Erdős Problem #120

Erdős Problem 120: avoiding infinite affine patterns

Statement and scope

For every infinite $A\subset\mathbb R$, is there a measurable set $E$ of positive Lebesgue measure containing no set $aA+b$ with $a\ne0$? This is the Erdős similarity problem. We prove two classes of infinite patterns can be avoided, explain why finite patterns cannot, and reduce the difficult case to sequences converging to zero. The general question is unresolved here.

The bounded and nowhere-dense constructions retain the valid parts of the earlier GPT-5.2 attempt; the finite-pattern proof and reduction are included to identify the gap precisely. Source statement: https://www.erdosproblems.com/120. No new resolution is claimed.

Two complete avoidance cases

If $A$ is unbounded, every $aA+b$ with $a\ne0$ is unbounded, so $E=[0,1]$ avoids it.

Suppose the closure of $A$ contains a nondegenerate interval. Enumerate the rationals of $[0,1]$ as $q_1,q_2,\ldots$ and remove the open intervals of radii $2^{-j-4}$ about them. The remaining set \[ E=[0,1]\setminus\bigcup_{j\ge1}(q_j-2^{-j-4},q_j+2^{-j-4}) \] is closed, has empty interior, and has measure at least $1-\sum_{j\ge1}2^{-j-3}=7/8$. If $aA+b\subset E$, closedness gives $a\overline A+b\subset E$, contradicting its empty interior. Thus this $E$ avoids every affine copy of $A$.

Why the infinite hypothesis matters

Every measurable set $E$ of positive measure contains an affine copy of every finite set $F=\{u_1,\ldots,u_k\}$. Choose a bounded measurable $E_0\subset E$ of finite positive measure. Translations are continuous in $L^1$: \[ \mu((E_0-h)\mathbin{\triangle}E_0)\longrightarrow0\quad(h\to0). \] One proof approximates $E_0$ in measure by a finite union of bounded intervals, for which the assertion follows by adding the changes at finitely many endpoints.

Choose a sufficiently small nonzero $a$ so that $\mu(E_0\setminus(E_0-au_i))<\mu(E_0)/(2k)$ for every $i$. The union bound gives \[ \mu\left(E_0\cap\bigcap_{i=1}^k(E_0-au_i)\right)\ge\mu(E_0)/2>0. \] Any $b$ in this intersection satisfies $b+aF\subset E$. The same estimate cannot be applied to an infinite collection with a fixed nonzero $a$.

A reduction of the remaining case

Every infinite bounded subset of $\mathbb R$ has a sequence of distinct points converging to a limit $c$. Passing to points on one side of $c$ and then to a subsequence, their distances from $c$ can be made strictly decreasing to zero. Translation and, if necessary, reflection turn that subsequence into \[ b_1>b_2>\cdots>0,\qquad b_j\to0. \] If some positive-measure set avoids every affine copy of this normalized sequence, it avoids every affine copy of the original $A$: any such copy of $A$ would contain the corresponding copy of the subsequence. Together with the unbounded case, this shows that it suffices to settle all strictly decreasing positive sequences tending to zero.

Remaining gap

A convergent sequence can lie in a closed nowhere-dense set, so the second construction supplies no contradiction for this reduced case. The finite-pattern argument gives potentially different, shrinking scales for different prefixes; it supplies no common nonzero scale for an infinite pattern. Even the geometric-sequence case is not established here.

COMPLETION ESTIMATE

COMPLETION: 30%

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).