User blog comment:Ikosarakt1/Coding strings as ordinals (for Friedman's n(k))/@comment-5529393-20130715142426/@comment-5529393-20130715163509

Yes, there is a natural characterization of the ordinals below epsilon-zero using sequences of letters. With one letter we get the ordinals below omega again, and, given that n letters correspond to ordinals below omega^^n, we show how n+1 letters correspond to ordinals below omega^^(n+1). Let A be the 1st letter. A sequence using n+1 letters will be of the form

S = S_1 A S_2 A S_3 A ... A S_m

where the S_i are sequences using the last n letters, and hence correspond to ordinals a_i below omega^^n by considering them as strings with the letters shifted down by one letter. S then corresponds to

omega^a_1 # omega^a_2 # ... # omega^a_m

where # means natural sum which means we rearrange the terms so that they are in weakly decreasing order then add them. This will lead to some terms corresponding to the same ordinal - for example, BA and AB both correspond to omega + 1 - but we can resolve this by ordering sequences with the same ordinal based on lexicographic order. (So BA will be greater than AB since BA comes after AB in the dictionary.)  This will lead to a new ordinal correspondences, but the order type is still epsilon-zero.

So, for example,

BBABA

has substrings

BB, B, and {}

which we shift down by one letter to get

AA, A, and {}

which have ordinals

2, 1, and 0

so BBABA corresponds to

w^2 + w + 1.

Similarly,

CCBCBABBA

corresponds to

w^(w^2 + w + 1) + w^2 + 1

This ordinal representation is useful for example in determining the order type of Beklemishev's worms (which LittlePeng9 used in one of his Turing machines). But it doesn't help with Friedman's n(k), since it doesn't obey the subsequence property: for example, BABA has ordinal w*2 + 1, whereas BBA has ordinal w^2 + 1, so BBA is greater, but it is a subsequence if BABA. In fact, we know that we cannot get an ordinal correspondence up to epsilon_0 that obeys the subsequence property: if we could, we could use it to find a lower bound for F(k) of approximately H_{epsilon_0}(k). But Friedman proved that F(k) is dominated by H_{w^(w^w + 1)} (k), so we know we can't get an ordinal correspondence of more than w^(w^w + 1). (It seems highly unlikely that the limit would be between w^w^w and w^(w^w + 1), so the limit is likely w^w^w.)