Begin with the empty set then construct the set of the empty set, then construct the set of all subsets of the latter set, then at each level of construction construct the next level as the set all subsets of that level that are definable using formulas restricted to that level, i.e. follow Godel's construction of stages, then continue that iterative construction, lets have a first limit level among levels of that construction, lets continue this construction transfinitely as long as at each level $L_{\alpha}$ there exists a set $x$ constructed at a prior level such that there is a well ordering on the class of all levels from the empty set till $L_{\alpha}$, that is isomorphic to $x$, i.e. the height of each level corresponds to a well ordering that is a set formed at a prior level.

Question: what is the limit to this construction?

Is it $L_{\omega_1}$? or is it some fixed countable point, like $L_{\omega^{\omega^{\omega^{.^{.^{.}}}}}}$?

I tend to think it is $L_{\omega^{\omega^{\omega^{.^{.^{.}}}}}}$, and that this (I suppose) would make it of the strength of $PA$?

thewell-ordering"? Both when you talk of the well-ordering class and the well-ordering of some set. I'm afraid I don't understand yet what you are trying to say. $\endgroup$class$\{L_0, L_1,....,L_{\alpha}\}$, in other words suppose you have a well ordering $s$ that is a set (i.e. an element of some level $L_{\alpha}$, then there is a subclass $\{L_0,L_1...,L_{\kappa}\}$ (that is a hierarchy) of the class of all levels of the hierarchy that is isomorphic to $s$. For example the stage $L_{\omega_1}$ is not reachable from below, because no stage $L_{\alpha}$ for a countable $\alpha$ would contain a set that is a well ordering that is isomorphic to $\endgroup$4more comments