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Source. Two write-ups of the same construction: E. Glazer, Erdős Problem 501 after adding $\omega_2$ random reals, draft rev10, Section 6, physical p. 8, held by Glazer (2026); and S. Lee, Relative independence of Erdős problem #501, second version dated 2026-06-01, Appendix A, physical pp. 5--6, held by Lee (2026). Both attribute the result to S. H. Hechler, On two problems in combinatorial set theory, Bull. Acad. Polon. Sci. 20 (1972), 429--431, which is not held; the problem page records the open attribution question.

Standing. This is an author-recorded reconstruction. It is not an independent review and changes no status and assigns no tier.

Statement

Assume CH. There is a family $(A_y)_{y\in\mathbb R}$ such that every $A_y$ is countable (so $\lambda^*(A_y)=0<1$) and bounded, and no infinite $X\subseteq\mathbb R$ satisfies $x\notin A_y$ for all distinct $x,y\in X$. Hence CH implies $\neg P$, where $P$ is the positive assertion of the first question of Problem 501.

Proof

By CH, $|\mathbb R|=\aleph_1$; fix an enumeration $\mathbb R=\{r_\alpha:\alpha<\omega_1\}$ without repetition, and let $\prec$ be the induced well-ordering: $r_\alpha\prec r_\beta$ if and only if $\alpha<\beta$. For $y=r_\beta$ define

$$A_y=\{r_\alpha:\alpha<\beta,\ |r_\alpha|\le|y|+1\}.$$

Each $A_y$ is a subset of the countable set $\{r_\alpha:\alpha<\beta\}$, so it is countable and therefore Lebesgue null: $\lambda^*(A_y)=0$. Each $A_y$ is contained in $[-(|y|+1),|y|+1]$, so it is bounded.

Suppose $X\subseteq\mathbb R$ is infinite and independent. Since $\prec$ well-orders $\mathbb R$ and $X$ is infinite, $X$ contains a strictly increasing sequence $x_0\prec x_1\prec x_2\prec\cdots$ (its first $\omega$ elements in the order $\prec$). Let $i<j$ and write $x_i=r_\alpha$, $x_j=r_\beta$, so $\alpha<\beta$. Independence gives $x_i\notin A_{x_j}$. By the definition of $A_{x_j}$, a point $r_\alpha$ with $\alpha<\beta$ lies outside $A_{x_j}$ only when $|r_\alpha|>|x_j|+1$. Hence

$$|x_i|>|x_j|+1\qquad(i<j).$$

Applying this to consecutive indices, $|x_0|>|x_1|+1>|x_2|+2>\cdots$, so $|x_n|<|x_0|-n$ for every $n\ge1$. For an integer $n>|x_0|$ this gives $|x_n|<0$, which is impossible. So no infinite independent set exists.

Boundary. The sets are null, so the construction refutes even the variant of the first question with "outer measure below one" replaced by "null"; the problem page records the same construction along a well-ordering of order type $\mathfrak c$ under Martin's axiom. Both theorem pages, Glazer Theorem 1.1 and Lee Theorem 1.1, use this page for the negative half of their corollaries.