Applying a function ordinal number of times

Question

I'm taking an introductory topology course and the idea of the ordinal numbers was very briefly introduced in class. I was wondering what determines whether function application can be done an ordinal number of times.

More formally, define the nesting function $N(f,x,a)$ -- where $f:X\rightarrow X$, $x\in X$ and $a$ is an ordinal number -- as follows:

• If $a=0$, then $N(f,x,a) = x$.
• If $a$ has an immediate predecessor, $N(f,x,a) = f(N(f,x,\mathrm{pred}(a)))$
• Otherwise (i.e., if $a$ is a limit ordinal) then how do you sensibly define $N(f,x,a)$?

What would be the necessary and sufficient conditions so that one can make a "sensible" definition for $N$? For example, I would naively expect that for $X=\{0,1\}$ and $f(0)=1, f(1)=0$, one might not be able to make such a definition (what would $N(f,0,\omega)$ be?). On the other hand, if $X$ is the set of all ordinals and $f$ is the usual successor function, I think that $N$ can probably be suitably defined.

Answer 1

I think the key property you're looking for is convergence in an appropriate sense: if $\lambda$ is a limit ordinal and $a_\alpha$ is a sequence in some space $X$ for $\alpha<\lambda$, say $(a_\alpha)_{\alpha<\lambda}\rightarrow b$ ($b\in X$) if for every open set $U\ni b$, there is some $\beta<\lambda$ such that $a_\alpha\in U$ for every $\alpha>\beta$. If such a $b$ exists and is unique (if our space is Hausdorff we don't need to specify uniqueness - exercise!), say the sequence $(a_\alpha)_{\alpha<\lambda}$ is convergent.

For example, the $\omega$-sequence $0, 1, 0, 1, 0, 1, . . .$ is not convergent (at least, assuming the usual topology on $\{0, 1\}$). Meanwhile, looking at the ordinals with their usual topology (the order topology), we see that the sequence $a_\alpha=\alpha$ ($\alpha<\lambda$) converges to $\lambda$; this meets the intuition that if I apply successor an ordinal number of times to $0$, I should get that ordinal back.

Some caution is needed, however: consider the maps $f:\alpha\mapsto \alpha$ (identity) and $g:\alpha\mapsto 2\alpha$ (doubling) on the ordinals. Then we have $f^\omega(0)=\omega=g^\omega(0)$, even though as $n\rightarrow\omega$ the "gap" between $f^n(0)$ and $g^n(0)$ gets larger and larger! So our intuitions aren't always great at figuring out how ordinally-iterated maps behave.

In fact, there's a more general theorem lurking here: if $h$ is an increasing continuous function from ordinals to ordinals, then there are many ordinals $\alpha$ such that $h(\alpha)=\alpha$. A keyword here is club (= closed unbounded set of ordinals), and as you go further you'll also wind up caring about stationary sets of ordinals ($A$ is stationary if $A\cap C\not=\emptyset$ for every club $C$; think of clubs as being "big," and stationary sets as being "not small").