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Statement

A map $f:\mathbb R\to\mathbb R$ is bi-Lipschitz if for some constant $L\ge1$, $L^{-1}|x-y|\le|f(x)-f(y)|\le L|x-y|$ for all $x,y$; a set $A$ is bi-Lipschitz measure universal if for every measurable set $E$ of positive Lebesgue measure some bi-Lipschitz $f$ has $f(A)\subset E$ (p. 5).

Theorem 2.1 (Feng--Lai--Xiong; p. 5). Let $A=(a_n)_{n=1}^\infty$ be a strictly decreasing sequence converging to $0$, and let $E$ be a measurable set of positive Lebesgue measure on $\mathbb R^1$.

1. If $\lim_{n\to\infty}a_{n+1}/a_n=1$, then $A$ is not bi-Lipschitz measure universal.
2. If $\limsup_{n\to\infty}a_{n+1}/a_n<1$, then there is a bi-Lipschitz map $f:\mathbb R\to\mathbb R$ with $f(A)\subset E$.

An affine map with nonzero slope is bi-Lipschitz, so part (1) gives a new proof of Theorem 1.3 (p. 5). The survey adds that with more care the map in part (2) can be chosen with $f'(0)=1$ (p. 8).

Source. Yeonwook Jung, Chun-Kit Lai and Yuveshen Mooroogen, Fifty years of the Erdős similarity conjecture, arXiv:2412.11062v2 (1 January 2025), whose labels and page numbers are cited here; the edition is identified on the source card. The theorem is from De-Jun Feng, Chun-Kit Lai and Ying Xiong, Erdős similarity problem via bi-Lipschitz embedding, Int. Math. Res. Not. IMRN (2024), no. 17, 12327--12342.

Read depth. Claims checked: the statement and definitions were read clause by clause on p. 5, and the survey's proofs (pp. 5--8) were followed through their displayed estimates; nothing here is independently reviewed, and the paper of Feng, Lai and Xiong was not read.

Proof pointer

Part (1), pp. 5--7. Lemma 2.2 (p. 5) passes to a subsequence that is still sublacunary and whose consecutive gaps are, up to a factor $2$, nonincreasing. Choosing indices $n_k$ where the relative gap is at most $k^{-2}4^{-k}$, the proof removes from $[0,1]$ about $a_{n_k}^{-1}k$ evenly spaced gaps of length $\delta_k=k(a_{n_k}-a_{n_k+1})$ at level $k$; the intersection $E$ has measure at least $1/3$. A bi-Lipschitz image of the tail of the sequence moves in steps shorter than $\delta_k$ for $k>2L$, so it cannot cross a level-$k$ gap and stays in one component of length below $a_{n_k}/k$, while its distance to the limit point is at least $a_{n_k}/L$, a contradiction.

Part (2), pp. 7--8. Translate a density point of $E$ to $0$, take $\delta<1$ bounding the ratios $a_{n+1}/a_n$ (display (2.9)) and fix $\eta\in(\delta,1)$; for large $n$ the disjoint intervals $[\eta a_n,a_n]$ meet $E$ by the density theorem, so a point $b_n$ of $E$ is chosen in each, and the piecewise linear map through $(a_n,b_n)$ has slopes bounded above and below.

Dependencies

Lemma 2.2 (p. 5) and the Lebesgue density theorem.

Bears on

• Problem 120: part (1) answers the question affirmatively for strictly decreasing sublacunary sequences, as Theorem 1.3 does. Part (2) concerns the weaker bi-Lipschitz embedding: it shows that this relaxation cannot avoid sequences with $\limsup a_{n+1}/a_n<1$, such as $2^{-n}$, and says nothing about whether a set of positive measure contains an affine copy of them. It does not settle the problem.