-3
$\begingroup$

Let's define the $V$-rank (also can be named as set \ true - rank) of a set $X$ as the minimal $\alpha$ such that $X \in V_{\alpha+1}$.

On the other hand, define the $L$-rank (also can be named as constructible-rank) of a constructible set $X$ as the minimal $\alpha$ such that $X \in L_{\alpha+1}$.

A set in $L$ is concordant if and only if its $V$ and $L$ ranks are equal. Formally: $$ \operatorname {conc}(X) \iff \rho^L(X)=\rho^V(X)$$

We know that $L$ does contain sets that are not concordant, an example is the diagonal of a constructible bijection between $L_\omega$ and $L_{\omega+1}$.

Now, I want to motivate the criterion of Quine's stratification in relation to this notion of concordance. So, the working assumption is that if we weaken the definition of $L$ to stratified-constructible sets (i.e., require an additional condition in defining stages of $L$, that of the defining formulas being also stratified), then the resulting hierarchy (call it stratified-$L$) can have all of its elements being concordant! That is, stratification can prevent constructible jumps.

Is there a clear counter-example to the above assumption?

$\endgroup$
15
  • $\begingroup$ Why do you think restricting attention to stratified formulas would prevent such jumps? Jumps of this sort are unavoidable as long as you want a reasonably rich universe built in the end. $\endgroup$ Commented May 26 at 23:02
  • $\begingroup$ @NoahSchweber, it depends on how you qualify "rich", for instance do you think the universe of NF[U] rich? $\endgroup$ Commented May 26 at 23:10
  • $\begingroup$ Note that NFU proves the existence of non-arithmetic sets. But non-arithmetic sets aren't in $L_{\omega+1}$, so they automatically lead to jumps of the sort you want(?) to avoid. $\endgroup$ Commented May 26 at 23:11
  • $\begingroup$ @NoahSchweber, I'm not sure of that. Can you give examples of NFU doing that? $\endgroup$ Commented May 26 at 23:35
  • 2
    $\begingroup$ Rank $\alpha$ should mean that $\alpha$ is least such that $x\in V_{\alpha+1}$, not $V_\alpha$, and similarly with $L_{\alpha+1}$, since these hierarchies are both continuous. $\endgroup$ Commented May 27 at 0:36

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.