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(Why is Kunen inconsistency at the top of Cantor's upper attic?) I have seen statements about the Kunen Inconsistency being thrown around like that it is the 'upper-bound' or 'limit' for how strong of a large cardinal you can create within ZFC (Kunen inconsistency cardinals are inconsistent with ZFC). This implies that these cardinals (Reinhardt, ω-huge cardinals, etc.) are too 'strong' for choice and that there mere implication 'break choice.' This may apply to Reinhardt and Berkeley cardinals, but what about ω-huge cardinals?

As seen from Definition of an $\omega$-huge cardinal , ω-huge cardinals are equivalent to the large cardinal axiom I1 in strength (they are also both constructed via a elementary embedding from some rank Vλ+1 to itself), below the likes of I0 (which is consistent with ZFC). Here is my question: Why are ω-huge cardinals below the choiceless cardinals (Reinhardt, etc.) when it is apart of the Kunen Inconsistency and are supposed to 'too strong' for choice?

My current theory (with my limited knowledge) rather is that the 'Kunen Inconsistency isn't a limit, but rather a inconsistency occurs when the schema for these large cardinals are found within ZFC (Like they are 'incompatible' with the axiom of choice). Is the Kunen inconsistency really a limit or not?

SI J
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  • The definition of $\omega$-huge in the Martin–Steel paper is different from what you'd expect nowadays. So you need to be very explicit in what you mean when you say $\omega$-huge. – Asaf Karagila Oct 12 '22 at 09:44
  • @AsafKaragila From my understanding the ω-huge cardinals defined from the Martin - steel paper is defined as (equivalent to) I2, while currently they are defined as I1? Nonetheless they are still inconsistent with choice and doesn't really matter. For this question just go with the current version. – SI J Oct 12 '22 at 09:58
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    None of the I(n) axioms are known to be inconsistent with ZFC. – Asaf Karagila Oct 12 '22 at 10:10
  • @AsafKaragila Yeah I am aware of that, but ω-huge cardinals are meant to be equivalent to I1 in strength (it is also constructed similarly, I myself do not know the difference only that is inconsistent with choice). – SI J Oct 12 '22 at 10:26
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    For context I had some conversations with SI J about this, and we agree that horizontal "barriers" drawn in large cardinal trees (the V=L line, the Kunen inconsistency, etc) are statements about only the axioms about large cardinals. E.g all the Kunen inconsistency line is stating is that ZFC+X is inconsistent when an LCA X is written above the line. (This contrasts how we can choose non-LCAs X where ZFC+X is inconsistent but is weaker than ZFC+Inacc.) ZFC+'ω-huge exists' seems to be an exception, this phenomenon happens where it's weaker than I0 but inconsistent, but this time it is an LCA. – C7X Oct 12 '22 at 23:09
  • @AsafKaragila Would you say it's possible construct a hypothetical large cardinal that is 'stronger' than that of Reinhardt cardinals within ZFC (such that ZFC + 'hypothetical large cardinal that is large than Reinhardt cardinal' is consistent with ZFC)? If this is so then wouldn't the Kunen Inconsistency not be a limit? Or perhaps it is a limit in another way? – SI J Oct 14 '22 at 15:12
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    This seems like a relevant question: https://mathoverflow.net/q/320368 – C7X Nov 17 '22 at 02:08

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