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Statement
Theorem (p. 2). The paper states it as "Every analytic real closed proper subfield of has dimension ", where "dimension" means Hausdorff dimension (p. 1).
Here a subset of is analytic when it is the continuous image of a Borel subset of , and an ordered field is real closed when each of its positive elements has a square root in it and each odd-degree polynomial in one variable with coefficients in it has a root in it (p. 1). Every Borel subset of is analytic (p. 1).
Stronger form proved (p. 2, proof pp. 3--4). If is analytic and , then is contained in no proper real closed subfield of . The paper restates this as: contains a transcendence base for , a maximal algebraically independent subset of (abstract and p. 2). The Theorem follows by taking .
What it gives for subfields (p. 2). A real closed subfield of that is a Borel set, or more generally an analytic set, has Hausdorff dimension or , and dimension only when it is itself.
The converse fails (p. 2). Dimension does not keep an analytic set inside a proper real closed subfield: there are compact of dimension whose sum set has interior, so lies in no proper additive subgroup of . The paper takes with from Falconer's Fractal geometry (1990), Example 7.8.
Source. G. A. Edgar and Chris Miller, Hausdorff dimension, analytic sets and transcendence, Real Anal. Exchange 27 (2001/02), no. 1, 335--339. Page numbers are those of the authors' four-page preprint identified on the source card; the journal edition was not compared.
Read depth. Claims checked: the statement, the stronger form, the four lemmas and the proof of the Theorem were read clause by clause. Lemma 4 is stated in the paper without proof, and the facts the lemmas cite from Mattila, Oxtoby, Edgar and real algebraic geometry were not re-derived. Nothing here is independently reviewed.
Proof pointer
Pages 2--4. The paper says the Theorem is immediate from four lemmas.
- Lemma 1 (p. 2): for compact with there are and an -linear with having interior in . One picks with , so , takes an orthogonal projection whose image of has positive Lebesgue measure, and uses that the difference set of a set of positive measure has interior; then .
- Lemma 2 (p. 2): the same conclusion for analytic with , since such contains a compact set of positive dimension. The Remark after it notes that if is also an additive subgroup then .
- Lemma 3 (p. 3): if is analytic, the smallest real closed subfield of containing is analytic. It is the union of the images over the countably many semialgebraic defined over , and cell decomposition makes each image a finite union of continuous images of analytic sets. The Remark after it, credited to R. Dougherty, says the lemma fails with "Borel" in place of "analytic", even when is a subring, while the real closure of a Borel subfield is Borel.
- Lemma 4 (p. 3), a special case of van den Dries, Dense pairs of o-minimal structures, Fund. Math. 157 (1998), Lemma 4.1, stated without proof: if are real closed subfields of , , and is semialgebraic and defined over , then has empty interior in .
- Proof of the Theorem (pp. 3--4): for analytic with , the smallest real closed field is analytic by Lemma 3 and has positive dimension; Lemma 2 gives a linear, hence semialgebraic, with having interior; Lemma 4 with forces .
Dependencies
Lemmas 1--4 of the same paper. Through them: Mattila's Geometry of sets and measures in Euclidean spaces (1995) for the product-dimension bound and the projection theorem, Oxtoby's Measure and category for the difference-set theorem, Edgar's Integral, probability and fractal measure (1998) for compact subsets of analytic sets, the cell decomposition theorem of real algebraic geometry, and van den Dries's Lemma 4.1 cited above.
Bears on
- Problem 1154, which asks whether every is the Hausdorff dimension of some ring or field in : the Theorem shows that no real closed subfield of that is an analytic set, in particular a Borel set, has dimension strictly between and . The problem does not restrict the ring or field to analytic sets or to real closed fields, so the Theorem does not answer it.