Timeline for Countable abelian group of exponent $2$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 16, 2019 at 19:05 | comment | added | Jan-Christoph Schlage-Puchta | @YCor: You can define $|X|$ to be the equivalence class of $X$ under bijections. This will be a proper class, so you have to be quite careful. And in general sets cannot be compared. It is messy, but it can be done. | |
| Jan 13, 2019 at 12:32 | comment | added | YCor | @Jan-ChristophSchlage-Puchta what do you call $|X|$ in general? the definition I've learnt is that $|X|$ is the smallest cardinal equipotent to $X$, provided it exists. In this way it's meaningful in ZF but not every set has a cardinal. | |
| Jan 13, 2019 at 12:30 | comment | added | YCor | @Jan-ChristophSchlage-Puchta No "uncountable" for "continuum" is just a mistake. What' frequent, and not technically a mistake, consisting in writing down a result asserting that something is uncountable, when the proof shows that it has at least continuum cardinality. | |
| Jan 13, 2019 at 12:08 | comment | added | Jan-Christoph Schlage-Puchta | @YCor: Without AC cardinality becomes incredibly messy, but that doesn't mean that $|G|$ does not make sense. Also whenever one talks about exotic sets, one should not use sloppy language. "Uncountable" is too often used as a synonym "continuum", and AC is assumed even if it is not needed. | |
| Jan 13, 2019 at 11:57 | comment | added | YCor | @Jan-ChristophSchlage-Puchta there's no reference to AC in the post, which implicitly means that AC is assumed. The answer by GH, which also uses implicitly AC, was accepted by the OP, which confirms this interpretation. (Under the negation of AC, $|G|$ doesn't make sense by the way, so one should say "$G$ is infinite countable" rather than "$|G|=\aleph_0$".) It's a any useful remark that AC is not needed here. | |
| Jan 13, 2019 at 11:43 | comment | added | Jan-Christoph Schlage-Puchta | @YCor : Whether this is a duplicate or not depends on whether one believes in AC or not. With AC the question is trivial (the MSE question a little less so, as the exponent is not prime). Without AC the answer given on MSE is false, even for exponent 2. However, in the countable case we don't need AC. A bijection between $G$ and $\mathbb{N}$ induces a well ordering on $G$. Thus we can recursively construct the lexicographically minimal basis of $G$, which induces an isomorphism $G\cong \{0,1\}^{<\omega}$. | |
| S Jan 7, 2019 at 15:59 | history | suggested | Alex Kruckman | CC BY-SA 4.0 |
Fixed a typo
|
| Jan 7, 2019 at 15:48 | review | Suggested edits | |||
| S Jan 7, 2019 at 15:59 | |||||
| Jan 7, 2019 at 15:23 | comment | added | YCor | Subduplicate of MathSE post: math.stackexchange.com/questions/1193556/… | |
| Jan 7, 2019 at 15:11 | history | edited | Neil Hoffman | CC BY-SA 4.0 |
Changed $x$ in the final question to $g$ which seems to agree with the qualifier for all $g \in G$.
|
| Jan 7, 2019 at 9:55 | review | Close votes | |||
| Jan 14, 2019 at 18:28 | |||||
| Jan 7, 2019 at 8:50 | vote | accept | Dominic van der Zypen | ||
| Jan 7, 2019 at 7:58 | history | edited | GH from MO | CC BY-SA 4.0 |
edited body; edited title
|
| Jan 7, 2019 at 7:52 | answer | added | GH from MO | timeline score: 7 | |
| Jan 7, 2019 at 6:58 | comment | added | Gerhard Paseman | It is tempting to think so, as one can look at a direct sum of Z2 with index set G, and "remove" those indices c where c=ab and (under some well ordering of G) a and b are before c. However, there may be some exotic behaviour being overlooked by this view point. Gerhard "Should Look Up Boolean Groups" Paseman, 2019.01.06. | |
| Jan 7, 2019 at 6:34 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |