1

First Question:

Let IC be the statement "There is an inaccessible cardinal."

I have read that one cannot prove (in ZFC) the relative consistency of ZFC + IC w.r.t. ZFC.

i.e. $ Con(ZFC) \rightarrow (ZFC \nvdash Con(ZFC) \rightarrow Con(ZFC + IC)) $

The argument is as follows:

Assume $ ZFC \vdash Con(ZFC) \rightarrow Con(ZFC + IC) $

since $ ZFC + IC \vdash Con(ZFC)$

it follows that $ ZFC + IC \vdash Con(ZFC + IC)$ which contradicts Gödel's 2nd Incompleteness Theorem.

But from Gödel's Theorem we only can conclude $ \neg Con(ZFC + IC) $, so why is this a contradiction?

Second Question:

I call an $\in $-formula $\phi$ independent, if we have: $ \Phi_{1,1} := Con(ZFC) \rightarrow Con(ZFC + \phi)$ and $ \Phi_{1,2} := Con(ZFC) \rightarrow Con(ZFC + \neg \phi)$

Now $ \Phi_{1,1}$ and $\Phi_{1,2}$ are again $\in $-formulas and could again be independent if we have:

$ \Phi_{2,1} := Con(ZFC) \rightarrow Con(ZFC + \Phi_{1,1})$ and $ \Phi_{2,2} := Con(ZFC) \rightarrow Con(ZFC + \neg \Phi_{1,1})$

In this case $\phi$ would not be decided by ZFC, but we could not know about this result. Can such a situation occur?

Community
  • 1
abc
  • 1,409
  • 1
    It is best to ask separate questions in separate threads. The second question has been asked before, see http://math.stackexchange.com/questions/65248/are-there-statements-that-are-undecidable-but-not-provably-undecidable and the links there, as well http://math.stackexchange.com/questions/17212/is-there-a-statement-whose-undecidability-is-undecidable – Asaf Karagila Dec 29 '14 at 07:17
  • Thank you very much :) Since Henning also answered in this thread, I'll accept his answer here. – abc Dec 31 '14 at 00:39

1 Answers1

3

The Second Incompleteness Theorem says that if $T$ is a recursively axiomatized, consistent, and "strong enough" theory, then $T\not\vdash \mathcal{Con}(T)$.

Since ZFC+IC is recursively axiomatized and strong enough, then the Second Incompleteness Theorem gives us $$ \tag{*} \mathcal{Con}(\mathsf{ZFC+IC}) \to \Bigl[ \mathsf{ZFC+IC}\not\vdash \mathcal{Con}(\mathsf{ZFC+IC}) \Bigr]$$

The argument you're quoting implicitly assumes that ZFC is not only consistent but true (at least as regards arithmetic), such that if ZFC proves $\mathcal{Con}(\mathsf{ZFC})\to\mathcal{Con}(\mathsf{ZFC+IC})$ then it must be that ZFC+IC is actually consistent. Thus by (*) we have $\mathsf{ZFC+IC}\not\vdash \mathcal{Con}(\mathsf{ZFC+IC})$ which contradics the earlier conclusion that $\mathsf{ZFC+IC} \vdash \mathcal{Con}(\mathsf{ZFC+IC})$.

hmakholm left over Monica
  • 291,611
  • @user202791: No, at least that's not how I meant it. When I say that ZFC is true, then I mean that every arithmetic statement ZFC proves is actually satisfied by the Platonic naturals. – hmakholm left over Monica Dec 23 '14 at 01:34
  • @user202791 In this context, we usually say "true" to mean "true in the standard model". The natural numbers of the standard model are "as expected". – Andrés E. Caicedo Dec 23 '14 at 01:39