gpt pro 5.2 attempt on Math-5955

FORMAL RESTATEMENT. Let $A\subseteq\mathbb{N}$ (here $\mathbb{N}=\{0,1,2,\dots\}$ or $\{1,2,3,\dots\}$; the distinction does not affect "all sufficiently large integers"). Say $A$ is an asymptotic additive basis of order $2$ if there exists $n_0$ such that every integer $n\ge n_0$ can be written as $n=a+a'$ with $a,a'\in A$. Let \[ r_A(n):=(1_A*1_A)(n)=\left|\{(a,a')\in A\times A: a+a'=n\}\right|, \] (the ordered representation function). A subset $B\subseteq A$ is a minimal asymptotic basis of order $2$ if $B$ is an asymptotic basis of order $2$ and for every $b\in B$, the set $(B\setminus\{b\})+(B\setminus\{b\})$ fails to contain infinitely many integers.

Question: If $A$ is an asymptotic basis of order $2$ and $r_A(n)\to\infty$ as $n\to\infty$, must $A$ contain a minimal asymptotic basis of order $2$? What if $r_A(n)\ge \varepsilon\log n$ for all large $n$?

QUICK LITERATURE/CONTEXT CHECK. The problem statement records:

• Erdős--Nathanson $1979$ proved the conclusion under the stronger hypothesis $r_A(n)>(\log(4/3))^{-1}\log n$ for all large $n$.
• Härtter $1956$ and Nathanson $1974$ constructed additive bases containing no minimal additive bases.
• Erdős--Nathanson $1989$ constructed, for any fixed $t$, bases $A$ with $r_A(n)\ge t$ for all large $n$ but containing no minimal asymptotic basis of order $2$.

(Per the project integrity constraints, I do not use or assert further literature beyond what is in the problem text.)

ATTACK PLAN.

• Prove basic implications of $r_A(n)\to\infty$, e.g. robustness under deleting finitely many elements.
• Try to formalize a Zorn/chain-minimization approach: show every descending chain of asymptotic bases inside $A$ has an intersection that is still an asymptotic basis, which would yield a minimal element.
• Disproof track: attempt to amplify known constructions with bounded $r_A(n)$ to force $r_A(n)\to\infty$ while still forbidding minimal subbases.

I only establish basic robustness lemmas; I do not resolve existence of a minimal subbasis.

WORK.

Lemma 868.1 (Finite deletions preserve the basis property under $r_A(n)\to\infty$). Assume $A$ is an asymptotic basis of order $2$ and $r_A(n)\to\infty$ as $n\to\infty$. Then for every finite set $F\subseteq A$, the set $A\setminus F$ is also an asymptotic basis of order $2$.

Proof. Fix a finite $F\subseteq A$ and put $m:=|F|$. For each integer $n$, let $R(n)$ be the set of ordered representations $(a,a')\in A\times A$ with $a+a'=n$, so $|R(n)|=r_A(n)$. Let $R_F(n)\subseteq R(n)$ be those representations that use at least one summand from $F$. If $f\in F$, then in an ordered representation $(a,a')$ of $n$ the condition $a=f$ forces $a'=n-f$ uniquely, and likewise $a'=f$ forces $a=n-f$ uniquely. Therefore, for each $n$, \[ |R_F(n)|\le 2m. \] Since $r_A(n)\to\infty$, there exists $n_1$ such that for all $n\ge n_1$, $r_A(n)>2m$. For such $n$, we have $|R(n)\setminus R_F(n)|\ge r_A(n)-2m\ge 1$, so there exists an ordered representation $(a,a')$ of $n$ with $a,a'\notin F$. Thus $n\in (A\setminus F)+(A\setminus F)$ for all $n\ge n_1$, i.e. $A\setminus F$ is an asymptotic basis of order $2$. $\square$

Lemma 868.2 (Minimality $\Leftrightarrow$ infinitely many "forced" representations). Let $B$ be an asymptotic basis of order $2$. Then $B$ is minimal (asymptotic) if and only if for every $b\in B$ there are infinitely many integers $n$ such that every representation of $n$ as $n=x+y$ with $x,y\in B$ satisfies $x=b$ or $y=b$.

Proof. Fix $b\in B$. If $B$ is minimal, then by definition $B\setminus\{b\}$ fails to represent infinitely many integers $n$. For each such $n$, every representation of $n$ using $B$ must involve $b$ (otherwise $n$ would still be representable after deleting $b$). Thus the "forced representation" property holds for infinitely many $n$.

Conversely, suppose the forced-representation property holds for $b$ on an infinite set $S\subseteq\mathbb{N}$. Then for each $n\in S$, there is no representation of $n$ using only $B\setminus\{b\}$, so $n\notin (B\setminus\{b\})+(B\setminus\{b\})$. Hence deleting $b$ creates infinitely many missing integers, so $B$ is minimal. $\square$

FAST REALITY CHECK (sanity examples).

• If $A=\mathbb{N}$, then $r_A(n)=n+1\to\infty$, and Lemma 868.1 correctly says deleting finitely many elements preserves being an asymptotic basis.
• Lemma 868.2 matches the intuition: in a minimal basis, each element must be essential for infinitely many target integers.

VERIFICATION.

• Lemma 868.1: the only nontrivial estimate is $|R_F(n)|\le 2|F|$, justified by the uniqueness of a partner for a fixed summand in an ordered sum representation.
• Lemma 868.2: both implications are direct unpacking of the definition, with no hidden compactness assumptions.

FINAL. UNRESOLVED

(i) Strongest proved partial result. If $A$ is an asymptotic basis of order 2 and $r_A(n)\to\infty$, then $A$ is robust under deleting any finite subset (Lemma 868.1). Minimality can be characterized by having infinitely many integers whose representations are forced to use each given element (Lemma 868.2).

(ii) First gap (crisp). Show that the family of asymptotic bases contained in $A$ has a minimal element under inclusion, e.g. by proving that the intersection of any descending chain of such bases is still an asymptotic basis. This is exactly where a Zorn-type approach would need a verified chain-intersection property.

(iii) Top 3 next moves.

• Attempt to prove a chain-intersection lemma under quantitative growth (e.g. $r_A(n)\ge \varepsilon\log n$): if $A=A_0\supseteq A_1\supseteq A_2\supseteq\cdots$ is a nested sequence of asymptotic bases with each $A_{m+1}=A_m\setminus\{a_m\}$, show $\bigcap_m A_m$ still represents all large $n$.
• Try to adapt (or reprove from scratch) the Erdős--Nathanson threshold $(\log(4/3))^{-1}\log n$ as an explicit "sufficient slack" for iterative deletions.
• Construct candidate counterexamples by building $A$ in blocks so that $r_A(n)$ grows slowly, but every purported subbasis can be thinned further without losing the basis property.

(iv) Minimal counterexample structure. A counterexample would be an asymptotic basis $A$ with $r_A(n)\to\infty$ but such that for every $B\subseteq A$ which is an asymptotic basis, there exists $b\in B$ with $B\setminus\{b\}$ still an asymptotic basis (so no minimal element). Such an $A$ would need representations to be plentiful yet "spread" so that no single element is forced infinitely often inside any subbasis.