User blog:Nayuta Ito/A suggestion to the extension of Psi's Letter Notation

Now I'm going to extend Psi's Letter Notation to ε_0. The concept is a bit complicated (and unique), so please get ready.

Superordinals
This concept is for expressing fractional values used in the notation.

First, consider a pair of an ordinal $$\alpha$$ and a non-negative real number $$x$$: $$(\alpha,x)$$. Now consider the following equivalence relation:

$$(\alpha,x)\sim(\beta,y)\Leftrightarrow (\exists n\in\mathbb{N}, \alpha+n=\beta \and x=y+n) \or (\exists n\in\mathbb{N}, \alpha=\beta+n \and x+n=y)$$

For example:

$$(\omega,3)\sim(\omega+1,2)\sim(\omega+2,1)\sim(\omega+3,0)$$

$$(\omega^{\omega}+4,1.27)\sim(\omega^{\omega}+3,2.27)\sim(\omega^{\omega}+2,3.27)$$

Intuitively, this is the relationship that the "sum" of the ordinal and the real number is same (note that sometimes the sum is not defined - there is no such ordinal as $$\omega^{\omega}+5.27$$).

The construction of superordinals is easy in this relationship. Each equivalence class is called a superordinal. That's all.

For simplicity, I'm going to write $$\alpha+x$$ for the equivalent class of $$(\alpha,x)$$. This is consistent with ordinal addition because of the definition of the equivalence relation (note that $$\omega^{\omega}+5.27$$ is now well-defined - not an ordinal but a superordinal).

For future use, I will define these terms:

A superordinal $$(\alpha,x)$$ is normal if there is an ordinal $$\beta$$ such that $$(\alpha,x)\sim(\beta,0)$$. (In this case $$\alpha+x=\beta$$ holds under ordinal addition)

A superordinal is extended if it is not normal.

Hierarchy
To extend Psi's Letter Notation, I will use a hierarchy. The symbol for the hierarchy is $$\uparrow$$ because it has a lot to do with Arrow notation.

This hierarchy needs a number called "base". The base is 10 by default, but any integer bigger than 2 will work.

The hierarchy can take two forms:

$$\uparrow_{\alpha+x}(b) $$ ,where α+x is a superordinal (can be normal) and b is the base.

$$\uparrow_{\alpha}(x) $$ ,where α is a normal ordinal and x is a non-negative real number (can be base).

You can't use BOTH superordinal and non-integer at the same time. It must be either or none.

Restriction: For a normal ordinal $$\alpha$$ and base $$b$$, noth rule apply for $$\uparrow_{\alpha}(b) $$. The result MUST be the same whichever rule is used.