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ZFC can't prove that strongly inaccessible cardinals exist, or else it would prove that a model of itself exists and hence $Con(ZFC)$. So this leaves us with two options:

  1. ZFC proves there are no strongly inaccessible cardinals
  2. The existence of strongly inaccessible cardinals is independent from ZFC

I've heard, however, that you can't actually prove that it's #2. That is, you can't prove the independence of large cardinal axioms from ZFC. Why is this?

At first, I thought it was because proving independence would mean that ZFC proves the consistency of a larger theory that contains ZFC as a model. However, I don't think my reasoning was right. That wouldn't prove that a model exists - just that a model could exist. And isn't it proven that Con(ZFC) is independent from ZFC, and hence that ZFC + Con(ZFC) is consistent iff ZFC is?

So how do you show that you can't show the independence of large cardinal axioms?

Asaf Karagila
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Mike Battaglia
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  • You don't necessarily have to talk about models. Working with weak arithmetical theories that suffice to develop an internal formalization of first-order logic are usually enough. Then you don't have models, but you can talk about a proof of contradiction or the lack thereof. – Asaf Karagila Nov 10 '15 at 23:21

1 Answers1

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Well. In order to establish "true independence" you need to show that $\sf ZFC$ is consistent with both the statement and its negation, at least assuming that $\sf ZFC$ is consistent.

However just assuming that $\sf ZFC$ is consistent is not enough to establish that $\sf ZFC$+large cardinal axioms are consistent, because from large cardinal axioms we can prove the consistency of $\sf ZFC$; thus a proof of consistency from $\sf ZFC$ of large cardinal axioms will ultimately prove the consistency of $\sf ZFC$, and by the second incompleteness theorem this is impossible.

So you can only prove that $\sf ZFC$ does not prove the existence of large cardinals, not that it does not refute them.

And indeed every now and then you can find people claiming to have proofs that inaccessible cardinals are inconsistent with $\sf ZFC$.

(The situation is similar with $\sf ZFC+\operatorname{Con}(ZFC)$, which is also strictly stronger than just $\sf ZFC$ in terms of consistency. But I don't think anyone who seriously considers $\sf ZFC$ a reasonable theory thinks it is inconsistent, so this assumption gets questioned far less often than inaccessible cardinals.)

Caveat lector:

Relative consistency results about $\sf ZFC$ and assumptions stronger than $\sf ZFC$ itself will depend a lot on the meta-theory. If you work in a theory which assumes $\sf\operatorname{Con}(ZFC)+\lnot\operatorname{Con}(ZFC+LC)$, then $\sf ZFC$ will refute large cardinals in that meta-theory. Because there will be a "natural number" encoding a proof of contradiction in a way recognizable by $\sf ZFC$.

The point is that when you want to talk about the ability to refute something stronger than $\sf ZFC$ you need to ask yourself what sort of meta-theory you have. If inaccessible cardinals are refutable from $\sf ZFC$, that's great (or actually the other thing, sucky); if not, then the question if they are refutable becomes meta-theory dependent.

Asaf Karagila
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  • I think you're restating the same thing I said in different terms. For large cardinals to be independent, you'd have to prove that you can't prove their nonexistence. My understanding is that you can prove that you can't prove that you can't prove their nonexistence. Right? Why is that? (I think I worded that right...) – Mike Battaglia Nov 10 '15 at 23:26
  • No, I'm saying that you need to show it is consistent they exist and that it is consistent they don't. The latter is easy; the former is impossible, because you can't start just from $\sf ZFC$ and prove the consistency of $\sf ZFC+LC$. So $\sf ZFC$ cannot do the former, therefore it is impossible to establish independence. Only unprovability. And who knows, maybe inaccessible cardinals are inconsistent with $\sf ZFC$. – Asaf Karagila Nov 10 '15 at 23:28
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    OK, then in those terms, what I'm asking is: why can't ZFC prove the consistency of ZFC + LC? – Mike Battaglia Nov 10 '15 at 23:29
  • Because from the consistency of ZFC+LC you can prove the consistency of ZFC. So if ZFC could do it, it would prove its own consistency and thus be inconsistent, subjected to the second incompleteness theorem. – Asaf Karagila Nov 10 '15 at 23:30
  • OK, so assuming ZFC is consistent, it can't prove Con(ZFC). And assuming ZFC is consistent, then it can't prove ~Con(ZFC), but it also can't prove that it can't prove ~Con(ZFC). That is, if ZFC is consistent, ZFC must believe that it "might" be able to prove ~Con(ZFC). And since LC is much stronger than Con(ZFC), the same reasoning applies, which is why independence can't be proven for ZFC+LC from ZFC. Is that right? – Mike Battaglia Nov 10 '15 at 23:48
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    Well. Yes. But there is a huge caveat here about your metatheory. I've already went to bed, so this will have to wait a couple of hours until morning finds me. Sorry for that. But in a naive nutshell, this is the argument, yes. – Asaf Karagila Nov 10 '15 at 23:51
  • "ZFC must believe that it "might" be able to prove" This makes no sense as written. What do you mean by "must believe"? What do you mean by "might"? This is a formal theory, not a person. – Andrés E. Caicedo Nov 11 '15 at 00:34
  • @AndrésCaicedo: I mean that ZFC can't prove that it can't prove ~Con(ZFC), and it can't prove that it can prove ~Con(ZFC). That is, the decidability in ZFC of ~Con(ZFC) is undecidable in ZFC. – Mike Battaglia Nov 11 '15 at 02:40
  • @Mike: I've added the caveat. I hope it didn't confuse you too much. – Asaf Karagila Nov 11 '15 at 05:52