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Statement

Notation (p. 1). $\dim$ is Hausdorff dimension and $E^k$ is the $k$-fold Cartesian product of $E$. From the Dimension Inequality $\dim(A\times B)\ge\dim A+\dim B$ for Borel $A\subseteq\mathbb R^n$, $B\subseteq\mathbb R^m$, the paper gets $\dim(A^{n+m})\ge\dim(A^n)+\dim(A^m)$, so $(1/n)\dim(A^n)$ converges as $n\to\infty$ to $\sup_n(1/n)\dim(A^n)$; this limit is the Cartesian--Hausdorff (CH) dimension of $A$. The CH dimension of $A$ is $0$ exactly when $\dim(A^n)=0$ for every positive integer $n$, and then in particular $\dim A=0$.

Theorem 1 (p. 1). "Let $E\subseteq\mathbb R$ be a subring and a Borel set. Then either $E$ has CH dimension zero or $E=\mathbb R$."

Analytic sets (Remarks, pp. 3--4). The paper states that Theorem 1 holds for analytic sets as well: a subring $E\subseteq\mathbb R$ that is an analytic set has CH dimension zero or equals $\mathbb R$. It supports this by noting that the three Borel-set inputs of the proof extend to analytic sets: the Dimension Inequality (through compact subsets of nearly full dimension), the Projection Theorem (the same way), and the Borel measurability of the inverse of a Borel measurable bijection. It then draws a consequence: for an analytic set $X\subseteq\mathbb R$ the ring $\mathbb Z[X]$ it generates is analytic, and if $\dim X^k>0$ for some $k$ then $\mathbb Z[X]=\mathbb R$. If $X$ is moreover compact, the Baire Category Theorem gives an $n$ and an open interval $I$ such that every element of $I$ is a sum of at most $n$ terms, each plus or minus a product of at most $n$ elements of $X$ (empty sum $0$, empty product $1$).

Source. G. A. Edgar and Chris Miller, Borel subrings of the reals, Proc. Amer. Math. Soc. 131 (2003), no. 4, 1121--1129, DOI 10.1090/S0002-9939-02-06653-4. Theorem 1 and the definition of CH dimension on p. 1, Lemmas 1.1--1.4 on pp. 2--3, the proof of Theorem 1 on p. 3, the Remarks on pp. 3--4. Pages and labels are those of the authors' nine-page preprint identified on the source card; the journal edition was not compared.

Read depth. Claims checked: the definition, the theorem, the statements of Lemmas 1.1--1.4 and the Remarks were read clause by clause on the page images. The proofs of the lemmas were read for structure only; nothing here is independently reviewed.

Proof pointer

Pages 2--3. Suppose $E$ has nonzero CH dimension. Then some power has positive dimension, and by the Dimension Inequality some $E^k$ has dimension greater than $1$. Lemma 1.1 (p. 2), a special case of the Projection Theorem, says a Borel $A\subseteq\mathbb R^k$ with $\dim A>1$ has image of positive Lebesgue measure under almost every linear functional $\mathbb R^k\to\mathbb R$; the image of $E^k$ is then an additive subgroup of positive measure, and Steinhaus's theorem makes it all of $\mathbb R$ (Lemma 1.2, p. 2, for a Borel additive subgroup). Lemma 1.3 (pp. 2--3) uses the ring structure: taking $k$ least, a nontrivial relation $\sum b_jr_j=0$ with $b_j\in E$ among the images $r_j$ of the coordinate vectors would let one coordinate be dropped, so the functional is injective on $E^k$. Lemma 1.4 (p. 3): if a linear functional maps $E^k$ bijectively onto $\mathbb R$ for a Borel additive subgroup $E$, its inverse is Borel measurable, so the first coordinate of the inverse is a Borel measurable additive map $\mathbb R\to\mathbb R$, hence $x\mapsto cx$ with $c\ne0$; it cannot vanish at the image of a second coordinate vector, so $k=1$ and $E=\mathbb R$. The proof of Theorem 1 (p. 3) chains Lemmas 1.2, 1.3 and 1.4.

Dependencies

The Dimension Inequality (Mattila, Geometry of sets and measures in euclidean spaces, Thm. 8.10; Falconer, Fractal geometry, 7.2); the Projection Theorem (Mattila, Cor. 9.8); Steinhaus's theorem on difference sets of sets of positive measure; the Borel measurability of the inverse of a Borel measurable bijection (Cohn, Measure theory, Prop. 8.3.5; Kechris, Classical descriptive set theory, 15.2); and the linearity of Borel measurable additive maps $\mathbb R\to\mathbb R$ (Kechris, 9.10, among the paper's references). The analytic extension cites Davies, Subsets of finite measure in analytic sets (1952), for compact subsets of nearly full dimension, and Cohn, Prop. 8.6.2, for Borel isomorphism.

Bears on

• Problem 1154, which asks whether each $\alpha\in[0,1]$ is the Hausdorff dimension of some ring or field in $\mathbb R$: by the theorem and its analytic extension, a subring of $\mathbb R$ that is Borel or analytic, and so in particular such a subfield, has Hausdorff dimension $0$ or is $\mathbb R$. No witness for $0<\alpha<1$ can therefore be Borel or analytic. The theorem says nothing about rings outside these classes, and it decides no $\alpha$ in $(0,1)$. The problem page's account of a proof claim on the site invokes the generated-ring consequence of the Remarks.