User blog:Simply Beautiful Art/Veblen style ordinal collapsing function

I came up with an ordinal collapsing function which behaves extremely similarly to that of the Veblen function. It works similarly to that of Feferman's $$\theta$$ but goes much further and is more general. I define it as follows, where Greek letters are ordinals or functions $${\rm Ord}^n\mapsto\rm Ord$$ and capital English letters are sets of ordinals.

Definition:
$${\rm cl}(A)=A\cup\{\sup(B)~|~B\subset A\}$$

$$C_0(\alpha,\xi,\pi)=\{0\}\cup\xi$$

$$C_{n+1}(\alpha,\xi,\pi)=\{\gamma+\delta,\Omega_\mu,\vartheta_\gamma(\eta)~|~\gamma,\delta,\mu,\eta\in C_n(\alpha,\xi,\pi)\land\pi\in\mu\land\eta\in\alpha\}$$

$$C(\alpha,\xi,\pi)=\bigcup_{n\in\omega}C_n(\alpha,\xi,\pi)$$

$$\vartheta_\pi(\alpha)=C(\alpha,\min\{\xi~|~\alpha\in{\rm cl}(C(\alpha,\xi,\pi))\},\pi)\cap\Omega_{\pi+1}$$

Explanation:
$$\vartheta_\pi(0)$$ is equal to the set of all ordinals less than $$\Omega_{\pi+1}$$ that can be formed by repeatedly including $$\gamma+\delta$$ and $$\Omega_\mu$$ into $$C_{n+1}(0,0,0)$$, where $$\gamma,\delta,\mu$$ are in $$C_n(0,0,0)$$ and $$\mu>\pi$$ and $$C_0(0,0,0)=\{0\}$$. Trivially this leaves us with $$\vartheta_\pi(0)=\{0\}=1$$.

We do the same for $$\vartheta_\pi(1)$$ except this time we may include $$\vartheta_\pi(0)$$, and by repeatedly adding finite numbers, we reach $$\vartheta_\pi(1)=\omega$$.

Generally, for small enough ordinals, we have $$\vartheta_\pi(\alpha)=\omega^\alpha=\varphi(\alpha)$$.

The first fixed point: $$\vartheta_\pi(\varepsilon_0)=\varepsilon_0=\omega^{\omega^{\omega^{\dots}}}$$

To go beyond this, my OCF then includes the least $$\xi$$ needed so that $$\alpha\in{\rm cl}(C(\alpha,\xi,\pi))$$ when all ordinals less than $$\xi$$ are included into $$C_0(\alpha,\xi,\pi)$$. For $$\varepsilon_0+1$$ to be in that set, we must have $$C_0(\varepsilon_0+1,\xi,\pi)=\{0,1,2,3,\dots,\varepsilon_0\}=\varepsilon_0+1$$ and thus we have $$\xi=\varepsilon_0+1$$, and furthermore $$\vartheta_\pi(\varepsilon_0+1)=\omega^{\varepsilon_0+1}$$.

So on and so forth my function continues until we reach $$\alpha=\Omega_1$$, which is the first ordinal where the subscript makes a difference. Since $$\pi=0$$ is the only $$\pi$$ such that $$\Omega_\mu$$ may be included for $$\mu=1$$, it follows that $$\Omega_1\in{\rm cl}(C(\Omega_1,0,\pi))$$ for $$\pi=0$$. Otherwise we need to include all ordinals less than $$\Omega_1$$ in the same fashion that we reached $$\varepsilon_0+1$$.

That being said, $$\vartheta_{\pi>0}(\Omega_1)=\omega^{\Omega_1}=\Omega_1$$, while $$\vartheta_0(\Omega_1)$$ is equal to the set of all ordinals less than $$\Omega_1$$ and in $$C$$. This happens to be all the ordinals less than $$\sup\{\vartheta_0(0),\vartheta_0(\vartheta_0(0)),\vartheta_0(\vartheta_0(\vartheta_0(0))),\dots\}=\varepsilon_0$$, and so $$\vartheta_0(\Omega_1)=\varepsilon_0$$.

Next we have $$\vartheta_0(\Omega_1+1)$$. Note that since $$\Omega_1\in C_n(\Omega_1+1,0,0)\cap(\Omega_1+1)$$, then $$\varepsilon_0=\vartheta_0(\Omega_1)\in C_{n+1}(\Omega_1)$$. From there we may add one to this ordinal and then repeatedly apply $$\vartheta_0(\vartheta_0(\dots))$$ to it. This gives us $$\vartheta_0(\Omega_1+1)=\sup\{\vartheta_0(\vartheta_0(\Omega_1)+1),\vartheta_0(\vartheta_0(\vartheta_0(\Omega_1)+1)),\vartheta_0(\vartheta_0(\vartheta_0(\vartheta_0(\Omega_1)+1))),\dots\}=\varepsilon_1$$.

Continuing the pattern, we get $$\vartheta_0(\Omega_1+2)=\varepsilon_2,\vartheta_0(\Omega_1+3)=\varepsilon_3,\dots$$. Generally $$\vartheta_0(\Omega_1+\alpha)=\varepsilon_\alpha=\varphi(1,\alpha)$$.

All the way until we hit $$\vartheta_0(\Omega_1+\zeta_0+1)$$. It is here where we must include $$\zeta_0\in C_0(\zeta_0+1,\xi,0)$$ to progress further (by taking $$\xi=\zeta_0+1$$) and after doing so, we should find $$\vartheta_0(\Omega_1+\zeta_0+1)=\varepsilon_{\zeta_0+1}$$.

And so, for all $$\alpha<\Omega_1$$, we have $$\vartheta_0(\Omega_1+\alpha)=\varepsilon_\alpha$$, increasing $$\xi$$ as needed.

And then we hit $$\vartheta_0(\Omega_12)$$, whereupon we reset $$\xi$$ back to zero. We then have $$\vartheta_0(\Omega_12)=\sup\{\vartheta_0(\Omega_1),\vartheta_0(\Omega_1+\vartheta_0(\Omega_1)),\vartheta_0(\Omega_1+\vartheta_0(\Omega_1+\vartheta_0(\Omega_1))),\dots\}=\varepsilon_{\varepsilon_{\varepsilon_{\ddots}}}=\zeta_0$$.

Perhaps you are seeing the pattern? We end up having the nice relationship to the Veblen function:

$$\vartheta_0(\Omega_1^n\gamma_n+\Omega_1^{n-1}\gamma_{n-1}+\dots+\Omega_1\gamma_1+\gamma_0)=\varphi(\gamma_n,\gamma_{n-1},\dots,\gamma_1,\gamma_0)$$

For $$\gamma_k<\Omega_1$$.

And this pattern ought to continue higher and higher, up to the Veblen function with transfinitely many arguments and beyond that too.

Another note here is that we cannot construct $$\Omega_1\omega$$ using repeated addition, but this is why we have $$\vartheta_1(\Omega_1+1)=\omega^{\Omega_1+1}=\Omega_1\omega$$. In general, each $$\vartheta_\pi$$ shares the same relationship to the Veblen function for arguments less than $$\Omega_{\pi+1}$$.