_2, ⋯, is a sequence of points in a closed bounded set Γ_1 … this sequence must have at least one limit point, say x_0, in Γ_1". Γ_1 = {x | f_A(x) ≥ M/2} need not be closed, so x_0 may lie outside Γ_1; the lemma's conclusion still holds.
- p. 12 Proof of Lemma (maximal grade essentially attained)
is a sequence of points in a closed bounded set Γ_1. By the Bolzano-Weierstrass theorem, this sequence must have at least one limit point, say x_0, in Γ_1
Only the authors can say what is right here.
Judge's note: Checked pp.348–349 (PDF p11–12). Γ_n = {x | f_A(x) ≥ M − M/(n+1)} is defined on p.348. The paper only requires boundedness, and membership functions are arbitrary maps into [0,1], so the level sets need not be closed. Counterexample I checked: in E^1, f_A(x)=x on (0,1) and 0 elsewhere. Γ_α=[α,1) is bounded for every α>0, so A is bounded, and M=1. Then Γ_1=[1/2,1) is not closed. x_n ∈ Γ_n=[1−1/(n+1),1) converges to 1, which is not in Γ_1 (f_A(1)=0). Bolzano–Weierstrass still gives a limit point in the closure of Γ_1. The rest of the argument (every neighbourhood of x_0 contains x_N, x_{N+1}, … in Q(ε)) does not use x_0 ∈ Γ_1, so the lemma stands. Step faulty, result stands: S3.
Second check: another model agreed
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Typo The captions of Fig. 2 and Fig. 3 print "simultating" for "simulating": "Parallel and series connection of sieves simultating ∪ and ∩" (PDF p6) and "A network of sieves simultating {[f1(x) ∨ f2(x)] ∧ f3(x)} ∨ f4(x)" (PDF p7).
- p. 6 Fig. 2 caption
connection of sieves simultating
- p. 6 Fig. 2 caption
sieves simultating
- p. 7 Fig. 3 caption
A network of sieves simultating
Judge's note: Read both captions on the crops and the full page images. The extra 't' is printed in both captions, so it is not an extraction artefact. Spelling only; meaning unaffected.
Second check: none (judge only)
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Typo "If A is a bounded set, then for each ε > 0 then exists a hyperplane H such that fA(x) ≦ ε ...": the second "then" should be "there".
- p. 11 Boundedness, second paragraph
then for each ε > 0 then exists a hyperplane H
- p. 11 Paragraph after the definition of Boundedness
then for each
- p. 11 Same sentence
then exists a hyperplane H
- p. 11
for each e > 0 then exists a hyperplane H
Judge's note: Read on the crop and the full PDF p11 image: the printed word is "then exists", not an extraction artefact. The evident fix is "there exists". Meaning unaffected.
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Typo "To the end, let P be a cylinder of radius ε with the line passing through x0 and x1 as its axis.": probably meant "To this end".
- p. 12 Proof of the theorem that the core is convex, last paragraph of page
To the end, let P be a cylinder of radius
- p. 12 Proof of the theorem that the core of a convex fuzzy set is convex, last paragraph of the page
To the end, let P be a cylinder
Judge's note: Read on the crop and the full PDF p12 image: printed as "To the end". The sentence introduces the construction that proves the claim in the preceding paragraph, where the usual idiom is "To this end"; no "end" has been named. Cosmetic, and the printed form could be the author's phrasing, so filed as an observation.
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Typo "... such that fA(x0) ≠ fB(x0), then their exists a hyperplane H such that ...": "their" should be "there".
- p. 13 Shadow of a fuzzy set, paragraph 'To prove this assertion'
then their exists a hyperplane
Judge's note: Read on the crop and the full PDF p13 image: "their exists" is printed, not an extraction artefact. The evident fix is "there exists". Meaning unaffected.
Second check: none (judge only)