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:


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


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:




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.

  • $\begingroup$ 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. $\endgroup$ – Simply Beautiful Art Sep 21 '17 at 1:33
  • $\begingroup$ 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! $\endgroup$ – Deedlit Sep 21 '17 at 2:01
  • $\begingroup$ The last paragraph seems faulty. Shouldn't the operation $\alpha\mapsto\alpha\cdot2$ carry $\omega_1$ to $\omega_1\cdot\omega$? $\endgroup$ – Simply Beautiful Art Oct 23 '17 at 20:28
  • $\begingroup$ You are correct; the $\omega^{\omega_1 + 1}$ is correct, but not the $\omega_1^2$. Fixing. $\endgroup$ – Deedlit Oct 24 '17 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.