GPT 6 Astra Ultra attempt on Math-733

Problem 254: correcting the interval bootstrap and isolating its input

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. I read the complete 254.tex. Its target is completeness of distinct-element subset sums under growing dyadic block counts and the problem's fractional-part divergence assumption. The supplied attempt obtains a conditional interval bootstrap but does not produce its starting interval. I audit that bootstrap and make its exact sufficient input explicit.

2) WORK

Let $F$ be finite, with $[L,U]\cap\mathbb Z\subseteq\Sigma(F)$, and put $W=U-L+1$. List the unused elements increasingly as $b_1,b_2,\ldots$. If \[ b_i\le W+\sum_{h<i}b_h\qquad(i\ge1), \] then induction gives \[ \left[L,U+\sum_{h\le i}b_h\right]\cap\mathbb Z \subseteq\Sigma(F\cup\{b_1,\ldots,b_i\}). \] Indeed, adjoining $b_i$ adds a translate of the preceding interval, and the displayed inequality says the two integer intervals overlap or are adjacent. This proves completeness if the $b_i$ are infinite. For this particular union-of-translates construction the first failed inequality leaves an actual gap between the intervals arising from subsets of the preceding elements and those using $b_i$; additional subset sums of $F$ outside $[L,U]$ may still fill it, so failure is not a converse for completeness.

Here is a corrected dyadic corollary. If $F=A\cap[1,2^J]$ contains an interval of subset sums of length at least $2^{J+1}$, and \[ |A\cap(2^j,2^{j+1}]|\ge2\qquad\text{for every }j\ge J, \] then $A$ is complete. At the beginning of block $j$, maintain width at least $2^{j+1}$. Every member of that block is at most that width, so it can be absorbed. Two members increase the width by more than $2^{j+1}$, giving the invariant for block $j+1$.

The supplied statement began its count hypothesis at $j\ge J+1$, but its induction tried to absorb the block $(2^J,2^{J+1}]$. That block has no guaranteed two elements under the stated hypothesis. Starting the count hypothesis at $J$ repairs the proof; equivalently, supply the wider starter needed to jump directly to the next controlled block.

3) ADVERSARIAL VERIFICATION

Growing dyadic counts and $\gcd(A)=1$ alone cannot produce the missing starter. For $A=3\mathbb N\cup\{1\}$, both hold, but every subset sum is $0$ or $1$ modulo $3$. At $\theta=1/3$, the fractional-part sum is only $1/3$, so this example does not satisfy the additional divergence assumption. It illustrates precisely why that assumption still needs to enter the proof.

4) FINAL

UNRESOLVED. The propagation step is rigorous after an indexing correction. Deriving a sufficiently long interval from the original hypotheses remains entirely necessary; no implication from the fractional-part condition to that interval has been established here.

5) SOURCES CHECKED

The complete supplied 254.tex, read 2026-09-23. The interval proof and modular obstruction above are self-contained. An attempted official-page retrieval was unavailable, so no additional external formulation is asserted.

6) COMPLETION ESTIMATE

Subjective progress toward proving completeness under all original hypotheses. COMPLETION: 10%