How to avoid a compact set
classification
🧮 math.LO
math.MG
keywords
dimensionmathbbeverycompacthausdorffclosedoperationssubset
read the original abstract
A first-order expansion of the $\mathbb{R}$-vector space structure on $\mathbb{R}$ does not define every compact subset of every $\mathbb{R}^n$ if and only if topological and Hausdorff dimension coincide on all closed definable sets. Equivalently, if $A \subseteq \mathbb{R}^k$ is closed and the Hausdorff dimension of $A$ exceeds the topological dimension of $A$, then every compact subset of every $\mathbb{R}^n$ can be constructed from $A$ using finitely many boolean operations, cartesian products, and linear operations. The same statement fails when Hausdorff dimension is replaced by packing dimension.
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