Abstract
We prove that superclub implies s=ω1. More generally, if κ is weakly compact then superclub implies sκ=κ+. Based on this statement, we separate tiltan from superclub at successors of supercompact cardinals. We also use the Galvin property in order to separate tiltan from superclub at both successors of regular and successors of singular cardinals.
Original language | English |
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Journal | Periodica Mathematica Hungarica |
DOIs | |
State | Accepted/In press - 2025 |
Bibliographical note
Publisher Copyright:© The Author(s) 2025.
Keywords
- Splitting number
- Superclub
- The Galvin property
- Tiltan
- Weakly compact cardinals