Mathworks (Not the main chat!) - 2017-10-24 (page 2 of 3)

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8:01 PM

@chronolegends You should really get towards seeing the footnotes on my TREE(3) program though.

$\Omega[n,\Omega] =\omega, \psi(\omega), \psi^2(\omega), \psi^3(\omega), \psi^4(\omega),...$

$= \omega, \epsilon_{\omega},\epsilon_{\epsilon_{\omega}},\epsilon_{\epsilon_{\epsilon_{\omega}}},\epsilon_{\epsilon_{\epsilon_{\epsilon_{\omega}}}},...$

So: $\psi (\Omega) = \zeta_0$

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@Secret :P Who need's $\zeta_0$ when you have $\psi(\Omega)$? amirite?

i need it

It still bothers me that any OCF need the first uncountable ordinal to reach $\zeta_0$

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@chronolegends need what?

@Secret not any

8:02 PM

z_0

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@chronolegends Oh.

@Secret The OCF in my program has $\psi(\Omega)>\Gamma_0$

Actually.... lemme calculate $\psi(\Omega)$ in my program real quick...

(brb...)

lol

@Secret it depends on the closure operations allowed

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In terms of the Veblen function

my program has $\psi(\Omega)=\varphi(1,1,0)$

@chronolegends what he said

psi(W) = psi(1,1,0) is a result of having Gamma_alpha in there ... (?)

That translates to : $\Gamma_0\text{ something } \epsilon_0$

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8:06 PM

@chronolegends The second one is a phi.

right

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$\psi(\alpha\cdot\omega)=\Gamma_\alpha$ for small $\alpha$

So yeah, more or less.

ah nope. $\varphi (1,1,0)$ is $\Gamma_{\epsilon_0}$ I think...

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@Secret nope

psi(1,1,0) = limit to T_T_T....

i mean phi

dammit. that hiatus really set me back

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8:08 PM

$\varphi(1,1,0)=1$-st fixed point of $x=\Gamma_x$

@chronolegends lots of tears were required to reach φ(1,0,0)

Btw, you know you can copy the symbols from the MathJax and it'll give you the unicode characters?

i need to get my ordinal-fu back to its prime

$\varphi (1,1,0)=\Gamma_{\Gamma_{\Gamma_{\ddots}}}$?

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lmao

and thanks i didnt know that

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@Secret yup

8:10 PM

Yeah

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But back to our functions here

I'll explain mine later.

=P

I think I started to have some idea on what the second argument is doing: It is needed to vary the argument at the bottom of the nesting. Since we have $\Omega=\omega_1$ that second argument allow us to access the uncountably many fundemental sequence in increasing order to reach higher ordinals before $\Omega$

How do i make it show me the jax code?

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@chronolegends right click -> show math as -> TeX commands

Or refresh the page

I.e.
$\Omega[n,\alpha] =\omega, \psi(\alpha), \psi^2(\alpha), \psi^3(\alpha), \psi^4(\alpha),...$

8:12 PM

Cool

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@Secret Noppie doodles

$\Omega[1,1]=\psi(1[0,1])=\psi(0)\ne\psi(1)$

Get's a little bit tricky

You'll never reach the expression $\Omega[1,1]$, but it was the quickest counterexample that came to mind.

0\ne\1

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@chronolegends Put dollar signs around it to make math

$0\ne\1$

I see

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Yeah, also no idea why you're doing a \1

8:15 PM

$0\ne1$

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:P

Climbing the $\psi(\Omega)$:

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@Secret The second term should actually be $\psi(\alpha[0,\alpha])$

And the third term is not always $\psi^2(\alpha)$

In fact, $\psi(\Omega)$ is the only time we'll be able to use iteration notation so nicely.

@chronolegends Say, do you program by any chance?

Not outside of minor golfing

@Secret have you skimmed the wikipedia article on ordinal collapsse?

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@chronolegends Probably

8:19 PM

Ordinal collapsing function

In mathematical logic and set theory, an ordinal collapsing function (or projection function) is a technique for defining (notations for) certain recursive large countable ordinals, whose principle is to give names to certain ordinals much larger than the one being defined, perhaps even large cardinals (though they can be replaced with recursively large ordinals at the cost of extra technical difficulty), and then “collapse” them down to a system of notations for the sought-after ordinal. For this reason, ordinal collapsing functions are described as an impredicative manner of naming ordinals....

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But Secret has also found all those technical definitions a bit too much.

@chronolegends I have, but the non recursive version is so annoying to follow. This one is MUCH BETTER

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lol

I'll have to agree though, I like this recursive form much more.

I have two rules in maths life:

1. Higher dimensions and columns makes things easy
2. Operators makes things easy

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You can actually make an ordinal collapsing function in a program by asking it to repeat a set of operations infinitely and find the smallest ordinal that's not inside of it @chronolegends

8:20 PM

um.

okay

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Would probably result in a program that never terminates.

If you're to do it like the definitions prescribe

23

Q: Output the simplified Goodstein sequence

A number is in base-b simplified Goodstein form if it is written as b + b + ... + b + c, 0 < c ≤ b The simplified Goodstein sequence of a number starts with writing the number in base-1 simplified Goodstein form, then replacing all 1's with 2's and subtracting 1. Rewrite the result in base-2...

code-golf math sequence

@chronolegends This challenge might be easy enough and still of interest to you

Anyways...

@Secret So can you see that $\psi(\Omega+1)=~^\omega(\psi(\Omega)) =\varepsilon_{\zeta_0+1}$

$\psi(\Omega+\alpha)=~^\omega(\psi(\Omega)) =\varepsilon_{\zeta_0+\alpha}$

um, exclude the middle part

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:P

19 mins ago, by chronolegends

i need to get my ordinal-fu back to its prime

You really do.

\psi(\Omega+1) =\varepsilon_{\zeta_0+1}

$\psi(\Omega+\alpha) = \varepsilon_{\zeta_0+\alpha}$

ok better

so then we

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But yeah, $\psi(\Omega+\alpha)=\varepsilon_{\zeta_0+\alpha}$ for reasonably small $\alpha$.

Wonder where @Secret went.

@chronolegends and I totally cut you off! xD Sorry.

8:31 PM

$\psi(\Omega+\psi(\Omega)) =\varepsilon_{\zeta_0+\varepsilon_{\zeta_0}}$

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@chronolegends D:

Stop stealing my light man

i meant $ \psi(\Omega+\psi(\Omega)) =\varepsilon_{\zeta_0+\varepsilon_{\zeta_0+1}} $

but since limit ordinals additively left cancel from the left, we have

effectively $\psi(\Omega+\psi(\Omega)) =\varepsilon_{\varepsilon_{\zeta_0+1}} $

\begin{align}
\psi (\Omega)[n,\Omega] & = \psi^{n+1} (\omega) = \zeta_0\\
\psi (\Omega+1)[n,\Omega] & = {}^n\psi(\Omega) = \epsilon_{\zeta_0+1}\\
\psi (\Omega+m)[n,\Omega] & = {}^{n^m}\psi(\Omega) = \epsilon_{\zeta_0+m}\\
\psi (\Omega (m+1))[n,\Omega] & = \psi (\Omega m+ \Omega [n,\Omega]) = \psi (\Omega m + \psi^{n+1}(\omega)) = \psi(\Omega m + \zeta_0) = \text{I need to do it on paper}
\end{align}

you can take the torch @SimplyBeautifulArt to $\zeta_1$ if you want :p

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@Secret Whoa whoa whoa

8:37 PM

not so fast cowboy :D

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We don't jump from $\Omega+m$ to $\Omega m$

Dats a bit jump

Back down to $\Omega2$

Or rather, $\Omega+\Omega$

$$\psi(\Omega2)=~?$$

Do this one.

Also, no skipping steps

The intuition from $\psi(\Omega)$ may be your downfall, pressing you to make jumps that are not valid.

holds @Secret 's hand

$\psi (\Omega 2)[n,\Omega] = \psi (\Omega +\Omega)[n,\Omega] = \psi (\Omega + \Omega [n, \Omega]) = \psi (\Omega + \psi (\Omega [n-1,\Omega]))$

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@Secret yeup.

R.I.P. because it's not being very nice for iteration notation

$=\psi (\Omega + \psi^2 (\Omega [n-2,\Omega])) = \psi (\Omega + \psi^3 (\Omega [n-3,\Omega])) = \cdots = \psi (\Omega + \psi^{n+1}(\omega)) = \psi (\Omega + \zeta_0) = ???$

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8:42 PM

Nope

Oh

x.x you messed up the $z$

You should be having $\Omega2[n,\Omega2]$

Not $\Omega2[n,\Omega]$

$$\psi (\Omega 2)[n,\Omega 2] = \psi (\Omega +\Omega)[n,\Omega 2] = \psi (\Omega + \Omega [n, \Omega 2]) = \psi ( \Omega + \psi ( \Omega 2 [n-1,\Omega 2])) = \psi (\Omega + \psi (\Omega +\Omega [n-1,\Omega 2])) = \psi (\Omega + \psi (\Omega +\psi (\Omega [n-2,\Omega 2]))) = \cdots = [\psi (\Omega + ]^{n+1} (\omega) = [\psi (\Omega + ]^{n} (\psi (\Omega+\zeta_0)) = ???$$

no, tripped again

$\psi(\Omega2)=~?$

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@chronolegends what?

@Secret well, you shouldn't be getting any $\zeta_0$ at the end

rest looks fine to me.

So we more or less are looking at:

$$[\psi(\Omega+]^n\omega)$$

What a broken notation lol

$\psi(\Omega2)=\psi(\Omega+\Omega)=\psi(\Omega+0),\psi(\Omega+\psi(\Omega+0)), \psi(\Omega+\psi(\Omega+\psi(\Omega+0))) ...$

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@chronolegends For some reason I made it $\omega$'s instead of zeroes because that's how my program went.

$$\underbrace{\psi(\Omega +\psi(\Omega +\psi(\Omega +\dots}_n\omega\dots)))$$

But we can agree on this expression more or less

Maybe there's $n+1$, but we get the idea.

8:53 PM

$$\psi (\Omega 2)[n,\Omega 2] = \psi (\Omega +\Omega)[n,\Omega 2] = \psi (\Omega + \Omega [n, \Omega 2]) = \psi ( \Omega + \psi ( \Omega 2 [n-1,\Omega 2])) = \psi (\Omega + \psi (\Omega +\Omega [n-1,\Omega 2])) = \psi (\Omega + \psi (\Omega +\psi (\Omega 2[n-2,\Omega 2]))) = \cdots = [\psi (\Omega + ]^{n+1} (\omega) = ???$$

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Now, recall that we said that:

$ = \zeta_0, \epsilon_{\zeta_0+1},\epsilon_{\epsilon_{\zeta_0+1}}, ... = \zeta_1$

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$$\psi(\Omega+\alpha)=\varepsilon_{\zeta_0+\alpha}$$

@chronolegends jeez man, I'm trying to teach what I wrote, not that normal stuff

:D

brb sorry lol trying to warm up back to a working standard

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My FS are non-standard FYI @chronolegends

8:55 PM

i know i'm cramping your style in the process, i apologize :p

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@Secret Anyways, we get:$$\psi(\Omega2) =\varepsilon_{\zeta_0 +\varepsilon_{\zeta_0 +\varepsilon_{\zeta_0 +\varepsilon_{\zeta_0 +\varepsilon_\ddots}}}}$$

That's even worse then continued fractions

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X'D

Realize that $\zeta_0$ is smaller than the limit ordinal to it's right.

That is, $\alpha+\beta=\beta$ if $\beta$ is a limit ordinal and $\beta>\alpha$

So it reduces down to$$\psi(\Omega2)=\varepsilon_{ \varepsilon_{ \varepsilon_{ \varepsilon_{ \ddots_{\zeta_0+\omega}}}}}$$

Anyways, you should recognize this as $\zeta_1$.

I always thought $\zeta_1$ is $\zeta_{\zeta_{\zeta_{\cdots}}}(\zeta_0)$

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@Secret Uh, that doesn't even make sense?

9:07 PM

subscript towers of $\zeta_0$

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$\zeta_{x+1}$ is basically infinitely many $\varepsilon$'s with a $\zeta_x$ at the bottom (but with a +1 or something to avoid collapsing due to fixed-points)

@Secret That's $\eta_0$.

$\varepsilon$ enumerates over exponentiation.
$\zeta$ enumerates over $\varepsilon$'s.
$\eta$ enumerates over $\zeta$'s.

But it all works out much much nicer once you develop a good understanding of OCF's and polynomials of $\Omega$'s.

@Secret $\zeta_{\zeta_...} = \eta_0$

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@chronolegends >.> I already said that mate.

i was like agreeing with u on that

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>.>

9:15 PM

btw if there is infinitely many ε's then there can't be a ζx at the bottom :p

since there is no bottom :D

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:P

@Secret Done for the day, or moving on?

Tidying stuff in my brain atm

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Fair warning, its getting to the point where we may want to switch to the Veblen function.

take it easy bro

one ordinal at a time :D

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@chronolegends Too slow man. Never would've reached $\psi(\Omega)$ at that rate. Sometimes you gotta jump and ask questions afterwards.

9:23 PM

You are the better teacher so i concede you might be right

so should we jump to $\psi(\Omega^2)$

grin

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@chronolegends >:(

$\psi(\Omega3) = \psi(\Omega2+\Omega) = \psi(\Omega2), \psi(\Omega2+\psi(\Omega2)), \psi(\Omega2+\psi(\Omega2+\psi(\Omega2))) ...$

$= \zeta_1 , \epsilon_{\zeta_1+1}, \epsilon_{..._{\zeta_1+1}} = \zeta_2$

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@chronolegends Nah

In here, it shall be:
$\psi(\Omega2+\omega),\psi(\Omega2+\psi(\Omega2+\omega)),\dots$

$\psi(\Omega( 1+\alpha)) = \zeta_\alpha$

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Uh no

Parenthesis sir.

9:37 PM

there ;D

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=P

ah feels good doing the baby steps again. thanks for inviting me to the chat lol, brb

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@chronolegends I really like how you can interpret the Veblen function as $\psi(P(\Omega))$, where $P(\Omega)$ is a polynomial of $\Omega$'s with countable coefficients.

ugh I am trying to rederive $\psi (\epsilon_{\omega})$ using that supremum rule so I can see why it is acting an epsilon

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@Secret ew, don't write it like dat.

$$\psi(\psi(\omega))$$

No $\Omega$'s in this expression, so it's safe to use [n] instead of [n,z]

\begin{align}\psi(\psi(\omega))[n]&\to \psi(\psi(\omega)[n])\\&\to\psi(\psi(\omega[n]))\\&\to\psi(\psi(n))\end{align}

9:42 PM

Now how to apply the supremum rule and thus conclude that it is $\epsilon_{\epsilon_{\omega}}$. I am always losing track...

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Uh...

Well it just follows from $\varepsilon_\alpha=\sup\{\varepsilon_\beta, \beta<\alpha\}$ for all $\alpha\in\Bbb{Lim}$

The rule aligns with the rules of the OCF is all you need to really verify.

@SimplyBeautifulArt true, that principle also applies to extending up to and above hyper inaccessibles

but replacing psi with the appropiate function

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Yeah ofc

and phi too :D

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@Secret I like to imagine $\psi$ as multiple different functions. You have the function $\psi(x)$, then $\psi(\Omega+x)$, then $\psi(\Omega2+x)$, ...

Then maybe a generalized $\psi(\Omega y+x)$

And you'll notice something peculiar here....

@Secret How would you describe the nature of what the $\Omega$ is doing so far?

9:54 PM

It keeps nesting $\zeta_0$ terms at increasingly higher sophistication controlled by $y$

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@Secret Nope. What does $\Omega$ do in general.

$\psi(\Omega)$ never nested any $\zeta_0$'s, for example.

Other than to get past $\zeta_0$ I have no idea yet

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What did $\psi(\Omega)$ expand into?

(in terms of more $\psi$'s, but not reducing to anything else)

What did $\psi(\Omega+\Omega)$ expand into?

What do you think $\psi(\Omega2+\Omega)$ will expand into?

Remember. Only write your expansions in terms of $0,1,2,\dots,\omega,\Omega,\psi,+,\times$, and if necessary, exponentiation.

@chronolegends Say, have you been on Discord?

*::Patiently waits for the messages to stop sliding to the right::*

no i haven't

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Okay, just checking

10:02 PM

i'm too ashamed by my lack of practice to go back

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lol

Hm

@chronolegends you know how to write TREE(n)'s fgh approximation in terms of these OCF's?

which one ? :p

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$ϑ(Ω^ωω)+1$

Hold on, i think i may have a way if i just look it up

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I'm pretty confident it should translate directly into $\psi$ where that $\vartheta$ is.

But I just wanna check in on it.

10:07 PM

$\theta(\alpha, \beta) = \vartheta (\Omega * \alpha + \beta)$

$\theta (\alpha, \beta) = \psi (\Omega^{\alpha} (1 + \beta))$

so um. it should follow that

What is $\sup (\epsilon_n +1 | n < \omega)$?

$ \vartheta (\Omega * \alpha + \beta) = \psi (\Omega^{\alpha} (1 + \beta)) $

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Ah, okay

@Secret $\epsilon_\omega$

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@Secret $\varepsilon_n<\varepsilon_n+1<\varepsilon_{n+1}$

10:10 PM

Hmm...

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What do the upper and lower bounds go to as $n\to\omega$?

$\omega$

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Gonna go eat.

cya

$\psi(\psi(\omega)) = \sup ({}^{\omega}\psi (\psi (\omega)[n,\omega]|n < \omega)) = \sup ({}^{\omega}\psi (\psi (n)|n < \omega)) = .........................$

$\psi(\psi(\omega)) = \epsilon_{\epsilon_\omega} $

10:13 PM

no way I can churn a $\epsilon_{\epsilon_{\omega}}$ from here... using just the sup

I don't know about @SimplyBeautifulArt 's method but

How can you show that with the $\psi$ fully unravelled into sups?

We established earlier that for $\alpha < \Omega$ , $\psi(\alpha) = \epsilon_\alpha$

I am not very certain, I think I don't really understand how to establish that

Alright, in the closure operations we allow + x and ^

and for c_0 we allow 0 , 1 and w (using the wikipedia definition)

so the smallest ordinal we can't write as a finite expression using those is $\epsilon_0$

thus $\psi(0)=\epsilon_0$ , next since $\epsilon_0 has been constructed, it can be used, so the next ordinal we can't write as a finite expression is now

$\epsilon_0^{\epsilon_0^...} = \epsilon_1$

so $\psi(1) = \epsilon_1$

Good so far? :D

10:20 PM

yup

well, to figure out what $\psi(\omega)$ we consider all the $\epsilon_n$ for $n<\omega$ are constructed, so

$ \epsilon_0^{\epsilon_0^...} = \epsilon_1 $ is constructed, $ \epsilon_1^{\epsilon_1^...} = \epsilon_2 $ is constructed, $ \epsilon_{(1+n)}^{\epsilon_{(1+n)}^...} = \epsilon_n $ is constructed ...

so the smallest ordinal not in the set c(ω) is clearly $\epsilon_\omega$

ergo $\psi(\omega) = \epsilon_\omega$

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I'm back.

*hides his ordinals beneath the table *

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@chronolegends You really need to get back into your ordinal-fu

i wasn't doing anything i swear

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10:28 PM

13 mins ago, by chronolegends

We established earlier that for $\alpha < \Omega$ , $\psi(\alpha) = \epsilon_\alpha$

Pft.

What's this?

lies

17 mins ago, by Secret

$\psi(\psi(\omega)) = \sup ({}^{\omega}\psi (\psi (\omega)[n,\omega]|n < \omega)) = \sup ({}^{\omega}\psi (\psi (n)|n < \omega)) = .........................$

BSOD

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You really believe that $\psi(\alpha)$ outputs every epsilon number from $\varepsilon_0$ to $\Omega$?

@Secret :o new acronym.

not up to but below

Blue Screen of Death

A stop error, better known as a Blue Screen of Death (also known as a blue screen or BSoD) is an error screen displayed on a Windows computer system after a fatal system error, also known as a system crash: when the operating system reaches a condition where it can no longer operate safely. == History == BSoDs have been present in Windows NT 3.1 (the first version of the Windows NT family, released in 1993) and all Windows operating systems released afterwards. (See History of Microsoft Windows.) BSoDs can be caused by poorly written device drivers or malfunctioning hardware, such as faulty memory...

My brain is BSOD in trying to unravel $\psi(\psi(\omega))$

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10:30 PM

@Secret :o

I am trying to justify why it is $\epsilon_{\epsilon_{\omega}}$

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@chronolegends So you believe it outputs all of those ordinals?

@Secret Easier to use induction

Yeah

I hope i was of some help @Secret

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Suppose that $\psi(y)=\varepsilon_y$ for every $y<x$.

@chronolegends :| You were using C.

Secret finds the C(..) to be a bit confusing.

Hence why I'm using this approach.

Back to the induction

But you haven't introduced me to your inductive ways

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10:33 PM

Suppose that $\psi(y)=\varepsilon_y$ for every $y<x$.

so i don't know any better :P

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Suppose that $x=y+1$ for one of those $y$'s.

Prove that $\psi(x)=~^\omega(\psi(y))$ for that $x=y+1$.

Otherwise, $x$ is a limit (and not involving $\Omega$'s).

Then prove that $\psi(x)=\sup\{\psi(y):y<x\}$

me ?

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(not really a question. These are all from definitions.)

(No not you you silly)

(Hello @Hans)

@Secret If you prove those statements, you'll have proven that $\psi(x)=\varepsilon_x$ for every $x$ not involving $\Omega$'s.

$\psi (y+1)[n,z] = \sup ({}^{\omega}\psi (y+1[n,z])) = \sup ({}^{\omega}\psi (y))$

$\psi (x) [n,z] = \sup({}^{\omega}\psi(x[n,z])) = \sup ({}^{\omega} \psi (x[n])) = \sup ({}^{\omega} \psi (y_n))$

10:39 PM

@SimplyBeautifulArt regarding your earlier question, i pm'd deedlit for help, in terms of psi it's $\psi(\Omega^{(\Omega^\omega\omega)})$

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@chronolegends M'kay

Oh, a full nother exponent thingy

Looks much baller doesn't it

and I guess we get the base case from $\epsilon_0$

join the dark c(...) side

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@chronolegends :|

Honestly much better over here.

10:42 PM

Climbing the $\Omega$ (Again!)

yay

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Well

Use $\psi (\Omega +1)[n,\Omega +1]$ or $\psi (\Omega +1)[n,\Omega]$?

Can you link me to your recursive psi definition

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(in terms of more natural numbers, $\omega$, $\Omega$ and $\psi$'s, but not reducing to anything else)
What did $\psi(\Omega)$ expand into?
What did $\psi(\Omega+\Omega)$ expand into?
What do you think $\psi(\Omega2+\Omega)$ will expand into?

@chronolegends Uh, its kinda a mess.

10:45 PM

iight

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You can kinda find it on the TREE(3) program

I'll give @Secret my cheaty way of dealing with $\Omega$

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(just replace a(↑b)c with addition, multiplication, and exponentiation)

@chronolegends did you know Deedlit is online?

Steps problem: $\psi (\Omega +1)[n,\Omega +1] = \psi(\Omega +1[n,\Omega+1])$ or $\psi (\Omega +1)[n,\Omega +1] = {}^{\omega}\psi(\Omega)[n,\Omega+1])$?

I didn't

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10:48 PM

@Secret $\psi(Ω+(1[n,Ω+1]))$

@chronolegends he's in my realm.

but that will become $\psi (\Omega+0)$?

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@Secret $=ψ(Ω)=\zeta_0$, yes.

$\psi(x+1)$ is supposed to reduce straight down to $\psi(x)$, and then you take an infinite tower.

But that will mean $\psi (\Omega+m) = \psi (\Omega)$?

Simply Beautiful Art

No

@Secret Whenever you see a $*\Omega$ in an expression where * is some operation, make a sequence, the first step is $*0$, the second step is $*first step$, the third step is $*second step$ and so on

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10:51 PM

You said it yourself, $(m+1)[n,z]=m+(1[n,z])=m$

It just drops to the next lower natural number.

@chronolegends \# and \text{...}

fixied

ok I see

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@chronolegends >:( No, the first step is $\omega$

Alright :D

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Anyways, I'm gonna go now.

We spent a lot of hours talking today lol

Don't you have, like, idk, a life to get back to @Secret ?

10:54 PM

again, thanks for ropeing me back in

:)

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I'm sick at home more or less, so I have an excuse :P

@chronolegends anytime mate

feels good annoying you on a regular basis

I can push a bit more, but only a bit more, then I must get back to my chemistry

Also I did not sleep so far...

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@chronolegends -__-

lol jk @SimplyBeautifulArt i hope you feel better soon , friend :)

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10:55 PM

:P

@Secret I can help you with psi stuff but, i only know the usual c(...) style

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