# Explanation: Building the Successors of Ordinal Numbers

(related to Part: Ordinal Numbers)

### Finite Ordinal Numbers

Given an ordinal $\alpha$, we can use the formula to build the successor of an ordinal number to create a fascinating chain of ordered ordinals.

We start with finite ordinals, i.e. the natural numbers. $$\begin{array}{rcl}0&:=&\{\emptyset\}\\ 1&:=&0\cup\{0\}\\ 2&:=&1\cup\{1\}\\ &\vdots&\\ \end{array}$$

This way, we will come up with the set of all finite ordinals (i.e. natural numbers), which we want to denote by $$\omega=\{0,1,2,\ldots\}.$$ ### Countably Infinite Ordinal Numbers

$\omega$ can be perceived as the supremum of all finite natural numbers $$\omega=\sup\{k\mid k\in\mathbb N\}.$$ $\omega$ has infinitely many elements, however only countably many infinitely many. Moreover, $\omega$ is the first limit ordinal, since it has no direct predecessor, although it is greater than all of its predecessors.

Now, we can continue to create the successor of $\omega$ by setting

$$\begin{array}{rcl}\omega+1&:=&\omega\cup\{\omega\}\\ \omega+2&:=&\omega+1\cup\{\omega+1\}\\ &\vdots&\\ \end{array}$$

We can continue this process to create again an infinite, but countable number of consecutive successors, until we reach the set $$\omega+\omega=2\omega:=\{0,1,2,3,4,\ldots,\omega,\omega+1,\omega+2,\omega+3,\omega+4,\ldots\},$$ which is the second limit ordinal without a direct predecessor.  But wait, we can continue the process of finding new limit ordinals $k\omega$, $k\in\mathbb \omega$ infinitely (but only countably many) times once again to get a new limit ordinal, for which there is no direct predecessor,

$$\omega\cdot\omega=\omega^2:=\sup\{k\omega\mid k\in\omega\}$$   We can continue the process of building consecutive ordinal numbers even further. Now, starting with $\omega^2+1:=\omega^2\cup\{\omega^2\},$ leading to the limit ordinals $\omega+k\omega$, for all $k\in\omega.$ It is now clear that the supremum of all theses sets is the set $2\omega^2.$ It should be also clear, how this process can be continued until we reach further limit ordinals $$\begin{array}{rcl}\omega\omega^2&=&\omega^3,\\ \omega\omega^3&=&\omega^4,\\\omega\omega^4&=&\omega^5,\\ &\vdots\end{array}.$$

### Uncountably Infinite Ordinal Numbers

Now, we have to hold on. The new limit ordinal would be $\omega ^\omega,$ which is defined as the following supremum:

$$\omega^\omega=\sup\{\omega^k\mid k\in\omega\}.$$

The special thing about this $\omega ^\omega$ is that it is the first limit ordinal which has uncountably many elements, although all of its predecessors have only countably many elements! The proof of this claim is similar to the proof that the set of real numbers is uncountable, using a diagonal argument.

The counting of ordinals we have done so far up to $\omega^\omega$ was a transfinite one. Each time we have created a new limit ordinal, we have conducted mentally a transition to a new infinite limit. It is, however hard to imagine and even to visualize the transfinite counting process up to $\omega^\omega.$ Note that above we only provided figures up to $\omega^2.$ The unimaginable transfinite counting process up to $\omega^\omega$ can be formulated as follows:

In each of the countably infinitely many steps, we have counted a countably infinite set countably infinitely many times.

But this process can be continued even further, ad infinitum. The next limit numbers are (these are just a few examples) \begin{align} \omega^\omega +1, \ldots, \omega^\omega +k,\ldots, \omega^\omega +k\omega,\ldots,\nonumber\\ \omega^\omega +k\omega^2,\ldots,k\omega^\omega,\ldots,\omega^{\omega +k},\ldots,\nonumber\\ \omega^{k\omega},\ldots,\omega^{\omega^{\omega}},\ldots,\omega^{\omega^{\omega^{\omega}}},\ldots,\omega^{\omega^{\omega^{\omega^{\omega}}}},\ldots\nonumber\\ \end{align}

Now, let us denote by $\omega^{\uparrow k}$ the ordinal number, in which $k$ times we reach a new exponent power, i.e. $\omega^\omega=\omega^{\uparrow 1},$ $\omega^{\omega^{\omega}}=\omega^{\uparrow 2},$ $\omega^{\omega^{\omega^{\omega}}}=\omega^{\uparrow 3},$ etc. These all are only countably many new limit numbers with uncountably many elements. Therefore, a new interesting question arises: Is there a transfinite number such that it is the limit number $$\Omega:=\sup\{\omega^{\uparrow k}\mid k\in\omega\}?$$

It turns out that there is no such ordinal number $\Omega,$ its existence would cause a contradiction, which became known as the Burali-Forti paradox.

Github: ### References

#### Bibliography

1. Hoffmann, Dirk W.: "Grenzen der Mathematik - Eine Reise durch die Kerngebiete der mathematischen Logik", Spektrum Akademischer Verlag, 2011