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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 $\aleph_{\alpha}=\alpha$ such as the union of $\aleph_0, \aleph_{\aleph_0},\aleph_{\aleph_{\aleph_0}}\dots$

[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

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    $\begingroup$ Find whoever wrote $\aleph_{\aleph_0}$ and tell them they are a bad person for doing that. $\endgroup$ – Asaf Karagila♦ Mar 13 '18 at 14:57
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    $\begingroup$ Also, math.stackexchange.com/questions/1747465, math.stackexchange.com/questions/2405052, math.stackexchange.com/questions/633605, math.stackexchange.com/questions/37543, math.stackexchange.com/questions/644784, math.stackexchange.com/questions/772104 $\endgroup$ – Asaf Karagila♦ Mar 13 '18 at 15:00
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    $\begingroup$ 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. :) $\endgroup$ – Asaf Karagila♦ Mar 13 '18 at 15:32
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    $\begingroup$ 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. $\endgroup$ – Dave L. Renfro Mar 13 '18 at 16:44
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    $\begingroup$ 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. $\endgroup$ – Noah Schweber Mar 18 '18 at 21:20
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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.

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