## Wednesday, March 21, 2012

### Infinity: Impredicative Ordinals

This is the tenth post of the Infinity Series. For the first, see here. For all posts, see the Infinity Series Portal.

At the conclusion of the previous post, Γ0, the first impredicative ordinal, and the first defined with a "second-order" definition, was introduced. The remainder of the Γ-sequence was also defined. To find higher countable ordinals, one precedes in a similar manner to previously, by finding fixed points.

For example, the smallest fixed point of α → Γα is also the supremum of the set {Γ0Γ0ΓΓ0,...}, and is known as the Ackermann ordinal, named after Wilhelm Ackermann, known for his contributions to computation theory.

Again, it becomes useful to have a standard notation for these ordinals, and a generalized form of the Veblen hierarchy is quite useful. To extend the Veblen notation, first we replace what was previously denoted as φβ(α) with φ(α,β), namely a function of two arguments, the first being what was before the argument, and the second that which was the subscript. To extend the Veblen functions beyond what was already considered, we add a third argument γ, and consider the ordinals defined by expressions of the form φ(α,β,γ). These expressions are evaluated in the following way:

First, φ(α,β,0) = φ(α,β). Then, φ(0,0,1) is defined as the first ordinal that is inaccessible through the expressions φ(α,β), constructing from 0 upward. Since this definition is identical to the one for the Feferman-Schutte orindal, φ(0,0,1) = Γ0. To become more familiar with the notation, consider these further examples:

φ(0,1,1) = φΓ0 + 1(0), or the first fixed point of the first Veblen function after φΓ0, φ(0,0,2) is equal to Γ1, or the second fixed point of φα(0) → α. Every increase here in the second argument of the generalized Veblen function represents listing fixed points. Each increase in the third represents another Γ value. For example, φ(0,0,2) = sup{φ(0,0,1),φ(0,φ(0,0,1),1),φ(0,φ(0,φ(0,0,1),1),1),...}, with the second argument β assuming the value of the previous member in each element of the set, until Γ1 is obtained. It is easy to see this is consistent with the previously cited value of Γ1. All other φ(0,0,γ) can be defined this way, and, continuing from the discussion of Γ0, which was defined in a second-order fashion, these ordinals have second-order definitions, and are all impredicative.

The smallest ordinal inaccessible from the standard operations on all the ordinals φ(0,0,1), φ(0,0,2), φ(0,0,3), etc. of finite third argument γ, yields a new γ value: ω, the ordinal being written φ(0,0,ω). Of course other values for α and β are also possible, such as φ(1,5,ω) or φ(ε03,ω). But of more concern is the third argument, which can increase through more countable values until reaching nested Veblen functions such as φ(0,0,φ(0,0,1)), defining a sequence in which the γ assumes the value of the previous member of the sequence. Having gone as far as possible with three arguments, a fourth is introduced in the following way:

sup(φ(0,0,1),φ(0,0,φ(0,0,1)),φ(0,0,(φ(0,0,φ(0,0,1))),...} = φ(0,0,0,1)

Since φ(0,0,1) = Γ0, the above equation is exactly the same as sup{Γ0Γ0ΓΓ0,...} = φ(0,0,0,1). Additionally, φ(0,0,0,1) is the first fixed point of Γα → α, and, as said above, is also known as the Ackermann ordinal. This ordinal has a "third-order" definition, and is also impredicative by most definitions.

One can go on to define φ(α,β,γ,...,δ) for any finite number of arguments, all of which are limits similar to the example above, and although inaccessible from previous operations, are countable for all countable α,β,γ,...,δ. It becomes inconvenient to deal with so many arguments however, so there is an alternate method to create large countable ordinals with a function of only one argument.

As stated above, it is possible to go still further, by the use of a new kind of function, called an ordinal collapsing function. These functions again use the idea of inaccessibility in their definition, but there is a crucial difference between these and Veblen functions: ordinal collapsing functions assume the existence of an ordinal Ω that is beyond any ordinals that can be constructed with ordinal collapsing functions. This ordinal does not have to be defined in any precise way, but simply is assumed to exist as an upper limit to any ordinal constructed. However, since ordinals generated by ordinal collapsing functions make use of Ω, this definition is impredicative, as it assumes the existence of a larger ordinal to define a smaller.

To construct this function, first consider the set of ordinals S = {0,1,ω,Ω}. The set C(0) is then defined as containing all ordinals that can be constructed through a finite process from 0,1,ω, and Ω using addition, multiplication and exponentiation, including 1, 2, 3, ω, ωω, and ωωω. C(0) also includes Ω, Ω + 1, ΩΩ, and others of this type, but these are not as important for the time being.

Now, we denote the smallest ordinal that is not in C(0) as ψ(0) (with the symbol psi, rather than phi for Veblen functions). ψ(0) is then equal to ε0, the first ordinal that requires infinite steps from ω to construct. Next, C(1) is the set of ordinals constructible from S with addition, multiplication, and exponentiation, and the function ψ|1, or the ψ function restricted to 1. Such a function can only take as an argument values that are members of 1, namely, only 0 (all ordinals are sets of all previous ordinals). Hence the ordinal ψ(0) = ε0 is a member of C(1), as well as sequences of the normal three operations on it, such as ε0 + 1, ε0ε0, and so on.

The smallest ordinal not in C(1) is written ψ(1), and is equal to ε1, as no finite sequence of operations from ε0 yields this value. Similarly, C(2) is the set of ordinals accessible from members of S through the normal operations plus ψ|2. Since 1 is a member of 2, ψ(1) = ε1 is a member of C(2), as is ε0 and any operations on these two ordinals with the other members of S. Preceding recursively, C(α) has as a member all ordinals defined through the three normal operations, and ψ|α, which has a domain of all ordinals less than α.
ψ(α) is defined as before, and is seen to equal εα for all finite α.

But what about when α = ω? C(ω) includes all values taken by ψ|ω, namely ε0, ε1, ε2,..., εα, and so on for all finite α, as well as operations upon them. The corresponding ψ(ω) is the clearly εω as this transcends all previous epsilon values and operations on them. Thus the equivalence of ψ(α) and εα continues for some infinite values of α. In fact, this equality holds true as long as α is less then or equal to εα.

However, in the case of α = ζ0 (for which εα = α), C0) includes all of the epsilon sequence below ζ0, but does not contain the ordinal itself, as even applying the function ψ|ζ0 to 0 any finite number of times (producing the sequence ε0, εε0,...) cannot yield ζ0, and ψ(ζ0) = ζ0. α = ζ0 + 1 presents an interesting case. One would think that the pattern would continue, generating εζ0 + 1 as the value for ψ(ζ0 + 1). However, for this to happen, ζ0 would have to be a member of C0), since if this were not the case, εζ0 + 1 would not be the smallest ordinal not in C0). But, beginning with the set S, it is impossible to construct ζ0 with a finite number of operations. Even though one of the operations available for C0) is ψ|ζ0 + 1, which has ζ0 in its domain and would produce ζ0, no number of steps from S can produce this value in the first place to plug it into the function! There is therefore no way to obtain ζ0, and
ψ(ζ0 + 1) = ζ0.

By the same logic, ψ(ζ0 + 2), ψ(ζ0 + ω), and others are still equal to ζ0. It may seem that the function is "stuck" at ζ0, and can produce no higher value. This is where Ω comes in. Consider the set C(Ω). This set is no different from previous ordinals in that it ψ(Ω) = ζ0. The difference appears when considering C(Ω + 1), as ordinals in this set can be constructed from S using ψ|Ω + 1 along with the other three operations. Since Ω is in S and in the domain of ψ|Ω + 1, ψ(Ω), that is, ζ0, is in C(Ω + 1). If not for the addition of Ω to S, this function would cease to increase, and this is the importance of Ω. It also explains the name "ordinal collapsing function", as it takes large ordinals such as Ω, and using a function, "collapses" them into smaller ordinals, in this case ζ0.

It has been established that ψ(Ω + 1) is larger than ζ0, but what is it? C(Ω + 1) also includes the usual operations on ζ0, including iterated exponents such as ζ0ζ0ζ0. Clearly the smallest ordinal that cannot be accessed in this way is εζ0 + 1, and this is equal to ψ(Ω + 1). The correlation between the ε and ψ functions continues from there, with ψ(Ω + α) = εζ0 + α for all finite and some infinite values of α, including cases greater than ζ0. For example, ψ(Ω + ζ0*2) =
εζ0 + ζ0*2 = εζ0*3.

The above trend occurs up to ψ(Ω + ζ1), but the function ψ|Ω + ζ1 cannot equal ζ1 for any ordinal in its domain, nor can iterated exponentiation generate this value. Thus, ψ(Ω + ζ1) = ζ1. The next problem occurs when one reaches ψ(Ω + ζ1), as the set C(Ω + ζ1) has the operation ψ|Ω + ζ1 + 1, which gives ζ1, but only for the input Ω + ζ1, and this input cannot be constructed with other operations to plug into the function. Thus, ψ(Ω + ζ1) = ζ1. Hence the function is stuck once again, with ψ(Ω + α) being equal to ζ1 for α greater than or equal to ζ1, but only up to a certain point.

Again, Ω bails us out, or, more precisely, Ω*2. Though ψ(Ω*2) still is ζ1, Ω*2 can be constructed easily from S, and the set C(Ω*2 + 1) has available the operation ψ|Ω*2 + 1, which has Ω*2 in its domain, and can produce ζ1. Through similar logic as above, ψ(Ω*2 + 1) = εζ1 + 1. The general idea continues, with each ψ(Ω*(1 + α) + 1) transcending another ζ ordinal, and yielding εζα + 1 (if this formula is unclear, try substituting 0 and 1 for α, which give known values, and serve as examples).

As it turns out, this repetition of multiples of Ω "unsticking" the function continues all the way up to ψ(Ω*η0), with η0 being the first fixed point of
α → ζα. ψ(Ω*η0) is equal to η0, as finite iterations of α → ζα from 0 give ordinals strictly less then this value. However, ψ(Ω*η0 + 1) is also η0, and the function is stuck again, this time all the way until ψ(Ω2), with ψ(Ω2 + 1) equal to εη0 + 1, the first epsilon ordinal greater than η0. In fact, further powers of Ω have ψ values corresponding roughly to Veblen functions of two arguments, up to
ψ(ΩΓ0) = ψ(ΩΩ) = Γ0, with the function stuck at Γ0 for all ψ(Ωα) for Γ0 ≤ α ≤ Ω. ψ(ΩΩ) is then the same as φ(0,0,1) = Γ0. Multiplying by Ω corresponds to raising the second argument of the Veblen function, and multiplying the exponent of Ω by Ω raises the third argument. For example, ψ(ΩΩ + 1) is
φ(0,0,2), or Γ1. More values include:

ψ(ΩΩ2) = φ(0,0,0,1), or the Ackermann ordinal, ψ(ΩΩ3) = φ(0,0,0,0,1), and so on, with each ψ(ΩΩα) equal to the smallest ordinal requiring α + 2 arguments to represent. Alternatively, each of these is the supremum of all ordinals requiring at most α + 1 arguments to represent. However, the ordinal collapsing function has only yielded ordinals expressible as Veblen functions of finite arguments. Even higher countable ordinals accessible through this function are discussed in the next post.

Sources: http://www.ams.org/journals/tran/1908-009-03/S0002-9947-1908-1500814-9/S0002-9947-1908-1500814-9.pdf, Ordinal Collapsing Function - Wikipedia

#### 1 comment:

Kevin Stefan said...

You did a good job giving accessible explanations up to this point, but I got totally lost somewhere in the ordinal collapsing function. And the following pages only get harder... Guess this is all I'll ever be able to understand without a formal background. :)