# “Nested independence” of $\mathsf{ZFC}$? [duplicate]

I have a question about undecidable statements in $\mathsf{ZFC}$.

I know there are true statements like: $$X\text{ is independent of }\mathsf{ZFC}.$$

But is it also possible that such a statement exists?

$$\text{"}X\text{ is independent of }\mathsf{ZFC}\text{" is independent of }\mathsf{ZFC}$$

Or even worse:

$$\text{""}X\text{ is independent of }\mathsf{ZFC}\text{" is independent of }\mathsf{ZFC}\text{" is independent of }\mathsf{ZFC}$$

I think you see what I mean by "nested independence of $\mathsf{ZFC}$". Is it possible that such a true statement exists? If it is possible: What does it mean? And is there an example for such a true statement? Unless such a true statement can exist: Why not?

Another funny case would be is the depth of this nested statement goes to infinity:

\begin {align} \mathrm{statement}_0 &= X \\ \mathrm{statement}_{n+1} &= \mathrm{statement}_n\text{ is independent of }\mathsf{ZFC} \\ \mathrm{specialstatement} &= \lim_{n\to\infty}\mathrm{statement}_n \end{align}

Does such a $\mathrm{specialstatement}$ exist? And what is the meaning of it?

• thx for the link:) – Kevin Meier Dec 25 '15 at 0:24
• Milo's answer to the 'duplicate' question is wrong because it does not deal with provability correctly. Specifically, you need to state provability over what theory. Independence likewise. Also, ZFC cannot even prove that any one sentence is independent of ZFC, otherwise ZFC is inconsistent! Thus your initial statement is already wrong, unless your meta-system is stronger than ZFC, in which case your question is not well-defined because you didn't specify what exactly your meta-system is. – user21820 May 2 '16 at 3:05