I heard that $\omega^n$ (for any positive real, n) is an omnific integer, but dose this property also extend to numbers such as $\sqrt[\omega]{\omega}$ (aka $\omega^{1/\omega}$)?
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1Quick beginner guide for asking a well-received question + How to ask a good question – terran Sep 06 '23 at 10:53
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Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 06 '23 at 10:55
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2Not every operation making sense for real numbers also makes sense for infinite ordinal numbers. – Peter Sep 06 '23 at 12:39
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2@Peter This is taking place in the surreal numbers, per the tag, where "$\sqrt[\omega]{\omega}$" actually is meaningful. Remember that every ordinal is also a surreal number. – Noah Schweber Sep 06 '23 at 17:55
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2I would remind voters that just because you don't understand a question does not necessarily mean it is unclear. This question is perfectly clear to anyone familiar with the surreal numbers. – Eric Wofsey Sep 07 '23 at 13:31
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2What more info could be added to the question? – Gabriel Tellez Sep 07 '23 at 15:46
1 Answers
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Assuming you mean to define $\omega^{1/\omega}$ by the definition of powers of $\omega$ in Conway's writings that is featured on Wikipedia, then yes, $\omega^{1/\omega}$ is an omnific integer. By definition, $\omega^{1/\omega}=\{\mathbb R^+\mid r\omega^{1/n}\}$ where $r$ ranges over the positive reals and $n$ ranges over the positive naturals (or the powers of $2$, say). Subtracting $1$ or adding $1$ won't go beyond those bounds (the same is true of any positive power of $\omega$), so $\omega^{1/\omega}$ is indeed an omnific integer by the simplicity theorem.
Mark S.
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