## Hmm, Ordinal Numbers...

In the last post, Why Ordinal Numbers, I had intuitively guessed that I would be left with a perfect set after enough application of derived sets on $${S}$$. By the end, I gave an example of a set that had a nonempty “infinite” derived set, $${S^{(\infty)},}$$ that was not perfect but a singleton. Rather than disprove my hypothesis, this shows that $${\infty}$$, as we'd used it, was not enough but $${\left(S^{(\infty)}\right)' = \emptyset}$$: the empty set is perfect by definition. This leads to very novel ideas and the answer to our problem will lie in a completely unexpected area. The following definitions follow naturally from our previous discussions.:

Definition 1. For any set $${S \subset \mathbb R}$$, let $${S^{(\infty + n)} = \left(S^{(\infty)}\right)^{(n)}}$$ for $${n\in \mathbb N}$$, \begin{aligned} S^{(\infty \cdot 2)} &= S^{(\infty+\infty)} = \left(S^{(\infty)}\right)^{(\infty)} \end{aligned}

I could easily define $${S^{(\infty \cdot n)}, ~ S^{(\infty \cdot \infty)}, ~ S^{(n + \infty)}, }$$ and even $${S^{(n \cdot \infty)}}$$. This suggests a sort of arithmetic with these “numbers”, including $${\infty}$$. The arithmetic has some interesting properties; for instance, the operations are not commutative: $${n + \infty = \infty = n \cdot \infty}$$ when $${n \in {\mathbb N}}$$.

To avoid confusion and the repeated use of “infinite” with quotations, I will proceed by using the symbol $${\omega}$$ instead of $${\infty}$$. This highlights that: 1) $${\omega}$$ is a number that can be treated with arithmetic as any other; 2) $${\omega}$$ has a successor, namely $${\omega + 1}$$. Let me illustrate further on why the second point is important. Using $${\infty}$$ in my first definition was not arbitrary. It was based on my (false) assumption that finite derived sets and the intersection of all finite derived set were the only interesting derived sets. I did not expect to learn anything useful past $${S^{(\infty)}}$$ hence my use of the symbol. There is nothing wrong in continuing to use it but I (and Cantor over a century ago) chose to change it in order to separate the two notions.

Now this is where Cantor's genius truly shines. He recognized that the numbers we defined, $${1, 2, 3, \ldots,\omega, \ldots}$$, represented something special: they represented order. Even the familiar $${1, 2, 3, \ldots}$$ were not acting as natural numbers or signifiers of size (cardinal numbers) but as signifiers of order, specifically well-orders of countable sets. Let's look at $${\omega}$$ for instance. I can see that it represents the natural order of natural numbers since that's the order I used to define it. But that's not all of it. I could have started with $${S^{(3)} = S'}$$ and, in this case, $${\omega}$$ would represent the natural order of all integers greater than 2. But aren't these two orderings essentially the same? Lastly, I was never required to use just natural numbers. So $${\omega}$$ could represent the well-order of any countable set where every element has a successor and only one element is not a successor of any number. What about $${\omega \cdot 2}$$? That would be the well-order where every element has a successor and only two elements are not successors of any number. Two such well-orders in natural numbers would be: $${1 \prec 3 \prec 5 \prec \cdots \prec 2 \prec 4 \prec \cdots}$$ or $${3 \prec 6 \prec 9 \prec \cdots \prec 1 \prec 2 \prec 4 \prec \cdots}$$. Lastly, $${\omega^\omega}$$ could represent my counterexample set $${S}$$ ordered by $${>}$$.

All in all, whenever I used a number in $${S^{(n)}}$$, what really mattered was the order $${n}$$ represents. In retrospect, my initial definition for $${S^{(\infty)}}$$ was very naive for there are many more orders available. So what does it mean for orders “to be essentially the same” or “represented by the same number”? The following definition will help:

Definition 2. Two (linearly or partially) ordered sets $${A}$$ and $${B}$$ are said to be order isomorphic if there exists a bijection $${f:A \rightarrow B}$$ such that for every $${x,y \in A}$$, $x \prec y \iff f(x) \prec f(y)$ $${f}$$ is also known as an order-preserving function.

Note: While this investigation started with a problem in topology, we are now about to momentarily leave that topic and delve into set-theory, naive for now.

Naive Definition. An ordinal number is the unique mathematical object that represents order-isomorphic well-orderings of sets. A countable/uncountable ordinal is an ordinal number that represents order-isomorphic well-orderings of countable/uncountable sets, respectively.

This definition is anything but rigorous for current math standards but it is basically what Cantor started with. It allowed him to define objects like “a set of all countable ordinal numbers.” However, since ordinal numbers are still in use in modern mathematics, there must be a rigorous definition that uses axiomatic set theory. I suppose one could define countable ordinal numbers in terms of equivalence classes of well-orderings of $${{\mathbb N}}$$ but I don't know if that's enough to account for the existence of an uncountable ordinal. More on this in the next post; for now, the naive approach will do.

Let's return to $${\omega}$$ and $${\omega \cdot 2}$$ and the well-orders they represent: \begin{aligned} \omega &: \qquad 1 \prec 2 \prec 3 \prec \cdots \\ \omega \cdot 2 &: \qquad 1 \prec 3 \prec 5 \prec \cdots \prec 2 \prec 4 \prec \cdots \end{aligned}

It seems like $${\omega \cdot 2}$$ contains $${\omega}$$ -- whatever “contains” means. Cantor gave the following definition:

Definition 4. A well-order on set $${A}$$ is contained in another well-order on set $${B}$$ if there exists an order-preserving injective function $${f:A \rightarrow B}$$ but no such function $${g:B \rightarrow A}$$. Furthermore, if $${\alpha}$$ and $${\beta}$$ represented the well-ordered sets $${A}$$ and $${B}$$, then we say $${\alpha \in \beta}$$.

Note: Cantor had already developed the idea that any mathematical object can be a set and $${\in}$$ is just a relation on sets. He thus defined the relation $${\in}$$ on the newly defined ordinals. Axiomatic set theorist would later take his ideas and build a proper foundation: everything was made a set and axioms dictated how the relation $${\in}$$ could be properly defined on sets.

The following is an illustration using finite sets. If I have sets $${A = {3,7,8}}$$ and $${B = {a,b}}$$, then clearly the well-order $${a \prec b}$$ is contained in $${7 \prec 3 \prec 8}$$. Since $${2}$$ and $${3}$$, as ordinal numbers, represent the two well-orders, respectively, then I can write $${2 \in 3}$$. In fact, $${1 \in 3}$$ and $${0 \in 3}$$, where $${0}$$ would represent the well-order on an empty set.

Cantor then proved the important facts: ordinal numbers are well-ordered by $${\in}$$; ordinals were uniquely determined by the ordinals contained in them. Thus one could say that the $${3 = {0,1,2}}$$ and, at the same time, this ordinal represented its own well-ordering: $${0 \in 1 \in 2}$$. He went further and define the set $${\omega_1 = {\text{all countable ordinals}}}$$. This is a well-defined set since countable ordinals represent well-orderings of $${{\mathbb N}}$$. Furthermore, since $${\omega_1}$$ is well-ordered by $${\in}$$ then it must be represented by an ordinal number I'll, incidentally, call $${\omega_1}$$.

Definition 5. A nonzero ordinal number is called a limit ordinal if it is not a direct successor of any ordinal number. Otherwise, the ordinal is a successor ordinal.

Finally, I will briefly return to the original problem: can we make any set $${S}$$ perfect if we derive it enough times? This is now evidently a misguided question since "enough times" assumed magnitude mattered but what we should care about is order. The counterexample at the end of the last post showed that one could derive a set countably infinite times and get different results if the order is changed. Thus $${S^{(\omega)}}$$ was a singleton but $${S^{(\omega+1)}}$$ was a perfect empty set. So the problem should be rephrased to "can we always make $${S}$$ perfect if we derive it in the right order?"

These new definitions of derived sets using ordinals will prepare you for the last step:

Definition 6. For any set $${S \subset \mathbb R}$$ and ordinal $${\alpha}$$, let $${S^{(0)}=S}$$, $${S^{(\alpha+1)} = \left(S^{(\alpha)}\right)',}$$ and $S^{(\alpha)} = \bigcap_{\beta \in \alpha} S^{(\beta)} \quad \text{ if } \alpha \text{ is a limit ordinal}$

To be continued...