User blog:Simply Beautiful Art/Recursively closing OCF

For this blog post, I'm going to introduce my recursively closing OCF. The simple version is defined as follows:

$$\begin{align}&C(\alpha,0)=\{0,1,\omega,\Omega\}\\&C(\alpha,\varphi)=\left.\begin{cases}\gamma+\delta,\gamma\cdot\delta,\gamma^\delta,\psi(\eta)|\zeta\in\varphi\land\gamma,\delta\in C(\alpha,\zeta)\land\eta\in(C(\alpha,\zeta)\cap\alpha)\\\sup A|\xi\in\alpha\land\pi\in(C(\xi)\cap\varphi)\land\pi\in C(\alpha,\pi)\land A\subset C(\alpha,\pi)\end{cases}\!\!\!\!\!\right\}\\&C(\alpha)=\bigcup_{\varphi\in\rm Ord}C(\alpha,\varphi)\\&\psi(\alpha)=C(\alpha)\cap\Omega\end{align}$$

Let's start with $$C(0)$$. Note that there are no $$\xi\in0$$, so we don't need to worry about the second line in the recursive step.

Without it, $$C(0,\varphi)$$ is basically Madore's OCF. Since we can't construct any ordinals between $$\varepsilon_0$$ and $$\Omega$$, we find that $$C(0,\varphi)$$ doesn't change for $$\varphi\ge\varepsilon_0$$. Likewise, we easily find that $$\varepsilon_0=C(0)\cap\Omega$$, and thus, $$\psi(0)=\varepsilon_0$$.

Now comes the fun part. In particular, we may now use $$\sup A$$, which will increase the growth of $$C(1,\varphi)$$. For starters, $$A\subset C(\alpha,\pi\in\omega)$$ is a finite set, so taking $$\sup A$$ doesn't contribute anything.

Just like normal, $$\varepsilon_1=C(1,\omega)\cap\Omega$$, so again, nothing of particular interest here.

But now we reach $$C(1,\omega+1)$$. Note that we have the following conditions satisfied:

$$0\in1\land\omega\in(C(0)\cap(\omega+1))\land\omega\in C(1,\omega)\land\varepsilon_1\subset C(1,\omega)$$

Therefore, we may put $$\varepsilon_1=\sup\varepsilon_1\in C(1,\omega+1)$$, and from there we reach $$\varepsilon_2=C(1,\omega2)\cap\Omega$$.

By repeating this process, I believe we come to the conclusion that $$\psi(1)=\varepsilon_{\varepsilon_0}$$.

Furthermore, I believe that

$$\psi(2)=\varepsilon_{\varepsilon_{\varepsilon_0}}$$

$$\psi(3)=\varepsilon_{\varepsilon_{\varepsilon_{\varepsilon_0}}}$$

$$\vdots$$

$$\psi(\omega)=\zeta_0=\varepsilon_{\zeta_0}$$

$$\psi(\omega+1)=\varepsilon_{\zeta_0+\varepsilon_{\zeta_0}}$$

$$\psi(\omega+2)=\varepsilon_{\zeta_0+\varepsilon_{\zeta_0+\varepsilon_{\zeta_0}}}$$

etc.

This OCF does something peculiarly funky for large ordinals as well. For example, consider $$x=\sup\{0,\psi(0),\psi(\psi(0)),\dots\}$$. You will find that $$\psi(x)>x$$! Perhaps contrary to expectations, and furthermore you will find that $$\psi(\Omega)$$ exists and is likely larger than the Bachmann-Howard ordinal, though these are stories for a later blog.