## FANDOM

10,818 Pages

Work in progress!!!

I was thinking of making some strong ordinal notation to use it to compare functions.I was gonna define it using simpe recursion,but then I thought of Taranovsky's notation which is very strong BECAUSE it's not defined recursively.So I decided to make a very strong using a different aproach than what I usually do.If you're about to comment,please bare in mind that this is just an early version and the reason why I am making this post is because I want to know if the way I described it is well-defined and if it's really what I'm trying to do.Also we should note that it is defined by recursion,so it's limit must be equal to or below $$\omega^{\text{CK}}_1$$ and it is probably weaker than Taranovsky's notation.

I wasn't sure how to define it but now I realized it doesn't take a lot of definitions.

Let $$\sigma(0,\alpha) = \alpha+1$$ and $$\sigma(\alpha)$$ will be short for $$\sigma(\alpha,0)$$.

Let $$\sigma(\alpha+1) = \text{Sup}\{\sigma(\alpha,\sigma(\alpha)),\sigma(\alpha,\sigma(\alpha,\sigma(\alpha))),\sigma(\alpha,\sigma(\alpha,\sigma(\alpha,\sigma(\alpha)))),........\}$$ (if $$\alpha$$ is not a supremum to $$L(\beta)$$)

Now let (if $$\alpha$$ is not a supremum to $$L(\gamma)$$) $$\sigma(\alpha +1,\beta +1) = \\ \text{Sup}\{\sigma(\alpha,\sigma(\alpha+1,\beta)+1),\sigma(\alpha,\sigma(\alpha,\sigma(\alpha+1,\beta)+1)),\sigma(\alpha, \sigma(\alpha,\sigma(\alpha,\sigma(\alpha+1,\beta)+1)))........\}$$.

If $$\alpha$$ is a limit ordinal (that is not supremum of $$L(\beta)$$) then $$\sigma(\alpha) = \text{Sup}\{\sigma(\gamma),\sigma(\gamma,\sigma(\gamma)),\sigma(\gamma,\sigma(\gamma,\sigma(\gamma)))........\}\forall \gamma (\gamma < \alpha)$$

And $$\sigma(\alpha,\beta+1) = \text{Sup}\{\sigma(\gamma,\sigma(\alpha,\beta)+1),\sigma(\gamma,\sigma(\gamma,\sigma(\alpha,\beta)+1)).........\}\forall \gamma (\gamma < \alpha)$$

And if $$\beta$$ is a limit ordinal then $$\sigma(\alpha,\beta) = \text{Sup}\{\sigma(\alpha,\gamma)\}\forall \gamma (\gamma < \beta)$$

(if $$\alpha$$ is not a supremum to $$L(\gamma)$$)

Let $$L(\alpha)$$ be the set of all ordinals $$\beta$$ such that $$\beta$$ is definable in $$\sigma()$$ using $$\sigma$$ no more than $$\alpha$$ times in the definition of $$\beta$$ and such that it uses no ordinal $$\gamma$$ if $$\gamma \notin L(\delta) \forall \delta (\delta +1 \geq \alpha)$$.Now let $$L(\alpha) = \text{Sup}\{L(\beta)\}\forall \beta (\beta < \alpha)$$ if $$\alpha$$ is a limit ordinal

Here comes the hard part.

If we do have an ordinal that is a supremum to $$L(\gamma)$$ as an initial input aka if $$\beta$$ is the last term in the fundamental sequence of $$\alpha$$ (aka the last part of the ordinal's normal form,such that it's broken down first when looking at the ordinal's fundamental sequence),which we denote as $$\alpha_\beta$$,and $$\beta \notin L(\gamma)$$ for some $$\gamma$$,then:

$$\forall \beta (\beta \notin L(\gamma)) : \sigma(\alpha_\beta +1) = \sigma(\alpha_\beta , \sigma(\alpha_\delta +1)) \forall \delta (\beta \in L(\epsilon) \land \delta \notin L(\epsilon) \Rightarrow \gamma < \epsilon \vdash \beta < \delta)$$

$$\forall \alpha_1,\alpha_2 (\alpha_1,\alpha_2 \in L(\beta)) :\alpha_1 < \alpha_2 \Leftrightarrow \exists \gamma (\gamma \in L(\delta)\forall \delta < \gamma) \land (\alpha_1 \leq \sigma(\gamma) < \alpha_2 \lor \alpha_1 < \sigma(\gamma) \leq \alpha_2)$$

It should be clear that $$\forall \alpha \forall \beta (\alpha < \beta) L(\alpha) \subset L(\beta)$$ and that this definition of $$\sigma(\alpha)$$ only works $$\forall \alpha (\alpha < \Omega)$$ (for now).

I will leave the analysis for another blog.