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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.

Alec Rhea
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  • Seeing how the question is *really* not about cardinal characteristics of $\omega$, it might be a good idea to change the title so it fits the question. – Asaf Karagila Apr 01 '19 at 10:42
  • @AsafKaragila I am open to suggestions or if you would like to edit feel free, but I still don’t quite see why cardinal characteristics is such a bad term for it — I agree that it’s not perfect, but I can’t think of a concise title that expresses the difference. ‘Cardinals of amorphous sets which satisfy properties of finite sets that fail for $\omega$’ seems like a mouthful :^). – Alec Rhea Apr 01 '19 at 10:59
  • But $\omega$ is *really* not the focal point here. You are not asking about CCC of $\omega$ at all. – Asaf Karagila Apr 01 '19 at 11:00
  • @AsafKaragila The Blass paper above doesn’t mention partial orders or CCC’s in the introduction or the parts I’ve gotten through (but I haven’t gotten too far), it characterizes cardinal characteristics of the continuum as I have above for $\omega$ — cardinals which satisfy properties true at $\omega$ but false at the continuum. This is the sense in which I thought this title was appropriate; is there an alternative that occurs to you? – Alec Rhea Apr 01 '19 at 11:05
  • My gripe about the title is that it seems to ask about CCCs *OF $\boldsymbol{\omega}$*. You are asking about amorphous sets. I'm not sure that this line of discussion is productive. So if you'll excuse me, I'll go about my day now. – Asaf Karagila Apr 01 '19 at 11:06
  • @AsafKaragila Fair enough, thank you for the input thus far. I will change it to ‘cardinal characteristics of amorphous sets’ or something like this if nothing better occurs to me soon. – Alec Rhea Apr 01 '19 at 11:09
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    @YCor Itlooks like the tag will change or be removed. Better wait. – Andrés E. Caicedo Apr 01 '19 at 11:32
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    @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. – Andrés E. Caicedo Apr 01 '19 at 11:35
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    It does not makes sense to say that CH is undecidable in a universe. – Andrés E. Caicedo Apr 01 '19 at 11:37
  • @AndrésE.Caicedo Got it, much appreciated -- I'll remove that comment as well. I think I'm conflating axioms and models of those axioms again. – Alec Rhea Apr 01 '19 at 11:38
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    (Thanks for the new title. It makes more sense.) – Andrés E. Caicedo Apr 01 '19 at 11:40
  • @AndrésE.Caicedo Thank you for correcting me again Andrés, I am green behind the ears in these matters. – Alec Rhea Apr 01 '19 at 11:46
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    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 – Andrés E. Caicedo Apr 01 '19 at 11:46
  • @AndrésE.Caicedo I am quoting directly from Andreas Blass's paper I linked above: "The cardinal characteristics are simply the smallest cardinals for which various results, true for $\aleph_0$, become false" (P. 3 second sentence). How do I reconcile this with what you and Asaf are telling me? I obviously don't doubt that either of you know far better than me, but you seem to be saying that Andreas is incorrect above? – Alec Rhea Apr 01 '19 at 11:50
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    @Alec You are misunderstanding the intent of the comment. You have something that is true of, say, the reals, and false of $\omega$. What is the size of the smallest set for which it is true? That is the opposite of "what is the largest size for which it is false?" In fact, that largest size may not exist. Again, it is not at all about properties of $\omega$. – Andrés E. Caicedo Apr 01 '19 at 11:55
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    For instance, $\mathbb N$ has measure 0. The natural question is not "for which sizes $\le\mathfrak c$ we have sets of that size and measure 0?" That is not so interesting, as the answer is "all". The interesting question is "$\mathbb R$ does not have measure 0. What is the smallest size of a nonmeasure 0 set?" – Andrés E. Caicedo Apr 01 '19 at 11:59
  • @AndrésE.Caicedo But he lists properties like 'if countably many sets each have Lebesgue measure $0$ then so does their union' which is true at $\omega$ and false at $|\mathbb{R}|$, which seems to indicate the interpretation I had in mind no? – Alec Rhea Apr 01 '19 at 11:59
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    No, not really. You are not asking for the property at size $\omega$ to still hold (which is what you wrote in the question). You are asking for the first size where it fails. Again, it is not about persistence. – Andrés E. Caicedo Apr 01 '19 at 12:02
  • @AndrésE.Caicedo I am missing something then and require further reflection; for now I will edit with a pointer to the fact that I'm wrong, and hopefully tomorrow will bring clarity and a better edit. On second thought, I'll just delete my incorrect interpretation and leave the link up -- if I understand things better at a later juncture I'll try again, thanks again for your patience Andrés. – Alec Rhea Apr 01 '19 at 12:03
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    And maybe in intuitionistic logic we can go even further and define cardinal characteristics of subsets of a singleton. (Ha, ha, only serious.) – Gro-Tsen Apr 01 '19 at 12:20

1 Answers1

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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.

Asaf Karagila
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  • 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. – Alec Rhea Apr 01 '19 at 09:31
  • 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 $\omega$. – Alec Rhea Apr 01 '19 at 09:39
  • 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. – Alec Rhea Apr 01 '19 at 09:53
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    Amorphous sets cannot even be linearly ordered, let alone carry an ordered field structure. – Asaf Karagila Apr 01 '19 at 10:36
  • I think that you're looking at cardinal characteristics in the wrong way. We're not "constructing" anything, and we're not "missing" anything. The universe is given to us, and we look at some very natural partial orders arising from it (e.g. $\omega^\omega/\equiv$ with $\leq^*$) and we are asking what are the properties of these partial orders. We do this, because it turns out that there are questions outside of set theory whose answers depend on the relationship between these cardinals, and we are interested in understanding how they interact with each other. – Asaf Karagila Apr 01 '19 at 10:38
  • They wouldn’t be fields or linearly ordered themselves, only members of a linearly ordered field, but the field contains $\omega$ so the linear ordering on it still prevents any amorphous cardinality from being a member. – Alec Rhea Apr 01 '19 at 10:38
  • In that case, I really don't see the point. Given any set $x$ we can define the surreals "of $x$", where we replace one of the surreal numbers by $x$ itself. What does that give us? Nothing. There's a reason why category theory is very successful in the study of algebraic structures. The elements of the field do not matter, the structure does. And set theory agrees with that. For all that we care, $0$ is an amorphous set, and there's no contradiction with anything. – Asaf Karagila Apr 01 '19 at 10:40
  • I don’t quite follow your last comment; what would be the point of replacing a surreal with a set (and how would we do it)? I was originally thinking that amorphous cardinals were ‘smaller’ than $\omega$ in the cardinal hierarchy, and all ordinals naturally correspond to surreal numbers (no replacing surreals or anything) so I thought these new cardinals might also naturally correspond to surreals below $\omega$ in a universe without choice. You’ve corrected my naïveté on the first issue and thank you for that, but I don’t understand your comment above. – Alec Rhea Apr 01 '19 at 10:46
  • Here, for example, $R_x=\Bbb R\setminus{\pi}\cup{x}$. Now transport the structure of $\Bbb R$ to $R_x$ using the obvious bijection $f(u)=u$ for $u\neq\pi$ and $f(\pi)=x$. Saying that you think that a set might be an element of the surreal numbers tells you nothing about that set, especially since the surreals do not form a transitive class. You could ask if the usual construction of the surreal numbers ever yields non-well ordered sets, but I'd be surprised if that is the case. And I'd be even more surprised if sets which cannot be linearly ordered would pop up there. – Asaf Karagila Apr 01 '19 at 10:49
  • I follow your construction now, but it’s not what I’m suggesting and you have my motivation reversed; I ask about these cardinals because they might tell me something about the surreals in a universe without choice, not because the surreals might tell me something about them. The surreals naturally contain a copy of the ordinals and thusly the cardinals in a universe with choice, and I was wondering if they still contained a copy of all the cardinals in a universe without choice; it looks like the answer is no for the reasons you’ve given here. – Alec Rhea Apr 01 '19 at 10:52
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    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. – Asaf Karagila Apr 01 '19 at 10:57
  • Very true regarding the arithmetic, but I think their relationship still naturally engenders questions about ordinals/cardinals corresponding to certain surreals (of course the cardinals of amorphous sets aren’t ordinals and thus have no Cantor normal form to manipulate for Hessenberg summing). – Alec Rhea Apr 01 '19 at 11:14