@user21820 Deedlit says it's probably Bachmann-Howard ordinal

@SimplyBeautifulArt Oh really. Thanks for asking! I can't quite believe it hahaha..

xP

Also if it interests you, I have a neat ordinal collapsing function involving the Mahlo cardinal(s) with decently nice fundamental sequences that can be made.

@SimplyBeautifulArt Can you convince me that a Mahlo exists first? =P

=P

I can probably convince you that Mahlo cardinals are a natural extension for regular ordinals in the context of ordinal collapsing functions

Ok let's hear it.

So the "big" ordinals used in ordinal collapsing function need to be sufficiently "big" that the ordinal collapsing function never outputs it.

More precisely, it should always stop growing before it reaches Ω or whatever

Yes.

Uncountable regular ordinals are the ordinals α where every function f : α → α has some ordinal β < α that is closed under f (γ < β implies f(γ) < β).

You can relate this back to cofinality by thinking of it as something like α cannot have cofinality β for β < α.

So ordinal collapsing functions can be extended by adding in more uncountable regular ordinals.

Beyond this, you can make an ordinal collapsing function which makes these regular ordinals.

The desired property for the "big ordinal" used in that ordinal collapsing function should be:

An ordinal α where every function f : α → α has some regular ordinal β < α that is closed under f.

which are Mahlos

@SimplyBeautifulArt: Sorry I was away for a bit. Reading what you wrote now.

no problem

Hmm.

I vaguely get the idea, but isn't the bottleneck still the existence of a Mahlo cardinal?

If you don't have such a cardinal, the "big ordinal" in that OCF wouldn't be able to serve as an upper bound.

Of course. My point was though that Mahlos are a natural next step for OCFs.

I see.

You can also do recursive analogues by restricting the function f to be "recursive" in some sense. And this has a fairly obvious generalization to OCFs which return Mahlos and so on.

Well here is how I see things. When we use a single Ω for the OCF, we do not need it to actually be an ordinal. We can just view it as a position placeholder and still sort of grasp how the recursion is well-founded. In fact, I think we can get by with Ω representing the first uncomputable ordinal δ, which is readily understood as well-defined as long as we understand programs and accept the well-definedness of the halting problem. But can we do the same with OCFs based on a Mahlo cardinal?

Yeah you can quite nicely actually

How?

I'm not talking about the syntax.

The idea is to have ψ[x](y) return ordinals less than x.

Then whenever the "last" part of y has some regular ordinal z ≥ x, then we replace it with ψ[z](~) like you do with yours.

If the "last" part of y is some limit ordinal z < x, then we simply pass the limit along as usual.

For ψ[x](y+1), the behavior depends on what x is. Supposing our OCF iterates over addition, then you can do something like ψ[x](y+1) = sup{ψ[x](y)*n | n < w} when x = ψ[M](z) and the last part of z is less than M.

@SimplyBeautifulArt What is M?

A Mahlo cardinal.

Conceptually, ψ[Mahlo] returns regulars, and ψ[regular but not Mahlo] returns limit ordinals.

If x = ψ[M](z) and the last part of z is ≥ M, then ψ[x](y+1) starts with ψ[x](y)+1 and iterates over ψ[M](~).

This isn't completely precise, but should explain the gist of it.

You mean you can have arbitrary nesting in the x in ψ[x](y)?

So it's like the Veblen kind of parameter?

No, you don't nest into the x, you nest into the y's last part

But you said "ψ[x](y+1) = sup{ψ[x](y)*n | n < w} when x = ψ[M](z)".

Which means you could have ψ[ψ[M][z]](...), or more nested.

Uh no?

ψ[Ω](1) = sup{ψ[Ω](0)*0, ψ[Ω](0)*1, ψ[Ω](0)*2, ...}

Like this for example

I'm just saying that's how you approach successor arguments here

You can replace it with repeatedly exponentiating or whatever if you want.

Then what you wrote cannot be correct. Please check again the statement I quoted?

It is equivalent to "ψ[ψ[M](z)](y+1) = sup{ψ[ψ[M](z)](y)*n | n < w}".

If that isn't what you meant, then maybe you misplaced an x somewhere.

Then I must've misinterpreted what you meant, because that is fine, as long as z doesn't end with M or similar.

How did you notate the nestings again?

like how would you have written φ(3, 2)?

For the simple version with finite parameters, it would be just φ([3],2), and Γ = φ([1,0],0).

I mean the fundamental sequence/limit with sup

Oh. For my own ease of reference let me copy my rules here again:

φ(3,2) encoded as φ([[0,3]],2) would follow the last rule.

so {a | f(~)} = {a, f(a), f(f(a)), ...}?

Yup.

aight

It's my laziness showing in syntax.

Er, I gtg but I'll be back lol

and try and write out the rules for the Mahlo OCF there.

Ok no hurry. =)

@user21820 This should work.

Oh yeah, and of course treat multiplication as repeated addition (we only have multiplication by finite numbers here).