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?
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> 
> Is it $L\_{\omega\_1}$? or is it some fixed countable point, like $L\_{\omega^{\omega^{\omega^{.^{.^{.}}}}}}$?
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> 
> 


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



What is the strength of this strict constructible iterative hierarchy?
If I've correctly understood the question, then I think the answer is $\omega\_1^{CK}$, the first non-recursive ordinal. 


On the one hand, as Nik Weaver pointed out in a comment, all recursive well-orderings of $\omega$ are in $L\_{\omega+1}$, so your construction certainly goes through at least all the recursive ordinals. 


On the other hand, every well-ordering of $\omega$ that is an element of $L\_{\omega\_1^{CK}}$ is hyperarithmetical. (This has nothing much to do with well-ordering; it's true for all relations on $\omega$ that are in $L\_{\omega\_1^{CK}}$.) And the lengths of hyperarithmetical well-orderings of $\omega$ are only the recursive ordinals. So $L\_{\omega\_1^{CK}}$ contains no well-orders of $\omega$ of length $\omega\_1^{CK}$ (or more), and your construction ends at that stage.


