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Aug 27 at 19:55 vote accept cody
Aug 27 at 18:16 history edited Noah Schweber
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Aug 27 at 13:20 history became hot network question
Aug 27 at 12:52 answer added Emil Jeřábek timeline score: 27
Aug 27 at 7:45 comment added Emil Jeřábek The theory of ordinals with $+$ is decidable. I do not remember off-hand what happens with $\omega^-$, but anyway, you should look at Ehrenfeucht's paper first.
Aug 27 at 2:18 history edited Alex Kruckman
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Aug 26 at 22:09 history edited cody CC BY-SA 4.0
Fix notation to more standard one.
Aug 26 at 22:08 comment added cody Both fair comments. I'll probably keep the title for continuity though (and I wouldn't mind an answer to the other question as well).
Aug 26 at 20:46 comment added Joel David Hamkins Regarding the title, Cantor normal form applies to all ordinals, not just ordinals up to $\varepsilon_0$. Every ordinal can be uniquely expressed as a finite sum $\omega^{\alpha_n}+\cdots+\omega^{\alpha_0}$ with $\alpha_n\geq\cdots\geq\alpha_0$. Below $\varepsilon_0$, the interesting thing is that you can get a nice terminating finite hereditary representation, with the exponents also represented this way. At $\varepsilon_0$, however, the hereditary representation starts not to work, since the representation $\varepsilon_0=\omega^{\omega^{\omega^{\cdot^{\cdot}}}}$ doesn't terminate.
Aug 26 at 20:35 comment added Joel David Hamkins If you intend to ask about the structure whose domain is the ordinals below $\varepsilon_0$, then you should write $(\varepsilon_0,\ldots)$, since $\text{Ord}^{<\varepsilon_0}$ would usually denote the proper class of all sequences of (arbitrary) ordinals, with length less than $\varepsilon_0$. Similarly, $\omega$ is the set of finite ordinals, whereas $\text{Ord}^{<\omega}$ is the class of all finite sequences of ordinals.
Aug 26 at 19:40 history asked cody CC BY-SA 4.0