Difference between revisions of "Madore's ψ function"

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Now we are introducing the Veblen function, which is explained in [[Diagonalization]].
 
Now we are introducing the Veblen function, which is explained in [[Diagonalization]].
  
\begin{eqnarray*} \psi(\Omega^3 \varphi_5(0)) &=& \varphi_5(0) \\ \psi(\Omega^4) &=& \varphi_5(0) \\ \psi(\Omega^n) &=& \varphi_{n+1}(0) \\ \psi(\Omega^{\Gamma_0}) &=& \Gamma_0 \\ \psi(\Omega^\Omega) &=& \Gamma_0 \psi(\Omega^\Omega+1) &=& \varepsilon_{\Gamma_0+1} \end{eqnarray*}
+
\begin{eqnarray*} \psi(\Omega^3 \varphi_5(0)) &=& \varphi_5(0) \\ \psi(\Omega^4) &=& \varphi_5(0) \\ \psi(\Omega^n) &=& \varphi_{n+1}(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_{n+1}(\Gamma_0+1) \end{eqnarray*}

Revision as of 18:44, 12 February 2017

Madore's \(\psi\) function is a function for expressing extremely large countable ordinals.

Defination

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

\(S=\{0,1,\omega,\Omega\}\) is our tool box.

\(C(0)=\)all of the things you can make from S and finite applications of \(+\times\text{^}\) in S.

\(C(n+1)=\) 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).

\(\psi(n)=\) the first countable ordinal not in \(C(n)\).

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.

\begin{eqnarray*} \psi(\Omega^3 \varphi_5(0)) &=& \varphi_5(0) \\ \psi(\Omega^4) &=& \varphi_5(0) \\ \psi(\Omega^n) &=& \varphi_{n+1}(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_{n+1}(\Gamma_0+1) \end{eqnarray*}