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Completeness number of ultrafilters
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In what I write below, by "ultrafilter" I mean a non-principal ultrafilter.
Given an ultrafilter $U$ on some set $S$, let $mu$ be the least cardinal such that $U$ is $mu$-complete but not $mu^+$-complete. Call this number the completeness number of $U$. It is easy to check that this must always be an infinite regular cardinal.
The question is which cardinals can appear as completeness numbers for ultrafilters. Stated otherwise: for which cardinals $mu$ we can find examples of ultrafilters that have completeness number $mu$.
For instance, if an ultrafilter is not countably complete, then its completeness number is $aleph_0$ (and if there are no measurable cardinals then in fact every ultrafilter has completeness number $aleph_0$). This means that $aleph_0$ can be completeness number.
What about $aleph_1$? or $aleph_n$? or any other regular or even large cardinal $mu$?
Perhaps this is well-known but I cannot find a reference.
Thanks.
set-theory large-cardinals ultrafilters
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In what I write below, by "ultrafilter" I mean a non-principal ultrafilter.
Given an ultrafilter $U$ on some set $S$, let $mu$ be the least cardinal such that $U$ is $mu$-complete but not $mu^+$-complete. Call this number the completeness number of $U$. It is easy to check that this must always be an infinite regular cardinal.
The question is which cardinals can appear as completeness numbers for ultrafilters. Stated otherwise: for which cardinals $mu$ we can find examples of ultrafilters that have completeness number $mu$.
For instance, if an ultrafilter is not countably complete, then its completeness number is $aleph_0$ (and if there are no measurable cardinals then in fact every ultrafilter has completeness number $aleph_0$). This means that $aleph_0$ can be completeness number.
What about $aleph_1$? or $aleph_n$? or any other regular or even large cardinal $mu$?
Perhaps this is well-known but I cannot find a reference.
Thanks.
set-theory large-cardinals ultrafilters
asked 2 hours ago
almost nowhere
111
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almost nowhere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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If there are no measurable cardinals, then all ultrafilters must have completeness $aleph_0$. So it's kinda hard to give you a straight answer.
– Asaf Karagila
2 hours ago
The possible completeness numbers of non-principal ultrafilters are $aleph_0$ and the measurable cardinals. I"ll write some details as an answer.
– Andreas Blass
57 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In what I write below, by "ultrafilter" I mean a non-principal ultrafilter.
Given an ultrafilter $U$ on some set $S$, let $mu$ be the least cardinal such that $U$ is $mu$-complete but not $mu^+$-complete. Call this number the completeness number of $U$. It is easy to check that this must always be an infinite regular cardinal.
The question is which cardinals can appear as completeness numbers for ultrafilters. Stated otherwise: for which cardinals $mu$ we can find examples of ultrafilters that have completeness number $mu$.
For instance, if an ultrafilter is not countably complete, then its completeness number is $aleph_0$ (and if there are no measurable cardinals then in fact every ultrafilter has completeness number $aleph_0$). This means that $aleph_0$ can be completeness number.
What about $aleph_1$? or $aleph_n$? or any other regular or even large cardinal $mu$?
Perhaps this is well-known but I cannot find a reference.
Thanks.
set-theory large-cardinals ultrafilters
asked 2 hours ago
almost nowhere
111
New contributor
almost nowhere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In what I write below, by "ultrafilter" I mean a non-principal ultrafilter.
Given an ultrafilter $U$ on some set $S$, let $mu$ be the least cardinal such that $U$ is $mu$-complete but not $mu^+$-complete. Call this number the completeness number of $U$. It is easy to check that this must always be an infinite regular cardinal.
The question is which cardinals can appear as completeness numbers for ultrafilters. Stated otherwise: for which cardinals $mu$ we can find examples of ultrafilters that have completeness number $mu$.
For instance, if an ultrafilter is not countably complete, then its completeness number is $aleph_0$ (and if there are no measurable cardinals then in fact every ultrafilter has completeness number $aleph_0$). This means that $aleph_0$ can be completeness number.
What about $aleph_1$? or $aleph_n$? or any other regular or even large cardinal $mu$?
Perhaps this is well-known but I cannot find a reference.
Thanks.
set-theory large-cardinals ultrafilters
set-theory large-cardinals ultrafilters
asked 2 hours ago
almost nowhere
111
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almost nowhere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 hours ago
almost nowhere
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asked 2 hours ago
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asked 2 hours ago
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asked 2 hours ago
almost nowhere
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111
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If there are no measurable cardinals, then all ultrafilters must have completeness $aleph_0$. So it's kinda hard to give you a straight answer.
– Asaf Karagila
2 hours ago
The possible completeness numbers of non-principal ultrafilters are $aleph_0$ and the measurable cardinals. I"ll write some details as an answer.
– Andreas Blass
57 mins ago
add a comment |Â
If there are no measurable cardinals, then all ultrafilters must have completeness $aleph_0$. So it's kinda hard to give you a straight answer.
– Asaf Karagila
2 hours ago
The possible completeness numbers of non-principal ultrafilters are $aleph_0$ and the measurable cardinals. I"ll write some details as an answer.
– Andreas Blass
57 mins ago
If there are no measurable cardinals, then all ultrafilters must have completeness $aleph_0$. So it's kinda hard to give you a straight answer.
– Asaf Karagila
2 hours ago
If there are no measurable cardinals, then all ultrafilters must have completeness $aleph_0$. So it's kinda hard to give you a straight answer.
– Asaf Karagila
2 hours ago
The possible completeness numbers of non-principal ultrafilters are $aleph_0$ and the measurable cardinals. I"ll write some details as an answer.
– Andreas Blass
57 mins ago
The possible completeness numbers of non-principal ultrafilters are $aleph_0$ and the measurable cardinals. I"ll write some details as an answer.
– Andreas Blass
57 mins ago
add a comment |Â
1 Answer
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Any countably incomplete ultrafilter has completeness number $aleph_0$, and if $kappa$ is measurable then any $kappa$-complete non-principal ultrafilter on $kappa$ has completeness number $kappa$. I claim that these are the only possible completeness numbers.
To prove it, suppose $U$ is a non-principal ultrafilter on some set $A$ and that its completeness number is an uncountable cardinal $kappa$. I'll prove that $kappa$ is measurable by producing a $kappa$-complete non-principal ultrafilter on $kappa$.
Since $U$ is not $kappa^+$-complete, we can fix $kappa$ sets $X_alphain U$ (where $alpha$ ranges over ordinals $<kappa$) such that then intersection $bigcap_alpha<kappaX_alphanotin U$. By subtracting this intersection from each $X_alpha$, we can assume, without loss of generality, that $bigcap_alpha<kappaX_alpha=varnothing$. So we can define a function $f:Atokappa$ by letting $f(p)$ (for any $pin A$) be the smallest $alpha$ such that $pnotin X_alpha$. Then let $V$ be the image of $U$ under this map $f$; that is, $V=Ysubseteqkappa:f^-1[Y]in U$. Since $f^-1$ preserves all Boolean operations, including infinitary ones, it is immediate that $V$ is, like $U$, a $kappa$-complete ultrafilter.
It remains to check that $V$ is non-principal, but this is easy. For any $alpha<kappa$, the definition of $f$ implies that $f^-1[alpha]$ is disjoint from $X_alpha$ and is therefore not in $U$. So $f^-1[alpha]$ is not in $V$.
answered 45 mins ago
Andreas Blass
55.8k7132213
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Any countably incomplete ultrafilter has completeness number $aleph_0$, and if $kappa$ is measurable then any $kappa$-complete non-principal ultrafilter on $kappa$ has completeness number $kappa$. I claim that these are the only possible completeness numbers.
To prove it, suppose $U$ is a non-principal ultrafilter on some set $A$ and that its completeness number is an uncountable cardinal $kappa$. I'll prove that $kappa$ is measurable by producing a $kappa$-complete non-principal ultrafilter on $kappa$.
Since $U$ is not $kappa^+$-complete, we can fix $kappa$ sets $X_alphain U$ (where $alpha$ ranges over ordinals $<kappa$) such that then intersection $bigcap_alpha<kappaX_alphanotin U$. By subtracting this intersection from each $X_alpha$, we can assume, without loss of generality, that $bigcap_alpha<kappaX_alpha=varnothing$. So we can define a function $f:Atokappa$ by letting $f(p)$ (for any $pin A$) be the smallest $alpha$ such that $pnotin X_alpha$. Then let $V$ be the image of $U$ under this map $f$; that is, $V=Ysubseteqkappa:f^-1[Y]in U$. Since $f^-1$ preserves all Boolean operations, including infinitary ones, it is immediate that $V$ is, like $U$, a $kappa$-complete ultrafilter.
It remains to check that $V$ is non-principal, but this is easy. For any $alpha<kappa$, the definition of $f$ implies that $f^-1[alpha]$ is disjoint from $X_alpha$ and is therefore not in $U$. So $f^-1[alpha]$ is not in $V$.
answered 45 mins ago
Andreas Blass
55.8k7132213
add a comment |Â
up vote
3
down vote
Any countably incomplete ultrafilter has completeness number $aleph_0$, and if $kappa$ is measurable then any $kappa$-complete non-principal ultrafilter on $kappa$ has completeness number $kappa$. I claim that these are the only possible completeness numbers.
To prove it, suppose $U$ is a non-principal ultrafilter on some set $A$ and that its completeness number is an uncountable cardinal $kappa$. I'll prove that $kappa$ is measurable by producing a $kappa$-complete non-principal ultrafilter on $kappa$.
Since $U$ is not $kappa^+$-complete, we can fix $kappa$ sets $X_alphain U$ (where $alpha$ ranges over ordinals $<kappa$) such that then intersection $bigcap_alpha<kappaX_alphanotin U$. By subtracting this intersection from each $X_alpha$, we can assume, without loss of generality, that $bigcap_alpha<kappaX_alpha=varnothing$. So we can define a function $f:Atokappa$ by letting $f(p)$ (for any $pin A$) be the smallest $alpha$ such that $pnotin X_alpha$. Then let $V$ be the image of $U$ under this map $f$; that is, $V=Ysubseteqkappa:f^-1[Y]in U$. Since $f^-1$ preserves all Boolean operations, including infinitary ones, it is immediate that $V$ is, like $U$, a $kappa$-complete ultrafilter.
It remains to check that $V$ is non-principal, but this is easy. For any $alpha<kappa$, the definition of $f$ implies that $f^-1[alpha]$ is disjoint from $X_alpha$ and is therefore not in $U$. So $f^-1[alpha]$ is not in $V$.
answered 45 mins ago
Andreas Blass
55.8k7132213
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Any countably incomplete ultrafilter has completeness number $aleph_0$, and if $kappa$ is measurable then any $kappa$-complete non-principal ultrafilter on $kappa$ has completeness number $kappa$. I claim that these are the only possible completeness numbers.
To prove it, suppose $U$ is a non-principal ultrafilter on some set $A$ and that its completeness number is an uncountable cardinal $kappa$. I'll prove that $kappa$ is measurable by producing a $kappa$-complete non-principal ultrafilter on $kappa$.
Since $U$ is not $kappa^+$-complete, we can fix $kappa$ sets $X_alphain U$ (where $alpha$ ranges over ordinals $<kappa$) such that then intersection $bigcap_alpha<kappaX_alphanotin U$. By subtracting this intersection from each $X_alpha$, we can assume, without loss of generality, that $bigcap_alpha<kappaX_alpha=varnothing$. So we can define a function $f:Atokappa$ by letting $f(p)$ (for any $pin A$) be the smallest $alpha$ such that $pnotin X_alpha$. Then let $V$ be the image of $U$ under this map $f$; that is, $V=Ysubseteqkappa:f^-1[Y]in U$. Since $f^-1$ preserves all Boolean operations, including infinitary ones, it is immediate that $V$ is, like $U$, a $kappa$-complete ultrafilter.
It remains to check that $V$ is non-principal, but this is easy. For any $alpha<kappa$, the definition of $f$ implies that $f^-1[alpha]$ is disjoint from $X_alpha$ and is therefore not in $U$. So $f^-1[alpha]$ is not in $V$.
answered 45 mins ago
Andreas Blass
55.8k7132213
Any countably incomplete ultrafilter has completeness number $aleph_0$, and if $kappa$ is measurable then any $kappa$-complete non-principal ultrafilter on $kappa$ has completeness number $kappa$. I claim that these are the only possible completeness numbers.
To prove it, suppose $U$ is a non-principal ultrafilter on some set $A$ and that its completeness number is an uncountable cardinal $kappa$. I'll prove that $kappa$ is measurable by producing a $kappa$-complete non-principal ultrafilter on $kappa$.
Since $U$ is not $kappa^+$-complete, we can fix $kappa$ sets $X_alphain U$ (where $alpha$ ranges over ordinals $<kappa$) such that then intersection $bigcap_alpha<kappaX_alphanotin U$. By subtracting this intersection from each $X_alpha$, we can assume, without loss of generality, that $bigcap_alpha<kappaX_alpha=varnothing$. So we can define a function $f:Atokappa$ by letting $f(p)$ (for any $pin A$) be the smallest $alpha$ such that $pnotin X_alpha$. Then let $V$ be the image of $U$ under this map $f$; that is, $V=Ysubseteqkappa:f^-1[Y]in U$. Since $f^-1$ preserves all Boolean operations, including infinitary ones, it is immediate that $V$ is, like $U$, a $kappa$-complete ultrafilter.
It remains to check that $V$ is non-principal, but this is easy. For any $alpha<kappa$, the definition of $f$ implies that $f^-1[alpha]$ is disjoint from $X_alpha$ and is therefore not in $U$. So $f^-1[alpha]$ is not in $V$.
answered 45 mins ago
Andreas Blass
55.8k7132213
answered 45 mins ago
Andreas Blass
55.8k7132213
answered 45 mins ago
Andreas Blass
55.8k7132213
answered 45 mins ago
Andreas Blass
55.8k7132213
55.8k7132213
add a comment |Â
add a comment |Â
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almost nowhere is a new contributor. Be nice, and check out our Code of Conduct.
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If there are no measurable cardinals, then all ultrafilters must have completeness $aleph_0$. So it's kinda hard to give you a straight answer.
– Asaf Karagila
2 hours ago
The possible completeness numbers of non-principal ultrafilters are $aleph_0$ and the measurable cardinals. I"ll write some details as an answer.
– Andreas Blass
57 mins ago