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I've been reading about set theory and the difference between small and large cardinals. since taking the power set of small cardinals (alephs) allows us to create larger cardinals/alephs I know that an inaccessible cardinal is equivalent to the regular aleph fixed point, and cannot be reached by taking the power sets of alephs. So my questions are:

  • what would the power set of an inaccessible cardinal be?
  • what is the regular fixed point of the first inaccessible cardinal? (basically, a cardinal that is larger than the first inaccessible cardinal to the same degree the first inaccessible cardinal is larger than the first small cardinals like aleph 0)
Adithya
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    What is the power set of anything? It's just the set of all subsets of that thing. And what does the second question even mean (syntactically, I can't parse it, what is a "fixed point of a cardinal")? – Asaf Karagila Jun 02 '23 at 08:55
  • @Asaf Karagila: what is a "fixed point of a cardinal" -- Probably the intended meaning is what Joel David Hamkins describes in this mathoverflow question. I don't know much about them, but my (and others) comments to Veblen function with uncountable ordinals & beyond may be of help, as well as the mathoverflow questions Mahlo cardinal and hyper k-inaccessible cardinal AND Limit of Mahlo cardinals. – Dave L. Renfro Jun 02 '23 at 09:14
  • @AsafKaragila i meant the aleph fixed point, something outlined in this thread https://math.stackexchange.com/questions/3089308/is-the-least-inaccessible-cardinal-equivalent-to-the-first-aleph-fixed-point would it not be an inaccessible cardinal? – Adithya Jun 03 '23 at 05:10
  • If $\kappa$ is an inaccessible cardinal, it is the $\kappa$th fixed point. – Asaf Karagila Jun 03 '23 at 05:34
  • @Adithya: "would it not be an inaccessible cardinal" -- You should look more thoroughly at the mathoverflow links I gave. In particular, see my comments to the question Veblen function with uncountable ordinals & beyond -- none of the least cardinals in any of the sets $A^{\beta}$ (or the first countably many such cardinals, or any of the first $\aleph_{\omega_{17}}$ such cardinals, etc.) are inaccessible as long as $\beta$ itself is chosen to be less than the smallest inaccessible. Similarly for their diagonal intersections. – Dave L. Renfro Jun 03 '23 at 13:14
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    “A cardinal that’s larger than the first inaccessible to the same degree that the first inaccessible is larger than $\aleph_0$”… if that means anything I would think it refers to the second inaccessible. – spaceisdarkgreen Jun 03 '23 at 16:04
  • what would that be? Ive seen someone do the math and argued it would be a 1-inaccessible cardinal, as it would act as a strong limit of inaccessible cardinals or as a fixed point – Adithya Jun 03 '23 at 19:32
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    The second inaccessible is the second least inaccessible… the next one. It is “inaccessible” from the least inaccessible in exactly the same manner as the least inaccessible is “inaccessible” from $\aleph_0$. A 1-inaccessible (i.e. a fixed point in the enumeration of inaccessibles) is much larger. – spaceisdarkgreen Jun 03 '23 at 23:10
  • so its just called the “second inaccessible cardinal” ? is there any other notation for it, and how do you define it mathematically – Adithya Jun 05 '23 at 09:36
  • @Adithya if you’re replying in the comments you need to address a person with the “@“ sign or they won’t be notified. Sure, no special notation. It’s defined mathematically as the second least inaccessible cardinal… that’s a perfectly precise specification (provided at least two inaccessible cardinals exist). – spaceisdarkgreen Jun 06 '23 at 20:04
  • @spaceisdarkgreen will do from now on. thanks for the explanation! – Adithya Jun 08 '23 at 13:04
  • @spaceisdarkgreen although, if the second inaccessible cardinal transcends the first the same way the first transcends aleph 0, what would transcend the first inaccessible the same way that aleph 1 transcends aleph 0? basically achieving higher cardinals via powersets? basically my original question of whether taking the power set of an inaccessible cardinal k or 2^k yields a higher cardinal – Adithya Jun 08 '23 at 19:32
  • @Adithya That's not at all how $\aleph_1$ transcends $\aleph_0$... that has nothing to do with power sets. There is a cardinal $\kappa^+$ right after $\kappa$ that is generated from $\kappa$ the same way $\aleph_1$ is generated from $\aleph_0.$ It is not inaccessible, since it is a successor cardinal. $2^\kappa$ is the cardinality of $P(\kappa)$ and is larger than $\kappa$ by Cantor's theorem... it could consistently be $\kappa^+$ or could consistently be strictly larger than $\kappa^+$ just like $2^{\aleph_0}$ could be equal to or strictly larger than $\aleph_1.$ – spaceisdarkgreen Jun 08 '23 at 19:48
  • @spaceisdarkgreen isnt p(k) a power set? which generates a cardinal strictly greater than k and is the next cardinal. so if you take the power set of an inaccessible cardinal k it would be a cardinal greater than k the same way aleph 1 is greater than aleph 0? because the successor cardinal is generated via the same operation? idk if im confused or not. but either way, then what would the power set of the first inaccessible cardinal be? – Adithya Jun 13 '23 at 16:36
  • @spaceisdarkgreen ah i misread your reply. but isnt 2^N0 equal to aleph 1? i know alephs arent inaccessible to their predecessors. – Adithya Jun 13 '23 at 16:45
  • @Adithya No, $2^{\aleph_0}=\aleph_1$ is the continuum hypothesis and the generalized continuum hypothesis is the statement that cardinal successor is the same thing as power set. You keep asking what the power set is and we keep answering “the power set” in various ways… that’s the answer and I don’t know what else you’re expecting. You need to understand the basics of how cardinals work before you can productively ask questions about hyperinaccessibility and the like. – spaceisdarkgreen Jun 13 '23 at 17:54
  • @Adithya To clarify, when I say “no, that’s the continuum hypothesis”, I say “no” because the continuum hypothesis is not necessarily true… it is consistent that it it true and consistent that it is false. – spaceisdarkgreen Jun 13 '23 at 18:09

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