2016
DOI: 10.14712/1213-7243.2015.159
|View full text |Cite
|
Sign up to set email alerts
|

On preimages of ultrafilters in ZF

Abstract: Institute of Mathematics of the Czech Academy of Sciences provides access to digitized documents strictly for personal use. Each copy of any part of this document must contain these Terms of use.

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1
1

Citation Types

0
4
0

Year Published

2017
2017
2019
2019

Publication Types

Select...
1
1

Relationship

0
2

Authors

Journals

citations
Cited by 2 publications
(4 citation statements)
references
References 5 publications
0
4
0
Order By: Relevance
“…It is not hard to verify, however, that ifẋ ∈ HS and 1 HSẋ ∈Ȧ, then {n | 1 ẋ =ȧn} is finite, and by an easy density argument, 1 does not decide the value of0 ∈ȧn for any finitely many reals. 4 This leads to the sometimes additional requirement that κ is uncountable in the definitions of measurable and strongly compact cardinals. For our purposes, however, it is better to allow ℵ 0 to be considered as measurable or strongly compact.…”
Section: Definition 22mentioning
confidence: 99%
See 1 more Smart Citation
“…It is not hard to verify, however, that ifẋ ∈ HS and 1 HSẋ ∈Ȧ, then {n | 1 ẋ =ȧn} is finite, and by an easy density argument, 1 does not decide the value of0 ∈ȧn for any finitely many reals. 4 This leads to the sometimes additional requirement that κ is uncountable in the definitions of measurable and strongly compact cardinals. For our purposes, however, it is better to allow ℵ 0 to be considered as measurable or strongly compact.…”
Section: Definition 22mentioning
confidence: 99%
“…In a recent paper [4] by Horst Herrlich, Paul Howard, and Eleftherios Tachtsis, the authors point out that it is open whether or not it is possible that there are ultrafilters on ω 1 , but there are no uniform ultrafilters on ω 1 . This can be done using a slight generalization of Feferman's argument from [3], 5 as we will show in this section.…”
Section: Generalization Of Feferman's Proofmentioning
confidence: 99%
“…It is not hard to verify, however, that if ẋ ∈ HS and 1 HS ẋ ∈ Ȧ, then {n | 1 ẋ = ȧn} is finite, and by an easy density argument, 1 does not decide the value of 0 ∈ ȧn for any finitely many reals. 4 This leads to the sometimes additional requirement that κ is uncountable in the definitions of measurable and strongly compact cardinals. For our purposes, however, it is better to allow ℵ 0 to be considered as measurable or strongly compact.…”
Section: Similarly We Say Thatmentioning
confidence: 99%
“…In a recent paper [4] by Horst Herrlich, Paul Howard, and Eleftherios Tachtsis, the authors point out that it is open whether or not it is possible that there are ultrafilters on ω 1 , but there are no uniform ultrafilters on ω 1 . This can be done using a slight generalization of Feferman's argument from [3], 5 as we will show in this section.…”
Section: Generalization Of Feferman's Proofmentioning
confidence: 99%