Abstract
We prove that if S is a commutative semigroup with well-founded universal semilattice or a solvable inverse semigroup with well-founded semilattice of idempotents, then every strongly productive ultrafilter on S is idempotent. Moreover we show that any very strongly productive ultrafilter on the free semigroup with countably many generators is sparse, answering a question of Hindman and Legette Jones.
Original language | English |
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Pages (from-to) | 242-257 |
Number of pages | 16 |
Journal | Semigroup Forum |
Volume | 92 |
Issue number | 1 |
DOIs | |
State | Published - 1 Feb 2016 |
Externally published | Yes |
Keywords
- Commutative semigroups
- Finite products
- Solvable groups
- Solvable inverse semigroups
- Strongly productive ultrafilters
- Ultrafilters
- Čech–Stone compactification