Using this definition of "closeness", how close is $L$ to $V$?

Question

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)

Answer 1

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 $x_r\notin {\mathcal T}_\delta(X)$.