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This is from a set of notes in a section on weakly inaccessible cardinals:

Although ZFC cannot prove the existence of weakly inaccessible cardinals, it can prove the existence of fixed points $\omega_{\alpha}=\alpha$ such as the union of $\omega, \omega_{\omega},\omega_{\omega_{\omega}},\dots$

[I know there is plenty of discussion regarding the notation as quoted. I does come from someone highly qualified.]

I know that for a weakly inaccessible cardinal how to show using cofinality relations that $\aleph_{\alpha}=\alpha$ (but this seems irrelevant here?).

My questions are: what is the union mentioned above, and what is the fixed point and how to derive it?

Thanks

Lucenaposition
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    Find whoever wrote $\aleph_{\aleph_0}$ and tell them they are a bad person for doing that. – Asaf Karagila Mar 13 '18 at 14:57
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    Also, https://math.stackexchange.com/questions/1747465/, https://math.stackexchange.com/questions/2405052/, https://math.stackexchange.com/questions/633605/, https://math.stackexchange.com/questions/37543/, https://math.stackexchange.com/questions/644784/, https://math.stackexchange.com/questions/772104/ – Asaf Karagila Mar 13 '18 at 15:00
  • @AsafKaragila You're fast (as well as better than good). I have my work cut out for me with the links. Taking a quick look, do any of them address my probably beginner's question as to what is the union? P.S. I assure you the credentials of the author are despite your characterization :) impeccable - or as Henning M said in a remark when I was looking (obviously not to effectively) for similar questions: someone who knows what they are talking about. Thanks as always, –  Mar 13 '18 at 15:23
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    I'm not sure about whether or not it will help you. The point is that the union is just a cardinal which is a fixed point, and there's no real way of describing it other than an $\aleph$ fixed point. As for the characterization, I don't know how impeccable this person can be if they write $\aleph_{\aleph_0}$... nobody is perfect. :) – Asaf Karagila Mar 13 '18 at 15:32
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    I'm pretty sure "union" means the usual union of sets, when the cardinals are replaced with sets representing their cardinality, or the cardinals are identified with certain sets (the initial ordinals in the Von Neumann ordinals) in the cumulative hierarchy. For the usual ordering on cardinal numbers, such a union gives you the least upper bound of the cardinal numbers the union is taken over. – Dave L. Renfro Mar 13 '18 at 16:44
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    @Andrew There is a very good reason not to write something like "$\aleph_{\aleph_0}$." Namely, what would you write for the next cardinal? You might be tempted to write "$\aleph_{\aleph_0+1},$" but the use of the cardinal notation "$\aleph_0$" in the subscript, as opposed to the ordinal notation "$\omega$, indicates that "$+$" refers to cardinal addition. In this sense, $\aleph_0+1=\aleph_0$, so we have a problem. You might think the successor cardinal notation saves us by letting us write "$\aleph_{\aleph_0}^+$" but... – Noah Schweber Mar 18 '18 at 21:17
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    this turns out to be a false hope: how should we denote (say) the "$\aleph_0$th" (better: $\omega$th!) cardinal after $\aleph_{\aleph_0}$? And so on. The problem is that cardinal arithmetic is completely unsuited to this task. By contrast, ordinal arithmetic behaves much better for our purposes: "$\aleph_{\omega+1}$" is clearly different from "$\aleph_\omega$," since the use of ordinal notation indicates that ordinal addition is meant and in the sense of ordinal addition $\alpha+1\not=\alpha$ for any ordinal $\alpha$. Similarly, expressions like "$\aleph_{\omega^2+17}$" make perfect sense. – Noah Schweber Mar 18 '18 at 21:20
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    Of course, this would be fixed if we just used different symbols for ordinal addition and cardinal addition rather than having the meaning depend on context: since $\aleph_0$ is literally the same set as $\omega$ (we just use two different notations to indicate context), if we wrote "$+{ord}$" for ordinal addition we could write "$\aleph{\aleph_0+{ord}1}$" for "$\aleph{\omega+1}$" without confusion. Unfortunately, that's not the way it turned out, and so you have to use the right notation corresponding to your particular context. – Noah Schweber Mar 18 '18 at 21:22
  • @NoahSchweber You always go above and beyond. Changed title so your comments are highlighted. –  Mar 18 '18 at 23:46
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    No, it doesn't. :) – Asaf Karagila Apr 10 '18 at 19:28

1 Answers1

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The union in your quote should probably be $$ \omega_0 \cup \omega_{\omega_0} \cup \omega_{\omega_{\omega_0}} \cup \cdots $$ is the smallest solution to the equation $\alpha=\omega_\alpha$, which is also mentioned in your quote.

I don't think that particular set has any nicer description than the two you already have, so asking "what it is" cannot really produce any answer other than the definitions you already have. It's a certain initial ordinal, with the properties you see before you.

ZFC can prove that a set with these properties exists, by defining a function from naturals to ordinals as $$ F(0) = \omega \\ F(n+1) = \omega_{F(n)} $$ Then applying the Axiom of Replacement to $\omega$ and this function gives you $$ \{ \omega_0, \omega_{\omega_0}, \omega_{\omega_{\omega_0}}, \ldots \} $$ and the Axiom of Unions applied to this gives $$ \alpha = \omega_0 \cup \omega_{\omega_0} \cup \omega_{\omega_{\omega_0}} \cup \cdots $$

To see that this set satisfies $\alpha = \omega_\alpha$, note that it is clearly a limit ordinal, and for limit ordinals $\omega_\alpha$ is defined as the smallest ordinal that is $\ge$ $\omega_\beta$ for all $\beta<\alpha$ -- which by well-known properties of ordinals is the union of all these ordinals.

Since $\alpha$ is the union of a sequence of ordinals, and $\omega_-$ of each of those ordinals is also in the sequence, this means that $\alpha$ and $\omega_\alpha$ are both the union of essentially the same increasing sequence of ordinals; therefore they have to be equal.

hmakholm left over Monica
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