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

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 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

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