A question on the ultrafilter numberSurjective Maps onto $aleph$-numbersMaximal chains in a quasi-order of...
A question on the ultrafilter number
Surjective Maps onto $aleph$-numbersMaximal chains in a quasi-order of linear order typesIs there a modification of Martin's Axiom which admits non-measurable sets of size less than continuum?What sort of large cardinal can continuum be?Is there a locally compact, $omega_1$-compact, not $sigma$-countably compact space of size $aleph_1$?Is there nontrivial structure to forcing axioms?The minimum cardinality of an almost disjoint reaping familyThe “strong” measure numberTranslates of measure zero setThe Parovichenko cardinal, is it equal to $max{aleph_2,mathfrak p}$?
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Let $mathfrak{u}$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcal{P}(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbb{N}$. Clearly $aleph_1leq frak{u}leq 2^{aleph_0}$, so it is only interesting to study $frak{u}$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $frak{u}=aleph_1$. Martin's axiom implies that $frak{u}=2^{aleph_0}$.
Is it consistent that $aleph_1<frak{u}<2^{aleph_0}$? If so, can I please have a reference?
set-theory lo.logic
asked 5 hours ago
IsaacIsaac
384
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add a comment |
$begingroup$
Let $mathfrak{u}$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcal{P}(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbb{N}$. Clearly $aleph_1leq frak{u}leq 2^{aleph_0}$, so it is only interesting to study $frak{u}$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $frak{u}=aleph_1$. Martin's axiom implies that $frak{u}=2^{aleph_0}$.
Is it consistent that $aleph_1<frak{u}<2^{aleph_0}$? If so, can I please have a reference?
set-theory lo.logic
asked 5 hours ago
IsaacIsaac
384
$endgroup$
add a comment |
$begingroup$
Let $mathfrak{u}$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcal{P}(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbb{N}$. Clearly $aleph_1leq frak{u}leq 2^{aleph_0}$, so it is only interesting to study $frak{u}$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $frak{u}=aleph_1$. Martin's axiom implies that $frak{u}=2^{aleph_0}$.
Is it consistent that $aleph_1<frak{u}<2^{aleph_0}$? If so, can I please have a reference?
set-theory lo.logic
asked 5 hours ago
IsaacIsaac
384
$endgroup$
Let $mathfrak{u}$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $mathcal{P}(mathbb N)$ which is a base for a nonprincipal ultrafilter on $mathbb{N}$. Clearly $aleph_1leq frak{u}leq 2^{aleph_0}$, so it is only interesting to study $frak{u}$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $frak{u}=aleph_1$. Martin's axiom implies that $frak{u}=2^{aleph_0}$.
Is it consistent that $aleph_1<frak{u}<2^{aleph_0}$? If so, can I please have a reference?
set-theory lo.logic
set-theory lo.logic
asked 5 hours ago
IsaacIsaac
384
asked 5 hours ago
IsaacIsaac
384
asked 5 hours ago
IsaacIsaac
384
asked 5 hours ago
IsaacIsaac
384
asked 5 hours ago
IsaacIsaac
384
384
add a comment |
add a comment |
1 Answer
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The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 3 hours ago
Andreas BlassAndreas Blass
58k7138224
$endgroup$
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
3 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 3 hours ago
Andreas BlassAndreas Blass
58k7138224
$endgroup$
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
3 hours ago
add a comment |
$begingroup$
The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 3 hours ago
Andreas BlassAndreas Blass
58k7138224
$endgroup$
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
3 hours ago
add a comment |
$begingroup$
The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 3 hours ago
Andreas BlassAndreas Blass
58k7138224
$endgroup$
The answer to your question is yes. In fact, one can force to make $mathfrak u$ equal to any prescribed uncountable regular cardinal while making the cardinal of the continuum equal to any larger prescribed uncountable regular cardinal. This is proved in and old paper by Shelah and me:
Blass, Andreas(1-PAS); Shelah, Saharon(1-RTG)
Ultrafilters with small generating sets.
Israel J. Math. 65 (1989), no. 3, 259–271.
answered 3 hours ago
Andreas BlassAndreas Blass
58k7138224
answered 3 hours ago
Andreas BlassAndreas Blass
58k7138224
answered 3 hours ago
Andreas BlassAndreas Blass
58k7138224
answered 3 hours ago
Andreas BlassAndreas Blass
58k7138224
58k7138224
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
3 hours ago
add a comment |
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
3 hours ago
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
3 hours ago
$begingroup$
Thanks Andreas.
$endgroup$
– Isaac
3 hours ago
add a comment |
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