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Can we have a number system $S$ of cardinality continuum such that for every $x \in S$, there is a unique $y \in S$, such that for all $z>x$ in S, $x<y\le z$ holds?

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I have added the tag [ordinals] because I wrote some expository paragraph on ordinal arithmetics, and I think it is worth having this under that tag for future users. – Asaf Karagila Oct 11 '11 at 18:27
up vote 5 down vote accepted

Assuming the axiom of choice we can define a well ordering of the real numbers. That is there is some $\aleph_\alpha$ whose cardinality is continuum.

Consider now the least ordinal of this cardinality, denoted as $\omega_\alpha$, equipped with the usual ordinal arithmetics. It is closed under addition, multiplication and exponentiation (of ordinals), as well there is a natural ordering on the elements of $\omega_\alpha$.

Now consider $x\in\omega_\alpha$, then there is $y=x+1$ for which every $z\in\omega_\alpha$ has the property $x<z\rightarrow y\le z$.

Under the continuum hypothesis $\alpha=1$, that is $2^{\aleph_0}=\aleph_1$. On the other hand, without the axiom of choice it may be that we cannot well order the continuum, implying that there is no ordinal in bijection with the continuum.

It occurs to me that you may not be familiar with the ordinal arithmetics. I'll write a bit about that:

Ordinals can be seen as generalization of the natural numbers. It is a number system in which every element has a successor, and every non-empty set has a minimal element. However there are ordinals which are not successor ordinals. They are the limit of smaller ordinals, and so they are called limit ordinals.

I will not go into the actual definitions, instead I will use the fact that these are uniquely defined ordered sets, one of each distinct order type, which behave very nicely.

Let us skip to the part where the addition and multiplication are defined.


When adding two ordinals $\alpha+\beta$, we concatenate the ordinals and adding a copy of $\beta$ to be over $\alpha$. Formally we define this in recursion:

  1. $\alpha+0 = \alpha$;
  2. $\alpha+(\beta + 1) = (\alpha+\beta) + 1$;
  3. for limit $\lambda$, $\alpha+\lambda=\sup\{\alpha+\beta\mid\beta<\lambda\}$

In a nutshell, $\alpha+1$ is the order type of $\alpha$ with a new maximum element, and so we have that in many times the addition is not commutative. For example, if $\lambda$ is a limit ordinal then $\lambda+1$ is the successor of $\lambda$. However $1+\lambda$ is the order type of adding a new minimal element to $\lambda$, which yields the ordinal $\lambda$ again.


Where as addition was a form of concatenating two ordinals, in the multiplication of $\alpha\cdot\beta$ we replace each point in $\alpha$ with a copy of $\beta$. Formally defined, again, in recursion:

  1. $\alpha\cdot 0 = 0$;
  2. $\alpha\cdot (\beta+1) = \alpha\cdot\beta + \alpha$;
  3. for limit $\lambda$, $\alpha\cdot\lambda=\sup\{\alpha\cdot\beta\mid\beta<\lambda\}$.

Here too we have that multiplication is not commutative, and again we can consider some limit ordinal $\lambda$, then $\lambda\cdot 2=\lambda + \lambda$, while $2\cdot\lambda=\lambda$.

Exercise: Define ordinal exponentiation based on the above definitions. (The solution below with a spoiler tag, hover to see it.)

1. $\alpha^0 = 1$; 2. $\alpha^{\beta+1} = \alpha^\beta\cdot\alpha$; 3. for a limit ordinal $\lambda$, $\alpha^\lambda=\sup\{\alpha^\beta\mid\beta<\lambda\}$.
Note that under this definition $2^\omega=\omega$, where $\omega$ is the least infinite ordinal. So once again peculiar behavior is expected.

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You might consider adding the intuitive descriptions of $\alpha+\beta$ and $\alpha\cdot\beta$ directly in terms of the orders: $\alpha+\beta$ is obtained by appending $\beta$ to $\alpha$, and $\alpha\cdot\beta$ is obtained by replacing each element of $\beta$ with a copy of $\alpha$. – Brian M. Scott Oct 11 '11 at 21:39
Thanks, @Brian. – Asaf Karagila Oct 11 '11 at 21:44
I realize this is a very old question, but in case anyone else stumbles across it as I did: a much simpler (though less interesting) solution is the lexicographical order on $\mathbb{R} \times \mathbb{Z}$. – Robin Saunders Oct 19 '15 at 12:42

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