For a class $W$:  



Let $\mathcal{T}\_0(W)=W$.  

Let $\mathcal{T}\_{\alpha+1}(W)=\{x\in V:x\subseteq\mathcal{T}\_\alpha(W)\}\cup\mathcal{T}\_\alpha(W)$.  

Let $\mathcal{T}\_{\beta}(W)=\bigcup\_{\alpha\in\beta}\mathcal{T}\_\alpha(W)$ for limit ordinals $\beta$.


Let $\mathrm{K}(W)=\min\{\alpha:\mathcal{T}\_\alpha(W)=V\}$ (if such a minimum exists). The smaller this value is, the "closer" $W$ is to $V$.


With this definition, $\mathrm{K}(L)$ is guaranteed to be either a limit ordinal or $0$. Quite clearly $\mathrm{K}(L)=0\Leftrightarrow V=L$. 


However, assuming $V\neq L$, is it possible to prove what $\mathrm{K}(L)$ is? Are there any models of ZFC in which it does not exist? (I.E. $M\models\forall\alpha\in\mathrm{Ord}(\mathcal{T}\_\alpha(L)\neq V)$)


A couple facts that may help:


* $\mathcal{T}\_\alpha(L)$ is transitive for every $\alpha$ (Proven by transfinite induction)
* $V\_{\omega+\alpha}\in\mathcal{T}\_\alpha(L)$ (Proven also by transfinite induction)
* $\mathrm{K}(L)\neq\alpha+1$ for any $\alpha$ (Transfinite induction)
* $V\setminus L$ is either the empty set or a proper class (Deduction using the fact that $L$ is transitive and a model of the axiom of pairing)



Using this definition of "closeness", how close is $L$ to $V$?
Assume $V\not=L$. I claim that $K(L)$ does not exist. 


There is a set $X\subset Ord$ with $X\notin L$.


For any ordinal $\alpha$, let $R\_\alpha$ be a well-founded tree of rank $\alpha$, and let $\rho:R\_\alpha\to \alpha$ be its rank function. For every $r\in R\_\alpha$ define a set $x\_r$ as follows:


* $x\_r = X$ if $\rho(r)=0$.
* $x\_r = \{ x\_t: t \in succ(r)\}$ if $\rho(r)>0$. Here, $succ(r) $ is the set of direct successors of $r$ in $R\_\alpha$.


I claim that $x\_r\notin {\mathcal T}\_{\rho(r)}$ for all $r$. 


This is clear if $\rho(r)=0$, as $x\_r=X\notin L={\mathcal T}\_0(L)$.


If $\rho(r)=\beta+1$, and $t\in succ(r)$ has rank $\beta$, then $x\_t\notin {\mathcal T}\_\beta(L)$, so $x\_r\nsubseteq {\mathcal T}\_\beta(L)$. 


If $\rho(r)=\delta$ is a limit, then for each $\beta<\delta$ we can find a successor $t$ of $r$ of rank $>\beta$. Since $x\_t\in x\_r$, we have $x\_r\nsubseteq {\mathcal T}\_\beta(X)$, so also $x\_r\notin {\mathcal T}\_\beta(X)$. So $x\_r\notin {\mathcal T}\_\delta(X)$.


