Reading Eight Erdős Problems Together

Will Cook

Cross-problem paper

Reading Eight Erdős Problems Together

Eight Erdős problems read together By Will Cook 22 September 2026 Authorship, AI use and citation

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Précis. One account of the mathematics developed across the programmes: an exact tail-capacity criterion under eventual congruences, its sharp factorial support-gap threshold, rational and irrational Lambert subsums across bases, and the limits of finite greedy tests and other irrationality methods. Full arguments, exact counterexamples, unsuccessful routes and attribution are retained together. The principal proofs are ordinary mathematics; cited Lean ingredients have their own stated scope. Historical novelty and independent expert review are not established.

This paper owns the synthesis exposition and ordinary proofs across the covered problems: capacity and congruence constructions, Lambert subsums, rational-point counts, method limits and their research record.

It is not authority for a solution to any original Erdős target, historical novelty, independent expert review, or a full Lean proof of the analytic capacity criterion or Lambert-chain theorem.

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In this paper

Base two

Removed intervals and finite translations

For the remainder of this subsection, take $t = 2$ : $w_{n} = (2^{n} - 1)^{- 1}$ , $R_{N} = \sum_{n > N} w_{n}$ , $g_{n} = w_{n} - R_{n} > 0$ , $X_{F} = \sum_{n \in F} w_{n}$ for finite $F$ , and $A$ is the set of all subsums.

Lemma 6.8 (the removed intervals). $[0, E] ∖ A$ is the disjoint union, over finite nonempty $F$ , of the open intervals $(X_{F} - g_{max F}, X_{F})$ . Their total length is $\sum_{n} 2^{n - 1} g_{n} = E - 1$ , and $g_{n} = \sum_{j \geq 2} \frac{2^{j} - 2}{2^{j} - 1} 2^{- j n} = \frac{2}{3} 4^{- n} + \frac{6}{7} 8^{- n} + \dots$ .

Proof. Fix the digits $ε_{1}, \dots, ε_{n - 1}$ and let $s = \sum_{i < n} ε_{i} w_{i}$ . The values with these digits lie in $[s, s + R_{n - 1}]$ , and they split into $[s, s + R_{n}]$ and $[s + w_{n}, s + w_{n} + R_{n}]$ . The interval between them is $(s + R_{n}, s + w_{n})$ . Its right end is $X_{F}$ with $F = {i < n : ε_{i} = 1} \cup {n}$ and its length is $g_{n}$ . Every finite nonempty $F$ arises once. The total length is $E$ minus the measure of $A$ , which is $1$ [33]. The series for $g_{n}$ follows from $w_{n} = \sum_{j \geq 1} 2^{- j n}$ and $R_{n} = \sum_{j \geq 1} 2^{- j n} / (2^{j} - 1)$ . ◻

Lemma 6.9 (translation by a finite subsum). Let $F$ be finite with largest element $n$ and let $0 \leq x \leq R_{n}$ . The greedy rule applied to $X_{F} + x$ selects exactly $F$ among the indices up to $n$ and then agrees with the greedy rule applied to $x$ . In particular $X_{F} + x \in A$ if and only if $x \in A$ , and $X_{F} + x$ is rejected at a step $m > n$ exactly when $x$ is.

Proof. At an index $i \leq n$ the remainder is $X_{F \cap [i, n]} + x$ . If $i \in F$ it is at least $w_{i}$ and the index is taken. If $i \notin F$ it is at most $R_{i} - R_{n} + x \leq R_{i} < w_{i}$ , so the index is skipped and nothing is rejected. After index $n$ the remainder is $x$ . ◻

Two consequences. The map $x \mapsto x + 1$ preserves reduced denominators, so at every step $n \geq 2$ the number of fractions of height at most $Q$ rejected at that step is even; every saved table has this property. The condition is $x \leq R_{max F}$ . It is not enough that $X_{F} + x \leq E$ : $x = 1 / 2$ is not rejected through step $160$ , while $1 / 2 + 1 / 3 = 5 / 6$ lies in $(R_{1}, 1)$ and is rejected at step $1$ . Counts of rejections should therefore be taken modulo these translations. At height $200$ the four fractions rejected at step $12$ are $46 / 183$ and its translates by $1 / 3$ , $1$ and $4 / 3$ , one event.

A finite subsum has odd denominator, since $\prod_{n \in F} (2^{n} - 1)$ is odd. So a rational with even denominator is never a finite subsum, and if it is a subsum at all its set $S$ is infinite.

Fixed-depth rational counts

Proof of Theorem 6.4. Fixing the first $N$ digits gives $2^{N}$ closed intervals of length $R_{N}$ . They are pairwise disjoint because $w_{n} > R_{n}$ , and a real in $[0, E]$ is not rejected in the first $N$ steps exactly when it lies in their union $K_{N}$ , which has measure $2^{N} R_{N}$ . For an interval $I$ of length $ℓ$ , the number of integers $p$ coprime to $q$ with $p / q \in I$ is $φ (q) ℓ + O (2^{ω (q)})$ by inclusion and exclusion. Summing over $q \leq Q$ with $\sum_{q \leq Q} φ (q) = 3 Q^{2} / π^{2} + O (Q \log Q)$ and $\sum_{q \leq Q} 2^{ω (q)} = O (Q \log Q)$ gives $3 Q^{2} ℓ / π^{2} + O (Q \log Q)$ . Apply this to the $2^{N}$ intervals of $K_{N}$ and to $(0, E]$ and divide. The second statement follows because every subsum lies in every $K_{N}$ and $2^{N} R_{N} \to 1$ . ◻

The intervals removed at step $n$ are explicit. Put $g_{n} = w_{n} - R_{n}$ , so that $g_{n} = \frac{2}{3} 4^{- n} + O (8^{- n})$ . For a finite nonempty $F$ with largest element $n$ , the interval $(X_{F} (2) - g_{n}, X_{F} (2))$ is removed at step $n$ , and

[0, E] ∖ A = ⨆_{F \neq \emptyset} (X_{F} (2) - g_{max F}, X_{F} (2)), \sum_{n \geq 1} 2^{n - 1} g_{n} = E - 1,

where $A$ is the set of subsums. By Theorem 6.2(b) a rational with an infinite $S$ is exactly a counterexample to #257 at base $2$ . So #257 at base $2$ holds if and only if every rational in $[0, E]$ that is not a finite subsum lies strictly between $X_{F} (2) - g_{max F}$ and $X_{F} (2)$ for some finite nonempty $F$ . This is a one-sided question of approximation by the countable set of finite subsums, with an error that shrinks like $4^{- max F}$ .

Under #257 the only fractions of height at most $Q$ that are subsums are the finite subsums, $40$ of the $19,653$ fractions with $2 \leq q \leq 200$ . A model that treats later remainders as equidistributed gives the opposite extreme, a proportion tending to $1 / E$ , because the shares $2^{n - 1} g_{n} / E$ of the removed intervals are summable. That model asserts that #257 fails for a positive proportion of all rationals. The fixed-depth limiting proportion does not distinguish this claim from its negation. An arithmetic argument, or a count with separately justified estimates as both depth and height grow, is needed. Measure does not decide either. Boes, Darst and Erdős construct symmetric Cantor sets of every measure in $[0, 1)$ that contain essentially no rationals [26].

The exact computation in [36] agrees with Theorem 6.4 step by step. Among the $19,653$ reduced fractions with $2 \leq q \leq 200$ , the numbers rejected at steps $1$ , $2$ and $7$ are $4809$ , $1470$ and $32$ , against $4811$ , $1467$ and $32$ from the measures of the removed intervals. The measure-based main term for the number rejected at step $n$ is $2^{n - 1} g_{n} \sum_{2 \leq q \leq Q} φ (q)$ , asymptotically $(3 Q^{2} / π^{2}) 2^{n - 1} g_{n}$ , which falls below $1$ near $n = 2 \log_{2} Q - 2 \log_{2} π$ . This is not a deterministic cutoff: the error in Theorem 6.4 does not justify such an extrapolation. For example, $189 / 388$ is first rejected at step $17$ , beyond this scale for $Q = 388$ . Its selected indices before rejection are $F = {2, 3, 7, 9, 10, 14, 15, 16}$ , and exact arithmetic gives

R_{17} \leq \frac{196609}{25769803776} < \frac{189}{388} - X_{F} (2) = \frac{9291822600689}{1217890317075045460} < \frac{1}{131071} = w_{17} .

Section 9 supplies the earlier-step checks. Deeper computation can therefore produce new exclusion certificates; absence of a later rejection still does not prove membership. No rational is known to have an infinite $S$ at base $2$ .

Problem 6.10. Decide whether $1 / 2$ is a subsum of $\sum (2^{n} - 1)^{- 1}$ . By [33] this holds if and only if the integer remainders of the greedy rule fail to increase at infinitely many steps.

The rational-point counting papers examined here concern null Cantor sets such as the middle-third set [27][28]. We did not locate a theorem settling the present positive-measure subsum problem. Problem 6.10 asks about one explicit rational point.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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