Timeline for Naturally occuring groups with cardinality greater than the reals.
Current License: CC BY-SA 2.5
19 events
| when toggle format | what | by | license | comment | |
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| May 30, 2015 at 9:03 | comment | added | few_reps | @PeteL.Clark : ok. Allow me try to figure out what you are asking for. Such a "natural group" (whatever "natural" means) would give a "natural set" by forgetting the group structure. Can you give an example of such a "natural set" (that isn't naturally endowed with a group structure -- so would not be an answer to your question) ? This would help to capture the notion of naturality you're interested in. | |
| May 30, 2015 at 7:35 | comment | added | Pete L. Clark | @few__reps: Certainly; this kind of cardinality construction is pursued in more generality in math.uga.edu/~pete/settheorypart4.pdf. The point is that it trivially works for all infinite cardinals, so it is precisely what I mean to exclude when I asked (not so precisely) for a "natural example". | |
| May 30, 2015 at 7:32 | comment | added | David Roberts♦ | @few_reps thanks, I was not sure, so this is quite handy to know (and I guess I should have been able to figure it out with a bit of thought!) | |
| May 30, 2015 at 3:05 | comment | added | few_reps | @DavidRoberts In fact such functions $S^1\to G$ are characterized by their values on, say, the set $\{e^{i\pi r }\}$ with $r$ rational (because they are continuous) ... so $\Omega G$ has the same cardinality as $G$ | |
| May 30, 2015 at 2:07 | comment | added | few_reps | @PeteL.Clark Take a set $S$ of infinite cardinal. Take the free abelian group $F(S)=\mathbf Z^{(S)}$ on this set. Then it has the same cardinal as $S$ ... right ? This would give a group of cardinal $\aleph_1$, whatever $\aleph_1$ is ... | |
| May 29, 2015 at 22:48 | comment | added | Douglas Gray | I have a more practical question. Are there any examples in applied mathematics dealing with any sets, objects, or spaces where one actually uses something with cardinality greater than that of R? | |
| Mar 10, 2011 at 4:09 | comment | added | David Roberts♦ | How about the group $\Omega G$ of (continuous) loops in a Lie group $G$? I'm only putting it in a comment because I'm not sure of the cardinality, but my guess (for what it's worth) is that it is bigger than the continuum. | |
| Mar 1, 2011 at 19:27 | comment | added | Andrés E. Caicedo | Regardless of "naturality": Shelah proved ("On a problem of Kurosh, Jónsson groups, and applications", in Word problems, II (Conf. on Decision Problems in Algebra, Oxford, 1976), pp. 373–394, Stud. Logic Foundations Math., 95, North-Holland, Amsterdam-New York, 1980) that (without any assumptions on cardinal arithmetic) there is a Jonsson group of size omega_1, that is, a group of size omega_1 such that any proper subgroup is countable. (On the other hand, I do not think it is known whether ZFC proves that there are Jonsson groups of size continuum.) | |
| Mar 1, 2011 at 18:26 | answer | added | Henno Brandsma | timeline score: 5 | |
| Aug 13, 2010 at 12:01 | vote | accept | Daniel Miller | ||
| Aug 13, 2010 at 3:48 | comment | added | Joel David Hamkins | You can push it up to $\Sigma^1_2$ using the Mansfield Solovay theorem, and if you assume PD, then there are no projective sets of size $\aleph_1$. In those models, there are arguably no natural examples of sets of reals of size exactly $\aleph_1$. | |
| Aug 13, 2010 at 1:31 | answer | added | Tracy Hall | timeline score: 8 | |
| Aug 13, 2010 at 1:05 | comment | added | Joel David Hamkins | Pete, very good question! Here is one answer: it is consistent with ZFC that there are no Borel sets, and even no analytic sets, of size $\aleph_1$. In such a model of set theory, the only groups (built out of real numbers) of size $\aleph_1$ would have high descriptive set-theoretic complexity. This could be taken as a negative answer to your question. But a positive answer could still arise by builiding a group directly out of the countable ordinals, rather than by using reals. | |
| Aug 13, 2010 at 0:55 | comment | added | Pete L. Clark | Turning this remark around, can one give a "natural" example of a group which has cardinality $\aleph_1$, independent of CH? | |
| Aug 13, 2010 at 0:54 | comment | added | BCnrd | What gave you the impression that the single most important piece of information about a group is its cardinality? It misses nearly all of the richness of group theory. For instance, many interesting classes of finite simple groups can be studied using a lot of the same tools as are used in the study of simple complex Lie algebras and semisimple linear algebraic groups. | |
| Aug 13, 2010 at 0:53 | answer | added | Pete L. Clark | timeline score: 30 | |
| Aug 13, 2010 at 0:45 | comment | added | François G. Dorais | The cardinality of $\mathbb{R}$ is not necessarily $\aleph_1$. | |
| Aug 13, 2010 at 0:44 | answer | added | Joel David Hamkins | timeline score: 11 | |
| Aug 13, 2010 at 0:26 | history | asked | Daniel Miller | CC BY-SA 2.5 |