Timeline for Is the power set axiom essential for constructing L?
Current License: CC BY-SA 4.0
10 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Aug 27, 2023 at 15:02 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
deleted 194 characters in body
|
Aug 27, 2023 at 14:16 | comment | added | Joel David Hamkins | The cardinal order relation on the set of cardinals is the same as the ordinal relation on the set of cardinals. But you are using the cardinality order relation on the set of ordinals, which is weird. No need to edit, though, since the question already has answers, and it will perhaps confuse the discussion to change it. | |
Aug 27, 2023 at 14:12 | comment | added | Zuhair Al-Johar | Ok, then I'll re-write it, I thought if I just say "cardinal" smaller than or bigger relation, then I'm done, and I thought it was clear. Thanks. | |
Aug 27, 2023 at 14:09 | comment | added | Joel David Hamkins | Your $\leq$ notation is very bad, since you will end up asserting things like $\omega+5\leq\omega$ and so forth. For ordinals $\alpha,\beta$, the notation $\alpha\leq\beta$ is universally understood to mean the ordinal order, and you invite needless confusion by redefining this totally standard relation. It is needless because you could just write $|\alpha|\leq\beta$ for your intended meaning. But it is better to just formulate your intended axiom by saying: every cardinal $\kappa$ has a successor cardinal $\kappa^+$. | |
Aug 27, 2023 at 14:00 | comment | added | Joel David Hamkins | That is very idiosyncratic notation. I had thought you had meant to restrict to cardinals $\alpha$, in which case the order relation is the same as for ordinals. But if you meant $|\alpha|\leq\kappa$, then you are indeed asserting that $\kappa^+$ exists. But why not say it that way? Every cardinal has a successor cardinal. It would be much clearer. | |
Aug 27, 2023 at 13:39 | comment | added | Zuhair Al-Johar | But $\leq$ refers to cardinal smaller than or equal relation and not to ordinal smaller than or equal relation. So, I think the axiom as I've stated it does capture successor cardinals and cannot be captured by successor ordinals. | |
Aug 27, 2023 at 12:46 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 22 characters in body
|
Aug 27, 2023 at 12:16 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 848 characters in body
|
Aug 27, 2023 at 12:08 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 848 characters in body
|
Aug 27, 2023 at 12:02 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |