REVIEW 3 major objections 3 minor 7 references
A counterexample to the Karvatskyi--Pratsiovytyi conjecture concerning the achievement set of an intermediate series
T0 review · 3 major / 3 minor · reviewed 2026-08-12 · deepseek-v4-flash
Pith's one-line read Three explicit multigeometric sequences satisfy the Karvatskyi–Pratsiovytyi conjecture's inequalities yet produce achievement sets of different topological types, refuting the conjecture.
desk verdict A real counterexample, currently obscured by two fixable typos that make Section 2 self-contradictory; the math holds up once they are corrected. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The central object is the achievement set $E(u_n)=\{\sum_{n=1}^\infty \varepsilon_n u_n : \varepsilon_n\in\{0,1\}\}$, the set of all subsums of a positive absolutely summable sequence. For multigeometric sequences, the achievement set is a self-similar set $K(\Sigma;q)=\{\sum_{n=0}^\infty d_n q^n : d_n\in\Sigma\}$, where $\Sigma$ is the finite digit set of one-period partial sums. The decisive tool is the criterion from Theorem 1.3(e): $K(\Sigma;q)$ is a Cantor-type set if $q^n<1/|\Sigma_n|$ for some n. The paper uses this criterion at different depths for $a$ and $b$, while the digit set for $c$ produces overlapping intervals that yield the Cantorval structure.
What would settle it
A reader could directly compute the fourteen values in Σ_2(b_n) to confirm the count and verify that (1/4)^2 < 1/14, then approximate E(c) by finite sums to check that it contains intervals. If E(c) turns out to be Cantor-type rather than a Cantorval, or if either E(a) or E(b) contains an interval, the counterexample would fail.
Extended reading notes
Core claim
The central claim is that the Karvatskyi–Pratsiovytyi conjecture, stated in the paper's introduction, is false. For every k ≥ 1, set $a_{2k-1}=a_{2k}=\alpha/4^k$ with $\alpha=1.95$, $b_{2k-1}=4\beta/4^k$ and $b_{2k}=3\beta/4^k$ with $\beta=0.8$, and $c_{2k-1}=3/4^k$, $c_{2k}=2/4^k$. The paper proves that $a_n \le c_n \le b_n$ and that the tail inequalities $b_n \le r^a_n$ and $r^b_n < a_n$ hold for all n. Using the known representation of multigeometric achievement sets as self-similar sets $K(\Sigma;q)$ with $q=1/4$, it applies Theorem 1.3(e) from Banakh et al. to conclude that $E(a)$ and $E(b)$ are Cantor-type sets, because $|\Sigma(a)|=3$ and $|\Sigma_2(b)|=14$ with $(1/4)^2<1/14$. However, $E(c)$ is the Guthrie–Nymann achievement set, a Cantorval. Thus the conclusion of the conjecture—that all three achievement sets have the same topological type—fails, even though the hypotheses hold completely.
Load-bearing premise
The counterexample depends on the external theorem that K(Σ;q) is Cantor-type when q^n < 1/|Σ_n| for some n, and on the exact count |Σ_2(b_n)| = 14; if that theorem is misstated, misapplied, or the count is wrong, the counterexample loses its ground.
Editorial extensions
If this is right
- The squeeze-theorem analogy does not carry over to achievement sets: termwise sandwiching plus tail inequalities does not force equal topological type.
- The explicit counterexample provides a concrete boundary case that any corrected version of the conjecture must exclude or handle.
- The proposed improved conjecture adds the condition $\lim_{n\to\infty} b_n/a_n = 1$, and the paper shows the counterexample does not refute that weaker statement.
- The construction demonstrates that multigeometric sequences with the same ratio $q=1/4$ can realize different achievement-set types despite satisfying identical interleaving and tail inequalities.
Reading between the lines
- The counterexample suggests that the ratio condition $\lim b_n/a_n = 1$ may be essential; one testable extension is to search for additional counterexamples that violate only that ratio condition but satisfy all other inequalities.
- The self-similar digit-set viewpoint used here could be turned into a systematic search method: vary the digit sets $\Sigma(a), \Sigma(b), \Sigma(c)$ with $q=1/4$ to find other triples where the theorem criterion applies to two sequences but not the third.
- If the improved conjecture is true, it would imply a genuine squeezing result for achievement sets under asymptotic closeness of the sandwiching sequences, which would be a useful classification tool for intermediate series.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper claims to refute the Karvatskyi–Pratsiovytyi conjecture by presenting three multigeometric sequences a, b, c that satisfy the conjecture's interleaving and remainder conditions (1), while E(a) and E(b) are Cantor-type sets and E(c) is a Cantorval. The intended counterexample shows that two outer sequences sharing the same achievement-set type do not force the middle sequence to have that type. The paper also proposes an improved conjecture adding the condition lim_{n→infty} b_n/a_n = 1, which it says the counterexample does not refute.
Significance. If the intended corrections are made, the paper provides a simple, explicit counterexample to a published conjecture, with hand-checkable constants and no parameter fitting. The proof relies legitimately on a known classification theorem for self-similar sets and on the standard identification of multigeometric achievement sets with K(Σ;q). The counterexample is narrow but sufficient to falsify the conjecture as stated, and the proposed refined conjecture is a reasonable direction for future work.
major comments (3)
- [Section 2, Proposition 1] The even case is proved by showing r^b_n < b_n, but condition (1) requires r^b_n < a_n. Since a_n ≤ b_n, the displayed inequality does not imply the required one. The intended inequality is true: for n = 2k, r^b_{2k} = 7β/(3·4^k) < α/4^k = a_{2k} because 7β/3 = 1.866… < α = 1.95; the proof needs this one-letter correction.
- [Section 2, final paragraph] The sentence 'the achievement set E(b_n) is Cantorval' directly contradicts Proposition 3, which proved E(b_n) is a Cantor-type set. The intended statement is that E(c_n) is a Cantorval; indeed c is exactly the Guthrie–Nymann sequence. Without this correction the counterexample is internally inconsistent and cannot be checked from the manuscript as written.
- [Section 3, introductory paragraph] The prose says 'we suggest adding a condition lim_{n→∞} b_n/a_n ≠ 1', but the stated improved conjecture in (2) and the preceding explanation require lim_{n→∞} b_n/a_n = 1. The displayed '≠' is a typo that should be corrected to '='.
minor comments (3)
- [Section 2, Proposition 3] The set Σ_2(b_n) is listed explicitly, but a short explanation of why |Σ_2(b_n)| = 14 would help the reader verify the crucial count on which the application of Theorem 1.3(e) depends.
- [Throughout] The phrase 'In [1, Theorem 1.3(e)] proved that...' is ungrammatical; it should read 'Theorem 1.3(e) in [1] proves that...' or similar.
- [Introduction] The definition of a Cantor-type set as 'homeomorphic to the Cantor set, which is the achievement set of the sequence (u_n) with u_n = 2/3^n' could be phrased more clearly by saying that the standard Cantor set is the achievement set of that representative sequence.
Circularity Check
No circularity: the counterexample rests on explicit constants, direct inequality checks, and external theorems; no step assumes its own conclusion.
full rationale
The derivation chain is self-contained with respect to circularity. The sequences (a_n), (b_n), and (c_n) are explicitly defined with numerical constants; Proposition 1 verifies the inequalities in (1) by direct computation. The type classifications of E(a_n) and E(b_n) invoke Theorem 1.3(e) of Banakh et al. [1], an external published theorem with no author overlap, together with explicit cardinalities |Sigma(a_n)| = 3 and |Sigma_2(b_n)| = 14; the Cantorval classification of E(c_n) invokes the Guthrie-Nymann result [4], also external. No parameter is fitted to force the conclusion, and the final claim does not appear among the assumptions. Therefore no step reduces to its own input by construction. The printed text contains apparent typographical or proof-checking slips: the even case of Proposition 1 states r^b_n < b_n instead of the required r^b_n < a_n, and the final sentence appears to say E(b_n) is a Cantorval after Proposition 3 classifies it as Cantor-type. These are correctness concerns for the manuscript as printed, not circularity, because fixing them does not change the evident intended derivation.
Assumptions & free parameters
free parameters (2)
- α =
1.95
- β =
0.8
assumptions (4)
- domain assumption Achievement sets of absolutely summable positive series classify into finite unions of intervals, Cantor-type sets, or Cantorvals.
- domain assumption Theorem 1.3(e) of [1]: K(Σ;q) is a Cantor-type set of zero Lebesgue measure if q^n < 1/|Σ_n| for some n.
- domain assumption For a multigeometric sequence with block sums Σ and ratio q, the achievement set equals K(Σ;q).
- domain assumption The achievement set of the sequence c with c_{2k-1}=3/4^k, c_{2k}=2/4^k is the Guthrie-Nymann Cantorval.
Cite this review
Pith. "Pith review of A counterexample to the Karvatskyi--Pratsiovytyi conjecture concerning the achievement set of an intermediate series." pith.science (2026). https://pith.science/paper/A7YRRAI6
@misc{pith2026241200042,
author = {Pith},
title = {Pith review of: A counterexample to the Karvatskyi--Pratsiovytyi conjecture concerning the achievement set of an intermediate series},
year = {2026},
howpublished = {\url{https://pith.science/paper/A7YRRAI6}},
note = {Machine review of arXiv:2412.00042}
}
read the original abstract
We found a counterexample to the conjecture of Karvatskyi and Pratsiovytyi concerning the topological type of the achievement set of an intermediate series (Proceedings of the International Geometry Center, 2023. https://doi.org/10.15673/pigc.v16i3.2519). This conjecture is based on an analogy with the squeeze theorem from calculus. We also proposed an improved version of the conjecture, which this counterexample does not refute.
Reference graph
Works this paper leans on
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[1]
T. Banakh, A. Bartoszewicz, E. Szymonik, and M. Filipcza k, Topological and measure properties of some self-similar se ts, Topological Methods in Nonlinear Analysis. 46:2 (2015), 1013–1028. https://doi.org/10.12775/TMNA.2015.075
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[2]
M. Banakiewicz and Fr. Prus-Wi´ sniowski, M-Cantorvals of Ferens type , Mathematica Slovaca. 67:4 (2017), 907–918. https://doi.org/10.1515/ms-2017-0019
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[3]
A. Bartoszewicz, M. Filipczak, and E. Szymonik, Multigeometric sequences and Cantorvals , Central European Journal of Math- ematics. 12:7 (2014), 1000–1007. https://doi.org/10.2478/s11533-013-0396-4
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[4]
J. A. Guthrie and J. E. Nymann, The topological structure of the set of subsums of an infinite series, Colloquium Mathematicae. 55:2 (1988), 323–327. https://doi.org/10.4064/CM-55-2-323-327
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Kakeya, On the partial sums of an infinite series , Tˆ ohoku Sci
S. Kakeya, On the partial sums of an infinite series , Tˆ ohoku Sci. Rep. 3 (1914), 159–164
work page 1914
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[6]
J. E. Nymann and R. A. S´ aenz, On the paper of Guthrie and Nymann on subsums of infinite serie s, Colloquium Mathematicae. 83:1 (2000), 1–4. https://doi.org/10.4064/cm-83-1-1-4
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[7]
M. Pratsiovytyi and D. Karvatskyi, Cantorvals as sets of subsums for a series related with trigo nometric functions , Proceedings of the International Geometry Center. 16:3-4 (2023), 262–271. https://doi.org/10.15673/pigc.v16i3.2519 M.Moroz: Department of Dynamical Systems and Fractal Analysis, Inst itute of Mathematics of NAS of Ukraine, Tereschenkivska ...
Reviewed August 12, 2026 · model on record in the stance chip above.
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