# Properties of exponantiation of ordinals

In the last post, we worked pretty hard to define ordinals exponentiation, however, This hard work revealed some great properties:

1. $\alpha^0 = 1$
2. If $\beta$ is a successor then: $\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$
3. If $\beta$ is a limit ordinal then; $\alpha^{\beta}=\sup\{\alpha^{\gamma}:\gamma<\beta\}$

Those properties are great – the first two are pretty intuitive, and the third is also a pretty comfortable definition of $\alpha^{\beta}$. We can now prove properties on the $\alpha^{\beta}$ without ‘going back’ to the orginial definition, which is less intuitive…

## The properties you would expect

I think the most famous properties of the action of exponantiation are:

\alpha^{\beta+\gamma}=\alpha^{\beta}\cdot\alpha^{\gamma}
(\alpha^{\beta})^{\gamma}=\alpha^{\beta\gamma}

Luckily, those properties apply to ordinal exponentiation as well!

However before I’ll prove those, I want to show a way to compare ordinals first:

Just a little remark: I am going to use transfinite induction a lot!

## Comparing powers

If $\alpha>1$ and $\beta<\gamma$ then $\alpha^{\beta}<\alpha^{\gamma}$.

Note that it is important that $\alpha>1$. Indeed, if $\alpha=0$ the statement is not always true (why? How many function are there to 0? What happens if $\alpha=1$?).

Let’s prove it with transfinite induction on $\gamma$:

#### Base case

Suppose that $\gamma=0$ then there is no such $\beta$, so the statement is vacuous truth…

#### Successor case

Supposet that the statement is true for $\gamma$, we shall prove it is true for $\gamma+1$ as well:

\alpha^{\beta}\overset{\text{induction}}{<}\alpha^{\gamma}=\alpha^{\gamma}\cdot1\overset{\alpha>1}{<}\alpha^{\gamma}\cdot\alpha=\alpha^{\gamma+1}

#### Limit case

Suppose that the statement is for every $\delta<\gamma$ and $\gamma$ is a limit ordinal. Then:

\alpha^{\gamma}=\sup\{\alpha^{\delta}:\delta<\gamma\}=\bigcup_{\delta<\gamma}\alpha^{\delta}

However, since $\gamma$ is a limit ordinal and $\beta<\gamma$, we know that $\beta+1<\gamma$ as well. Therefore:

\alpha^{\beta}\overset{\text{induction}}{<}\alpha^{\beta+1}<\bigcup_{\delta<\gamma}\alpha^{\delta} =\alpha^\gamma \\ \Downarrow \\
\alpha^{\beta}<\alpha^{\gamma}

And that’s it.

Now we are ready to prove the major properties:

### The first property : $\alpha^{\beta+\gamma}=\alpha^{\beta}\cdot\alpha^{\gamma}$ $\alpha^{\beta+\gamma}=\alpha^{\beta}\cdot\alpha^{\gamma}$

We shall prove it by transfinite induction on $\gamma$:

#### Base case

If $\gamma=0$ then:

\alpha^{\beta+0}{=}\alpha^{\beta}=\alpha^{\beta}\cdot1\overset{\alpha^0=1}{=}\alpha^{\beta}\cdot\alpha^{0}

#### Successor case

Suppose that the statement is true for $\gamma$, we shall prove it’s true for $\gamma+1$:

\alpha^{\beta+(\gamma+1)}\overset{\text{associativity}}{=}\alpha^{(\beta+\gamma)+1}\overset{(a^{b+1}=a^b\cdot a)}{=}\alpha^{\beta+\gamma}\cdot\alpha\overset{\text{induction}}{=}(\alpha^{\beta}\cdot\alpha^{\gamma})\cdot\alpha
\overset{\text{associativity}}{=}\alpha^{\beta}\cdot(\alpha^{\gamma}\cdot\alpha)\overset{(a^{b+1}=a^b\cdot a)}{=}\alpha^{\beta}\cdot\alpha^{\gamma+1}

#### Limit case

Before I’ll prove the limit case, I want to prove two sub-statements first:

##### Sub-statement 1

Note that since $\gamma$ is a limit ordinal, then so as $\beta+\gamma$. Why?

Suppose that’s not the case, then there exists some $\delta$ such that $\delta+1=\beta+\gamma$. However:

\delta<\delta+1=\beta+\gamma\overset{\gamma\text{ is limit}}{=}\beta+\sup\{\epsilon:\epsilon<\gamma\}\overset{\text{'continuous' property}}{=}\sup\{\beta+\epsilon:\epsilon<\gamma\} \\ \Downarrow \\
\delta\in\sup\{\beta+\epsilon:\epsilon<\gamma\}=\bigcup_{\epsilon<\gamma}(\beta+\epsilon)

So there exists some $\epsilon_0<\gamma$ such that $\delta\in\beta+\epsilon_0$ which is the same as $\delta<\beta+\epsilon_0$. Therefore $\delta+1\leq \beta+\epsilon_0$.

But $\gamma$ is a limit ordinal, so $\epsilon_0+1<\gamma$. Thus:

\delta+1\leq \beta+\epsilon_0<\beta+(\epsilon_0+1) \\
\Downarrow\\
\delta+1\in \bigcup_{\epsilon<\gamma}(\beta+\epsilon)=\delta+1

And that’s a contradiction, since an ordinal is not a member of itself!

##### Sub-statement 2
\sup\{\alpha^{\beta+\delta}:\delta<\gamma\}=\sup\{\alpha^{\delta}:\delta<\beta+\gamma\}

If $\alpha=0$ or $\alpha=1$, then the statment is obviously true (why?). So let’s assume that $\alpha>1$.

Using the previous sub-statement, we know that $\beta+\gamma$ is a limit ordinal. Therefore:

\sup\{\alpha^{\delta}:\delta<\beta+\gamma\}=\alpha^{\beta+\gamma}

Now, suppose that $\delta<\gamma$, so $\beta+\delta<\beta+\gamma$. Hence:

\alpha^{\beta+\delta}<\alpha^{\beta+\gamma}=\sup\{\alpha^{\delta}:\delta<\beta+\gamma\}

And that’s true for every $\delta<\gamma$, so we can conclude that:

\sup\{\alpha^{\beta+\delta}:\delta<\gamma\}\leq\sup\{\alpha^{\delta}:\delta<\beta+\gamma\}

On the other hand, if $x<\beta+\gamma$ then:

x<\beta+\gamma\overset{\text{'continuous'}}{=}\sup\{\beta+\delta:\delta<\gamma\}=\bigcup_{\delta<\gamma}(\beta+\delta)

So there exists some $\delta_0<\gamma$ such that $x<\beta+\delta_0$. So $\alpha^x<\alpha^{\beta+\delta_0}$. And that’s true for every $x<\beta+\gamma$, therefore:

\sup\{\alpha^{\beta+\delta}:\delta<\gamma\}\geq\sup\{\alpha^{\delta}:\delta<\beta+\gamma\}

In total, we get that:

\sup\{\alpha^{\beta+\delta}:\delta<\gamma\}=\sup\{\alpha^{\delta}:\delta<\beta+\gamma\}

As we wanted.

##### Back to the proof

Suppose that $\gamma$ is a limit ordinal and the statement is true for every $\delta<\gamma$. Then:

\alpha^{\beta}\cdot\alpha^{\gamma}\overset{\gamma\text{ is limit ordinal}}{=}\alpha^{\beta}\cdot\sup\{\alpha^{\delta}:\delta<\gamma\}\overset{\text{'continuous'}}{=}\sup\{\alpha^{\beta}\cdot\alpha^{\delta}:\delta<\gamma\}
\overset{\text{induction}}{=}\sup\{\alpha^{\beta+\delta}:\delta<\gamma\}\overset{\text{sub-statement 2}}{=}\sup\{\alpha^{\delta}:\delta<\beta+\gamma\}\overset{\text{sub-statement 1}}{=}\alpha^{\beta+\gamma}

And that’s it!

### The second property : $(\alpha^{\beta})^{\gamma}=\alpha^{\beta\gamma}$ $(\alpha^{\beta})^{\gamma}=\alpha^{\beta\gamma}$

This case is actually very similar, again, we shall prove it with induction on $\gamma$:

#### Base case

I $\gamma=0$ then:

(\alpha^\beta)^{0}=1=\alpha^{0}=\alpha^{\beta\cdot0}

#### Successor case

Suppose that the statement is true for $\gamma$, we shall prove it’s true for $\gamma+1$:

(\alpha^{\beta})^{\gamma+1}=(\alpha^{\beta})^{\gamma}\cdot\alpha^{\beta}\overset{\text{induction}}{=}\alpha^{\beta\gamma}\cdot\alpha^\beta\overset{\text{property 1}}{=}\alpha^{\beta\gamma+\beta}\overset{\text{distributive}}{=}\alpha^{\beta(\gamma+1)}

#### Limit case

With almost identical proofs we can prove that if $\gamma$ is a limit ordinal, then $\beta\gamma$ is a limit ordinal as well. And that:

\sup\{\alpha^{\beta\delta}:\delta<\gamma\}=\sup\{\alpha^{\delta}:\delta<\beta\gamma\}

Try to prove it yourself! As I said, the proofs are almost identical to those of the sub-statements

Now, suppose that $\gamma$ is a limit ordinal and the statement is true for every $\delta<\gamma$. So:

(\alpha^{\beta})^{\gamma}\overset{\gamma\text{ is limit}}{=}\sup\{(\alpha^{\beta})^{\delta}:\delta<\gamma\}\overset{\text{induction}}{=}\sup\{\alpha^{\beta\delta}:\delta<\gamma\}\overset{\text{sub-statement}}{=}\sup\{\alpha^{\delta}:\delta<\beta\gamma\}
\overset{\beta\gamma\text{ is limit}}{=}\alpha^{\beta\gamma}

That’s it.

## Some cool tricks

Ok, now that we are equipped with lots of great properties of the ordinals operations, I would like to present some pretty cool results:

Consider the ordinal $\omega+\omega^{2}$. We don’t really have an intuition to what it is exactly… However, maybe there is a ‘nicer’ way to represent it. Let’s see what we can do:

\omega+\omega^2=\omega+\omega^{1+1}\overset{\alpha^{\beta+1}=\alpha^{\beta}\cdot\alpha}{=}\omega+\omega\cdot\omega\overset{\text{distributive}}{=}\omega(1+\omega)

However, recall that $1+\omega=\omega$, therefore, we are left with:

\omega+\omega^2=\omega(1+\omega)=\omega\cdot\omega\overset{\alpha^{\beta}\cdot\alpha}{=}\omega^2

Great, $\omega^2$ is much more intuitive than $\omega+\omega^2$. It seems like the $\omega$ there is only “disturbing” the representation of the ordinal…

Let’s see another example: What about the ordinal $(\omega+1)\cdot\omega^2$? Well, let’s find out:

(\omega+1)\cdot\omega^2\overset{\alpha^{\beta}\cdot\alpha}{=}(\omega+1)\cdot(\omega\cdot\omega)\overset{\text{distributive}}{=}((\omega+1)\cdot\omega)\cdot\omega

Now let’s focus on $((\omega+1)\cdot\omega)$. Since $\omega$ is a limit ordinal:

(\omega+1)\cdot\omega=(\omega+1)\cdot\sup\{n:n<\omega\}\overset{\text{'continuous'}}{=}\sup\{(\omega+1)\cdot n:n<\omega\}

Note that:

(\omega+1)\cdot n=\overbrace{(\omega+1)+(\omega+1)+\cdots+(\omega+1)}^{n\text{ times}}

This can be easilly proved by induction (try it!), therefore:

\sup\{(\omega+1)\cdot n:n<\omega\}=\sup\{\overbrace{(\omega+1)+(\omega+1)+\cdots+(\omega+1)}^{n\text{ times}}:n<\omega\}
\overset{\text{associative}}{=}\sup\{\omega+\overbrace{(1+\omega)+\cdots+(1+\omega)}^{n-1\text{ times}}+1:n<\omega\}
\overset{1+\omega=\omega}{=}\sup\{\omega+\overbrace{\omega+\cdots+\omega}^{n-1\text{ times}}+1:n<\omega\}
\sup\{\overbrace{\omega+\omega+\cdots+\omega}^{n\text{ times}}+1:n<\omega\}=\sup\{\omega\cdot n+1:n<\omega\}

However, it’s not hard to see that:

\sup\{\omega\cdot n+1:n<\omega\}=\sup\{\omega\cdot n:n<\omega\}

Therefore:

(\omega+1)\cdot\omega=\sup\{(\omega+1)\cdot n:n<\omega\}=\sup\{\omega\cdot n:n<\omega\}\overset{\text{'continuous'}}{=}\omega\cdot\sup\{ n:n<\omega\}
=\omega\cdot\omega=\omega^2

So, in total we get that:

(\omega+1)\cdot\omega^2=((\omega+1)\cdot\omega)\cdot\omega=\omega^{2}\cdot\omega=\omega^3

And that’s it!

## Summary

So we’ve met some really great properties of exponantiation of ordinals, and we’ve also seen how all the operations behave together. That’s going to be it for now…