A remark on hereditarily nonparadoxical sets

详细信息    查看全文

• 作者：Péter Komjáth
• 关键词：Paradoxical decompositions ; Combinatorial set theory ; Forcing ; Primary 03E35 ; Secondary 03E05
• 刊名：Archive for Mathematical Logic
• 出版年：2016
• 出版时间：February 2016
• 年：2016
• 卷：55
• 期：1-2
• 页码：165-175
• 全文大小：395 KB
• 参考文献：1.Hajnal A.: Proof of a conjecture of Ruziewicz. Fund. Math. 50, 123–128 (1961)MathSciNet MATH
  2.Hajnal A., Juhász I., Shelah S.: Splitting strongly almost disjoint families. Trans. Am. Math. Soc. 295, 369–387 (1986)CrossRef MATH
  3.Komjáth, P., Totik, V.: Problems and Theorems in Classical Set Theory, Springer, Ch. 13, p. 15 (2006)
  4.Nowik A.: Hereditarily nonparadoxical sets revisited. Topol. Appl. 161, 377–385 (2014)CrossRef MathSciNet MATH
  5.Penconek M.: On nonparadoxical sets. Fund. Math. 139, 177–191 (1991)MathSciNet MATH
  6.Zakrzewski P.: Paradoxical decompositions and invariant measures. Proc. Am. Math. Soc 111, 533–539 (1991)CrossRef MathSciNet MATH
  
• 作者单位：Péter Komjáth (1)
  
  1. Institute of Mathematics, Eötvös University, P.O. Box 120, 1518, Budapest, Hungary
  
• 刊物类别：Mathematics and Statistics
• 刊物主题：Mathematics
  Mathematical Logic and Foundations
  Mathematics
  Algebra
  
• 出版者：Springer Berlin / Heidelberg
• ISSN：1432-0665

文摘

Call a set \({A \subseteq \mathbb {R}}\) paradoxical if there are disjoint \({A_0, A_1 \subseteq A}\) such that both \({A_0}\) and \({A_1}\) are equidecomposable with \({A}\) via countabbly many translations. \({X \subseteq \mathbb {R}}\) is hereditarily nonparadoxical if no uncountable subset of \({X}\) is paradoxical. Penconek raised the question if every hereditarily nonparadoxical set \({X \subseteq \mathbb {R}}\) is the union of countably many sets, each omitting nontrivial solutions of \({x - y = z - t}\). Nowik showed that the answer is ‘yes’, as long as \({|X| \leq \aleph_\omega}\). Here we show that consistently there exists a counterexample of cardinality \({\aleph_{\omega+1}}\) and it is also consistent that the continuum is arbitrarily large and Penconek’s statement holds for any \({X}\). Keywords Paradoxical decompositions Combinatorial set theory Forcing