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## Welcome

Hi, welcome to Googology Wiki! Thanks for your edit to the Talk:Goodstein sequence page.

## Strength of & operator

Hey, Deedlit. You probably thought that $$f_{\Gamma_0}(n) \approx \{n,n / 2\}$$ from the reason that:

$$f_\omega(n) \approx \{n,n,n\}$$ and $$f_{\varphi(\omega,0)}(n) \approx \{X,X,X\} \&\ n$$.

$$f_{\omega+1}(n) \approx \{n,n,1,2\}$$ and $$f_{\varphi(\omega+1,0)}(n)$$ therefore should be $$\{X,X,1,2\} \&\ n$$.

That seems reasonable, but we can look at that by the other way: $$f_\varphi(\alpha,0)(n)$$ is the:

$$\alpha+1$$-th hyper-operator applied to X's. (When $$\alpha$$ is a finite number)

$$\alpha$$-th hyper-operator applied to X's. (When $$\alpha$$ is a transfinite ordinal)

So f_{$$f_\varphi(n,0)$$}(n) corresponds to $$\{X,X,n+1\} \&\ n$$ and $$f_\varphi(\omega,0)$$

Next look at $$\varphi(\omega+1,0)$$. It is equal to $$f_\varphi(\omega,\varphi(\omega,\cdots,\varphi(\omega,0))...))$$ (with $$\omega\text{ }\varphi$$'s (in the more compact way, it is the first ordinal such that $$\alpha$$ = $$\varphi(1,\alpha)$$). Hence, it behaves in the same way as, say $$\varphi(2,0) = \alpha = \varphi(1,\alpha)$$. We can pretty confident that $$\varphi(\omega+1,0)$$ corresponds to $$\{X,X,X+1\} = \{X,p,X+1\} = \{X,\{X,p-1,X\},X+1\}$$ structure.

When the one structure represents the hyper-operator of the other, X times, we reach expandal arrays, which is surely corresponds to $$\varphi(1,0,0) = \Gamma_0$$.

More generally, $$\varphi$$ function with n entries behaves at the same as linear arrays with n-1 entries. Thus, $$\{X,X (1) 2\} \&\ n\} \approx f_{\vartheta(\Omega^\omega)}(n)$$. Chris Bird received the same results, and after more considerations you can see that limit ordinal of the non-legion arrays is the $$\vartheta(\Omega^\Omega)$$, the Large Veblen Ordinal.

To be clear, I just present the selection of some comparisons:

$$X \uparrow\uparrow X \&\ n \approx f_{\varepsilon_0}(n)$$

$$X \uparrow\uparrow\uparrow X \&\ n \approx f_{\zeta_0}(n)$$

$$X \uparrow\uparrow\uparrow\uparrow X \&\ n \approx f_{\eta_0}(n)$$

$$\{X,X,X\} \&\ n \approx f_{\varphi(\omega,0)}(n)$$

$$\{X,X,X+1\} \&\ n \approx f_{\varphi(\omega+1,0)}(n)$$

$$\{X,X,2X\} \&\ n \approx f_{\varphi(\omega 2,0)}(n)$$

$$\{X,X,X^2\} \&\ n \approx f_{\varphi(\omega^2,0)}(n)$$

$$\{X,X,X^X\} \&\ n \approx f_{\varphi(\omega^\omega,0)}(n)$$

$$\{X,X,X \uparrow\uparrow X\} \&\ n \approx f_{\varphi(\varepsilon_0,0)}(n)$$

$$\{X,X,\{X,X,X\}\} \&\ n \approx f_{\varphi(\varphi(\omega,0),0)}(n)$$

$$\{X,X,1,2\} \&\ n \approx f_{\varphi(1,0,0)}(n)$$

$$\{X,X,2,2\} \&\ n \approx f_{\varphi(1,1,0)}(n)$$

$$\{X,X,3,2\} \&\ n \approx f_{\varphi(1,2,0)}(n)$$

$$\{X,X,X,2\} \&\ n \approx f_{\varphi(1,\omega,0)}(n)$$

$$\{X,X,\{X,X,X\},2\} \&\ n \approx f_{\varphi(1,\varphi(\omega,0),0)}(n)$$

$$\{X,X,\{X,X,1,2\},2\} \&\ n \approx f_{\varphi(1,\varphi(1,0,0),0)}(n)$$

$$\{X,X,1,3\} \&\ n \approx f_{\varphi(2,0,0)}(n)$$

$$\{X,X,1,4\} \&\ n \approx f_{\varphi(3,0,0)}(n)$$

$$\{X,X,1,X\} \&\ n \approx f_{\varphi(\omega,0,0)}(n)$$

$$\{X,X,1,\{X,X,1,2\}\} \&\ n \approx f_{\varphi(\varphi(1,0,0),0,0)}(n)$$

$$\{X,X,1,1,2\} \&\ n \approx f_{\varphi(1,0,0,0)}(n)$$

$$\{X,X,1,1,1,2\} \&\ n \approx f_{\varphi(1,0,0,0,0)}(n)$$

$$\{X,X (1) 2\} \&\ n \approx f_{\vartheta(\Omega^\omega)}(n)$$

$$\{X,X,2 (1) 2\} \&\ n \approx f_{\vartheta(\Omega^\omega+1)}(n)$$

$$\{X,X,3 (1) 2\} \&\ n \approx f_{\vartheta(\Omega^\omega+2)}(n)$$

$$\{X,X,X (1) 2\} \&\ n \approx f_{\vartheta(\Omega^\omega+\omega)}(n)$$

$$\{X,X,1,2 (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)2)}(n)$$

$$\{X,X,1,3 (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)3)}(n)$$

$$\{X,X,1,X (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)\omega)}(n)$$

$$\{X,X,1,1,2 (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)\vartheta(\Omega))}(n)$$

$$\{X,X,1,1,1,2 (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)\vartheta(\Omega^2))}(n)$$

$$\{X,X (1) 3\} \&\ n \approx f_{\vartheta((\Omega^\omega)\vartheta(\Omega^\omega))}(n)$$

$$\{X,X (1) X\} \&\ n \approx f_{\vartheta(\Omega^{\omega+1})}(n)$$

Notice that all these comparisons looks very similar with the regular arrays, where $$\{n,n (1) n\}$$ represented by $$X+1$$ structure. In general, if we have A structure (in BEAF), then $$A \&\ n \approx f_{\vartheta(\Omega^\alpha)}(n)$$, where $$\alpha$$ is the ordinal that associated with A. So, continuing onwards:

$$\{X,X (1) X,X\} \&\ n \approx f_{\vartheta(\Omega^{\omega+2})}(n)$$

$$\{X,X (1) X,X,X\} \&\ n \approx f_{\vartheta(\Omega^{\omega+3})}(n)$$

$$\{X,X (1)(1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega 2})}(n)$$

$$\{X,X (1)(1)(1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega 3})}(n)$$

$$\{X,X (2) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^2})}(n)$$

$$\{X,X (3) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^3})}(n)$$

$$\{X,X (0,1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^\omega})}(n)$$

$$\{X,X (1,1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega+1}})}(n)$$

$$\{X,X (0,2) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega 2}})}(n)$$

$$\{X,X (0,0,1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega^2}})}(n)$$

$$\{X,X (0,0,0,1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega^3}})}(n)$$

$$\{X,X ((1) 1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega^\omega}})}(n)$$

$$\{X,X ((0,1) 1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega^{\omega^\omega}}})}(n)$$

$$X \uparrow\uparrow X \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(1)})}(n)$$

$$X \uparrow\uparrow\uparrow X \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(2)})}(n)$$

$$X \uparrow\uparrow\uparrow\uparrow X \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(3)})}(n)$$

$$\{X,X,X\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\omega)})}(n)$$

$$\{X,X (1) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^\omega)})}(n)$$

$$\{X,X (2) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^2})})}(n)$$

$$\{X,X (3) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^3})})}(n)$$

$$\{X,X (0,1) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^\omega})})}(n)$$

$$\{X,X ((1) 1) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^{\omega^\omega}})})}(n)$$

$$\{X,X ((0,1) 1) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^{\omega^{\omega^\omega}}})})}(n)$$

$$X \uparrow\uparrow X \&\ n \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\vartheta(1)})})}(n)$$

$$X \uparrow\uparrow X \&\ n \&\ n \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\vartheta(\Omega^{\vartheta(1)})})})}(n)$$

All that limits to the Large Veblen Ordinal, $$\vartheta(\Omega^\Omega)$$. Ikosarakt1 (talk ^ contribs) 20:28, March 24, 2013 (UTC)

If you are right, what ordinal may be limit of BEAF? If we reach LVO without even && operator, I wonder how big meameamealokkapoowa oompa may really be. LittlePeng9 (talk) 21:36, March 24, 2013 (UTC)

Yes, I have recently come to the same conclusion concerning $$\{X,X,1,2\}$$ and $$\Gamma_0$$. I haven't examined linear arrays and above yet, but your conclusions seem reasonable. Of course, we need a rigorous definition for Bowers' arrays to be sure.

TREE(3) will have to be moved down the chain, to around the Small Veblen ordinal. (We don't know the exact ordinal for TREE(3) but I guess we will assume that it is not much larger than the Small Veblen ordinal.) Deedlit11 (talk) 22:01, March 24, 2013 (UTC)

@LittlePeng9: Chris Bird analyzed Bowers' notation and concluded that it ends well below the Bachmann-Howard ordinal, and I think below $$\vartheta (\Omega^{\Omega^{\Omega}})$$. Deedlit11 (talk) 00:05, March 25, 2013 (UTC)

Some measurements for legion arrays:

If {n,n/2} is at the level of the LVO, then {n,n/3} is at the level of LVO*2, {n,n/4} is at the level of {n,n/4} is at the level of LVO*3, and {n,n/1+F} is at the level of LVO*alpha, where we take alpha to be the ordinal associated to F. So {n,n / 1 / 2} will be at the level of LVO^2, {n,n / 1 / 1 / 2} will be at the level of LVO^3, {n,n (/1) 2} will be at the level of LVO^omega, and {n,n (/F) 2} will be at the level of LVO^(omega^alpha). Thus n && n && n will be at the level of LVO^LVO, and {n,n // 2} will be at the level of epsilon_{LVO+1}. We go through the same hierarchy up with //'2 so we get that {n, n /// 2} is at the level of epsilon_{LVO+2}. In general, {L, X^F} = {n,n (F)/ 2} will be at the level of epsilon_{LVO + alpha}. It follows that {L, L} = epsilon_{epsilon_{LVO+1}}, and probably {L, L, X} = phi(2, LVO+1). It's not clear to me how higher arrays are defined in terms of L - it seems like very bad notation to me - but it seems reasonable that {L, L, ..., L, X} = phi(n, LVO+1). So L2 space is at the level of Gamma_{LVO+1} = phi(1, 0, LVO+1). L3 space will be at level phi(1, 0, LVO+2), LF space will be at level phi(1, 0, LVO + alpha). LLF space will perhaps be at level phi(1, 1, LVO + alpha) (but again, I would need a precise definition of what LF is), and perhaps (L^F)G will be at phi(1, alpha, LVO + beta). So (L^L)F would be at level phi(2, 0, LVO + alpha).

Alternatively, perhaps {L, L, ..., L} is at level phi (1, 0, ..., 0, LVO+1}, with the same number of terms in each. So L2-space will be the second ordinal fixed by the Schutte Klammersymbolen, i.e. $$\theta(\Omega^{\Omega},1)$$. In general, LF space will be the alphath ordinal fixed by the Schutte Kalmmersymbolen, i.e. $$\theta(\Omega^{\Omega}, \alpha)$$. So perhaps LLF space will be $$\theta(\Omega^{\Omega}+1, \alpha)$$, and in general (L^F)G space will be $$\theta(\Omega^{\Omega}+\alpha, \beta)$$, and thus the closure space (I guess we are calling this L^L) will be at the level of $$\theta (\Omega^{\Omega} + \Omega, 0)$$. So we are very far from $$\theta(\Omega^{\Omega^{\Omega}}, 0)$$. Deedlit11 (talk) 01:54, March 25, 2013 (UTC)

But in BEAF there are array spaces such absurdal as L & L. Also, is there any reading about Klammersymbolen? I understand general concept, but nothing more LittlePeng9 (talk) 06:33, March 25, 2013 (UTC)

L & L? My analysis went well past the "&" operator.
I couldn't find much on the web on the Klammersymbolen, except for:
http://www.cs.man.ac.uk/~hsimmons/ORDINAL-NOTATIONS/FromBelow.pdf
http://www.cs.swan.ac.uk/~csetzer/articles/ordsyscor010124.ps
Both of those papers generalize the Klammersymbolen so they are harder to read than need be.
But, the notation is not that complicated; it's just an extension of the Extended Veblen notation into the transfinite.
To do that, we need to explicitly name the place values of the variables. For example, instead of phi(8, 0, 0, 0, 3, 0, 5, 6, 4), represent it as (8 @ 8, 3 @ 4, 5 @ 2, 6 @ 1, 4 @ 0). We can represent all Extended Veblen notations this way, but we can also say (1 @ omega, c @ 0), which is defined as the cth ordinal that satisfies (x @ n, 0 @ 0) = x for all n < omega.
More generally, define (a_1 @ b_1, a_2 @ b_2, ..., a_n @ b_n, c @ 0) as the cth ordinal that satisfies the system of equalities (a_1 @ b_1, a_2 @ b_2, ..., a_{n-1} @ b_{n-1}, d @ b_n, x @ e) = x for all d < a_n, e < b_n. Deedlit11 (talk) 03:50, March 26, 2013 (UTC)

## Array of

We found out that the & is the array of. And i am sorry for the confusion that we caused in BEAF. $$a$$$$l$$$$t$$ 04:19, April 14, 2013 (UTC)

But you never said is array of... $$a$$$$l$$$$t$$ 04:24, April 14, 2013 (UTC)

Sorry, I thought you knew. Deedlit11 (talk) 06:00, April 14, 2013 (UTC)

## SDA forums?

Just curious, are you the same person as "Deedlit" on the Speed Demos Archive forums? --Ixfd64 (talk) 19:51, April 16, 2013 (UTC)

Yep, that's me. I speedran FF12 a few years back. I go by Deedlit in a bunch of places. —Preceding unsigned comment added by Deedlit11 (talkcontribs)
Nice! It's really a small Internet after all... --Ixfd64 (talk) 20:17, April 16, 2013 (UTC)
If I not miscalculated, even if you turn all 0's and 1's in the Internet to elementary particles and crumple them in one ball, this ball will be so small that you will barely able to see it with the naked eye! Internet contains roughly 4.2 zettabytes = 33.6 zettabits (that is, 0's and 1's). Human body contains about 64 octillion elementary particles, and then $${6.4 \times 10^{28} \over 3.36 \times 10^{22}} \approx 1904762$$. In other words, this ball would be roughly 1904762 times smaller than body of average human, and we're speaking about entire Internet! Ikosarakt1 (talk ^ contribs) 20:42, April 16, 2013 (UTC)
One of my favorite cartoonists computed that storing the Internet into a bunch of modern hard drives would at most fill up an oil tanker. FB100Ztalkcontribs 20:49, April 16, 2013 (UTC)
I guess it's not surprising that the internet would be about a drop of water (1/18 of a gram in atoms, definitely visible), since the internet has to be stored, and we aren't anywhere near 3D atomic level of storage. But of course this stuff is increasing exponentially - it won't take all that long before we surpass a human in size. (well, it should be within our lifetimes.) Deedlit11 (talk) 20:56, April 16, 2013 (UTC)
It has been calculated that if we take all electrons moving due to global Internet transfer, mass-energy we are giving them for the flow is equivalent to 50 grams! Nicely explained here. LittlePeng9 (talk) 05:23, April 17, 2013 (UTC)

## Template:wedge

Vote now! $$a$$$$l$$$$t$$ 12:09, April 26, 2013 (UTC)

You might want to check out the Coq proof assistant to familiarize yourself with the system. I tried my hand at figuring out CoC several times, and I still don't fully understand it. you're.so.pretty! 20:02, April 8, 2014 (UTC)
Actually, I don't understand the Calculus of Constructions that well either. I could try writing something on it, but I would have to base it on what I can glean from the internet. Perhaps it would be best to start the simple typed lambda calculus, then work our way up to System T, System F, then the CoC. Deedlit11 (talk) 20:37, April 9, 2014 (UTC)

## Computer Turn-offage

Please don't tell me you haven't turned off your computer in 10 days. King2218 (talk) 16:06, July 27, 2014 (UTC)

## I think I killed you

[15:13] <Wojowu> !attack Deedlit11
[15:13] <XappolBot> Wojowu creeps up behind Deedlit11 and swings at them with a bat!
[15:13] <XappolBot> Thud! Deedlit11 falls to the ground unconscious!
[15:14] == Deedlit11 [~Deedlit@ip68-111-90-23.oc.oc.cox.net] has quit [Ping timeout: 263 seconds]
[15:14] <Wojowu> Well

Sorry LittlePeng9 (talk) 17:14, September 10, 2014 (UTC)

ahahahahahaha you're.so.pretty! 21:26, September 10, 2014 (UTC)

## Ramsey numbers

I did it . LittlePeng9 (talk) 12:38, May 4, 2015 (UTC)

## Reminder

Just a reminder than you have a few solutions left to write down. LittlePeng9 (talk) 19:29, May 20, 2015 (UTC)

## Fundamental sequences for extended Veblen-function

Hello, Deedlit, I will be very grateful If you'd read and commented this my post. Did I correctly understand FS up to SVO? If there are not mistakes, I'd like to write an article about Veblen function.Denis Maksudov (talk) 13:21, February 25, 2017 (UTC)

I published my article about Veblen function.

I have a question: let p=(...,a@b,g-1@0) then for case a-limit and b-successor

here Veblen defines FS as(...,a@b,g@0)[n]=(...,a[n]@b,p+1@0)

(I translated Veblen to zero-indexced).

and here Ranzi defines (...,a@b,g@0)[n]=(...,a[n]@b,p+1@b-1)

This diffence would be for all possible limit cases for a and b

Are both definitions correct? --Denis Maksudov (talk) 20:01, March 18, 2017 (UTC)

Yes, both look fine. Deedlit11 (talk) 07:57, March 19, 2017 (UTC)

## Ordinal hierarchy

I studied your googology and want to write a popular article about this in Russian. Because in Russian in the Internet there is almost no material on googology. I want to use the most visual comparisons.

And I want to forgive you for help. Could you check the comparisons in this table?

If possible, write in which step I was wrong.

Thank you. And sorry for my English.

Scorcher007 (talk) 02:04, March 23, 2017 (UTC)

I would just like to point out it is not true that $$I_n$$ (the $$n$$-th inaccessible cardinal) is not the least $$\Pi^0_n$$-indescribable cardinal. LittlePeng9 (talk) 05:21, March 23, 2017 (UTC)
I'm suspicious of the claim that $$\Pi^1_0$$-indescribable is equivalent to being Mahlo, but I don't really know. Can someone find a reference confirming or denying this?
I'm not familiar with your usage of $$\Omega_{(\alpha_n,\ldots,\alpha_0)}$$, but if you mean that $$\Omega_{(1,\alpha)}$$ is the $$\alpha$$th fixed point of $$\beta \rightarrow \Omega_\beta$$, and so on, it looks fine.
The major problem occurs when you go from M to K. This is something that I guess I didn't make clear enough when I wrote Ordinal Notations VI; the weakly compact notation is actually more powerful than people are thinking. When we get to Mahlo cardinals, we can define Mahlo(0) to be the Mahlo cardinals, then define Mahlo(1) to be the Mahlo cardinals that are limits of Mahlo cardinals, then define Mahlo(1) to be Mahlo(1) cardinals that are limits of Mahlo(1) cardinals, and so on. Then we can define Mahlo(1,0) to be cardinals $$\kappa$$ that are in Mahlo($$\kappa$$), and continue the hierarchy in the usual fashion. Then we can define $$M(\alpha_n,\ldots,\alpha_0,\beta)$$ to be the $$\beta$$th cardinal in Mahlo($$\alpha_n,\ldots,\alpha_0)$$. Thus we get a hierarchy similar to the inaccessible hierarchy, just like you have in your list. But, the cardinal that diagonalizes over this is not K! Rather, it is the first 2-Mahlo cardinal. This is the key thing to notice about the weakly compact ordinal notation: there are now different levels of collapsing. In the collapsing function $$\Psi_\pi(\alpha,\beta)$$, $$\alpha$$ repesents which type of collapsing we are doing. If $$\alpha = 0$$, we collapse to a strongly critical ordinal, which is what we normally collapse to anyway, so this is our normal collapsing function. If $$\alpha = 1$$, we collapse to a regular cardinal, so this is where Mahlo cardinals collapse to get us a hierarchy if inaccessible cardinals. If $$\alpha = 2$$, we collapse to a Mahlo cardinal, so we can collapse 2-Mahlo cardinals to get a hierarchy of Mahlo cardinals. But we can also collapse a 2-Mahlo at the $$\alpha = 0$$ and $$\alpha = 1$$ levels, so we get finer and finer structures as $$\alpha$$ increases.
In short, what you call K should actually be $$\Xi(3)$$, the smallest 2-Mahlo cardinal, and what you call S(n) should actually be $$\Xi(n+2)$$, the smallest (n+1)-Mahlo cardinal. The cardinal K will then diagonalize over this grander hierarchy of hierachies. So it's bigger than you think!
I don't think I will find the time to go over all of your inequalities, but some of them are making me wonder, like 219 and 220; why are those true? But overall, it seems like you have the right general idea. (except the major probelm above) Deedlit11 (talk) 07:42, March 24, 2017 (UTC)
As for the 219 and 220 steps (now it's other step, I changed table), I came to this conclusion on the basis of the following comparisons:

As for the 2-Mahlo cardinal, etc., as well as the diagonalization of collapsing functions with K, I believe that in my notation this will look like this:

Scorcher007 (talk) 02:33, March 25, 2017 (UTC)
As for the $$\Pi^1_0$$-indescribable $$\Leftrightarrow$$ Mahlo claim, this is simply not true. In fact, $$\Pi^1_0$$-indescribability is equivalent to inaccessibility. To see this, we have the following implications chain: inaccessible $$\Rightarrow$$ $$\Sigma^1_1$$-indescribable $$\Rightarrow$$ $$\Pi^1_0$$-indescribable $$\Rightarrow$$ $$\Pi^0_2$$-indescribable $$\Rightarrow$$ inaccessible, where the first and the last implications are standard results (mentioned on Wikipedia), and the two implications in the middle are clear, since every $$\Pi^0_2$$ formula is $$\Pi^1_0$$, and every such is in turn $$\Sigma^1_1$$. (in fact, Wikipedia also claims that $$\Pi^1_n$$-inaccessibility is equivalent to $$\Sigma^1_{n+1}$$-indescribability, but looking at the proof I am unsure it works for $$n=0$$; the above chain of implications shows the equivalence in a different way).
Also, it's worth pointing out that there is a bit of impreciseness when it comes to what $$n$$-Mahlo means - Wikipedia doesn't explicitly claim whether Mahlo cardinals are 0-Mahlo or 1-Mahlo, but this section suggests Mahlo is 1-Mahlo. On the other hand, Cantor's Attic explicitly states that 1-Mahlo is already the condition stronger than Mahloness, so Mahlo would have to mean 0-Mahlo. LittlePeng9 (talk) 08:44, March 25, 2017 (UTC)

## Similarities between collapsing functions

Hello, Deedlit!

I noticed several similar expressions:

Why not to create common ordinal notation? Something in follow kind:

,

, (Let ),

(Let ),

,

,

th common fixed point of functions for all , where is a string of arguments , including , and is string of zeros,

, where .

Then, for example:

is equivalent of ,

is equivalent of ,

is equivalent of

and so on.

I don't know how to write common collapsing function for this notation. I see something in this kind:

,

,

,

the th ordinal in

What do you think Deedlit, does this idea have any perspectives?--Denis Maksudov (talk) 09:01, March 23, 2017 (UTC)

The problem with that approach is that those elements don't actually form a sequence, so there is no obvious next step. (so beyond a unification of the symbols, there is no gain) Chronolegends (talk) 17:44, March 23, 2017 (UTC)

Hello Denis! I have thought about something like this; it would be nice to organize varying levels of diagonalizers into an overall system. I have in mind something that I am calling "the diagonalizer hierarchy"; the main problem is that I am not sure whether my description of it is enough to uniquely specify the system. I will try to dredge it up and put it on a blog post sometime.
With regards to your specific suggestions, Chronolegends is right that I,M,K don't quite form a sequence, or at least not a trivially obvious one. After $$\Omega_\alpha$$, we use I as a diagonalizer for the fixed point hierarchy of $$\Phi(\alpha) = \Omega_\alpha$$. Then I(1,0) diagonalizes the fixed point hierarchy of $$\alpha \rightarrow I(\alpha)$$, I(2,0) diagonalizes the fixed point hierarchy of $$\alpha \rightarrow I(1,\alpha)$$, and so on. So we build up an inaccessible hiearachy of arbitrary arity, which is a different sort of thing then we had before, and it is this hierarchy that M diagonalizes. Note that we don't then go from M to K; after we build up a hierarchy of Mahlo cardinals, the next step is to go the the smallest 2-Mahlo cardinal (let's call it M_2) and it is this cardinal M_2 that diagonalizes over the hierarchy of 1-Mahlo cardinals. Then we go to 3-Mahlo cardinals, alpha-Mahlo cardinals, (1,0)-Mahlo cardianls, and so on, and it is this hierarchy of hierarchies that K diagonalizes over. So we do have a sequence I,M,M_2,M_3... where each one diagonalizes over the hierarchy of the previous one. And I think we can talk about the sequence M,K,... where the third member would diagonalize over a hierarchy of hierarchies of hierarchies, and so forth; the natural setting for this would be to use indescribable cardinals. I'm hoping that the diagonalizer hierarchy would extend beyond this, however. Deedlit11 (talk) 06:08, March 24, 2017 (UTC)

I am officially hyped for that :D the strength of such a thing would be absolutely bonkers Chronolegends (talk) 16:14, March 24, 2017 (UTC)

Thanks for answer, Deedlit. I meaned something like this:

$$\theta(\varphi^{1}(1,0))=\theta(\theta^{1}(1,0))=\theta(\Omega_{\Omega...})$$, where $$\varphi^1$$ is Veblen function on base $$\alpha \mapsto \Omega_\alpha$$,

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

$$\theta(\theta^{1}(\Omega^{(1)}_{\Omega^{(1)}...}))=$$$$=\theta(\theta^{1}(\theta^{2}(1,0)))=\theta(\theta^{1}(\varphi^{2}(1,0)))$$, where $$\varphi^2$$ is Veblen function on base $$\alpha \mapsto \Omega_\alpha^{(1)}$$,

$$\theta(\theta^{1}(\varphi^{2}(1,0,0)))=\theta(\theta^{1}(\theta^{2}(\Omega^{(2)})))=$$

and so on (about M and K my knowledge is so small that I will not even try to write any examples).--Denis Maksudov (talk) 19:44, March 24, 2017 (UTC)

Hello, i don't think veblen functions on alpha -> Omega_alpha can have power equivalent to enumeration of the inaccessibles. (let alone with phi^2 and Omega^(1) reaching a mahlo) (Chronolegends (talk) 17:13, March 25, 2017 (UTC))

I would say Omega^(k+1) is the smallest number, which is greater all possible constructions of Omegas^(k). And yes, strength of Veblen function on alpha -> Omega_alpha is not enough. I included it in examples to show work of hypothetical theta-function, and possibly it has sense to include Veblen functions on bases alpha -> Omega_alpha^(k) in the definition of this theta-function - in set C_(n+1). --Denis Maksudov (talk) 19:03, March 25, 2017 (UTC)

I've also scribbled extensions based on phi functions on the fixed points of alpha -> Omega_alpha, it did not get me very far, but if you are interested i'd be happy to share my thoughts with you over private message some time Chronolegends (talk) 04:45, March 26, 2017 (UTC)

Ok, you can write me googology@list.ru --Denis Maksudov (talk) 09:24, March 26, 2017 (UTC)

Denis: So if we define $$\varphi^1$$ to be the Velben function starting with base $$\alpha \mapsto \Omega_\alpha$$, and $$\Omega^{(1)}$$ to be the diagonalizer of $$\varphi^1$$, then $$\Omega^{(1)}$$ will serve a similar role to I. You haven't defined what $$\Omega^{(1)}_\alpha$$ is, but I assume they serve the same role as $$I(\alpha)$$. So then, if we define $$\varphi^2$$ to be the Veblen function function starting from the base $$\alpha \mapsto \Omega_\alpha^{(1)}$$, and $$\Omega^{(2)}$$ to be the diagonalizer of $$\varphi^2$$, then $$\Omega^{(2)}$$ will serve the same role as $$I(1,0)$$. In general, $$\Omega^{(\beta)}_\alpha$$ will be just $$I(\beta,\alpha)$$. So you have to go farther to get to M, and much farther to get to K. Deedlit11 (talk) 06:40, March 29, 2017 (UTC)

Thanks for analysis, Deedlit.--Denis Maksudov (talk) 19:03, March 29, 2017 (UTC)

## Buchholz OCF and FS

Dear Deedlit, I will be very glad if you will read this my post. I want to know am I correctly understand how Buchholz's function works, since I plan to write article about this function for this wikia.

Denis, I'm sorry that I haven't gotten around to reading your article. It will take some time to go through everything, but I definitely will get around to it at some point. Deedlit11 (talk) 04:55, April 18, 2017 (UTC)

Deedlit, Buchholz mentioned that his notation was extended by Jager in article (G. Jager, p-inaccessible ordinals, collapsing functions, and a recursive notation system, Archiv f.math. Logik und Grundlagenf. 24 (1984) 49-62.). I unsuccessfully tried to find, may be you know, was this article placed in some site in open access. --Denis Maksudov (talk) 09:00, April 23, 2017 (UTC)

Sorry for the late reply. I haven't been able to find this article at an open access site. I do have a copy of the paper though. The ordinal collapsing function has I(a,b) among its closure functions, which tells us its strength. Deedlit11 (talk) 05:41, April 29, 2017 (UTC)
Nevertheless, I'm grateful since you tried to find. Regards, --Denis Maksudov (talk) 11:07, April 29, 2017 (UTC)

Dear Deedlit, I decided to publish my article about Buchholz function in wiki main space and now you can discuss and edit article if you want.--Denis Maksudov (talk) 15:59, May 13, 2017 (UTC)

Dear, Deedlit, may be it will be interesting for you That's extension of your notation. Actually instead of two types i.e. () and [] in your notation, I just used the notation with arbitrary amount of types of parentheses ()_b. That must work as rulesets for fast-growing hierarchy and for fundamental sequences of ordinals generated by Buchholz's function.--Denis Maksudov (talk) 21:23, May 21, 2017 (UTC)

## Ordinal catching point

Ordinal collapsing functions can take uncomputable or uncountable ordinals as argument, and result "small" (computable and countable) ordinal. That's similar to FGH, HH and SGH, which can take transfinite ordinal numbers as argument, and result "small" (finite) numbers.

We can have at least 2 different ordinal collapsing hierarchies. One starts with cardinality (a) while the other starts with admissibility (b). The $$\Omega$$ in (a) corresponds to the $$\omega_1^\text{CK}$$ in (b), $$\Omega_\omega$$ in (a) corresponds to $$\omega_\omega^\text{CK}$$ in (b), first weakly inaccessible in (a) corresponds to first recursively inaccessible in (b), first weakly Mahlo in (a) corresponds to first recursively Mahlo in (b), first weakly compact in (a) corresponds to first recursively $$\Pi_3$$-reflecting in (b), first $$\Pi_n^1$$-indescribable in (a) corresponds to first recursively $$\Pi_{n+2}$$-reflecting in (b), and so on. It seems that (b) is stronger than (a) because (b) can use smaller ordinal as input to generate the same size of output as (a), just as FGH can use smaller ordinal as input to generate the same size of output number as SGH.

FGH and HH catch up at $$\varepsilon_0$$ when take this as argument. FGH and SGH catch up at $$\psi(\Omega_\omega)$$. At the catching point, two hierarchies use the same input to generate "comparable" output. So here comes a question: where do these two ordinal collapsing hierarchies, (a) and (b), catch up together? {hyp/^,cos} (talk) 05:40, August 3, 2017 (UTC)

Of course, the size of a collapsed ordinal depends not only on the ordinal being collapsed, but the ordinal collapsing function used. Having said this, the idea behind the correspondence between recursively large countable ordinals and large cardinals is that they will be used in equivalent ordinal collapsing functions and will collapse to the same ordinals. So $$\psi(\varepsilon_{\omega_1^\text{CK} + 1})$$ will equal the Bachmann-Howard ordinal using the appropriate ordinal collapsing function, $$\psi(\varepsilon_{M + 1})$$ for M the first recursively Mahlo ordinal will still be the proof theoretic ordinal of KPM, and so on. The reason that large cardinals are typically used, even though they require stronger set theoretic assumptions, is that they allow easier proofs for some theorems necessary for ordinal analysis, because you can make cardinality arguments. Deedlit11 (talk) 14:44, August 3, 2017 (UTC)
Hmm, after rereading your post I think I see what you are asking. The ordinal collapsing hierarchy (b) uses large countable ordinals, whereas (a) starts with the first uncountable ordinal - they don't really share territory. Deedlit11 (talk) 15:00, August 3, 2017 (UTC)
Taranovsky's notation started with admissibility, e.g. $$C(\Omega_2+C(\Omega_2,C(\Omega_22,0)),0)$$ corresponds with the least recursively inaccessible, $$C(\Omega_2+C(\Omega_2,C(\Omega_22,0))^{C(\Omega_2,C(\Omega_22,0))},0)$$ corresponds with the least $$\Pi_3$$-reflecting. Then Taranovsky conjectured that $$C(\Omega_2^2+C(\Omega_2^2+\Omega_2,0),0)$$ corresponds with the omega-fixed-point, $$C(\Omega_2^2+\Omega_2,0)$$ corresponds with the least weakly compact cardinal. So it's possible that (b) may use cardinals as input. Or was Taranovsky wrong at this point? {hyp/^,cos} (talk) 00:21, August 4, 2017 (UTC)
Taranovsky has constructed a very intriguing notation, but his claims about certain expressions equaling certain ordinals doesn't make sense to me. His notation is a countable collection of abstract expressions for which we define a comparison rule. Assuming this turns the collection into a well-ordered set, each expression would then correspond to a countable ordinal, and in fact a recursive ordinal since the notation is recursive. So I don't understand what he means when he claims that certain expressions are equal to certain nonrecursive ordinals. The best I can figure is that he means certain expressions perform the same function as nonrecursive ordinals like $$\Omega$$ do in ordinal collapsing functions, just like we do. But then $$\omega_1$$ and $$\omega_1^{\text{CK}}$$ perform the same function in their corresponding ordinal collapsing functions, whereas Taranovsky has all the uncountable ordinals after all the countable ordinals, so it still doesn't make sense to me.
Now, of course it is possible to construct an OCF that uses both large countable ordinals and large cardinals. But, how that would work would depend on the particulars of the OCF. I don't think such an OCF would bear any relation to the analogy between large countable ordinals and large cardinals that you described. Deedlit11 (talk) 00:53, August 4, 2017 (UTC)
I'm not only talking about Taranovsky's notation. I'm talking about "general" notations, and such possibility to reach uncountables from large countable ordinals. Taranovsky wrote that
The assignment of ordinals assumes that the notation system has gaps; otherwise, all ordinals would be recursive. In this section, we will set the gaps in the notation system in the canonical way.
Maybe the possibility to reach uncountables from large countable ordinals depends on what "canonical gaps" are. {hyp/^,cos} (talk) 01:49, August 4, 2017 (UTC)
It is not really possible to "reach" uncountable ordinals from countable ordinals, other than to simply leap there in a single jump. Similar to how you can't get to infinite ordinals from finite ordinals; you can't, unless you simply say that you do. So I don't see how to parse your question in a way that makes it nontrivial. You can of course have an OCF that uses both nonrecursive countable ordinals and uncountable ordinals. But such an OCF wouldn't fit with the analogy between large countable ordinals and large cardinals, so it doesn't really go with your question.
The problem with Taranovsky's statement is I don't see how there could possibly be a canonical definition of "canonical gaps". It certainly doesn't seem like something that one should just choose to omit like Taranovsky does. But I suppose one could ask him what he means. Deedlit11 (talk) 03:28, August 4, 2017 (UTC)

Do I understand correctly? $$\Pi_3$$-reflecting using $$m-\M_n$$ where n - Mahloness, m - Weakly inaccessiblity, and we have two types fixed-point. Similarly $$\Pi_4$$-reflecting using $$n'(m-\K_a$$) where a - Weakly compactnes, m - Weakly inaccessiblity, n - Mahloness of Weakly inaccessiblity, and we have three types fixed-point. etc. $$\Pi_5$$- reflecting using 4 types fixed-point and $$\Pi_n$$- reflecting using n-1 types fixed-point.Scorcher007 (talk) 05:22, August 4, 2017 (UTC)

Not quite. If we consider m-weakly inacessibility, along with the accompanying fixed-point hierarchy, as "the first thing", then the "second thing" would be weakly Mahlos, inaccessible limits of weakly Mahlos, and the accompanying fixed point hierarchy.
So for example, at the weakly inaccessible level we have I(a0) meaning the a0'th weakly inaccessible cardinal (or limit of weakly inaccessible cardinal; from this point on I won't keep saying "and their limits", take it to be understood.). We can make a fixed point hierarchy over this function, and then we can use I(1,0), the smallest 1-weakly inaccessible cardinal, as the diagonalizer over this hierarchy. We can then go to I(1,1), the second smallest 1-weakly inaccessible, as a higher diagonalizer over the fixed point hierarchy of I(a0), and continue this for I(1,a0). Then we can define a fixed point hierarchy over the function a0 -> I(1,a0), and then use I(2,0), the smallest 2-weakly inaccessible as the diagonalizer for that hierarchy. We can continue in this fashion for all I(a1,a0), then create a fixed point hierarchy for the function a1 -> I(a1,0), and use I(1,0,0), the smallest cardinal k that is k-weakly inacessible (note that this is _not_ the smallest fixed point of a1 -> I(a1,0), as that smallest fixed point has cofinality w, so is not weakly inaccessible), to diagonalize over that hierarchy. And so on for I(...,a4,a3,a2,a1,a0). We can extend this further by using M, the smallest weakly Mahlo cardinal, to diagonalize over this inaccessible hierarchy, meaning xi(M^3 a3 + M^2 a2 + M a1 + a0) = I(a3,a2,a1,a0), and so on for higher powers of M.
The same sort of thing occurs for weakly Mahlo cardinals. We can define M(a0) as the a0'th weakly Mahlo cardinal; then M(0) diagonalizes over the inaccessible hierarchy, M(1) is a higher diagonalizer, and so on. Then we can define a fixed point hierarchy over this function M(a0), and then use M(1,0), the smallest Mahlo limit of Mahlo cardinals, as the diagonalizer over this fixed point hierarchy. Then M(1,1) is a higher diagonalizer, and so on for M(1,a0); M(2,a0), the Mahlo limits of Mahlo limits of weakly Mahlo cardinals, are diagonalizers over the fixed point hierarchy of the function a0 -> M(1,a0), M(3,a0) are diagonalizers over the fixed point hierarchy of the function a0 -> M(2,a0), and so on for general M(a1,a0). Then we can create a fixed point hierarchy over the function a1 -> M(a1,0), and use M(1,0,0), the smallest cardinal k that is a level-k Mahlo limit of weakly Mahlo cardinals. And so on, for M(...,a4,a3,a2,a1,a0). Next, we want to create a diagonalizer over this Mahlo hierarchy of weakly Mahlo cardinals. The answer is to use the smallest 2-weakly Mahlo cardinal, which is the a regular cardinal k such that the set of weakly Mahlo cardinals less that it form a stationary subset of k. This is a much more powerful notion than taking Mahlo limits, so it can diagonalize over any recursively applied application of Mahlo limits.
So, notice that the second paragraph is completely analogous to the first paragraph, and so if the weakly inaccessibles form the first stage, then the weakly Mahlos form the second stage, without even talking about any higher levels of weakly Mahlo. We can then create an analogous third paragraph for the 2-weakly Mahlos, then a fourth paragraph for the 3-weakly Mahlos, and so on. Once we have defined a0-weakly Mahlo, we can define a (1,0)- weakly Mahlo cardinal as a cardinal k that is k-weakly Mahlo; then (1,1)-weakly Mahlo cardinals are cardinals k such that the set of (1,0)-weakly Mahlo cardinals below it form a stationary subset of k, and so on. Extending this to (...a4,a3,a2,a1,a0)-weakly Mahlo, the next step is to create a new ordinal and collapsing function that diagonalizes over this place hierarchy. For this we use the smallest weakly compact cardinal, K. So, rather than the sequence going weakly inaccessible, weakly Mahlo, weakly compact, it actually goes weakly inaccessible, weakly Mahlo, 2-weakly Mahlo, 3-weakly Mahlo,... and the weakly compact cardinal K diagonalizes over this greater hierarchy.
So that takes us to the ordinal for KP + Pi-3 reflection. We can of course continue to K(2), K(3), etc., create a fixed point hierarchy, move on to hierarchies of weakly compact limits of weakly compact cardinals, then go on to taking stationary subsets and so forth. We can then go to stronger principles: define T_a(S) for an ordinal a and set/class of ordinals S to be the set/class of ordinals k such that k is a-indescribable on S. This is a stronger notion than S being stationary in k, and the strength increases as a increases, so we get arbitrarily strong operations. So this takes us far beyond the weakly compact notation.
However, I suspect that if we construct an ordinal notation in the usual way using these stronger operations, we still don't reach the ordinal for KP + Pi-4 reflection. I can't say for sure, but after reading the journal article that gives an ordinal notation for KP + Pi-4 reflection, the ordinal notation given is much more complicated than I expected, even if it only goes up to the smallest Pi^1_2-indescribable ordinal. I suspect that they wouldn't go to such trouble if the obvious extension was sufficient, so I imagine that things really do get quite a bit more complicated when we reach Pi-4 reflection.
There are also papers that give ordinal notations for Pi-n reflection for n > 4, as well as Pi-w reflection and beyond, using more sophisticated machinery. But I haven't come close to understanding these notations. Deedlit11 (talk) 07:14, August 4, 2017 (UTC)
I understand that the main difficulty for Pi-n reflection in determining the hierarchies of large cardinals. My knowledge of large cardinals is not enough to talk about these. But if we assume that: I has a hierarchy I_a and I(1,..,0) ; M has a hierarchy M_a and M(1,..,0) and has a superhierarchy a-M and (1,..,0)-M ; Then K may have K_a and K(1,..,0) and has a superhierarchy a-K and (1,..,0)-K and supersuperhierarchy K'a and K'(1,..,0).
I determined the collapse of such structures by example of theta-function: http://lihachevss.ru/compration.html
So are the properties of large cardinals important for constructing notation, because we only need objects for collapsing? Can we take any unidentified cardinal and define it as very large, having arbitrary properties we need for hierarchies?Scorcher007 (talk) 07:36, August 5, 2017 (UTC)
Well, I wouldn't say that determining hierarchies of large cardinals are the main difficulty in creating ordinal notations for Pi-n reflection. The notations in the literature are extremely complex, and it looks like the main difficulty is figuring out everything that can happen when you apply the various reflection rules.
No, I we cannot just create an unidentified cardinal and have it satisfy arbitrary properties that we choose. This can be inconvenient, for example, to extend the hierarchy of alpha-indescribables that I described above, it would be convenient if we could take about cardinals kappa that are kappa-indescribable, but it appears that such cardinals do not exist. Something we can do is add formal symbols to our notation that we imagine to be very large, and define comparison rules for all the expressions we can generate with the formal symbols and the "collapsing functions" that we associate with them. However, there are some disadvantages to this. One, defining the appropriate comparison rules can be quite difficult; for example, try to define the comparison rules for the weakly compact ordinal notation, it's not easy. Second, we would have to prove that the expressions in our notation form a well-order, whereas with a normal ordinal collapsing function you know that your expressions are ordinary ordinals, so you get well-ordering for free. So we would like to have appropriate large cardinal notions for our stronger and stronger ordinal notations, but I personally don't know where to go after alpha-indescribables. (There are of course many large cardinals higher than alpha-indescribables, but we need something which can form an expansive hierarchy even greater than the alpha-indescribables.) Deedlit11 (talk) 08:29, August 5, 2017 (UTC)

How large are the PTO of "KP + 'there exists a weakly inaccessible cardinal'", "KP + 'there exists a weakly Mahlo cardinal'" and "KP + 'there exists a weakly compact cardinal'"? Are they the same as "KP + 'there exists a recursively inaccessible ordinal'", "KP + 'there exists a recursively Mahlo ordinal'" and "KP + 'there exists a $$\Pi_3$$-reflecting ordinal'" respectively? {hyp/^,cos} (talk) 00:19, August 9, 2017 (UTC)

As far as I know, the existence of weakly inaccessible cardinals, weakly Mahlo cardinals, etc. are independent from ZFC, while recursively inaccessible ordinals, recursively Mahlo ordinals, etc. can be proven to exist in ZFC. So the PTOs of "KP + 'there exists a weakly inaccessible cardinal'" and such are much larger than the PTOs of "KP + 'there exists a recursively inaccessible ordinal'" and such. -- ☁ I want more clouds! ⛅ 07:35, August 9, 2017 (UTC)
Sorry for the late reply, but I agree with Cloudy. The part that I am unsure of is what happens with you add a large cardinal axiom to a weak base theory like KP. I still think that should be much stronger than adding an axiom for a recursively large ordinal, but I can't be sure. Deedlit11 (talk) 01:26, August 29, 2017 (UTC)
In system (b), we can make an OCF that $$\psi(\varepsilon_{i+1})$$ is the PTO of "KP + 'there exists a recursively inaccessible ordinal'" (where $$i$$ is the recursively inaccessible ordinal), then another OCF that $$\psi(\varepsilon_{M+1})$$ is the PTO of "KP + 'there exists a recursively Mahlo ordinal'" (where $$M$$ is the recursively Mahlo ordinal), then another OCF that $$\psi(\varepsilon_{K+1})$$ is the PTO of "KP + 'there exists a recursively $$\Pi_3$$-reflecting ordinal'" (where $$K$$ is the recursively $$\Pi_3$$-reflecting ordinal), and so on. Then there might be a series of OCF compatible with KP plus a large ordinal axiom. Finally there will be an OCF such that $$\psi(\varepsilon_{I+1})$$ is the PTO of "KP + 'there exists a weakly inaccessible cardinal'" (where $$I$$ is the weakly inaccessible cardinal), and now system (b) can also use uncountable cardinals in some way. {hyp/^,cos} (talk) 03:09, August 29, 2017 (UTC)
Perhaps, but the devil is in the details. The belief among the mathematicians that do ordinal analysis seems to be that even second-order arithmetic is way out of reach of current methods, much less ZFC or ZFC + Inaccessible, Taranovsky's claims aside. So, there could be a notation such that $$\psi(\varepsilon_{I+1})$$ is the PTO of "KP + 'there exists a weakly inaccessible cardinal'", but the bigger question is what sort of monstrous notation gets you that far. Deedlit11 (talk) 05:01, August 29, 2017 (UTC)

### KP catching ZFC point

The "KP catching ZFC point" is such a large ordinal $$\rho$$ (may not be a cardinal) with special property that the PTO of (ZFC + "there exists a $$\rho$$") equals the PTO of (KP + "there exists a $$\rho$$"). Does this "KP catching ZFC point" exist? If so, how large is it? {hyp/^,cos} (talk) 07:48, December 23, 2017 (UTC)

## Pi-reflection fast-growing hierarchy

As I understand for transfinite n is undefined. And if further creation of collapsing functions becomes very complicated, then we can diagonalize the fast-growing function.

First, we can create a new fast-growing function:

first fixed point
second fixed point
and so on to:

And we get a super fast growing function:

etc...

After that, we can continue to diagonalize. How do you like this idea? —Preceding unsigned comment added by Scorcher007 (talkcontribs) 05:00, September 9, 2017 (UTC)

I'm not sure what you mean by a fixed point of . The right hand side is a natural number, and is only defined when is finite, so we can only be talking about natural numbers. But that function is strictly increasing for natural numbers.
Iterating things is not that strong: , for instance, and more generally. Deedlit11 (talk) 15:14, September 9, 2017 (UTC)

## Operations with inaccessible ordinals

If we are talking about operations on inaccessible ordinals, then we can write , but what does it mean? Similarly, we can introduce such an operation for Mahlo ordinal . Then will it make sense ? If so, why it is not used in the OCF, and why after immediately used ? Scorcher007 (talk) 03:49, September 14, 2017 (UTC)

Dear Scorcher,

is smallest ordinal which is greater than . In terms of functions collapsing -weakly inaccessible cardinals is smallest 0-inaccessible cardinal which is greater than

or in other words

.

Considering Veblen function's analogy it looks like

- smallest power of omega which is greater than .

Taking into account the rules for fundamental sequences assigned for those functions we can say that in this case

and

Although I did not assign fundamental sequences for functions collapsing Mahlo-cardinals but if I would did it I believe I would got analogical result and for

--Denis Maksudov (talk) 10:28, September 14, 2017 (UTC)

But do these expressions have meaning outside the collapsing function? I suppose as well as means fixed point of , and should mean with cardinality , should mean with cardinality or , and should mean -th inaccessible cardinal or , and should mean -th 2-inaccessible cardinal or .
Do I understand this correctly? And if this is so, why do not we then make a collapsing function, where after used , , etc.? This will be a much more powerful notation.
Scorcher007 (talk) 02:59, September 15, 2017 (UTC)

If you think that for all existing colapsing functions follows immediately after , , then it is not correctly.

As you can see, even for example follows not after

, , ...

but after

, , ...

And what to say about Mahlo-cardinals! Smallest of them is greater than any inaccessible cardinals same way as first uncountable ordinal is greater than any countable ordinal of Veblen function.

But I will not try to look smart saying about cardinals of Mahlo, since I only begin to understand them, nevertheless I will say something intuitively:

we can define collapsing function

such that

and we can define collapsing function

such that

Second seems stronger.

But!

Look at analogy: although collapsing function such that is stronger than another collapsing function such that but the second will catch the first at the level

Same way and for functions collapsing Mahlo: the first will catch the second at the level . Thus although the second function collapsing cardinals of Mahlo is stronger but not really much stronger meanwhile the first function collapsing cardinals of Mahlo is simplier for creating of common rules for fundamental sequences. --Denis Maksudov (talk) 21:17, September 15, 2017 (UTC)

@Denis: is not the smallest 0-inaccessible cardinal which is greater than , and so it is not true that . Rather, is simply the st cardinal, or the smallest cardinal greater than .
@Scorcher: , , and are the th cardinal, weakly inaccessible cardinal, and weakly Mahlo cardinal respectively. So they do have meaning outside of ordinal collapsing functions. As for why we don't use the I function in the later ordinal collapsing functions, the reason is that it is subsumed by the Mahlo hierarchy. In the ordinal collapsing function up to the weakly Mahlo cardinal M, we get by , then is the th 1-weakly inaccessible cardinal, and so on and so forth. Of course, this only goes up to M, but beyond M we can use and beyond to collapse. So we do get things like using the Mahlo hierarchy. For that matter, we don't really need the hierarchy either, but it is kept to better match up with previous ordinal notations.
I'm not sure what you mean by saying things like will lead to a more powerful notation. I(2,M+1) and cardinals like it are present in the last notation I presented (the one that went up to a weakly compact cardinal). Deedlit11 (talk) 21:40, September 20, 2017 (UTC)
@Deedlit: I used following definition:

An ordinal is weakly inaccessible if it is a regular limit cardinal larger than i.e. = the th cardinal in the set

An ordinal is -inaccessible if it is a regular cardinal and limit of -inaccessibles for all

Thus we can define binary function enumerating the -inaccessible ordinals the th ordinal in the set where denotes the set of all regular cardinals i.e.

In this case i.e. 0-inaccessible ordinals are exactly the regular cardinals

and

. So the 1-inaccessible ordinals are the inaccessible ordinals.

Hence --Denis Maksudov (talk) 11:41, September 21, 2017 (UTC)

Sorry thar i replied so late but would like to discuss this:

https://googology.wikia.com/wiki/User_blog:Ynought/Trying_to_make_the_SCG_function_even_quicker Ynought (talk) 20:16, January 28, 2019 (UTC)

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