# A slower ordinal collapsing function and the Bachmann-Howard ordinal.

Madore's ordinal collapsing function is often defined by

$$\psi(\alpha)=\min\{\lambda:\lambda\notin C(\alpha)\}\\C(\alpha)=\bigcup_{n<\omega}C(\alpha)_n\\C(\alpha)_{n+1}=C(\alpha)_n\cup\{\gamma+\delta,\gamma\cdot\delta,\gamma^\delta,\psi(\zeta):\gamma,\delta,\zeta\in C(\alpha)_n,\zeta<\alpha\}\\C(\alpha)_0=\{0,1,\omega,\omega_1\}$$

It's supremum is known as the Bachmann-Howard ordinal:

$$\text{BHO}=\sup\{\psi(\omega_1),\psi(\omega_1^{\omega_1}),\psi(\omega_1^{\omega_1^{\omega_1}}),\dots\}$$

Consider two variable variants:

$$\psi_1(\alpha,\beta)=\min\{\lambda:\lambda\notin C_1(\alpha),\omega_\beta<\lambda<\omega_{\beta+1},\forall\eta(\eta<\lambda\implies\eta\cdot\eta<\lambda)\}\\C_1(\alpha)=\bigcup_{n<\omega}C_1(\alpha)_n\\C_1(\alpha)_{n+1}=C_1(\alpha)_n\cup\{\gamma+\delta,\gamma\cdot\delta,\psi_1(\zeta,\delta),\omega_\gamma:\gamma,\delta,\zeta\in C_1(\alpha)_n,\zeta<\alpha\}\\C_1(\alpha)_0=\{0,1,\omega\}$$

$\psi_2$ and $C_2$ are identical, with the exception that

$$\psi_2(\alpha,\beta)=\min\{\lambda:\lambda\notin C_2(\alpha),\omega_\beta<\lambda<\omega_{\beta+1},\forall \eta(\eta<\lambda\implies\eta+\eta<\lambda)\}\\C_2(\alpha)_{n+1}=C_2(\alpha)_n\cup\{\gamma+\delta,\psi_2(\zeta,\delta),\omega_\gamma:\gamma,\delta,\zeta\in C_2(\alpha)_n,\zeta<\alpha\}$$

Their countable supremums are given by

$$S_{1,2}=\sup\{\psi_{1,2}(\omega,0),\psi_{1,2}(\omega_\omega,0),\psi_{1,2}(\omega_{\omega_\omega},0),\dots\}$$

And my question is:

Are these supremums greater than the Bachmann-Howard ordinal? If so, for what ordinal $\beta$ does $\text{BHO}=\psi_{1,2}(0,\beta)$?

I attempted to evaluate some of these values, however, I was not able to draw any strong conclusions:

$$\beta\le\varepsilon_0\implies\psi_2(0,\beta)=\omega^{2+\beta}$$

$$\beta\le\varepsilon_1\implies\psi_2(0,\omega_1+\beta)=\varepsilon_0\omega^\beta$$

$$\gamma<\omega,~\beta\le\varepsilon_{1+\gamma}\implies\psi_2(0,\omega_1\cdot(1+\gamma)+\beta)=\varepsilon_\gamma\omega^\beta$$

But this gets very tedious very fast, and the expansions become increasingly convoluated...

I will start by analyzing $\psi_1$.

Some early values:

$\psi_1(\alpha,0) = \omega^{\omega^{1+\alpha}}$ for $\alpha \le \varepsilon_0$.

$\psi_1(\omega_1 + \alpha, 0) = \omega^{\omega^{\varepsilon_0+\alpha}}$ for $\alpha \le \varepsilon_1$.

$\psi_1(\omega_1(1+ \alpha) + \beta,0) = \omega^{\omega^{\varepsilon_\alpha+\beta}}$ for $\beta \le \varepsilon_{\alpha+1}, \alpha < \varphi(2,0)$.

$\psi_1(\omega_1^2,0) = \varphi(2,0)$.

$\psi_1(\omega_1^\alpha,0) = \varphi(\alpha,0)$ for $\alpha \le \Gamma_0$.

By comparison, $\psi(\Omega^\alpha) = \varphi(1+\alpha,0)$ using the traditional notation for the Bachmann-Howard ordinal, so the $\psi_1$ notation "catches up" to the traditional notation at $\psi_1(\omega_1^\omega,0) = \psi(\Omega^\omega)$.

Beyond that point, we will have equivalence between the two notations at all multiples of $\omega_1^\omega$. So the question is how far we can continue with powers of $\omega_1$. To answer that, we examine $\psi_1(\alpha, 1)$.

$\psi_1(0,1)$ is the limit of $\omega_1$ under the operation $\alpha \to \alpha^2$; this will be $\omega_1^\omega = \omega^{\omega^{\omega_1+1}}$. More generally, $\psi_1(\alpha,1) = \omega^{\omega^{\omega_1+1+\alpha}}$ for all $\alpha$ up until the fixed point of this map, which will be $\varepsilon_{\omega_1+1}$. Beyond that point, $\psi_1(\alpha,1)$ will stay fixed until we reach an ordinal that can be added to $C_1(\alpha)$ some other way, namely $\omega_2$. So $\psi_1(\omega_2,1) = \varepsilon_{\omega_1+1}$, and $\psi_1(\omega_2,0)$ is the Bachmann-Howard ordinal.

For $\psi_2$, we have:

$\psi_2(\alpha,0) = \omega^{2+\alpha}$ for $\alpha \le \varepsilon_0$.

$\psi_2(\omega_1 + \alpha, 0) = {\omega^{\varepsilon_0+\alpha}}$ for $\alpha \le \varepsilon_1$.

$\psi_2(\omega_1(1+ \alpha) + \beta,0) = {\omega^{\varepsilon_\alpha+\beta}}$ for $\beta \le \varepsilon_{\alpha+1}, \alpha < \varphi(2,0)$.

$\psi_2(\omega_1^2,0) = \varphi(2,0)$.

$\psi_2(\omega_1^\alpha,0) = \varphi(\alpha,0)$ for $\alpha \le \Gamma_0$.

So we have equivalence between $\psi_1$ and $\psi_2$ at all multiples of $\omega_1$. So again, we want to see how high a power of $\omega_1$ we can get.

$\psi_2(0,1)$ is the limit of $\omega_1$ under the operation $\alpha \to \alpha 2$; this will be $\omega_1 \omega = {\omega^{\omega_1+1}}$. More generally, $\psi_1(\alpha,1) = {\omega^{\omega_1+1+\alpha}}$ for all $\alpha$ up until the fixed point of this map, which again will be $\varepsilon_{\omega_1+1}$. So again we will have $\psi_2(\omega_2,1) = \varepsilon_{\omega_1 + 1}$, and $\psi_2(\omega_2,0)$ will again be the Bachmann-Howard ordinal.

• Huh, interesting. But I'm not sure about your second line. You say that $\psi_1(1,1)=\omega_1+\omega^\omega$, but it does not seem that$$\forall\eta(\eta<\omega_1+\omega^\omega\not\Rightarrow\eta\cdot\eta<\omega_1+\omega^\omega)$$Take $\eta=\omega_1$ for example. Commented Sep 21, 2017 at 1:33
• Whoops! I noted that clause early on and then forgot about it later when I was analyzing higher $\beta$. That makes a rather large difference! Commented Sep 21, 2017 at 2:01
• The last paragraph seems faulty. Shouldn't the operation $\alpha\mapsto\alpha\cdot2$ carry $\omega_1$ to $\omega_1\cdot\omega$? Commented Oct 23, 2017 at 20:28
• You are correct; the $\omega^{\omega_1 + 1}$ is correct, but not the $\omega_1^2$. Fixing. Commented Oct 24, 2017 at 22:42