This (not-so-)short text is about ordinals. It may contain some maths, but don't worry, it won't hurt. Nonetheless, I think the following requirements should be fullfilled: you may want to know a little about what is a set, how we (usually) write them and what is an "abstract" function. Additionnaly, if you know the difference between a set and a (proper) class, that's good.
Do not hesitate to pause-and-ponder if you ever feel the need. It is a rather complicated subject and, depending on your knowledge and your intuition, it may be hard to grasp. Also, as I said, this is quite a long text, so take your time, you're not in a hurry.
Please note that this text is broken down into color-coded boxes:
those with quartz border are the main text;
those with red border are pictures and visual examples;
those with green border are sidenotes;
those with light cream border delve into harder topics, if you understand, that's good, if not, do not bother.
Also, mathematical statement are displayed like this: $e^{i\pi} + 1 = 0$ and sometimes you'll see this symbol:, hover or touch it to uncover some additionnal informations!
Hopefully, this page is responsive, i.e. you can browse it with whatever device you want! If, however, you ran into any issue, please contact me at: contact@timotheehuneau.fr.
Finally, before we start, I would like to thank Grant Sanderson, who gave me the impetus to start this project, as a part of the Summer of Math Exposition v2 (SoME2)
Orders!
When counting, one gives an order to things. There is the first thing, the second thing, the third thing, ... and sometimes, when there is nothing at all, there is a total of zero things. As you notice, counting in intrinsically bound to the idea of order.
Not any orders, in fact! When dealing with finite sets, only those which are total orders allows to count. By this, we mean orders on a set such that every two elements $a$ and $b$ are in relation: either $a < b$, $a > b$ or $a = b$. This may seem like an unnecessary remark, but the study of partial orders is particularly interesting (and we won't deal with them here).
One order is good, but two is better! Consider the usual ones on $A := \{0,1,2\}$ and $B := \{3,4,5\}$? They look different, but are pretty similar: they both have elements, both have a first, second and third element and both doesn't have a fourth element. In fact, they are said to be isomorphic, i.e. up to a change of label, they are identical. I will often use the symbol $a \mapsto b$ to tell "let's rename $a$ into $b$" and the symbol $A \simeq B$ to mean "$A$ and $B$ are isomorphic".
The idea of isomorphism is key in mathematics. It always means that some objects (set, well-orders, ring ...) are almost indistinguishable: they share the same structure, the only difference is the name of their elements. Here, in our example, we have the isomorphism $0 \mapsto 3$, $1 \mapsto 4$ and $2 \mapsto 5$.
Let's ask ourselves what we like when we count. If we are to generalize counting, we need to get rid of a few properties and, most importantly, to keep some. We first notice $\Nd$ has a least element: $0$. Also, when we count (downward) in the usual set $\Nd$ of the natural integer, we notice something: every strictly decreasing sequence of integer eventually comes down to $0$. So, we may want to go with that: every order we'll consider will have these properties: (1) having a least element and (2) every of decreasing sequence eventually comes down to it; and we will call them well-orders.
An equivalent property to being well-ordered for an ordered set $S$ is that every subset $T$ of $S$ has a least element (i.e. an element less than any other element in $T$). . Still using the $\Nd$ example, try and find a set of integers with no least element. If you can't, that's cool, because such a thing does not exists.
We may now try to find a classification of these order, and for each class, a quintessential order. In fact, we'll do better: first, some fidgeting with well-orders and then a classification.
Numbers?
Let's look at a few particular well-ordered sets. The zeroth one is $A_0 := \emptyset = \{\}$, the first one $A_1 := \{\emptyset\}$, the second $A_2 := \{\emptyset, \{\emptyset\}\}$ and the $(n+1)$-th is the set containing all the previous ones, that is $A_{n+1} := \{A_1, A_2, \dots, A_n\} = A_n \cup \{A_n\}$. You first notice that if $n < m$, then $A_n \in A_m$.
Sidenote here: I will often use the notation $a := b$ to say: "let's define $a$ by it being equal to $b$", where $b$ is often complex-looking. This differs from the usual $a = b$ being only a statement: "I state that $a$ and $b$ are equal".
Let's do the converse: let's define an order on the family of the $A_n$ by saying that $x < y$ if and only if $x \in y$. This may seem overly complex (and complex it is), but bear with me and look at what we've just done. Starting with the empty set, we created a family of thing that behave just like the plain good old natural numbers. So, rather than calling this family $A_0$, $A_1$, $A_2$, $A_3$, ..., let's say that these are the natural numbers, the ones usually called $0$, $1$, $2$, $3$, ...
A closer look at the process we used to create our natural numbers shows that $n+1 := n \cup \{n\}$. That is, $n+1$ is the union of the set $n$ (so it contains every element that $n$ contains) and of the set containing only $n$. Imagine taking a big box an tossing in it the content of $n$ and then $n$ itself. That big box is now $n+1$.
So, why would we do that? Why would we call these things "numbers"? Because saying what a number precisely is is complicated (try it for yourself: how would you define a number?) and here, we just have a nice recipe! To be extra precise, we often call them "ordinal numbers", or simply "ordinals".
Here I come to warn you: from now on, things will get weirder and weirder. For instance, with this definition, a number is both an element and a set; $3$ is a set with three ordered elements, but also the greatest element of $4$.This may look like a weaknes or an overly-complex thing to do, but it will have its use.
Comparisons
Imagine you have two well-orders $V$ and $W$ (whatever this means, we don't have much examples yet). What can you say about them? First, they could be equal, or, in a broader sense, isomorphic. But what if they are not ?
They could somehow still be related. What if a part (aka a subset) of one well-order (say $W$) is isomorphic to the other (say $V$)? That would mean $W$ is, in a certain way, bigger than $V$, since $W$ contains $V$. As an example, $A := \{ 0, 2 \}$, with $A \subset 4$ is isomorphic to the set $2$ (we have $A \simeq 2$), and so $4$ is bigger than $2$.
We now have a new mean of comparisons, not between elements in a well-orders, but among well-orders themselves. You can think of $2$ and $3$ as element of a bigger well-order (say, $4$), but also as well-orders and can be ordered without having to refer to a bigger set. In some weird way, $2 \preceq 3$ is an absolute statement, rather than one depending on the well order $4$, or $5$, ...
Here, I used the symbol $\preceq$ rather than $<$ in order to compare $2$ and $3$ as well-ordered sets. It is only to show explicitly that this is not the same thing, $<$ is to compare elements inside a well-order and $\preceq$ to compare two given well-orders. Again, if I put $W := \{a,b,c\}$ and $V := \{d,e\}$, ordered by $a < b < c$ and $d < e$, I can compare $W$ and $V$ through $\preceq$, but not through $<$.
Also, there is a little ambiguity, one you may not even have noticed: when using the symbol $<$, because it is relative to a well-order, what I state may not always be clear. Consider the sets $S := \{a,b,c\}$ and $T := \{a,b\}$. I could well-order $S$ with $a < b < c$ and $T$ with $b < a$ and by doing so, knowing wether $a$ is greater than $b$ depends on the well-order we put ourselves in, but we always have $S \preceq T$.
Before we move on, I shall just state one very important remark, although you may not feel like it is now: given two well-orders, one is always smaller than the other, or, in other words, embedded in the other. Note that I just said "smaller" and not "strictly smaller", because we always have $W \preceq W$ for every well-order $W$.
To infinity!
We had some fun with well-orders, but still, the only one we know are finite well-orders. This means that for now, all we've done is a formal definition of what is a natural integer, but we did't go beyond infinity. Every set we know is finite.
And here comes the deus ex machina, the axiom saving us: the collection of all natural numbers is a set. Let's call it $\omega$. Why does its existence matters ? Because now, we have a new well-order! One that we did not know previously existed, an infinite one!
Before we go further, we should notice that what I called $\omega$ may look like what is usually called $\Nd$. Why naming it differently? Because I swiped a few things under the rug, some weird technicalities, like that we cannot be sure that $\omega$ and $\Nd$ are exaclty the same object, that $\Nd$ do is a set and some other odd stuff. You shouldn't care too much about this, because a teacher once told us that "it is a matter of faith".
What does $\omega$ looks like? It looks like a ladder, like all the finite numbers we just (re-)created, but an infinite one. It never stops an goes on and on and on forever! Nonetheless, it goes like that in only one direction, only upward; it is solidely rooted in the ground at the rung $0$.
Now that we have a new "number", we can start to use it, try to extend the usual operations ($+$, $\times, ...) to it and see what it yields us. But because we still don't know what an infinite ordinal looks like (we have only one example), we shall do that on well-orders: we don't really know what they are in general, but at least we have a definition for them.
Addition
When adding $n$ and $m$, you count "one, two, ..., $n$, $n+1$, ..., $n+m$". In other words, you concatenated the order on $n$ with the one on $m$. Let's do the same with $V$ and $W$, two well-orders. For the sake of simplicity, we will assume that $V \cap W = \emptyset$. If the intersection of $V$ and $W$ is not empty, I will take $V'$ isomorphic to $V$ such that $V' \cap W = \empyset$.
How do we do it? We simply consider the set $V \cup W$, well-ordered as follows: if you compare two elements from $V$ or from $W$, you keep the old order and every element from $V$ is smaller than every element from $W$. We will call this new well-order $V+W$.
Let's try out with $\omega$. Put $V := \{z\} \simeq 1$ and $W := \omega$. Consider the set $V + W := \{z,0,1,2, \dots \}$, ordered with $z < 0 < 1 < 2 < \dots$. Well, if we do the following renaming : $z \mapsto 0$, $0 \mapsto 1$, $1 \mapsto 2$, $2 \mapsto 3$,... we have what I called sooner an isomorphism between $\{z\}+\omega$ and $\omega$. Because $V$ is isomorphic to $1$, we may have stumbled upon the fact that $1+ \omega = \omega$. Strange.
What if we do the opposite? Put $V := \omega$ and $W := \{z\} \simeq 1$? We find the set $\omega + \{z\} := \{0,1,2, \dots, z\}$. It looks just like $\omega$, but with a greater element: $z$. Thus, we have that $\omega + 1 \simeq V + W$ is something entirely new.
We could continue: put $V := \omega$ and $W := \{0',1'\} \simeq 2$, ordered by $0' < 1'$. We have here $\omega + 2 \simeq V + W = \{0,1,2,\dots, 0',1'\}$, ordered as written. It just looks like $\omega + 1$ but with an added greater element. You may realize that it looks like $(\omega + 1) + 1$. We could continue adding one, going always higher and higher: $\omega + 3$, $\omega + 4$, ..., $\omega + n$, ... until what?
Multiplication
We need multiplication. Again, lets look at a small exemple. Put $n := 2$, $m := 3$ and let's look at $n \times m$. The usual way of doing that is with rectangle, of height $n$ and width $m$. To represent it using sets, we'll do the cartesian product, or set multiplication, between $n$ and $m$; that is the set of all possible pairs. So we have $(n\times m) := \{(0;0), (1;0), (0;1), (1;1), (0;2), (1;2)\}$.
Again, we'll now need an order on this set! The convention is to give it the anti-lexicographic order. This barbaric piece of jargon means that in order to compare two pairs, you first compare the right-most coordinates of the two pairs and then, if they are equal, the coordinates to the left.
To be clear, with that mean of ordering in mind, we have $(10;2) > (20;1)$ because $2 > 1$ (and so here, we do not care about the order between $10$ and $20$). We also have $(10;3) < (20;3)$, because $3 = 3$ and then, $10 < 20$. Also, the word lexicographic means "like in a dictionnary", because you compare the first letters of two words and then the second ones, and so on.
In our previous example, we thus have the ordering $(0;0) < (1;0) < (0;1) < (1;1) < (0;2) < (1;2)$, which, by a clever renaming, can be seen as the order on $6 = \{0,1,2,3,4,5\}$. We've just seen that $2 \times 3 = 6$. Hurray! We could redo this example on $(3 \times 2) := \{(0;0), (1;0), (2;0), (0;1), (1;1), (2;1)\}$, but we wont.
Now, let's see what happens with $V := \omega$ and $W := 2$. We do have $(V \times W) = \{(0;0), (1;0), (2;0), \dots, (0;1), (1;1), (2;1), \dots , \}$. And about the order on that set ? Well, we have $(0;0) < (1;0)< (2;0) < \dots < (0;1) < (1;1) < (2;1) < \dots$. Here, you may start to wonder : "Have I ever seen that before?". Well, yes, because it looks like a sum, but no, because I skipped it. Try and add $\omega$ to itself: you fisrt have a copy of $\omega$, and then a second one. It looks like $\omega \times 2$ and $\omega + \omega$ are equivalent.
With the same argument, we find out what $\omega \times 3$ and, more generally, $\omega \times n$, for any $n$ finite, are: it is the concatenation of $n$ infinite ladders. Although it is fun, doing all of this purely by hand is tedious; it is time to power up.
Exponentiation!
No surprise here : first, let's see how it works on finite well-orders (aka natural numbers). Put $n := 2$, $m := 3$ and consider $n^m$. If your intuition whispers "$8$", you're correct. But how to see that ? Well, we consider the set of all triplets (also called $3$-uple) made out of elements of $2$. In other words, that is the set of elements of the form $(a;b;c)$ where $a$, $b$ and $c$ are elements of the set $2$. I could list it, but I won't.
To be extra clear, an element from $2^3$ can be seen as a function from the set $3$ to the set $2$: the element $(0;1;0)$ is the map $0 \mapsto 0$, $1 \mapsto 1$ and $2 \mapsto 0$. To keep the little sanity I still have, let's use notation $a[n]$ to represent the $n$-th element from the tuple $a$, that is to say what one would normally write $a(n)$ if we consider $a$ to be a full-on function.
Again, we're looking for ordinals, so we need an order on that $2^3$ thing. Let's choose the same as before: the anti-lexicographic ordering. So $(0;0;0)$ is the first element, $(1;0;0)$ the second one, then comes $(0;1;0)$, $(1;1;0)$, ... and finally, $(1;1;1)$.
Let's generalize! So, we take $V := \omega$, $W := 2$ and we consider the set of all functions $2 \to \omega$. And we give it the anti-lexicographic ordering. But what does it even mean? Well, you already know: here, such a function $f$ is wholly defined by its value on $0$ and on $1$; so it looks like the $2$-tuple $(f[0];f[1])$. In fact, the set $\omega ^ 2$ is the set of all pairs of numbers.
Anti-lexicographic still means "first compare the right-most coordinates, and then the one to its left (and so on)", so you know how to deal with that. And now, we can count up to $\omega^3$ and even $\omega^n$, with $n$ any finite ordinal.
Let's try our new operation on $V := \omega$ and $W := \omega$. So we consider $\omega^\omega$ the set of $\omega$-tuple made out of elements of $\omega$, also called "integer sequences", which we'll see as functions $\omega \to \omega$. Although here comes a trap: we cannot use our magical anti-lexicographic ordering in order to have a well-order. Consider the sequence of sequences $u_0 := (1;1;1;1;\dots)$, $u_1 := (0;1;1;1;\dots)$, $u_2 := (0;0;1;1;\dots)$, ... It is a stricly decreasing sequence, the kind we forbid early on.
We need a way out. We'll find it by restricting our functions to those with finite support. Here, the support of a given sequence (aka function, aka $\omega$-tuple) $f : \omega \to \omega$ is the set of all elements $x$ such that $f[x] \neq 0$. More generally, we can always call $0$ the least element of any well-order (and it always exists) and apply the same definition.
This way, elements like $u_0$ and the like are not taken into account and there is no strictly decraesing sequence. This proof is out of reach, so take it as granted.
The set of finitely-supported functions from $\omega$ to $\omega$ is often denoted $\omega^{(\omega)}$. More generally, the set of finitely-supported functions $W \to V$, with $V$ and $W$ being two well-orders, is denoted $V^{(W)}$.
You may want a drawing of $\omega^\omega$, but sadly, I don't know of any way to represent it. $\omega$ is a ladder, $\omega\times 2$ is two ladders, and so on, $\omega^2$ is a ladder upon which a ladder is attached to every step, $\omega^3$ is a ladder upon which a a copy of $\omega^2$ is attached to every step, and so on. So $\omega^\omega$ is huge. The image found on ordinal number's Wikipedia page is not as clear as it seems.
Though, a way to understand it is to see that the set $\Pc$ of all prime numbers (indexed as $(p_i)_i$), ordered by the with usual order, is a well-order isomorphic to $\omega$. So, one can identify a finitely-supported function $f : \omega \to \omega$ with $\phi := \prod_i p_i^{f[i]}$. It does give an order on $\Nd \setminus \{0\}$, which goes like $1,2,4,8,\dots$ and all the powers of $2$ and then $3,6,12,24,\dots$ and all the $3$ times a power of $2$, then all the $9$ times a power of $2$, and so on, and then when you're done with the power of $2$ and $3$, you start again with everything you've done times $5$, and then times $25$, and so on and on until you reach $7$, and you continue until you've done that for every prime that exists. Voilà !
More?
So here we are, with our newly made ordinal numbers, going past infinity. But how do they truly work? Well, first thing first, finite ordinals work like any usual finite number: you can add them up, multiply, exponentiate and compute things like $2 \times 3^{3^3+1}+3+1$ without thinking about the sets of function with some properties, order, or anything. This means that our generalization work has turned out well.
Here, I could add an extra layer to the already existing never-ending ladder of definition by stating some arithmetic properties of ordinals. I won't, because it would be totally uninteresting to type and to read. Also, it is a very good exercise to try and find some rules, to take a break from reading and start exercising. If you want, there is a whole Wikipedia page on ordinal arithmetic. You can always come back later!
Ordinals?
Now that we know how to handle well-orders, we want an application. Let's work on our fundamental mission: classifying them. Consider the following definition: an ordinal is a set $\alpha$ consisting only of ordinals and such that if $\beta \in \alpha$, then $\beta \subset \alpha$.
First, let's break it down. We want every ordinals to be a set consisting of ordinals. This means that if $0$, $1$ and $2$ are ordinals, then $\{0,1,2\}$ is a good candidate for being an ordinal, and we usually call it $3$, but I've already said that. Moreover, because $2 \in 3$, we need $2 \subset 3$. Is it true? Well, $2 = \{0,1\}$ and $3 = \{0,1,2\}$, so yes, it is an ordinal.
Let's look at $\omega$. Is it an ordinal? Every of its element are ordinals and for any integer $n \in \omega$, we have every smaller integer (the elements of $n$) in $\omega$. So $\omega$ is an ordinal, by our definition.
This way of giving instructions on how to build ordinals (through nested sets) gives our work is a bit more of constructivism than just existentialism. We don't only assert that things exists, we display them. What good does it do to us? We can build things on that; because an ordinal is a set, we can consider making unions or intersections of ordinals.
But how to work out the ordinality of $\omega \times 2$? And $\omega^2$? And even $\omega^\omega$? We can't do it by hand. So, let's do some theorem-proving. First, we will need our theorem: every well-order is isomorphic to an ordinal.
Before we prove it, we shall understand how it could work: put $W$ a well-order. Consider the function $i$ associating every element $x \in W$ to the subset of $W$ consisting of all the element strictly smaller than $x$ (it is called the initial segment up to $x$). Imagine you got the set $W := \{a,b,c\}$, ordered by $a < b < c$. Then $i(a) = \emptyset$, $i(b) = \{a\}$ and $i(c) = \{a,b\}$.
You notice that all these sets can be well ordered and associated to ordinals, respectively $i(a)$ to $0$, $i(b)$ to $1$ and $i(c)$ to $2$. You also notice that every of these well-orders are smaller than the whole $W$ (we indeed have that $i(a) \preceq i(b) \preceq i(c) \preceq W$). Also, $W$ can be associated to the ordinal $3$, aka the set $\{0,1,2\}$, through the isomorphism $a \mapsto 0$, $b \mapsto 1$ and $c \mapsto 2$.
How do we generalize and prove our theorem? Rather than doing this for every ordinals, we ask ourselves: "what if it weren't true?". In that case, there is a smallest well-order such that it isn't true, call it $W$. We consider the function $i$ associating every element of $W$ with its initial segment. All these subsets are well-orders smaller than $W$.
By hypothesis, all well-orders smaller than $W$ are associated to an ordinal. So, all these subsets $S$ can be associated to an ordinal $o(S)$. This means that $o(i(x))$ is an ordinal, associated to an element $x$ of $W$.
But then, the set $\alpha$ of all these ordinals (as in: $\alpha := \{o(i(x)): x \in W\}$) is an ordinal, and it can be associated to $W$. That is absurd, because we jut assumed that $W$ does not correspond to any ordinal. This means that the premise was faulty: such a $W$ does not exist.
What we've just done is a proof ad absurdum: by assuming something to be true, we show an inconsistency, which means that what we've just assumed isn't correct.
This means that, indeed, every well-order can be associated (aka is isomorphic) to an ordinal. Why does it matter? Because now, we have our quintessential collection of well-orders.
Induction!
Because we wanted to learn how to count past infinity, we copied some properties of $\Nd$ and transposed them to our new ordinal numbers. It turned out well. But what do we often do with natural numbers? I give it to you: we prove things via (strong) induction.
But what is an induction? Imagine a ladder (an infinite one). You may know that you can climb onto the first rung and that whatever the rung you're at, you can step up to the following one. This shows that you can climb as far as you want. In mathematical terms, if you have an infinite amount of statements $P(n)$ (indexed over $\Nd$) such that for every $n$, if $P(n)$ holds, then $P(n+1)$ does too and such that $P(0)$ also holds, then all of the $P(n)$ are true.
In this kind of proof, the part that needs proving is either that $P(n)$ implies $P(n+1)$ or that $P(0)$ holds (or both). An usual example is about the sum of integers below $n$: put $P(n) : 1 + 2 + \dots + n = \frac{n(n+1)}{2}$. How do you prove all $P(n)$ simultaneously? By showing $P(0)$ holds, and it does: $0=0$; and that, assuming $P(n)$ holds, we show that $P(n+1)$ does too.
Sadly, this way of proving things does not work well with infinte ordinals. Precisely, you get stuck at the finite ordinals only: in order to prove some property $P(\omega)$, you would need the previous property to hold, but no such thing exists: there is no ordinal $\alpha$ such that $\alpha + 1 = \omega$.
The way out is to consider the strong induction: rather than needing only the previous statement to hold, you ask for all statement indexed by smaller numbers to hold. So, given a collection of statement $P(n)$ indexed over $\Nd$, such that if for any integer $m$: for every $n < m$, $P(n)$ holds, then $P(m)$ holds; then $P(k)$ holds for any $k$.
In other words, it just means that "if all statements indexed by ordinals strictly smaller than $m$ are true, then $P(m)$ is too" implies that all $P(n)$ are true.
Here is an example in the finite case: put $P(n):$ there exists at least $n$ prime number; we can build $p_0$, $p_1$, ..., $p_n$ the sequence of the $n+1$ first ones. To prove it, we assume every $P(n)$ to hold up to some $m$ and consider $q := 1 + \prod_{n < m} p_n$. Because no prime number up to the $m-1$-th divides $q$, we conclude that either $q$ is prime or it has a prime factor (less than itself) that is not in our sequence of $p_n$. So the set $S_m := \{1, \dots, q\} \setminus \{p_n : n < m\}$ contains at least one prime number, it thus has a least prime number; we call it $p_m$.
We just proved that all $P(n)$ for $n < m$ together implies $P(m)$. This means that for any $n$, $P(n)$ is true. Here, we needed all the information up to $m$ to prove it, just knowing $P(m-1)$ wasn't sufficient.
And this, we can transpose to ordinals: you can always consider the set $s_\alpha := \{ \beta : \beta < \alpha \}$ of ordinals smaller than $\alpha$. So we are ready to state our transfinite induction theorem: given a collection of statements indexed by ordinals such that for any ordinal $\alpha$: if every statements indexed by ordinals strictly smaller than $\alpha$ holds, then the $\alpha$-th statement holds; then every of these statements hold.
To prove this theorem, we go ad absurdum again: consider such a collection $P(\alpha)_\alpha$ of statements and assume that one fails to hold (say the $\alpha$-th). Then, one can consider the set of all $\beta$ such that $\beta < \alpha$ and $P(\beta)$ fails. It is a subset of $\alpha$, so it has a least element, say $\gamma$. But then, for all $\beta < \gamma$, $P(\beta)$ holds. So $P(\gamma)$ does too. That is absurd. So, such a $\gamma$ does not exists, nor does $\alpha$.
Classy!
Here, you may start to wonder what is the collection of all ordinals: we indexed collection of statements using it. It looks like a well-order and even, to some sense, like an ordinal We call it $ON$ (or $Ord$ sometimes). Alas ! It is not a set: assume it is a set, then it is a well-order, containing only ordinals and such that if $\alpha \in ON$, then $\alpha \subset ON$. Thus it would be an ordinal, and so $ON \in ON$. You may know that it is absurd, given that no set can contain itself.
So, $ON$ is no set, but it is a well-ordered class. This means that every subset (and in fact, every subclass) of $ON$ has a least element. This will lead to very, very bizarre behaviours, as we shall see.
Consider a non-empty subclass $S$ of $ON$. It contains an element, say $\alpha_0$. Either $\alpha_0 \cap S = \emptyset$ and so $\alpha_0 = \min S$ or there exists some $\alpha_1 \in \alpha_0 \cap S$. In fact, one can construct a striclty decreasing sequence $(\alpha_n)_n$ of ordinals by putting $\alpha_{n+1} \in \alpha_n \cap S$. It must terminate at some point, and it terminates at the $n$-th step if and only if $\alpha_n \cap S = \emptyset$, aka $\alpha_n = \min S$. This shows that any set (or even class) of ordinal has a least element.
Limits?
We considered strictly decreasing sequences of ordinals, but they're not very interesting: they are all - by definition - finite. Let's now consider a strictly increasing sequence of ordinals; let's call it $(\alpha_\iota)_{\iota \in \omega}$. My claim is that it is bounded - in other words, there is a smallest ordinal (among $ON$) that is bigger than any element $\alpha_\iota$.
Indeed, consider the set $A := \cup_\iota \alpha_\iota$. It is an ordinal, and clearly, because it contains all the $\alpha_\iota$, it is greater than any of them.
In fact, given any ordinal-indexed $\alpha$-sequence $(\beta_\iota)_{\iota \in \alpha}$ (aka a function with domain $\alpha$ for some ordinal $\alpha$), it is bounded: you can alway consider $B := \cup_{\iota \in \alpha} \beta_\iota$; it will always be greater than all the $\beta_\iota$
This leads to another definition: an ordinal $\alpha$ is said to be a limit ordinal if $\alpha = \{\beta \in ON: \beta < \alpha\}$. Conversely, an ordinal that is not a limit one is a successor, equivalently, successor ordinals are of the form $\alpha := \beta +1$ for some $\beta$.
Up!
In our last attempt to reach great heights and numbers, we counted up to $\omega^\omega$ and then stopped, because I could not draw it. But then, even if we do not (truely) understand it, we may consider $\omega^{\omega^\omega}$, and so on, always piling up more $\omega$. In fact, we could define the sequence defined by $\alpha_0 := 1$ and $\omega_{n+1} := \omega^{\alpha_n}$.
Because it is a strictly increasing sequence, it has a limit: let's call it $\varepsilon_0$. What property does it have? A very weird one: we have $\omega^{\varepsilon_0} = \varepsilon_0$. What this means is that the function $\alpha \mapsto \omega^\alpha$ has a fixed point.
In fact, the function $f : \alpha \mapsto \omega^\alpha$ has more than one fixed point. The first one is $\varepsilon_0$, the second one is $\varepsilon_1$, the $n$-th one is $\varepsilon_n$. We can even define the function $\varepsilon : \alpha \mapsto \varepsilon_\alpha$ the function yielding the $\alpha$-th fixed point of $f$, and this epsilon function has fixed points, too. We could continue like that, enumerating fixed point of functions, but even I start to not understand what's going on.
Cardinality?
Finally, if you know what "counntable" means, here is the last blow to your little brain: every ordinal I mentionned up to this point is countable! Obviously, $\omega$, $\omega \times 2$, $\omega^2$ are countable. It is a bit harder to see, but ordinal exponentiation keeps the countability in check, so $\omega^\omega$ is too, and so is $\omega^{\omega^\omega}$, and so on. Moreover, because the limits we considered are in fact countable union of countable sets, they also are countable. So are $\varepsilon_0$, $\varepsilon_{\varepsilon_0}$ and all the other too.
In fact, because any limit of countable ordinal sequence is a countable union of countable sets, any ordinal made through addition, multiplication, exponentiation and limit and starting with only countable ordinals is countable. But do not despair, as they are some uncountable ordinals, the least of them being often called $\omega_1$ or $\Omega$. But, let' not dive into that whole other topic!
and me: the remaining ones all are homemade, handmade, free-to-use, public domain SVG files.
The fonts are ...
Ovo, by Nicole Fally, used for all the content texts;
Alegreya Sans, by Juan Pablo del Peral, for all the titles.
Also ...
... this page uses MathJax to render beautiful mathematics.
... and let's have a few words about me: I'm a French math student who loves to do things (including math, baking, biking, ...) and whose main webpage is mostly blank (and in French).
