User blog comment:LittlePeng9/Long hierarchies of functions (on my other blog)/@comment-11227630-20171101003439/@comment-5529393-20171101092955

Wojowu's argument can be modified to work for these other versions of "eventually outgrows" as well.

Let's say that "f really dominates g" if, for all k > 0, there is an N > 0 such that n > N implies f(n) > g(n^k).

Let f_0(n) = n, and pick some function F that really dominates f_0. (F(n) = n^n will work.) Then, given f_a(n), we will define f_{a+1} by f_{a+1}(n) = f_a(n^2). Clearly f_{a+1} "strictly eventually outgrows" f_a (for either definition of "strictly eventually outgrows"), and if f_a is really dominated by F, then f_{a+1} is also really dominated by F.

All that is left is to define f_a for limit a. Suppose a has a fundamental sequence a_1, a_2, ... .  We want to define a function f_a such that f_a strictly eventually outgrows f_a_i for all i, but is still really dominated by F.  We do this as follows:  We start by letting f_a(n) = f_a_1(n), until we reach an n such that for all m >= n, we have f_a_2(m) >= f_a_1(m), and F(m) >= f_a_2(m^2). We can do this, because we know that f_a_2 strictly eventually outgrows f_a_1, and that F really dominates f_a_2, by induction. At that n, we start letting f_a(n) = f_a_2(n), until we reach an n such that for all m >= n, we have f_a_3(m) >= f_a_2(m), and F(m) >= f_a_3(m^3). Again, we know this will eventually happen. At that point we start letting f_a(n) = f_a_3(n), and so on. The resulting f_a will strictly eventually outgrow f_a_i for any i, since by the way we defined it, f_a will eventually be greater than f_a_{i+1}, and we know that f_a_{i+1} strictly eventually outgrows f_a_i by induction. On the other hand, we know that F(n) really dominates f_a(n), since we defined f_a(n) in sections, and in the ith section we have F(n) >= f_a(n^i). So by induction, we can keep going through all the countable ordinals, and we will have a hierarchy f_a for 0 <= a < w_1 such that a < b implies f_b strictly eventually outgrows f_a, and F really dominates f_a for all a.

We can then apply Wojowu's proof by induction that we can define f_a for all 0 <= a < c for any c < w_2 such that a < b implies f_b strictly eventually outgrows f_a. To reiterate: We apply induction on c.  The above cases take care of when c is a successor ordinal or c is a limit ordinal of cofinality w, so the only remaining case is when c is a limit ordinal of cofinality w_1. In that case, we have a fundamental sequence c[a]: w_1 -> c. We know that we can define a hierarchy of functions g_a(n) for 0 <= a < w_1 as above; note that, if a < b, then g_{wa} is really dominated by g_{wb}, since g_{wa+i}(n) = g_{wa}(n^2^i), so since g_{w(a+1)} eventually dominates all g_{wa+i}, it must really dominate g_{wa}. We can then set f_{c[a]}(n) = g_{wa}(n), and then for each a, we know that we can fill in the remaining functions between f_{c[a]} and f_{c[a+1]} by induction, since c[a+1] - c[a] < c, and f_{c[a+1]} really dominates f_{c[a]}.

As for computable functions, the family of functions f_q(n) = Floor[qn] for positive rationals q work just fine for Sbiis's definition of "strictly eventually outgrows", since if 0 < a < b are rational numbers, then Floor[a(n+c)]+d is eventually dominated by Floor [bn]. It follows that we can define f_a(n) for 0 <= a < c for any countable ordinal c, since any countable ordinal has an order-preserving injection into the positive rationals. For Hyp Cos's definition, we can use f_q(n) = Floor[n^q] for positive rationals q. Then, if 0 < a < b, [(cn+d)^a] will be dominated by [n^b] for all c and d.