Why does $\epsilon_0 = \omega^{\epsilon_0}$

Question

I saw this on wikipedia and wasn't sure why. It seems obvious given the definition but how can we be sure?

Also, what would be $\epsilon_0 \bullet \epsilon_0$? Would it be $\epsilon_0$?

Thanks

Answer 1

This is an instance of a more general phenomenon.

Suppose I have a function on ordinals $f: ON\rightarrow ON$ such that

• $f(x)$ is nondecreasing, and

• $f$ is continuous (that is, $\sup \{f(x): x<\lambda\}=f(\lambda)$ for all limit ordinals $\lambda$).

Then $f$ has a fixed point, that is, an ordinal $\alpha$ such that $f(\alpha)=\alpha$. Actually, $f$ will have many fixed points. And if additionally $f$ sends countable ordinals to countable ordinals, $f$ will have many (in fact, uncountably many) countable fixed points.

Proof. Let $\beta_0=0, \beta_{i+1}=f(\beta_i)$; and let $\beta=\sup\{\beta_i: i\in\mathbb{N}\}$. Then by continuity, $$f(\beta)=\sup\{f(\beta_i): i\in\mathbb{N}\}=\sup\{\beta_{i+1}: i\in\mathbb{N}\}=\beta.$$

Now, it's not hard to show that the map $f(x)=\omega^x$ has the properties above; so, it has a countable fixed point. $\epsilon_0$ is defined as the least such fixed point - so $\omega^{\epsilon_0}=\epsilon_0$ by definition.

On the other hand, some natural functions on ordinals do not have fixed points - e.g. the map $x\mapsto x+1$. Why? Well, it's not continuous: $\omega+1\not=\sup\{n+1: n<\omega\}$. So we don't get fixed points "for free."

Finally, you asked about $\epsilon_0\cdot\epsilon_0$. HINT: show that $\epsilon_0+1\le\epsilon_0\cdot\epsilon_0$. Indeed, $\epsilon_0^2$ is much larger than $\epsilon_0$. This might be surprising - we tend to think of the map "$x\mapsto \omega^x$" as being much bigger than the map "$x\mapsto x^2$" on most inputs - but is due, again, to the failure of continuity.

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Note that I've skipped an important detail: showing that the map "$x\mapsto \omega^x$" is continuous! This is either immediate or somewhat tedious, depending on how you define ordinal exponentiation.

As another side note, if we look at the order topology on the class of ordinals, "continuous" does in fact mean "continuous" for nondecreasing functions.