Amorphous proper classes in MKWhat sort of structure can amorphous sets support?Splitting infinite setsFor...
Amorphous proper classes in MK
What sort of structure can amorphous sets support?Splitting infinite setsFor models of ZF, if for some $A$ we have $L[A] = L$, what can we deduce about $A$?What sort of structure can amorphous sets support?Some questions about Ackermann set theoryHartogs number and the three power setsCan $mathbb{R}$ be partitioned into dedekind-finite sets?How many Dedekind-finite sets can $mathbb{R}$ be partitioned into?Can ZFC be interpreted in a set theory having finitely many ranks?An axiom for collecting proper classesDo choice principles in all generic extensions imply AC in $V$?
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Working in $ZFC$ every cardinal is either finite or in bijection with a proper subset of itself (Dedekind infinite). Without Choice it is consistent that there are infinite sets which can't be partitioned into two infinite subsets (amorphous sets), so the above statement no longer holds since a bijection to a proper subset implies a partition into two disjoint infinite subsets as proven on the wiki -- all of this is discussed in the question and answers here much more succinctly.
Is it consistent in $MK$ without Global Choice that there are amorphous proper classes, meaning proper classes which can't be partitioned into two proper class sized subclasses?
Directly generalizing the argument given on the wiki article for amorphous sets seems to require a notion of transfinite function composition which can be defined in good categorical generality using colimits, but it is not immediately apparent how to generalize the recursive definition of the $S_i$'s for limit ordinal $i$ since the given definitions depend on immediate predecessor steps.
set-theory lo.logic axiom-of-choice
edited 57 mins ago
David Roberts
17.5k463177
asked 4 hours ago
Alec RheaAlec Rhea
1,3331819
$endgroup$
add a comment |
$begingroup$
Working in $ZFC$ every cardinal is either finite or in bijection with a proper subset of itself (Dedekind infinite). Without Choice it is consistent that there are infinite sets which can't be partitioned into two infinite subsets (amorphous sets), so the above statement no longer holds since a bijection to a proper subset implies a partition into two disjoint infinite subsets as proven on the wiki -- all of this is discussed in the question and answers here much more succinctly.
Is it consistent in $MK$ without Global Choice that there are amorphous proper classes, meaning proper classes which can't be partitioned into two proper class sized subclasses?
Directly generalizing the argument given on the wiki article for amorphous sets seems to require a notion of transfinite function composition which can be defined in good categorical generality using colimits, but it is not immediately apparent how to generalize the recursive definition of the $S_i$'s for limit ordinal $i$ since the given definitions depend on immediate predecessor steps.
set-theory lo.logic axiom-of-choice
edited 57 mins ago
David Roberts
17.5k463177
asked 4 hours ago
Alec RheaAlec Rhea
1,3331819
$endgroup$
add a comment |
$begingroup$
Working in $ZFC$ every cardinal is either finite or in bijection with a proper subset of itself (Dedekind infinite). Without Choice it is consistent that there are infinite sets which can't be partitioned into two infinite subsets (amorphous sets), so the above statement no longer holds since a bijection to a proper subset implies a partition into two disjoint infinite subsets as proven on the wiki -- all of this is discussed in the question and answers here much more succinctly.
Is it consistent in $MK$ without Global Choice that there are amorphous proper classes, meaning proper classes which can't be partitioned into two proper class sized subclasses?
Directly generalizing the argument given on the wiki article for amorphous sets seems to require a notion of transfinite function composition which can be defined in good categorical generality using colimits, but it is not immediately apparent how to generalize the recursive definition of the $S_i$'s for limit ordinal $i$ since the given definitions depend on immediate predecessor steps.
set-theory lo.logic axiom-of-choice
edited 57 mins ago
David Roberts
17.5k463177
asked 4 hours ago
Alec RheaAlec Rhea
1,3331819
$endgroup$
Working in $ZFC$ every cardinal is either finite or in bijection with a proper subset of itself (Dedekind infinite). Without Choice it is consistent that there are infinite sets which can't be partitioned into two infinite subsets (amorphous sets), so the above statement no longer holds since a bijection to a proper subset implies a partition into two disjoint infinite subsets as proven on the wiki -- all of this is discussed in the question and answers here much more succinctly.
Is it consistent in $MK$ without Global Choice that there are amorphous proper classes, meaning proper classes which can't be partitioned into two proper class sized subclasses?
Directly generalizing the argument given on the wiki article for amorphous sets seems to require a notion of transfinite function composition which can be defined in good categorical generality using colimits, but it is not immediately apparent how to generalize the recursive definition of the $S_i$'s for limit ordinal $i$ since the given definitions depend on immediate predecessor steps.
set-theory lo.logic axiom-of-choice
set-theory lo.logic axiom-of-choice
edited 57 mins ago
David Roberts
17.5k463177
asked 4 hours ago
Alec RheaAlec Rhea
1,3331819
edited 57 mins ago
David Roberts
17.5k463177
asked 4 hours ago
Alec RheaAlec Rhea
1,3331819
edited 57 mins ago
David Roberts
17.5k463177
edited 57 mins ago
David Roberts
17.5k463177
edited 57 mins ago
David Roberts
17.5k463177
17.5k463177
asked 4 hours ago
Alec RheaAlec Rhea
1,3331819
asked 4 hours ago
Alec RheaAlec Rhea
1,3331819
asked 4 hours ago
Alec RheaAlec Rhea
1,3331819
1,3331819
add a comment |
add a comment |
1 Answer
1
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oldest
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Unless I'm missing something, the answer is no: we have a surjection $s$ from a given proper class to the class of ordinals - sending each element to its rank and then "collapsing" appropriately - and this lets us partition the original class into two proper classes, for example $s^{-1}(limits)$ versus $s^{-1}(successors)$.
answered 3 hours ago
Noah SchweberNoah Schweber
19.5k349146
$endgroup$
$begingroup$
Very nice, you haven't missed anything -- if I create a moving target and ask about $MK-Foundation$ is the answer still trivially no?
$endgroup$
– Alec Rhea
3 hours ago
1
$begingroup$
@Alec: In that case the answer is positive. Just do Fraenkel's model over a proper class of atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
@Alec: That's the OG model for amorphous sets. Just remember that ZFA (or ZFU) is equivalent to ZF-Foundation with Quine atoms for the atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
Tsk tsk tsk. TSK. TSK. TSK.
$endgroup$
– Asaf Karagila
1 hour ago
3
$begingroup$
@Noah Asaf is calling you uncool for not knowing.
$endgroup$
– David Roberts
55 mins ago
|
show 5 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Unless I'm missing something, the answer is no: we have a surjection $s$ from a given proper class to the class of ordinals - sending each element to its rank and then "collapsing" appropriately - and this lets us partition the original class into two proper classes, for example $s^{-1}(limits)$ versus $s^{-1}(successors)$.
answered 3 hours ago
Noah SchweberNoah Schweber
19.5k349146
$endgroup$
$begingroup$
Very nice, you haven't missed anything -- if I create a moving target and ask about $MK-Foundation$ is the answer still trivially no?
$endgroup$
– Alec Rhea
3 hours ago
1
$begingroup$
@Alec: In that case the answer is positive. Just do Fraenkel's model over a proper class of atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
@Alec: That's the OG model for amorphous sets. Just remember that ZFA (or ZFU) is equivalent to ZF-Foundation with Quine atoms for the atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
Tsk tsk tsk. TSK. TSK. TSK.
$endgroup$
– Asaf Karagila
1 hour ago
3
$begingroup$
@Noah Asaf is calling you uncool for not knowing.
$endgroup$
– David Roberts
55 mins ago
|
show 5 more comments
$begingroup$
Unless I'm missing something, the answer is no: we have a surjection $s$ from a given proper class to the class of ordinals - sending each element to its rank and then "collapsing" appropriately - and this lets us partition the original class into two proper classes, for example $s^{-1}(limits)$ versus $s^{-1}(successors)$.
answered 3 hours ago
Noah SchweberNoah Schweber
19.5k349146
$endgroup$
$begingroup$
Very nice, you haven't missed anything -- if I create a moving target and ask about $MK-Foundation$ is the answer still trivially no?
$endgroup$
– Alec Rhea
3 hours ago
1
$begingroup$
@Alec: In that case the answer is positive. Just do Fraenkel's model over a proper class of atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
@Alec: That's the OG model for amorphous sets. Just remember that ZFA (or ZFU) is equivalent to ZF-Foundation with Quine atoms for the atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
Tsk tsk tsk. TSK. TSK. TSK.
$endgroup$
– Asaf Karagila
1 hour ago
3
$begingroup$
@Noah Asaf is calling you uncool for not knowing.
$endgroup$
– David Roberts
55 mins ago
|
show 5 more comments
$begingroup$
Unless I'm missing something, the answer is no: we have a surjection $s$ from a given proper class to the class of ordinals - sending each element to its rank and then "collapsing" appropriately - and this lets us partition the original class into two proper classes, for example $s^{-1}(limits)$ versus $s^{-1}(successors)$.
answered 3 hours ago
Noah SchweberNoah Schweber
19.5k349146
$endgroup$
Unless I'm missing something, the answer is no: we have a surjection $s$ from a given proper class to the class of ordinals - sending each element to its rank and then "collapsing" appropriately - and this lets us partition the original class into two proper classes, for example $s^{-1}(limits)$ versus $s^{-1}(successors)$.
answered 3 hours ago
Noah SchweberNoah Schweber
19.5k349146
answered 3 hours ago
Noah SchweberNoah Schweber
19.5k349146
answered 3 hours ago
Noah SchweberNoah Schweber
19.5k349146
answered 3 hours ago
Noah SchweberNoah Schweber
19.5k349146
19.5k349146
$begingroup$
Very nice, you haven't missed anything -- if I create a moving target and ask about $MK-Foundation$ is the answer still trivially no?
$endgroup$
– Alec Rhea
3 hours ago
1
$begingroup$
@Alec: In that case the answer is positive. Just do Fraenkel's model over a proper class of atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
@Alec: That's the OG model for amorphous sets. Just remember that ZFA (or ZFU) is equivalent to ZF-Foundation with Quine atoms for the atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
Tsk tsk tsk. TSK. TSK. TSK.
$endgroup$
– Asaf Karagila
1 hour ago
3
$begingroup$
@Noah Asaf is calling you uncool for not knowing.
$endgroup$
– David Roberts
55 mins ago
|
show 5 more comments
$begingroup$
Very nice, you haven't missed anything -- if I create a moving target and ask about $MK-Foundation$ is the answer still trivially no?
$endgroup$
– Alec Rhea
3 hours ago
1
$begingroup$
@Alec: In that case the answer is positive. Just do Fraenkel's model over a proper class of atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
@Alec: That's the OG model for amorphous sets. Just remember that ZFA (or ZFU) is equivalent to ZF-Foundation with Quine atoms for the atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
$begingroup$
Tsk tsk tsk. TSK. TSK. TSK.
$endgroup$
– Asaf Karagila
1 hour ago
3
$begingroup$
@Noah Asaf is calling you uncool for not knowing.
$endgroup$
– David Roberts
55 mins ago
$begingroup$
Very nice, you haven't missed anything -- if I create a moving target and ask about $MK-Foundation$ is the answer still trivially no?
$endgroup$
– Alec Rhea
3 hours ago
$begingroup$
Very nice, you haven't missed anything -- if I create a moving target and ask about $MK-Foundation$ is the answer still trivially no?
$endgroup$
– Alec Rhea
3 hours ago
1
1
$begingroup$
@Alec: In that case the answer is positive. Just do Fraenkel's model over a proper class of atoms.
$endgroup$
– Asaf Karagila
3 hours ago
$begingroup$
@Alec: In that case the answer is positive. Just do Fraenkel's model over a proper class of atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
1
$begingroup$
@Alec: That's the OG model for amorphous sets. Just remember that ZFA (or ZFU) is equivalent to ZF-Foundation with Quine atoms for the atoms.
$endgroup$
– Asaf Karagila
3 hours ago
$begingroup$
@Alec: That's the OG model for amorphous sets. Just remember that ZFA (or ZFU) is equivalent to ZF-Foundation with Quine atoms for the atoms.
$endgroup$
– Asaf Karagila
3 hours ago
1
1
$begingroup$
Tsk tsk tsk. TSK. TSK. TSK.
$endgroup$
– Asaf Karagila
1 hour ago
$begingroup$
Tsk tsk tsk. TSK. TSK. TSK.
$endgroup$
– Asaf Karagila
1 hour ago
3
3
$begingroup$
@Noah Asaf is calling you uncool for not knowing.
$endgroup$
– David Roberts
55 mins ago
$begingroup$
@Noah Asaf is calling you uncool for not knowing.
$endgroup$
– David Roberts
55 mins ago
|
show 5 more comments
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