On the Existence of Free Ultrafilters on ω and on Russell-sets in ZF

Eleftherios Tachtsis

Bulletin of the Polish Academy of Sciences. Mathematics (2015)

  • Volume: 63, Issue: 1, page 1-10
  • ISSN: 0239-7269

Abstract

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In ZF (i.e. Zermelo-Fraenkel set theory without the Axiom of Choice AC), we investigate the relationship between UF(ω) (there exists a free ultrafilter on ω) and the statements "there exists a free ultrafilter on every Russell-set" and "there exists a Russell-set A and a free ultrafilter ℱ on A". We establish the following results: 1. UF(ω) implies that there exists a free ultrafilter on every Russell-set. The implication is not reversible in ZF. 2. The statement there exists a free ultrafilter on every Russell-set" is not provable in ZF. 3. If there exists a Russell-set A and a free ultrafilter on A, then UF(ω) holds. The implication is not reversible in ZF. 4. If there exists a Russell-set A and a free ultrafilter on A, then there exists a free ultrafilter on every Russell-set. We also observe the following: (a) The statements BPI(ω) (every proper filter on ω can be extended to an ultrafilter on ω) and "there exists a Russell-set A and a free ultrafilter ℱ on A" are independent of each other in ZF. (b) The statement "there exists a Russell-set and there exists a free ultrafilter on every Russell-set" is, in ZF, equivalent to "there exists a Russell-set A and a free ultrafilter ℱ on A". Thus, "there exists a Russell-set and there exists a free ultrafilter on every Russell-set" is also relatively consistent with ZF.

How to cite

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Eleftherios Tachtsis. "On the Existence of Free Ultrafilters on ω and on Russell-sets in ZF." Bulletin of the Polish Academy of Sciences. Mathematics 63.1 (2015): 1-10. <http://eudml.org/doc/281273>.

@article{EleftheriosTachtsis2015,
abstract = { In ZF (i.e. Zermelo-Fraenkel set theory without the Axiom of Choice AC), we investigate the relationship between UF(ω) (there exists a free ultrafilter on ω) and the statements "there exists a free ultrafilter on every Russell-set" and "there exists a Russell-set A and a free ultrafilter ℱ on A". We establish the following results: 1. UF(ω) implies that there exists a free ultrafilter on every Russell-set. The implication is not reversible in ZF. 2. The statement there exists a free ultrafilter on every Russell-set" is not provable in ZF. 3. If there exists a Russell-set A and a free ultrafilter on A, then UF(ω) holds. The implication is not reversible in ZF. 4. If there exists a Russell-set A and a free ultrafilter on A, then there exists a free ultrafilter on every Russell-set. We also observe the following: (a) The statements BPI(ω) (every proper filter on ω can be extended to an ultrafilter on ω) and "there exists a Russell-set A and a free ultrafilter ℱ on A" are independent of each other in ZF. (b) The statement "there exists a Russell-set and there exists a free ultrafilter on every Russell-set" is, in ZF, equivalent to "there exists a Russell-set A and a free ultrafilter ℱ on A". Thus, "there exists a Russell-set and there exists a free ultrafilter on every Russell-set" is also relatively consistent with ZF. },
author = {Eleftherios Tachtsis},
journal = {Bulletin of the Polish Academy of Sciences. Mathematics},
keywords = {axiom of choice; free filter on a set; free ultrafilter on a set; Russell-set; symmetric models of ZF},
language = {eng},
number = {1},
pages = {1-10},
title = {On the Existence of Free Ultrafilters on ω and on Russell-sets in ZF},
url = {http://eudml.org/doc/281273},
volume = {63},
year = {2015},
}

TY - JOUR
AU - Eleftherios Tachtsis
TI - On the Existence of Free Ultrafilters on ω and on Russell-sets in ZF
JO - Bulletin of the Polish Academy of Sciences. Mathematics
PY - 2015
VL - 63
IS - 1
SP - 1
EP - 10
AB - In ZF (i.e. Zermelo-Fraenkel set theory without the Axiom of Choice AC), we investigate the relationship between UF(ω) (there exists a free ultrafilter on ω) and the statements "there exists a free ultrafilter on every Russell-set" and "there exists a Russell-set A and a free ultrafilter ℱ on A". We establish the following results: 1. UF(ω) implies that there exists a free ultrafilter on every Russell-set. The implication is not reversible in ZF. 2. The statement there exists a free ultrafilter on every Russell-set" is not provable in ZF. 3. If there exists a Russell-set A and a free ultrafilter on A, then UF(ω) holds. The implication is not reversible in ZF. 4. If there exists a Russell-set A and a free ultrafilter on A, then there exists a free ultrafilter on every Russell-set. We also observe the following: (a) The statements BPI(ω) (every proper filter on ω can be extended to an ultrafilter on ω) and "there exists a Russell-set A and a free ultrafilter ℱ on A" are independent of each other in ZF. (b) The statement "there exists a Russell-set and there exists a free ultrafilter on every Russell-set" is, in ZF, equivalent to "there exists a Russell-set A and a free ultrafilter ℱ on A". Thus, "there exists a Russell-set and there exists a free ultrafilter on every Russell-set" is also relatively consistent with ZF.
LA - eng
KW - axiom of choice; free filter on a set; free ultrafilter on a set; Russell-set; symmetric models of ZF
UR - http://eudml.org/doc/281273
ER -

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