The Hurewicz covering property and slaloms in the Baire space

According to a result of Kocinac and Scheepers, the Hurewicz covering property is equivalent to a somewhat simpler selection property: For each sequence of large open covers of the space one can choose finitely many elements from each cover to obtain a groupable cover of the space. We simplify the characterization further by omitting the need to consider sequences of covers: A set of reals $X$ satisfies the Hurewicz property if, and only if, each large open cover of $X$ contains a groupable subcover. This solves in the affirmative a problem of Scheepers. The proof uses a rigorously justified abuse of notation and a "structure" counterpart of a combinatorial characterization, in terms of slaloms, of the minimal cardinality b of an unbounded family of functions in the Baire space. In particular, we obtain a new characterization of \b.Comment: Small update

Topological diagonalizations and Hausdorff dimension

The Hausdorff dimension of a product XxY can be strictly greater than that of Y, even when the Hausdorff dimension of X is zero. But when X is countable, the Hausdorff dimensions of Y and XxY are the same. Diagonalizations of covers define a natural hierarchy of properties which are weaker than ``being countable'' and stronger than ``having Hausdorff dimension zero''. Fremlin asked whether it is enough for X to have the strongest property in this hierarchy (namely, being a gamma-set) in order to assure that the Hausdorff dimensions of Y and XxY are the same. We give a negative answer: Assuming CH, there exists a gamma-set of reals X and a set of reals Y with Hausdorff dimension zero, such that the Hausdorff dimension of X+Y (a Lipschitz image of XxY) is maximal, that is, 1. However, we show that for the notion of a_strong_ gamma-set the answer is positive. Some related problems remain open.Comment: Small update

Rothberger bounded groups and Ramsey theory

We show that: 1. Rothberger bounded subgroups of sigma-compact groups are characterized by Ramseyan partition relations. 2. For each uncountable cardinal $\kappa$ there is a ${\sf T}_0$ topological group of cardinality $\kappa$ such that ONE has a winning strategy in the point-open game on the group and the group is not a subspace of any sigma-compact space. 3. For each uncountable cardinal $\kappa$ there is a ${\sf T}_0$ topological group of cardinality $\kappa$ such that ONE has a winning strategy in the point-open game on the group and the group is \sigma-compact.Comment: 11 page

A note on unbounded strongly measure zero subgroups of the Baer–Specker group

AbstractWe show that it is consistent with ZFC that there exist:(1)An unbounded (with respect to ⩽∗) and strongly measure zero subgroup of ZN, but without the Menger property.(2)An unbounded (with respect to ⩽∗) and strongly measure zero subgroup of ZN with the Menger property which does not have the Rothberger property. This answers the last two problems which remained from a classification project of Machura and Tsaban

Strong measure zero and meager-additive sets through the prism of fractal measures

We develop a theory of \emph{sharp measure zero} sets that parallels Borel's \emph{strong measure zero}, and prove a theorem analogous to Galvin-Myscielski-Solovay Theorem, namely that a set of reals has sharp measure zero if and only if it is meager-additive. Some consequences: A subset of $2^\omega$ is meager-additive if and only if it is $\mathcal E$-additive; if $f:2^\omega\to2^\omega$ is continuous and $X$ is meager-additive, then so is $f(X)$.Comment: arXiv admin note: text overlap with arXiv:1208.552

Strong measure zero and meager-additive sets through the prism of fractal measures

summary:We develop a theory of sharp measure zero sets that parallels Borel's strong measure zero, and prove a theorem analogous to Galvin--Mycielski--Solovay theorem, namely that a set of reals has sharp measure zero if and only if it is meager-additive. Some consequences: A subset of $2^{\omega}$ is meager-additive if and only if it is $\mathcal E$-additive; if $f\colon 2^{\omega}\to 2^{\omega}$ is continuous and $X$ is meager-additive, then so is $f(X)$