This is a question that I always had in mind.

When it is said that the consistency of a theory $T$ requires the assumption of existence of some specified cardinal $\kappa$. Is that taken to mean that $T$ can prove the existence of all cardinals strictly smaller than $\kappa$?

A clarifying example, suppose its said that a theory $T$ requires existence of $0^\sharp$, one might get the impression that $T$ can interpret each theory ZFC+ $\kappa$, where $\kappa$ is a cardinal property such that ZFC + $\kappa$ is interpretable in ZFC+$0^\sharp$. So viewing the list of large cardinal properties, then one would for example expect that $T$ can interpret ZFC+ Mahlo cardinal exist. Is that impression correct?


This is a situation where there is a fair amount of abuse of language, unfortunately.

Generally statements like this are shorthand for a consistency strength claim over the base theory ZFC; e.g. when people say "X requires a measurable cardinal" what is meant is that ZFC+"There is a measurable cardinal" is consistent relative to ZFC+X (even more formally, that PRA proves "If ZFC+X is consistent, then ZFC+"There is a measurable cardinal" is consistent).

Sometimes a different base theory on one or the other side is meant; e.g. "AD requires a measurable" means that ZF+AD is consistent only if ZFC+"There is a measurable" is, and this is clear from context since ZFC+AD is outright inconsistent.

Note that this is all strictly weaker than saying that the large cardinal actually needs to exist; when we say that X requires a measurable cardinal we generally do not mean that ZFC proves "If X, then there is a measurable cardinal." In particular, there's a huge difference between "There is a measurable cardinal" and "There is an inner model with a measurable cardinal."

(Incidentally, it's not even clear to me what "$T$ can prove the existence of all cardinals strictly smaller than $\kappa$?" would mean in the first place.)

  • $\begingroup$ I've added a clarifying example addressing the last two lines of your answer. $\endgroup$ – Zuhair Apr 13 at 18:17
  • $\begingroup$ I think the implications in your answer are the converse. I mean if I say that X requires measurable cardinal then its taken to mean that PRA proves that if ZFC+measurable cardinal is consistent, then ZFC+X is consistent. $\endgroup$ – Zuhair Apr 13 at 18:25
  • $\begingroup$ @Zuhair Absolutely not - under that reading, "$2+2=4$" would require a measurable! "Requires" is analogous to "only if." $\endgroup$ – Noah Schweber Apr 13 at 18:33
  • $\begingroup$ I wouldn't call it abuse of language. It is standard jargon. $\endgroup$ – Andrés E. Caicedo Apr 13 at 18:47
  • $\begingroup$ @NoahSchweber, Oh yes of course, I was confused, I thought its the least cardinal in that direction. Anyhow i think its equivalent to the direction you've stated. $\endgroup$ – Zuhair Apr 13 at 18:56

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