gpt pro 5.2 attempt on Math-640, version 2

Round 2: Hypercube isoperimetry and a sharp general lower bound

1) Round-2 Objective

Path (A): proof attempt toward the conjectured bound $N\gg 2^n$. Building on the Round-1 gap---to prove or disprove the existence of an absolute $c>0$ such that $N\ge c\,2^n$ for all $n$---the most concrete promising route is an anti-concentration argument for subset sums. In this round we implement a sharp anti-concentration mechanism via the vertex-isoperimetric inequality on the hypercube, obtaining the best-known general lower bound \[ N\ge \binom{n}{\lfloor n/2\rfloor}. \] This still falls short of $N\gg 2^n$ by a factor $\Theta(\sqrt n)$, so the original conjecture remains unresolved.

2) Round-1 Foundation Used

We use the following Round-1 foundations.

1. The global hypothesis: $A\subseteq\{1,\dots,N\}$, $|A|=n$, and the subset-sum map $S\mapsto \sigma(S)$ is injective on $\mathcal P(A)$.
2. The standard identification of subsets with vertices of the hypercube $\{0,1\}^n$ and the fact that injectivity implies the $2^n$ values are distinct integers.

(We do not use Round-1 Lemmas 2--3 in this round.)

3) New Insight / Tool (Round-2)

New tool: hypercube vertex-isoperimetry (Harper) + a Lipschitz property.

Let $f:\{0,1\}^n\to\mathbb Z$ be the subset-sum function $f(x)=\sum_{i=1}^n x_i a_i$. If $x$ and $y$ differ in exactly one coordinate $i$, then $|f(x)-f(y)|=a_i\le N$, so $f$ is $N$-Lipschitz on the hypercube graph. Harper's theorem lower-bounds the vertex boundary of any family of size $2^{n-1}$ by $\binom{n}{\lfloor n/2\rfloor}$. Combining these forces at least $\binom{n}{\lfloor n/2\rfloor}$ distinct subset sums into an interval of length $N$, yielding $N\ge \binom{n}{\lfloor n/2\rfloor}$.

4) Attack Plan (Round-2)

Gap after Round-1: prove/disprove an absolute-constant bound $N\ge c\,2^n$.

Round-2 target claim: prove the strongest general lower bound accessible from a clean anti-concentration argument: \[ N\ge \binom{n}{\lfloor n/2\rfloor}. \]

Plan. Choose a median threshold among the $2^n$ distinct subset sums and take the corresponding "lower half" family $\mathcal F$ of vertices. Points in the external vertex boundary $\partial\mathcal F$ must have subset sums within $N$ of the median threshold (Lipschitz). Thus $|\partial\mathcal F|\le N$. Harper gives $|\partial\mathcal F|\ge \binom{n}{\lfloor n/2\rfloor}$, forcing the desired bound.

5) Work (Round-2)

5.1 Setup

Let $A=\{a_1,\dots,a_n\}\subseteq\{1,\dots,N\}$, and assume all subset sums are distinct. Identify each subset with its indicator vector $x\in\{0,1\}^n$ and define \[ f(x):=\sum_{i=1}^n x_i a_i\in\mathbb Z. \] Injectivity of subset sums is equivalent to $f$ being injective on $\{0,1\}^n$.

Let $Q_n$ be the hypercube graph on vertex set $\{0,1\}^n$, with $x\sim y$ if they differ in exactly one coordinate. For $\mathcal F\subseteq\{0,1\}^n$ define the external vertex boundary \[ \partial\mathcal F:=\{y\in\{0,1\}^n\setminus \mathcal F: \exists x\in\mathcal F\text{ with }x\sim y\}. \]

5.2 Lipschitz property

If $x\sim y$ and they differ in coordinate $i$, then \[ |f(x)-f(y)|=a_i\le N. \] Hence $f$ is $N$-Lipschitz on $Q_n$.

5.3 Harper's vertex-isoperimetric inequality (special case)

We use the following standard statement.

Harper (special case). If $\mathcal F\subseteq\{0,1\}^n$ has size $|\mathcal F|=2^{n-1}$, then \[ |\partial\mathcal F|\ge \binom{n}{\lfloor n/2\rfloor}. \]

5.4 Median split and localization of the boundary

Since $f$ is injective, its values on $\{0,1\}^n$ are $2^n$ distinct integers. List them in increasing order: \[ s_0<s_1<\cdots<s_{2^n-1}. \] Let \[ t:=s_{2^{n-1}-1}. \] Define the "lower half" family \[ \mathcal F:=\{x\in\{0,1\}^n: f(x)\le t\}. \] By definition of $t$ we have $|\mathcal F|=2^{n-1}$.

Now let $y\in\partial\mathcal F$. Then $y\notin\mathcal F$, so $f(y)>t$, and there exists $x\in\mathcal F$ with $x\sim y$. By the Lipschitz property, \[ f(y)-f(x)=a_i\le N \] for the flipped coordinate $i$. Since $f(x)\le t$ and $f(y)>t$, we obtain \[ t< f(y)\le t+N. \] As $f(y)$ is an integer, this implies \[ f(y)\in \{t+1,t+2,\dots,t+N\}, \] an interval containing exactly $N$ integers.

Because $f$ is injective, distinct $y\in\partial\mathcal F$ give distinct values $f(y)$. Therefore \[ |\partial\mathcal F|\le N. \]

5.5 Conclusion: the sharp general lower bound

By Harper's inequality applied to $\mathcal F$ (which has size $2^{n-1}$), \[ |\partial\mathcal F|\ge \binom{n}{\lfloor n/2\rfloor}. \] Combined with $|\partial\mathcal F|\le N$, we conclude \[ N\ge \binom{n}{\lfloor n/2\rfloor}. \]

5.6 Asymptotic form (optional corollary)

By Stirling's approximation, \[ \binom{n}{\lfloor n/2\rfloor}=\left(\sqrt{\frac{2}{\pi}}+o(1)\right)\frac{2^n}{\sqrt n}, \] so the bound implies \[ N\ge \left(\sqrt{\frac{2}{\pi}}+o(1)\right)\frac{2^n}{\sqrt n}. \]

6) Adversarial Verification

1. Small $n$ checks. For $n=1$, the bound gives $N\ge 1$; for $n=2$, it gives $N\ge 2$; for $n=3$, it gives $N\ge 3$. All are consistent with known minima and with Round-1 computations.
2. Boundary notion. The argument uses the external vertex boundary. The localization step crucially uses $y\notin\mathcal F$ (hence $f(y)>t$) and adjacency to an $x\in\mathcal F$ (hence $f(y)\le f(x)+N\le t+N$). This matches the standard formulation of Harper used here.
3. Quantifier/tie issues at the median. Injectivity of $f$ guarantees strict ordering $s_0<\cdots<s_{2^n-1}$ and ensures $|\mathcal F|=2^{n-1}$ with no ties.
4. Off-by-one in interval length. The set $\{t+1,\dots,t+N\}$ has exactly $N$ integers. Since $f$ is injective, $|\partial\mathcal F|\le N$ is correct.

7) Final Status (exactly one)

UNRESOLVED (but strictly advanced).

We proved the sharp general lower bound \[ N\ge \binom{n}{\lfloor n/2\rfloor}=\left(\sqrt{\frac{2}{\pi}}+o(1)\right)\frac{2^n}{\sqrt n}, \] via a median-splitting argument on the hypercube combined with Harper's vertex-isoperimetric inequality and the $N$-Lipschitz property of the subset-sum map. This strengthens the Round-1 lower bounds $N\ge (2^n-1)/n$ and $N\ge \binom{n}{2}+1$, but still does not prove the conjectured $N\gg 2^n$.

8) Completion Estimate

COMPLETION: 70%

9) References

References

• [BloomErdos1] T. F. Bloom, Erdős Problem #1: Distinct subset sums (online problem page).
• [DFX] Q. Dubroff, J. Fox, and M. W. Xu, A note on the Erdős distinct subset sums problem, arXiv:2006.12988.