The Natural numbers

What are the natural numbers? and yes, I do mean 1,2,3,\dots. What kind of question is it? Everyone knows what the natural numbers are, even my 6 years old brother, which is currently in the first grade, knows what the natural numbers are. However, how can you define them?

The natural numbers seem so fundemental and basic, how can define the most basic thing? Well, that’s what we are going to do today!

First of all, we need to ask ourselves – why do we use the natural numbers? What makes them so special?

For that question, there are two major answers:

  1. We use them to count. For example, 5 apples, 3 dogs, 10 pages… The natural numbers allows us to describe quantity.
  2. This is what we’re going to discuss about, considering what we’ve seen so far in the previous posts. They allow us to describe Order! For example, the first place in the race. The second student in the class. The tenth grade.

Considering what we dealt with so far, it’s pretty obvious that I want to talk about the second answer.

Everything we have done so far was related to Order. The main sets we delth with we’re ordinals and recall that they have some pretty amazing properties:

  • We can compare two ordinals
  • Their elements / transitive subsets are smaller ordinals than them.
  • From an existing ordinal \alpha , we can create a new ordinal: \alpha + 1=\alpha\cup\{\alpha\}.
  • Every ordinal is trully different from any other ordinal – Different ordinals can’t even be isomorphic to each other!

Now, those properties are actually properties we would like a natural number to fulfill. Of course we want to be able to compare two natural numbers, also, we would want to ‘add one’ to an existing natural number to get a new one.

This gives us a thick clue to what we are going to do – The most ‘natural’ way to define the naturals is by ordinals!

Ok, enough with the introduction, let’s dive into it:

Defining the naturals

What’s the smallest ordinal we know (recall that this is a valid question to ask, since we can compare any two ordinals) ? Well… The empty set. If you think about it, the empty set kind of acts like a number we already know:

\empty\cup A=A\ \ \ ,\ \ \ \empty\cap A=\empty

If we think about union as addition, and intersection as multiplication, it seems like the empty set is acting like zero.

So let’s make it official, we define:


Great! we have our first natural, although most people do not consider 0 as a natural, here, in set theory, it makes a lot of sense to define 0 as a natural.

What’s next? The only thing we are capale of doing now is ‘add one’ to an existing ordinal, This leads us to the definition of the number 1:


From now on, this is pretty clear how to define the next natural numbers:


And in general:


Do you remember the logic I’ve ‘invented’ In this post. I wrote 6 ordinals there:


So as it turns out – I was really on to something! Those are just the numbers 0,1,2,3,4,5. Ordinals allow us to define what we’ve thought until now to be the most basic and fundamental thing!

I want you to notice another cool result. By definition:


We can now use our knowledge on n as a natural number to get:

n+1=n\cup\{n\}=(n-1\cup\{n-1\})\cup n=n-1\cup\{n-1,n\}

Ok, I think we are on to something, let’s repeat this process:


We can keep doing this process until we get:


In my opinion, this is beautiful! Any natural number is actually a set made of the naturals before it! The order and rules behave so well here, it’s just makes perfect sense.

Expanding our view

After we’ve defined the natural numbers, I want to define \omega to be the set of all the naturals:


(Not a formal notation, but it gives better intuition…). What can we say about \omega? First of all, it is an ordinal itself:

Since \omega is a set of ordinals, It is indeed well ordered (and the minimal element of any subset A is \bigcap A).

We only need to prove that \omega is \in-transitive. Suppose that m\in n \in \omega. By the definition of \omega, we know that n is a natural ordinal, and we’ve just saw that an element of a natural ordinal is a natural ordinal itself, thus n\in\omega.


Since we defined the naturals formally, we can define size of sets – count element of a set. How? Well, we say that a set is of size n If there is a bijection between the set to n. That’s the perfect definition for size of a set, a bijection between two sets means that they are ‘the same’ sets – up to different names. We can now say that a set is finite if it is of size n for some n\in \omega.

What can we say now about \omega. It makes sense to think as \omega as the set of all the finite ordinals. However, is that true? are all the finite ordinals are natural ordinals? Let’s try to prove it:

First, by definition, all the elements of \omega are isomorphic to themselves, thus every element in \omega is finite.

On the other hand, suppose that there is some finite ordinal \alpha\notin \omega. Then we have two options: \alpha = \omega or \omega \in\alpha, thus, \omega\subseteq \alpha.

By our assumption, \alpha is finite, and there is a bijection from \alpha to some natural ordinal n. We can narrow down it’s domain to \omega (since \omega\subseteq \alpha) to get an injection from \omega to n.

I shall now consider the set:

D=\{n\in\omega|\text{ There exists an injection } \varphi:\omega\to n\}\sub\omega

And we have just proved that this is a non-empty set by finding such an injection. As a subset of \omega, which is well-ordered, we know it has a first element k\in D. Moreover, notice that \emptyset\notin D since there are no functions at all from \omega to the empty set. Therefore, we know that k=m+1=m\cup\{m\} for some m\in\omega.

What I’m going to do now, is to find an injection from \omega to m. By finding one, I will get a contradiction – according to our assumption, k is the first element in D, therefore, m can’t be in D as well (recall that m<k).

First, let’s denote the injection from \omega to k by \varphi. We know that k=m\cup\{m\}. There are two options now:

  1. m\notin\text{Im}(\varphi): Then \varphi is already an injection from \omega to m.
  2. m\in\text{Im}(\varphi). Then, there exists some x\in\omega such that \varphi(x)=m. notice that x is unique since \varphi is injective. Now we can define a new injection \Phi, where we just ‘skip’ m – for every y<x: \Phi(y)=\varphi(y) and for every y\geq x: \Phi(y)=\varphi(y+1) (This gives us: \Phi(x)=\varphi(x+1)\neq m).

From both of the options we were able to find an inection to m, and as I said before – this is a contradiction. Therefore, such an ordinal \alpha can’t exist.

Immediate corollaries

Equipped with the knowledge that \omega is the set of all finite ordinals while being an ordinal itself, we can conclude:

  • \omega is not a finite ordinal – Suppose it is, then \omega\in \omega and that’s a contradiction to \omega being an ordinal.
  • \omega is the first ordinal which is not finite – Notice that \omega=\omega|_\downarrow, and we’ve proved in this post that the generator of a lower set is the first element which is not in it.


So we now know what the natural numbers are, although we’ve known them since we were little kids, we have never defined them properly. Moreover, we’ve seen some interesting properties of the set of all the naturals – \omega. despite that it is an ordinal, it not finite…

In the next post I am going to start discussing about ordinals arithmetics – such as addition and multiplication.

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