A singleton obstruction to the Katz critical-set question

Definitions and mathematical statement

For a finite subset $X$ of an abelian group and integer $n\geq1$, write $nX=\{x_1+\cdots+x_n:x_i\in X\}$. The source, attributed to N. Katz, asks whether there exists finite $A\subseteq{\mathbb{Z}}$ such that \[ |2A|=k|A|,\qquad |nA'|\geq f(n,k)|A'|\quad\text{for every proper }A'\subsetneq A, \] where $f(n,k)$ is defined in the preceding Problem 3.16. That preceding definition is not included in this record. The supplied verification emphasizes that its defining doubling constraint is strict, whereas this question places $A$ on the equality boundary; the inequality for proper subsets is indeed $\geq$, not an extraction reversal.

The aid treats a literal reading with fixed integer $n\geq2$, real $k>1$, and $|A|\geq2$. To make one natural extremal reading explicit, define \[ f_{\mathrm{nat}}(n,k)=\sup\left\{\frac{|nX|}{|X|}:\varnothing\ne X\subseteq{\mathbb{Z}}\text{ finite},\ |2X|<k|X|\right\}. \] This displayed definition is a conditional reconstruction of the absent previous problem, not a verified quotation of it. Under this reading the reported singleton obstruction is \[ f_{\mathrm{nat}}(n,k)=1\quad\Longleftrightarrow\quad1<k\leq\tfrac32. \] Since every singleton is a proper subset of $A$ when $|A|\geq2$, the required inequality forces $f_{\mathrm{nat}}(n,k)\leq1$. The aid's reported classification for this reading is \[ \exists A\subseteq{\mathbb{Z}}\text{ finite},\ |A|\geq2,\ |2A|=k|A|, \ \forall\varnothing\ne A'\subsetneq A:\ |nA'|\geq f_{\mathrm{nat}}(n,k)|A'| \quad\Longleftrightarrow\quad k=\tfrac32. \] At $k=3/2$, precisely the two-element sets $A$ work. Thus, if $k$ was intended to be an integer at least two, this reading has a uniformly negative answer. This is not a classification for an unspecified different definition of $f$, for $n=1$, or for alternative subset conventions.

The aid also gives a cutoff obstruction independent of that missing definition. For every integer $n\geq1$, every finite $A$ in any abelian group, and every integer $1\leq r<|A|$, it reports \[ \min_{\substack{A'\subsetneq A\\|A'|\geq r}} \frac{|nA'|}{|A'|}\leq\frac1r\binom{n+r-1}{r-1}. \] Thus a repair requiring the inequality only for proper subsets of cardinality at least $r$ still cannot demand a larger universal lower-growth factor than this cap. The case $r=1$ recovers the singleton obstruction. The source's unqualified reference to every proper subset also includes the empty subset; with the usual $n\varnothing=\varnothing$ for $n\geq1$ and a finite factor, that case adds only $0\geq0$.

Short English statement

The preceding definition of $f(n,k)$ is missing. For the aid's natural strict-doubling extremal reading, singleton proper subsets collapse the desired lower bound, leaving exactly two-element sets at doubling ratio $3/2$. A separate cardinality-cutoff bound applies in every abelian group.

Sources

• Local registry: permanent UnsolvedMath ID $20001129$, exact public code catalogue entry; aim-combinatorics-notes.json, notes[253].
• AIM workshop "Recent trends in additive combinatorics", Problem 3.17, attributed to N. Katz: https://aimath.org/WWN/additivecomb/additivecomb.pdf. It refers to the function from Problem 3.16, whose full text is not supplied here. The literature field is empty.
• The supplied verification reports that the official source confirms the strict doubling inequality in 3.16, equality boundary in 3.17, and lower-bound direction $\geq$ for all proper subsets. This is a supplied verification claim, not independent PDF access by this draft.
• Research attempt 1: partial_result, status at run unknown, rigor label proved, novelty confidence low, classification PARTIAL-PROGRESS. It reports the classification for fixed $n\geq2$, real $k>1$, and the general abelian-group cutoff obstruction. No authoritative separate publication citation is supplied.