Let $K$ be the set of all total recursive functions of non-negative integers having only non-negative integers as values. Let $L$ be any well-ordered subset of $K$ in which the ordering $<$ is defined as follows. If $f(n),g(n)$ are elements of $L$, then $f(n)< g(n)$ just in case there exists a non-negative integer $h$ such that $f(n)$ is less than $g(n)$ whenever n is greater than $h$. Suppose that $L$ contains a maximum element. Is the ordinal number of the well-ordered set $L$ always a constructive ordinal number (in the sense of Church and Kleene)? Can arbitrarily large constructive ordinal numbers be represented in this way?



Questions about a possible way of representing construcive ordinal numbers
The answer to the first question is No and the second question Yes, because in fact every countable (successor) ordinal arises that way. (The successor part is only because you insisted that $L$ has a maximal element; otherwise we could say that every countable ordinal arises this way.)


The reason is that $K$ contains a countable dense linear order, and every countable linear order, including every countable well-order, can therefore be found as a suborder. 


In particular, we can find $L$ having order type larger than $\omega^{CK}\_1$ this way. 


