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It is known that if ZFC is consistent, ZFC+"no such large cardinal exists" is consistent. Then, is ZFC+"such large cardinal exists" known to be consistent? This would imply that proving large cardinal axiom is independent from ZFC...


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up vote 3 down vote accepted

ZFC cannot prove its own consistency, this is a result due to Godel's incompleteness theorems.

If, however, $\kappa$ is an inaccessible cardinal then $V_\kappa$ which is the collection of all sets whose von Neumann rank is $<\kappa$ is a model of ZFC. Due to the completeness theorem of Godel, this implies that ZFC is consistent.

So we cannot prove the existence of an inaccessible cardinal in ZFC.

Furthermore, suppose ZFC+There is an inaccessible cardinal is consistent, let $\kappa$ be the least inaccessible cardinal, then $V_\kappa$ is a model of ZFC but there are no inaccessible in $V_\kappa$.

(and almost all large cardinals are inaccessible, and if they are not inaccessible they imply that below them there are inaccessible cardinals, it might be that a cardinal is weakly inaccessible but "going down" to $L$ makes it a strongly inaccessible, and we are only interested in consistency results anyway)

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