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12:19 AM

ERROR STACK OVERFLOW

\begin{align}
\psi(\Omega^m\Omega)[n,\Omega^m\Omega] &= [\psi(\Omega^m]^{n+1}(\omega) = \varphi(n+2,0)\\
\psi(\Omega^{\Omega})[n,z] = \psi(\Omega^{\Omega[n,\Omega^{\Omega}]}) = \psi(\Omega^{\psi(\Omega^{\Omega[n-1,\Omega^{\Omega}]}}) & = [\psi(\Omega]^{n+1(\cdot)}(\omega) = \left(\mathop{\large{E}}_{k=1}^{n+1}\psi(\Omega\right)(\omega) = [\varphi(\omega,]^{n+1}(0) = \Gamma_0[n+1]\\
\psi(\Omega^{\Omega^{\Omega}})[n,z] = [\psi(\Omega^{\Omega}]^{n+1(\cdot)}(\omega) &= \text{Still Computing...}\\
\end{align}

$$Ω^{Ω^{Ω}} \to Ω^{Ω^{2}} \to Ω^{ΩΩ} \to Ω^{Ω2} \to Ω^{Ω+Ω} \to Ω^{Ω+2} = Ω^ΩΩΩ \to Ω^Ω2 = Ω^Ω + Ω^Ω \to Ω^Ω + ΩΩ \to Ω^Ω + Ω+Ω \to Ω^Ω + Ω+2 \to Ω^Ω + 2$$

Simply Beautiful Art

12:36 AM

lol

@Secret First line is definitely wrong

You drop the $m$ somewhere...

surely you do not think that $\psi(\Omega^m\Omega)[n,\Omega^m\Omega]$ is constant for all $m$?

Secret

n+1 nesting of $\varphi(m+1,0)$?

Simply Beautiful Art

Btw

Nvm

It ought to be $1+m$

(not $m+1$)

$n+1$ nestings of $x\mapsto\varphi(1+m,x)$

Secret

Some sort of pattern, not sure if reliable:

$\psi (\Omega) = \zeta_{\omega} = \varphi (1,0)$

$\psi (\Omega^{\Omega}) = \Gamma_{0} = \varphi (1,0,0)$

$\psi (\Omega^{\Omega^{\Omega}}) = ??? = \varphi (1,0,0,0)$?

ooops

should be $\zeta_{0} = \varphi (2,0)$

Simply Beautiful Art

12:58 AM

@Secret heh

Secret

Anyway, I think that's all for today, I am very sleepy right now

Simply Beautiful Art

$\psi(\Omega^{\Omega^0})=\varphi(2,0)$ (it's close enough to the pattern)
$\psi(\Omega^{\Omega^1})=\varphi(1,0,0)$
$\psi(\Omega^{\Omega^2})\stackrel?=\varphi(1,0,0,0)$
$\psi(\Omega^{\Omega^3})\stackrel?=\varphi(1,0,0,0,0)$
$\vdots$
$\psi(\Omega^{\Omega^\omega})\stackrel{!?}=\varphi(1,0,0,0,0,0,0,\dots)$
@Secret

So don't even get any ideas about $\psi(\Omega^{\Omega^\Omega})$ yet.

9 hours later…

Secret

10:26 AM

If you can prove that any predicative function applied to a countable ordinal generates a countable ordinal, then the answer is obviously not. Since $\omega_1$ is regular (at least assuming some bits of choice), any countable-to-countable function on ordinals would require $\omega_1$ iterations to reach $\omega_1$, which is exactly what being impredicative is all about. — Asaf Karagila 41 mins ago

Looks like even with types, we cannot predicatively reach $\omega_1$ because there is simply WAY TOO MANY STEPS

1 hour later…

Secret

11:44 AM

$\aleph_1$ and $\omega_1$ are the same object. The former is used to hint "we are talking about cardinals and cardinal arithmetic" and the latter denotes "we are talking about ordinals and order types". So $\aleph_1+\aleph_0=\aleph_1$, but $\omega_1+\omega\neq\omega_1$. Now. It is indeed consistent with ZF that $\omega_1$ is singular and thus the countable union of countable sets. But again, as far as predicativity, you would need to somehow argue that you can find a countable cofinal sequence which is obtained by iterating some predicative construction in a predicative way. [...] — Asaf Karagila 34 mins ago

[...] If you work under some reasonable assumptions that predicative definitions are absolute between transitive models of ZF, then this puts a big stick into the wheels of your idea, since $\omega_1$ while singular in your universe of ZF is not the $\omega_1$ that you find in L, an inner model of choice, thus the predicative construction cannot even surpass a very fairly specific countable ordinal as far as your universe is concerned. So the whole approach collapses. — Asaf Karagila 32 mins ago

Discussion here in this chat suggests the specific countable ordinal is somewhere just before Church Kleene ordinal

It is also good that the new OCF is a lot easier to use. That will mean should some time in the future the project to be focused in this room is something to do with proof theory, we can easily generate all those proof theoric ordinals in a systematic fashion

Simply Beautiful Art

@Secret ;P

Secret

Ψ₀(Ωω)

In mathematics, Ψ0(Ωω) is a large countable ordinal that is used to measure the proof-theoretic strength of some mathematical systems. In particular, it is the proof theoretic ordinal of the subsystem Π 1 1 {\displaystyle \Pi _{1}^{1}} -CA0 of second-order arithmetic; this is one of the "big five" subsystems studied in reverse mathematics (Simpson 1999). == Definition == Ω 0 = ...

This ordinal is somewhat related to user21860's second order arithmetic based mathematics foundation

It will be cool to see how it looks like under our $\psi$

Simply Beautiful Art

Oh, you want to go that far? We haven't even reached the Bachmann-Howard ordinal yet lol.

Secret

Don't worry, we will get there

Simply Beautiful Art

'tis a long road.

Secret

11:56 AM

nothing compared to uncountable sets

Simply Beautiful Art

:P Won't count on it.

Secret

obviously

Meanwhile, there is some progress in the new program coding in my chemistry PhD. Hopefully will get that done and knock out 500+ calculations in 20 clicks

Simply Beautiful Art

lol

Secret

btw, I am currently inactive because I am coding in the background

Simply Beautiful Art

same

4 hours later…

user21820

3:31 PM

@Secret Glad to hear that! Keep working regularly at your chem program! =)

@Secret I have considered asking on the main site what is the proof-theoretic ordinal of predicative higher-order arithmetic. If I ever do, I will let you know.

Simply Beautiful Art

@user21820 wonder how long it'll sit without answers

user21820

@SimplyBeautifulArt Lol.

I'm quite sure some logic expert knows, but right now I'm not feeling very welcome by some main site users who answer a lot of (introductory) logic questions.

Secret

4:12 PM

Momentarily back from coding, found a bug and then squished it. The code seemed fine for now. I will fill in the template files with content later on tomorrow. If all goes well, a few small tests should get it running and ready to go

Secret

4:33 PM

Side note: Now that we knew that both $\omega_1^{CK}$ and $\omega_1$ are impredicative. It will seemed that should anyone want to consider a predicative OCF, then $\Omega$ is more like a symbol that tells $\psi$ nesting is going to occur, rather than being the first uncountable ordinal

It will be interesting to see how far ACA and its relatives can go in terms of formalising the foundation of all predicative mathematics

user21820

@Secret That's exactly what I thought. I will have to think very carefully how exactly to justify the well-ordering of Madore's OCF predicatively. Next time! =)

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