Cardinal characteristics of amorphous setsIs the non-triviality of the algebraic dual of an...
Cardinal characteristics of amorphous sets
Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?Can one divide by the cardinal of an amorphous set?If Q is a subset of the plane of size less than continuum, then does every closed F in Q extend to a closed connected G in the plane with the same trace on Q? (Or is this independent of ZFC?)Are there large cardinals for $n$-elementarity?Surjective Maps onto $aleph$-numbersWhat are the known implications of “There exists a Reinhardt cardinal” in the theory “ZF + j”?Replacing Axiom of Choice with Axiom of Countable Choice“Largish” cardinalsHow much choice does a linear or well-order on cardinals imply?Brief history of Cardinal Characteristics of the Continuumcountable and uncountable subset of $P(omega)$“Antiforcing” - Is there a method to 'remove' sets from a model of ZF?
$begingroup$
In a universe where the continuum hypothesis ($CH$) fails we can ask about combinatorial cardinal characteristics of the continuum, but in a universe where $CH$ is true no such cardinals exist so this study becomes vacuous.
Does a similar phenomenon occur at the countable level in a universe without choice? Specifically, are there properties which are true for finite sets but false for $omega$ which are still true for the cardinal of an amorphous set, like divisibility as suggested here by François G. Dorais?
In a universe without choice we have the existence of amorphous sets and we can ask about their 'amorphous cardinals' which are incomparable with $omega$ (thank you Asaf for the correction) and may satisfy nice theorems, but in a universe with choice there are no infinite sets whose cardinality is incomparable with $omega$ so this study becomes vacuous in similar fashion to the uncountable case.
A possible candidate for characteristics smaller than $omega$ could come from theorems in finite group theory that become false for countable groups, since it is possible to have a group structure on an unbounded amorphous cardinal as constructed by Asaf Karagila here.
There is an article behind a paywall published in 2010 that appears to touch on these matters but I can't access it; if anyone is familiar with its contents and willing to give a brief exposition it would be greatly appreciated.
set-theory lo.logic axiom-of-choice
asked Apr 1 at 9:06
Alec RheaAlec Rhea
1,3771819
$endgroup$
|
show 15 more comments
$begingroup$
In a universe where the continuum hypothesis ($CH$) fails we can ask about combinatorial cardinal characteristics of the continuum, but in a universe where $CH$ is true no such cardinals exist so this study becomes vacuous.
Does a similar phenomenon occur at the countable level in a universe without choice? Specifically, are there properties which are true for finite sets but false for $omega$ which are still true for the cardinal of an amorphous set, like divisibility as suggested here by François G. Dorais?
In a universe without choice we have the existence of amorphous sets and we can ask about their 'amorphous cardinals' which are incomparable with $omega$ (thank you Asaf for the correction) and may satisfy nice theorems, but in a universe with choice there are no infinite sets whose cardinality is incomparable with $omega$ so this study becomes vacuous in similar fashion to the uncountable case.
A possible candidate for characteristics smaller than $omega$ could come from theorems in finite group theory that become false for countable groups, since it is possible to have a group structure on an unbounded amorphous cardinal as constructed by Asaf Karagila here.
There is an article behind a paywall published in 2010 that appears to touch on these matters but I can't access it; if anyone is familiar with its contents and willing to give a brief exposition it would be greatly appreciated.
set-theory lo.logic axiom-of-choice
asked Apr 1 at 9:06
Alec RheaAlec Rhea
1,3771819
$endgroup$
1
$begingroup$
@YCor Itlooks like the tag will change or be removed. Better wait.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:32
2
$begingroup$
@Alec Your proposal in the latest comment is a better title. In particular, the new title is misleading. You are not really talking about $omega$ or countability here.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:35
1
$begingroup$
It does not makes sense to say that CH is undecidable in a universe.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:37
1
$begingroup$
(Thanks for the new title. It makes more sense.)
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:40
2
$begingroup$
But I think your description of cardinal characteristics in the opening paragraph is still incorrect. We are not interested in persistence of properties of $omega$ or anything like that. You typically look at combinatorial (or topological or...) objects of size at most continuum and uncountable, not at $omega$ and any kind of persistence. In fact, inequalities between cardinal characteristics are usually established by arguments that make sense and carry nonvacuous information (they still tell you something about these objects), even if CH holds
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:46
|
show 15 more comments
$begingroup$
In a universe where the continuum hypothesis ($CH$) fails we can ask about combinatorial cardinal characteristics of the continuum, but in a universe where $CH$ is true no such cardinals exist so this study becomes vacuous.
Does a similar phenomenon occur at the countable level in a universe without choice? Specifically, are there properties which are true for finite sets but false for $omega$ which are still true for the cardinal of an amorphous set, like divisibility as suggested here by François G. Dorais?
In a universe without choice we have the existence of amorphous sets and we can ask about their 'amorphous cardinals' which are incomparable with $omega$ (thank you Asaf for the correction) and may satisfy nice theorems, but in a universe with choice there are no infinite sets whose cardinality is incomparable with $omega$ so this study becomes vacuous in similar fashion to the uncountable case.
A possible candidate for characteristics smaller than $omega$ could come from theorems in finite group theory that become false for countable groups, since it is possible to have a group structure on an unbounded amorphous cardinal as constructed by Asaf Karagila here.
There is an article behind a paywall published in 2010 that appears to touch on these matters but I can't access it; if anyone is familiar with its contents and willing to give a brief exposition it would be greatly appreciated.
set-theory lo.logic axiom-of-choice
asked Apr 1 at 9:06
Alec RheaAlec Rhea
1,3771819
$endgroup$
In a universe where the continuum hypothesis ($CH$) fails we can ask about combinatorial cardinal characteristics of the continuum, but in a universe where $CH$ is true no such cardinals exist so this study becomes vacuous.
Does a similar phenomenon occur at the countable level in a universe without choice? Specifically, are there properties which are true for finite sets but false for $omega$ which are still true for the cardinal of an amorphous set, like divisibility as suggested here by François G. Dorais?
In a universe without choice we have the existence of amorphous sets and we can ask about their 'amorphous cardinals' which are incomparable with $omega$ (thank you Asaf for the correction) and may satisfy nice theorems, but in a universe with choice there are no infinite sets whose cardinality is incomparable with $omega$ so this study becomes vacuous in similar fashion to the uncountable case.
A possible candidate for characteristics smaller than $omega$ could come from theorems in finite group theory that become false for countable groups, since it is possible to have a group structure on an unbounded amorphous cardinal as constructed by Asaf Karagila here.
There is an article behind a paywall published in 2010 that appears to touch on these matters but I can't access it; if anyone is familiar with its contents and willing to give a brief exposition it would be greatly appreciated.
set-theory lo.logic axiom-of-choice
set-theory lo.logic axiom-of-choice
asked Apr 1 at 9:06
Alec RheaAlec Rhea
1,3771819
asked Apr 1 at 9:06
Alec RheaAlec Rhea
1,3771819
edited Apr 1 at 12:06
Alec Rhea
asked Apr 1 at 9:06
Alec RheaAlec Rhea
1,3771819
asked Apr 1 at 9:06
Alec RheaAlec Rhea
1,3771819
asked Apr 1 at 9:06
Alec RheaAlec Rhea
1,3771819
1,3771819
1
$begingroup$
@YCor Itlooks like the tag will change or be removed. Better wait.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:32
2
$begingroup$
@Alec Your proposal in the latest comment is a better title. In particular, the new title is misleading. You are not really talking about $omega$ or countability here.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:35
1
$begingroup$
It does not makes sense to say that CH is undecidable in a universe.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:37
1
$begingroup$
(Thanks for the new title. It makes more sense.)
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:40
2
$begingroup$
But I think your description of cardinal characteristics in the opening paragraph is still incorrect. We are not interested in persistence of properties of $omega$ or anything like that. You typically look at combinatorial (or topological or...) objects of size at most continuum and uncountable, not at $omega$ and any kind of persistence. In fact, inequalities between cardinal characteristics are usually established by arguments that make sense and carry nonvacuous information (they still tell you something about these objects), even if CH holds
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:46
|
show 15 more comments
1
$begingroup$
@YCor Itlooks like the tag will change or be removed. Better wait.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:32
2
$begingroup$
@Alec Your proposal in the latest comment is a better title. In particular, the new title is misleading. You are not really talking about $omega$ or countability here.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:35
1
$begingroup$
It does not makes sense to say that CH is undecidable in a universe.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:37
1
$begingroup$
(Thanks for the new title. It makes more sense.)
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:40
2
$begingroup$
But I think your description of cardinal characteristics in the opening paragraph is still incorrect. We are not interested in persistence of properties of $omega$ or anything like that. You typically look at combinatorial (or topological or...) objects of size at most continuum and uncountable, not at $omega$ and any kind of persistence. In fact, inequalities between cardinal characteristics are usually established by arguments that make sense and carry nonvacuous information (they still tell you something about these objects), even if CH holds
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:46
1
1
$begingroup$
@YCor Itlooks like the tag will change or be removed. Better wait.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:32
$begingroup$
@YCor Itlooks like the tag will change or be removed. Better wait.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:32
2
2
$begingroup$
@Alec Your proposal in the latest comment is a better title. In particular, the new title is misleading. You are not really talking about $omega$ or countability here.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:35
$begingroup$
@Alec Your proposal in the latest comment is a better title. In particular, the new title is misleading. You are not really talking about $omega$ or countability here.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:35
1
1
$begingroup$
It does not makes sense to say that CH is undecidable in a universe.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:37
$begingroup$
It does not makes sense to say that CH is undecidable in a universe.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:37
1
1
$begingroup$
(Thanks for the new title. It makes more sense.)
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:40
$begingroup$
(Thanks for the new title. It makes more sense.)
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:40
2
2
$begingroup$
But I think your description of cardinal characteristics in the opening paragraph is still incorrect. We are not interested in persistence of properties of $omega$ or anything like that. You typically look at combinatorial (or topological or...) objects of size at most continuum and uncountable, not at $omega$ and any kind of persistence. In fact, inequalities between cardinal characteristics are usually established by arguments that make sense and carry nonvacuous information (they still tell you something about these objects), even if CH holds
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:46
$begingroup$
But I think your description of cardinal characteristics in the opening paragraph is still incorrect. We are not interested in persistence of properties of $omega$ or anything like that. You typically look at combinatorial (or topological or...) objects of size at most continuum and uncountable, not at $omega$ and any kind of persistence. In fact, inequalities between cardinal characteristics are usually established by arguments that make sense and carry nonvacuous information (they still tell you something about these objects), even if CH holds
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:46
|
show 15 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Any cardinal smaller than $aleph_0$ is finite. Amorphous sets are not "smaller", they are just incomparable with. They are very small, in some sense, for example we cannot even divide them into two infinite sets, but they are still infinite.
With respect to the article you linked, let me point out that amorphous sets cannot even be mapped onto $omega$, so they are definitely not the countable union of pairs.
Now, there are some combinatorial characteristics one can assign to general sets, which may be of interest in the case of amorphous sets. For example, if $A$ is amorphous, then any partition of $A$ is up to finitely many parts constant in size (i.e. all but finitely many parts are singletons, or pairs, or so on). We call this size the gauge of the partition, and we can ask what is the supremum of the gauges of possible partitions.
This can be $1$, or some finite $n$, or it can be "unbounded". We can prove, for example, that if $A$ is an amorphous set which can be made into a group, then it is unbounded. So it gives us some information.
But in general, this is not something too similar to cardinal characteristics in the traditional sense, and it is not something too helpful, since $omega$ is a very unique and a very concrete set, whereas amorphous sets can come in many different flavors, sizes, and support different structures.
answered Apr 1 at 9:18
Asaf KaragilaAsaf Karagila
21.7k681185
$endgroup$
$begingroup$
Thank you for the pointers Asaf, I was editing the question to add a link to your answer regarding a group structure on an unbounded amorphous set as a potential source of 'characteristics' but it seems I may have had things backwards. My very naive motivation for asking about this was the thought that 'amorphous cardinals' might correspond to surreal numbers in a universe without choice, but if their cardinals aren't comparable with $omega$ I don't see how they could be members of the surreals in any universe.
$endgroup$
– Alec Rhea
Apr 1 at 9:31
$begingroup$
I agree that $omega$ is a very concrete and special set, but in some sense with characteristics of the continuum we're asking if the way in which we move up the cardinal ladder from $omega$ using the powerset operation 'misses' any smaller cardinalities that could be defined if more care was taken, and I think we can ask exactly the same question with the infinitely iterated successor operation to move from finite cardinals to $omega$.
$endgroup$
– Alec Rhea
Apr 1 at 9:39
$begingroup$
Although I suppose it isn't exactly the same question since with the powerset operation we're asking if we have 'overshot' any smaller cardinals, and with the iterated successor operation we apparently can't overshoot any cardinals but can still ask if we've landed 'between' other incomparable cardinalities by taking the obvious route into the infinite.
$endgroup$
– Alec Rhea
Apr 1 at 9:53
1
$begingroup$
Amorphous sets cannot even be linearly ordered, let alone carry an ordered field structure.
$endgroup$
– Asaf Karagila
Apr 1 at 10:36
3
$begingroup$
The arithmetic of the surreals does not correspond to cardinal arithmetic. You could argue that it corresponds to natural arithmetic on the ordinals (i.e. Hessenberg sums), but the fact we need to specify the arithmetic tells us that it is not the arithmetic, so saying that the surreal numbers embed the ordinals is a bit of a stretch.
$endgroup$
– Asaf Karagila
Apr 1 at 10:57
|
show 7 more comments
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1 Answer
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1 Answer
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$begingroup$
Any cardinal smaller than $aleph_0$ is finite. Amorphous sets are not "smaller", they are just incomparable with. They are very small, in some sense, for example we cannot even divide them into two infinite sets, but they are still infinite.
With respect to the article you linked, let me point out that amorphous sets cannot even be mapped onto $omega$, so they are definitely not the countable union of pairs.
Now, there are some combinatorial characteristics one can assign to general sets, which may be of interest in the case of amorphous sets. For example, if $A$ is amorphous, then any partition of $A$ is up to finitely many parts constant in size (i.e. all but finitely many parts are singletons, or pairs, or so on). We call this size the gauge of the partition, and we can ask what is the supremum of the gauges of possible partitions.
This can be $1$, or some finite $n$, or it can be "unbounded". We can prove, for example, that if $A$ is an amorphous set which can be made into a group, then it is unbounded. So it gives us some information.
But in general, this is not something too similar to cardinal characteristics in the traditional sense, and it is not something too helpful, since $omega$ is a very unique and a very concrete set, whereas amorphous sets can come in many different flavors, sizes, and support different structures.
answered Apr 1 at 9:18
Asaf KaragilaAsaf Karagila
21.7k681185
$endgroup$
$begingroup$
Thank you for the pointers Asaf, I was editing the question to add a link to your answer regarding a group structure on an unbounded amorphous set as a potential source of 'characteristics' but it seems I may have had things backwards. My very naive motivation for asking about this was the thought that 'amorphous cardinals' might correspond to surreal numbers in a universe without choice, but if their cardinals aren't comparable with $omega$ I don't see how they could be members of the surreals in any universe.
$endgroup$
– Alec Rhea
Apr 1 at 9:31
$begingroup$
I agree that $omega$ is a very concrete and special set, but in some sense with characteristics of the continuum we're asking if the way in which we move up the cardinal ladder from $omega$ using the powerset operation 'misses' any smaller cardinalities that could be defined if more care was taken, and I think we can ask exactly the same question with the infinitely iterated successor operation to move from finite cardinals to $omega$.
$endgroup$
– Alec Rhea
Apr 1 at 9:39
$begingroup$
Although I suppose it isn't exactly the same question since with the powerset operation we're asking if we have 'overshot' any smaller cardinals, and with the iterated successor operation we apparently can't overshoot any cardinals but can still ask if we've landed 'between' other incomparable cardinalities by taking the obvious route into the infinite.
$endgroup$
– Alec Rhea
Apr 1 at 9:53
1
$begingroup$
Amorphous sets cannot even be linearly ordered, let alone carry an ordered field structure.
$endgroup$
– Asaf Karagila
Apr 1 at 10:36
3
$begingroup$
The arithmetic of the surreals does not correspond to cardinal arithmetic. You could argue that it corresponds to natural arithmetic on the ordinals (i.e. Hessenberg sums), but the fact we need to specify the arithmetic tells us that it is not the arithmetic, so saying that the surreal numbers embed the ordinals is a bit of a stretch.
$endgroup$
– Asaf Karagila
Apr 1 at 10:57
|
show 7 more comments
$begingroup$
Any cardinal smaller than $aleph_0$ is finite. Amorphous sets are not "smaller", they are just incomparable with. They are very small, in some sense, for example we cannot even divide them into two infinite sets, but they are still infinite.
With respect to the article you linked, let me point out that amorphous sets cannot even be mapped onto $omega$, so they are definitely not the countable union of pairs.
Now, there are some combinatorial characteristics one can assign to general sets, which may be of interest in the case of amorphous sets. For example, if $A$ is amorphous, then any partition of $A$ is up to finitely many parts constant in size (i.e. all but finitely many parts are singletons, or pairs, or so on). We call this size the gauge of the partition, and we can ask what is the supremum of the gauges of possible partitions.
This can be $1$, or some finite $n$, or it can be "unbounded". We can prove, for example, that if $A$ is an amorphous set which can be made into a group, then it is unbounded. So it gives us some information.
But in general, this is not something too similar to cardinal characteristics in the traditional sense, and it is not something too helpful, since $omega$ is a very unique and a very concrete set, whereas amorphous sets can come in many different flavors, sizes, and support different structures.
answered Apr 1 at 9:18
Asaf KaragilaAsaf Karagila
21.7k681185
$endgroup$
$begingroup$
Thank you for the pointers Asaf, I was editing the question to add a link to your answer regarding a group structure on an unbounded amorphous set as a potential source of 'characteristics' but it seems I may have had things backwards. My very naive motivation for asking about this was the thought that 'amorphous cardinals' might correspond to surreal numbers in a universe without choice, but if their cardinals aren't comparable with $omega$ I don't see how they could be members of the surreals in any universe.
$endgroup$
– Alec Rhea
Apr 1 at 9:31
$begingroup$
I agree that $omega$ is a very concrete and special set, but in some sense with characteristics of the continuum we're asking if the way in which we move up the cardinal ladder from $omega$ using the powerset operation 'misses' any smaller cardinalities that could be defined if more care was taken, and I think we can ask exactly the same question with the infinitely iterated successor operation to move from finite cardinals to $omega$.
$endgroup$
– Alec Rhea
Apr 1 at 9:39
$begingroup$
Although I suppose it isn't exactly the same question since with the powerset operation we're asking if we have 'overshot' any smaller cardinals, and with the iterated successor operation we apparently can't overshoot any cardinals but can still ask if we've landed 'between' other incomparable cardinalities by taking the obvious route into the infinite.
$endgroup$
– Alec Rhea
Apr 1 at 9:53
1
$begingroup$
Amorphous sets cannot even be linearly ordered, let alone carry an ordered field structure.
$endgroup$
– Asaf Karagila
Apr 1 at 10:36
3
$begingroup$
The arithmetic of the surreals does not correspond to cardinal arithmetic. You could argue that it corresponds to natural arithmetic on the ordinals (i.e. Hessenberg sums), but the fact we need to specify the arithmetic tells us that it is not the arithmetic, so saying that the surreal numbers embed the ordinals is a bit of a stretch.
$endgroup$
– Asaf Karagila
Apr 1 at 10:57
|
show 7 more comments
$begingroup$
Any cardinal smaller than $aleph_0$ is finite. Amorphous sets are not "smaller", they are just incomparable with. They are very small, in some sense, for example we cannot even divide them into two infinite sets, but they are still infinite.
With respect to the article you linked, let me point out that amorphous sets cannot even be mapped onto $omega$, so they are definitely not the countable union of pairs.
Now, there are some combinatorial characteristics one can assign to general sets, which may be of interest in the case of amorphous sets. For example, if $A$ is amorphous, then any partition of $A$ is up to finitely many parts constant in size (i.e. all but finitely many parts are singletons, or pairs, or so on). We call this size the gauge of the partition, and we can ask what is the supremum of the gauges of possible partitions.
This can be $1$, or some finite $n$, or it can be "unbounded". We can prove, for example, that if $A$ is an amorphous set which can be made into a group, then it is unbounded. So it gives us some information.
But in general, this is not something too similar to cardinal characteristics in the traditional sense, and it is not something too helpful, since $omega$ is a very unique and a very concrete set, whereas amorphous sets can come in many different flavors, sizes, and support different structures.
answered Apr 1 at 9:18
Asaf KaragilaAsaf Karagila
21.7k681185
$endgroup$
Any cardinal smaller than $aleph_0$ is finite. Amorphous sets are not "smaller", they are just incomparable with. They are very small, in some sense, for example we cannot even divide them into two infinite sets, but they are still infinite.
With respect to the article you linked, let me point out that amorphous sets cannot even be mapped onto $omega$, so they are definitely not the countable union of pairs.
Now, there are some combinatorial characteristics one can assign to general sets, which may be of interest in the case of amorphous sets. For example, if $A$ is amorphous, then any partition of $A$ is up to finitely many parts constant in size (i.e. all but finitely many parts are singletons, or pairs, or so on). We call this size the gauge of the partition, and we can ask what is the supremum of the gauges of possible partitions.
This can be $1$, or some finite $n$, or it can be "unbounded". We can prove, for example, that if $A$ is an amorphous set which can be made into a group, then it is unbounded. So it gives us some information.
But in general, this is not something too similar to cardinal characteristics in the traditional sense, and it is not something too helpful, since $omega$ is a very unique and a very concrete set, whereas amorphous sets can come in many different flavors, sizes, and support different structures.
answered Apr 1 at 9:18
Asaf KaragilaAsaf Karagila
21.7k681185
answered Apr 1 at 9:18
Asaf KaragilaAsaf Karagila
21.7k681185
answered Apr 1 at 9:18
Asaf KaragilaAsaf Karagila
21.7k681185
answered Apr 1 at 9:18
Asaf KaragilaAsaf Karagila
21.7k681185
21.7k681185
$begingroup$
Thank you for the pointers Asaf, I was editing the question to add a link to your answer regarding a group structure on an unbounded amorphous set as a potential source of 'characteristics' but it seems I may have had things backwards. My very naive motivation for asking about this was the thought that 'amorphous cardinals' might correspond to surreal numbers in a universe without choice, but if their cardinals aren't comparable with $omega$ I don't see how they could be members of the surreals in any universe.
$endgroup$
– Alec Rhea
Apr 1 at 9:31
$begingroup$
I agree that $omega$ is a very concrete and special set, but in some sense with characteristics of the continuum we're asking if the way in which we move up the cardinal ladder from $omega$ using the powerset operation 'misses' any smaller cardinalities that could be defined if more care was taken, and I think we can ask exactly the same question with the infinitely iterated successor operation to move from finite cardinals to $omega$.
$endgroup$
– Alec Rhea
Apr 1 at 9:39
$begingroup$
Although I suppose it isn't exactly the same question since with the powerset operation we're asking if we have 'overshot' any smaller cardinals, and with the iterated successor operation we apparently can't overshoot any cardinals but can still ask if we've landed 'between' other incomparable cardinalities by taking the obvious route into the infinite.
$endgroup$
– Alec Rhea
Apr 1 at 9:53
1
$begingroup$
Amorphous sets cannot even be linearly ordered, let alone carry an ordered field structure.
$endgroup$
– Asaf Karagila
Apr 1 at 10:36
3
$begingroup$
The arithmetic of the surreals does not correspond to cardinal arithmetic. You could argue that it corresponds to natural arithmetic on the ordinals (i.e. Hessenberg sums), but the fact we need to specify the arithmetic tells us that it is not the arithmetic, so saying that the surreal numbers embed the ordinals is a bit of a stretch.
$endgroup$
– Asaf Karagila
Apr 1 at 10:57
|
show 7 more comments
$begingroup$
Thank you for the pointers Asaf, I was editing the question to add a link to your answer regarding a group structure on an unbounded amorphous set as a potential source of 'characteristics' but it seems I may have had things backwards. My very naive motivation for asking about this was the thought that 'amorphous cardinals' might correspond to surreal numbers in a universe without choice, but if their cardinals aren't comparable with $omega$ I don't see how they could be members of the surreals in any universe.
$endgroup$
– Alec Rhea
Apr 1 at 9:31
$begingroup$
I agree that $omega$ is a very concrete and special set, but in some sense with characteristics of the continuum we're asking if the way in which we move up the cardinal ladder from $omega$ using the powerset operation 'misses' any smaller cardinalities that could be defined if more care was taken, and I think we can ask exactly the same question with the infinitely iterated successor operation to move from finite cardinals to $omega$.
$endgroup$
– Alec Rhea
Apr 1 at 9:39
$begingroup$
Although I suppose it isn't exactly the same question since with the powerset operation we're asking if we have 'overshot' any smaller cardinals, and with the iterated successor operation we apparently can't overshoot any cardinals but can still ask if we've landed 'between' other incomparable cardinalities by taking the obvious route into the infinite.
$endgroup$
– Alec Rhea
Apr 1 at 9:53
1
$begingroup$
Amorphous sets cannot even be linearly ordered, let alone carry an ordered field structure.
$endgroup$
– Asaf Karagila
Apr 1 at 10:36
3
$begingroup$
The arithmetic of the surreals does not correspond to cardinal arithmetic. You could argue that it corresponds to natural arithmetic on the ordinals (i.e. Hessenberg sums), but the fact we need to specify the arithmetic tells us that it is not the arithmetic, so saying that the surreal numbers embed the ordinals is a bit of a stretch.
$endgroup$
– Asaf Karagila
Apr 1 at 10:57
$begingroup$
Thank you for the pointers Asaf, I was editing the question to add a link to your answer regarding a group structure on an unbounded amorphous set as a potential source of 'characteristics' but it seems I may have had things backwards. My very naive motivation for asking about this was the thought that 'amorphous cardinals' might correspond to surreal numbers in a universe without choice, but if their cardinals aren't comparable with $omega$ I don't see how they could be members of the surreals in any universe.
$endgroup$
– Alec Rhea
Apr 1 at 9:31
$begingroup$
Thank you for the pointers Asaf, I was editing the question to add a link to your answer regarding a group structure on an unbounded amorphous set as a potential source of 'characteristics' but it seems I may have had things backwards. My very naive motivation for asking about this was the thought that 'amorphous cardinals' might correspond to surreal numbers in a universe without choice, but if their cardinals aren't comparable with $omega$ I don't see how they could be members of the surreals in any universe.
$endgroup$
– Alec Rhea
Apr 1 at 9:31
$begingroup$
I agree that $omega$ is a very concrete and special set, but in some sense with characteristics of the continuum we're asking if the way in which we move up the cardinal ladder from $omega$ using the powerset operation 'misses' any smaller cardinalities that could be defined if more care was taken, and I think we can ask exactly the same question with the infinitely iterated successor operation to move from finite cardinals to $omega$.
$endgroup$
– Alec Rhea
Apr 1 at 9:39
$begingroup$
I agree that $omega$ is a very concrete and special set, but in some sense with characteristics of the continuum we're asking if the way in which we move up the cardinal ladder from $omega$ using the powerset operation 'misses' any smaller cardinalities that could be defined if more care was taken, and I think we can ask exactly the same question with the infinitely iterated successor operation to move from finite cardinals to $omega$.
$endgroup$
– Alec Rhea
Apr 1 at 9:39
$begingroup$
Although I suppose it isn't exactly the same question since with the powerset operation we're asking if we have 'overshot' any smaller cardinals, and with the iterated successor operation we apparently can't overshoot any cardinals but can still ask if we've landed 'between' other incomparable cardinalities by taking the obvious route into the infinite.
$endgroup$
– Alec Rhea
Apr 1 at 9:53
$begingroup$
Although I suppose it isn't exactly the same question since with the powerset operation we're asking if we have 'overshot' any smaller cardinals, and with the iterated successor operation we apparently can't overshoot any cardinals but can still ask if we've landed 'between' other incomparable cardinalities by taking the obvious route into the infinite.
$endgroup$
– Alec Rhea
Apr 1 at 9:53
1
1
$begingroup$
Amorphous sets cannot even be linearly ordered, let alone carry an ordered field structure.
$endgroup$
– Asaf Karagila
Apr 1 at 10:36
$begingroup$
Amorphous sets cannot even be linearly ordered, let alone carry an ordered field structure.
$endgroup$
– Asaf Karagila
Apr 1 at 10:36
3
3
$begingroup$
The arithmetic of the surreals does not correspond to cardinal arithmetic. You could argue that it corresponds to natural arithmetic on the ordinals (i.e. Hessenberg sums), but the fact we need to specify the arithmetic tells us that it is not the arithmetic, so saying that the surreal numbers embed the ordinals is a bit of a stretch.
$endgroup$
– Asaf Karagila
Apr 1 at 10:57
$begingroup$
The arithmetic of the surreals does not correspond to cardinal arithmetic. You could argue that it corresponds to natural arithmetic on the ordinals (i.e. Hessenberg sums), but the fact we need to specify the arithmetic tells us that it is not the arithmetic, so saying that the surreal numbers embed the ordinals is a bit of a stretch.
$endgroup$
– Asaf Karagila
Apr 1 at 10:57
|
show 7 more comments
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1
$begingroup$
@YCor Itlooks like the tag will change or be removed. Better wait.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:32
2
$begingroup$
@Alec Your proposal in the latest comment is a better title. In particular, the new title is misleading. You are not really talking about $omega$ or countability here.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:35
1
$begingroup$
It does not makes sense to say that CH is undecidable in a universe.
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:37
1
$begingroup$
(Thanks for the new title. It makes more sense.)
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:40
2
$begingroup$
But I think your description of cardinal characteristics in the opening paragraph is still incorrect. We are not interested in persistence of properties of $omega$ or anything like that. You typically look at combinatorial (or topological or...) objects of size at most continuum and uncountable, not at $omega$ and any kind of persistence. In fact, inequalities between cardinal characteristics are usually established by arguments that make sense and carry nonvacuous information (they still tell you something about these objects), even if CH holds
$endgroup$
– Andrés E. Caicedo
Apr 1 at 11:46