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Math-682-A2 · revision r1 · on Math-682

gpt pro 5.2 attempt on Math-682

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #120
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK
  4. 3) ATTACK PLAN
  5. 4) WORK
  6. Lemma 120.1 (unbounded $A$ case is trivial).
  7. Lemma 120.2 (dense-in-interval case via a fat Cantor set).
  8. 5) VERIFICATION (FAST REALITY CHECK)
  9. 6) FINAL

Erdős Problem #120

1) FORMAL RESTATEMENT

Let $A\subseteq\mathbb R$ be an infinite set. Does there exist a measurable set $E\subseteq\mathbb R$ of positive Lebesgue measure such that $E$ contains no set of the form \[ aA+b:=\{aa'+b:\ a'\in A\} \] for any real numbers $a\neq0$ and $b$?

Equivalently: must every infinite $A$ admit a positive-measure set $E$ that avoids every similar copy of $A$ (scaling and translation)?

2) QUICK LITERATURE/CONTEXT CHECK

From the problem text:
  • The conjecture is true if $A$ is unbounded or dense in some interval.
  • It suffices to treat $A$ a decreasing sequence tending to $0$.
  • It is false for finite $A$ (Steinhaus).
  • The case $A=\{1,1/2,1/4,\dots\}$ is open.
No further sources are used here.

3) ATTACK PLAN

Prove the easy cases (unbounded $A$; $A$ dense in an interval) in a fully self-contained way, as sanity checks. The remaining difficult case is when $A$ is countable and accumulates only at $0$.

4) WORK

Lemma 120.1 (unbounded $A$ case is trivial).
If $A$ is unbounded, then there exists a positive-measure set $E$ containing no copy $aA+b$.

Proof. Take any bounded measurable set $E$ of positive measure, e.g. $E=[0,1]$. For any $a\neq0$ and $b$, if $A$ is unbounded then $aA+b$ is also unbounded (since $|aa'+b|\to\infty$ along some sequence $a'\in A$). Hence $aA+b$ cannot be a subset of the bounded set $E$. Therefore $E$ avoids every $aA+b$. $\square$

Lemma 120.2 (dense-in-interval case via a fat Cantor set).
Suppose $A$ is dense in some interval $I\subset\mathbb R$ with nonempty interior. Then there exists a measurable set $E$ of positive measure that contains no copy $aA+b$.

Proof. Fix such an interval $I$. For any $a\neq0$ and $b$, the set $aA+b$ is dense in the interval $aI+b$ because the map $x\mapsto ax+b$ is a homeomorphism and carries a dense set to a dense set.

Therefore, if a set $E$ contains $aA+b$, then $\overline{E}$ (the closure) contains the interval $aI+b$, hence $\overline{E}$ has nonempty interior.

Choose $E$ to be a fat Cantor set: a closed nowhere dense subset of $[0,1]$ with positive Lebesgue measure. Such a set can be constructed by iteratively removing open middle intervals of total removed length $<1/2$, leaving a closed set of measure $>1/2$ but with empty interior.

Since $E$ is nowhere dense, its closure is itself and has empty interior, so it cannot contain any interval. In particular, $E$ cannot contain any dense subset of a nontrivial interval. Therefore $E$ cannot contain any set $aA+b$ (each such set is dense in $aI+b$). Thus this $E$ has positive measure and avoids all similar copies of $A$. $\square$

5) VERIFICATION (FAST REALITY CHECK)

  • Lemma 120.1: if $A$ unbounded and $a\neq0$, then $aA+b$ unbounded; bounded $E$ excludes it.
  • Lemma 120.2: the key implication is "$aA+b$ dense in an interval $\Rightarrow$ any containing set has closure with interior". This is immediate from density.
  • Existence of fat Cantor sets is standard; the described removal process gives a closed set of positive measure with empty interior.

6) FINAL

UNRESOLVED.

(i) Strongest fully proved partial result obtained here. A complete proof of the conjecture in the two easy regimes explicitly mentioned in the problem text: $A$ unbounded (Lemma 120.1) and $A$ dense in an interval (Lemma 120.2).

(ii) Exact first gap. Handle the "hard" remaining case: $A$ countable, strictly monotone, and converging to $0$ (e.g. geometric sequences like $\{1,1/2,1/4,\dots\}$).

(iii) Top 3 next moves.

  1. For $A$ decreasing to $0$, try to construct $E$ by a scale-by-scale Cantor-type removal that blocks copies at each scale $a$.
  2. Identify quantitative density properties of $A$ (e.g. multiplicative structure) that force copies $aA+b$ to intersect most positive-measure sets.
  3. Explore Fourier-analytic criteria for a set $E$ to avoid dilates/translates of a given sequence.

(iv) Minimal counterexample structure. A counterexample to the conjecture would be an infinite set $A$ such that every positive-measure set $E$ contains some similar copy $aA+b$. By the reductions in the problem text, a minimal such $A$ would likely be a decreasing sequence tending to $0$ with strong self-similarity (e.g. geometric progressions).