# Difference between revisions of "Madore's ψ function"

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

− | == | + | == Definition == |

Madore's \(\psi\) function is defined as follows: | 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 can not be obtained from the set \(\{0, 1, \omega, \Omega\}\) using addition, multiplication, exponentiation and all \(\psi(\eta)\) with \(\eta < \alpha\). | ||

− | |||

== Values == | == Values == |

## Revision as of 10:43, 17 May 2018

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

## Contents

## 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 can not be obtained from the set \(\{0, 1, \omega, \Omega\}\) using addition, multiplication, exponentiation and all \(\psi(\eta)\) 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

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