Difference between revisions of "Madore's ψ function"

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[[Category:Lower attic]]
 
[[Category:Lower attic]]
Madore's \(\psi\) function is a function for expressing extremely large countable ordinals.  
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Madore's \(\psi\) function is an ordinal collapsing function introduced by David Madore.  
  
== Defination ==
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== Definition ==
 
Madore's \(\psi\) function is defined as follows:
 
Madore's \(\psi\) function is defined as follows:
  
\(S=\{0,1,\omega,\Omega\}\) is our tool box.
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Let \(\omega\) be the first transfinite ordinal and \(\Omega\) be the first uncountable ordinal. Then,
  
\(C(0)=\)all of the things you can make from S and finite applications of \(+\times\text{^}\) in S.
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\(C_0(\alpha) = \{0, 1, \omega, \Omega\}\)
  
\(C(n+1)=\) all of the things you can make from \(C(n)\) and finite applications of \(+\times\text{^}\) \(\psi\) in \(C(n)\)(only if \(\psi(n)\) is defined).
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\(C_{n+1}(\alpha) = \{\gamma + \delta, \gamma\delta, \gamma^{\delta}, \psi(\eta) | \gamma, \delta, \eta \in C_n (\alpha); \eta < \alpha\} \)
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\(C(\alpha) = \bigcup_{n < \omega} C_n (\alpha) \)
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\(\psi(\alpha) = \min\{\beta < \Omega|\beta \notin C(\alpha)\} \)
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In other words \(\psi(\alpha)\) is the least ordinal number less than \(\Omega\) which cannot be generated from ordinals \(0, 1, \omega, \Omega\) by  applying of addition, multiplication, exponentiation and the function \(\psi(\eta)\) with \(\eta < \alpha\).
  
\(\psi(n)=\) the first countable ordinal not in \(C(n)\).
 
  
 
== Values ==
 
== Values ==

Revision as of 11:13, 17 May 2018

Madore's \(\psi\) function is an ordinal collapsing function introduced by David Madore.

Definition

Madore's \(\psi\) function is defined as follows:

Let \(\omega\) be the first transfinite ordinal and \(\Omega\) be the first uncountable ordinal. Then,

\(C_0(\alpha) = \{0, 1, \omega, \Omega\}\)

\(C_{n+1}(\alpha) = \{\gamma + \delta, \gamma\delta, \gamma^{\delta}, \psi(\eta) | \gamma, \delta, \eta \in C_n (\alpha); \eta < \alpha\} \)

\(C(\alpha) = \bigcup_{n < \omega} C_n (\alpha) \)

\(\psi(\alpha) = \min\{\beta < \Omega|\beta \notin C(\alpha)\} \)

In other words \(\psi(\alpha)\) is the least ordinal number less than \(\Omega\) which cannot be generated from ordinals \(0, 1, \omega, \Omega\) by applying of addition, multiplication, exponentiation and the function \(\psi(\eta)\) with \(\eta < \alpha\).


Values

\begin{eqnarray*} \psi(0) &=& \varepsilon_0 \\ \psi(1) &=& \varepsilon_1 \\ \psi(2) &=& \varepsilon_2 \\ \psi(n) &=& \varepsilon_n \\ \psi(\zeta_0) &=& \zeta_0 \\ \psi(\zeta_0+1) &=& \zeta_0 \end{eqnarray*}

That seems strange. Shouldn't \(\psi(\zeta_0+1)=\varepsilon_{\zeta_0+1}\)? No. Look at \(C(\zeta_0+1)\). It has all the things from \(C(\zeta_0)\) and \(+\times\text{^}\) \(\psi\) of it. But in order to get \(\zeta_0\), you have to have \(\zeta_0\)in your list. So you are never going to have \(\zeta_0\) in your list. Or are you \(\cdots\cdots\)

\begin{eqnarray*} \psi(\Omega) &=& \zeta_0 \\ \psi(\Omega+1) &=& \varepsilon_{\zeta_0+1} \\ \psi(\Omega+n) &=& \varepsilon_{\zeta_0+n} \\ \psi(\Omega+\zeta_1) &=& \varepsilon_{\zeta_0+\zeta_1} &=& \zeta_1 \\ \psi(\Omega+\zeta_1+1) &=& \zeta_1 \end{eqnarray*}

We see that this \(\psi\) function got stuck at \(\zeta_0\). But when we arrive at \(\Omega+1\), then we are allowed to use the \(\Omega\) in \(C(\Omega)\) to create bigger ordinals. We then arrive at \(\psi(\Omega+\zeta_1)\) which is signaling that this function is stuck again until \(\Omega2\).

\begin{eqnarray*} \psi(\Omega2) &=& \zeta_1 \\ \psi(\Omega2+1) &=& \varepsilon_{\zeta_1+1} \\ \psi(\Omega2+n) &=& \varepsilon_{\zeta_1+n} \\ \psi(\Omega2+\zeta_2) &=& \varepsilon_{\zeta_1+\zeta_2} &=& \zeta_2 \\ \psi(\Omega2+\zeta_2+1) &=& \zeta_2 \end{eqnarray*}

Stuck again. We are going to fast forward now.

\begin{eqnarray*} \psi(\Omega3) &=& \zeta_2 \ \psi(\Omega n) &=& \zeta_{n-1} \\ \psi(\Omega \eta_0) &=& \eta_0 \\ \psi(\Omega \eta_0+1) &=& \eta_0 \end{eqnarray*}

This function is stuck again until \(\psi(\Omega^2)\) because \(C(\Omega \eta_0)\) contains all countable ordinals up to but not incliding \(\eta_0\), but there is no ordinal called \(\Omega \eta_0\) in \(C(\Omega \eta_0)\). So if you want to have \(\eta_0\) in your list, you need \(\Omega \eta_0\), and therefore, \(\eta_0\).

\begin{eqnarray*} \psi(\Omega^2) &=& \eta_0 \\ \psi(\Omega^2+1) &=& \varepsilon_{\eta_0+1} \\ \psi(\Omega^2+n) &=& \varepsilon_{\eta_0+n} \\ \psi(\Omega^2+\Omega) &=& \zeta_{\eta_0+1} \\ \psi(\Omega^2+\Omega2) &=& \zeta_{\eta_0+2} \\ \psi(\Omega^2+\Omega n) &=& \zeta_{\eta_0+n} \\ \psi(\Omega^2+\Omega\eta_1) &=& \eta_1 \\ \psi(\Omega^2 2) &=& \eta_1 \\ \psi(\Omega^2 n) &=& \eta_{n-1} \\ \psi(\Omega^2 \varphi_4(0)) &=& \varphi_4(0) \\ \psi(\Omega^3) &=& \varphi_4(0) \end{eqnarray*}

Now we are introducing the Veblen function, which is explained in Diagonalization, and also the Extended Veblen function.

\begin{eqnarray*} \psi(\Omega^3 \varphi_5(0)) &=& \varphi_5(0) \\ \psi(\Omega^4) &=& \varphi_5(0) \\ \psi(\Omega^n) &=& \varphi_{1+n}(0) \\ \psi(\Omega^{\Gamma_0}) &=& \Gamma_0 \\ \psi(\Omega^\Omega) &=& \Gamma_0 \\ \psi(\Omega^\Omega+1) &=& \varepsilon_{\Gamma_0+1} \\ \psi(\Omega^\Omega+\Omega) &=& \zeta_{\Gamma_0+1} \\ \psi(\Omega^\Omega+\Omega^n) &=& \varphi_{1+n}(\Gamma_0+1) \\ \psi(\Omega^\Omega+\Omega^{\Gamma_1}) &=& \Gamma_1 \\ \psi(\Omega^\Omega2) &=& \Gamma_1 \\ \psi(\Omega^\Omega n) &=& \Gamma_{n-1} \\ \psi(\Omega^{\Omega+1}) &=& \varphi(1,1,0) \\ \psi(\Omega^{\Omega+1}2) &=& \varphi(1,1,1) \\ \psi(\Omega^{\Omega+2}) &=& \varphi(1,2,0) \\ \psi(\Omega^{\Omega2}) &=& \varphi(2,0,0) \\ \psi(\Omega^{\Omega2+1}) &=& \varphi(2,1,0) \\ \psi(\Omega^{\Omega3}) &=& \varphi(3,0,0) \\ \psi(\Omega^{\Omega n}) &=& \varphi(n,0,0) \\ \psi(\Omega^{\Omega^2}) &=& \varphi(1,0,0,0) \\ \psi(\Omega^{\Omega^3}) &=& \varphi(1,0,0,0,0) \end{eqnarray*}

Small Veblen ordinal

The small veblen ordinal is defined as \(\psi(\Omega^{\Omega^\omega}) = \varphi(1,\underbrace{0,\cdots,0}_\omega)\). But it's only small compared to...

Large Veblen ordinal

The large veblen ordinal is defined as \(\psi(\Omega^{\Omega^\Omega}) = \psi(\Omega^{\Omega^{\psi(\Omega^{\Omega^{\psi (\cdots)}})}}) = \varphi(1,\underbrace{0,\cdots,0}_{\varphi(1,\underbrace{0,\cdots,0}_{\varphi(1,\underbrace{0,\cdots,0}_{\varphi(\cdots)})})})\). But even that's nothing compared to...

Bachmann-Howard ordinal

\(BHO = \psi(\varepsilon_{\Omega+1}) = \psi(\underbrace{\Omega^{\Omega^{\cdots^\Omega}}}_\omega)\)