I'm sure I don't quite understand this question, since my impression is that transfinite induction and ordinals are needed all the time, and often lengths are countable in practical cases.

Nevertheless, I can't resist mentioning one of my favorite results in symbolic dynamics, namely a theorem from "Structural aspects of tilings" by Ballier-Durand-Jeandel. It is one where I do not know a nice way to remove the use of countable ordinals. By basic topology, every $\mathbb{Z}^2$-subshift (closed shift-invariant subset of $A^{\mathbb{Z}^2}$ for $A$ finite set) is either finite, countably infinite or has the cardinality of the continuum. They prove an interesting result about the countable case.

Some definitions first. If $A$ is a finite set, we say $X \subset A^{\mathbb{Z}^2}$ is a **subshift of finite type** or **SFT** if there exists a clopen set $C \subset A^{\mathbb{Z}^2}$ such that
$$ X = \{x \in A^{\mathbb{Z}^2} \;|\; \forall \vec v \in \mathbb{Z}^2: \sigma^{\vec v}(x) \notin C \} $$
where $\sigma^{\vec v}(x)_{\vec u} = x_{\vec v + \vec u}$ is the shift action. (It's the same as saying it's defined by finitely many forbidden patterns.) An SFT is **countably infinite** if it has countably infinitely many configurations. We call elements $x \in X$ **configurations**. A configuration $x \in X$ is **singly-periodic** if the point stabilizer $\{\vec v \in \mathbb{Z}^2 \;|\; \sigma^{\vec v}(x) = x\} \leq \mathbb{Z}^2$ is nontrivial but not of finite index.

Let $X \subset A^{\mathbb{Z}^2}$ be a subshift of finite type which is countably infinite. Then $X$ contains a singly-periodic configuration.

The proof is quite interesting and I'll outline it as I remember it; there's lots of steps so this may be too quick to follow, but at least one sees from the summary that you really talk about ordinals and their successor relation on top of the Cantor-Bendixson argument. You can find the details in the Ballier-Durand-Jeandel paper.

First, you define the Cantor-Bendixson derivatives $X^{(\gamma)}$ for all ordinals $\gamma$ in the usual way. Since $X$ is countable and compact you have $X^{(\gamma)} = \emptyset$ for some $\gamma$ (or you find a perfect subset contradicting countability), and since the topology is second-countable, this happens for a countable ordinal $\gamma$.

Now let us analyze $\gamma$. It must be a successor ordinal, since otherwise $\emptyset$ is an intersection of nonempty sets contradicting compactness. So $\gamma = \beta + 1$ for some $\beta$. Since $X^{(\beta)}$ has empty Cantor-Bendixson derivative, it has to be finite. But it is classical that a finite subshift is a subshift of finite type, and it is also classical that SFTs have the "compactness" property that an SFT cannot be the intersection of a strictly descending chain of subshifts (it's a "finitely-generated group cannot be a strictly increasing union of subgroups" style argument). From this we deduce that also $\beta = \alpha + 1$ must be a successor ordinal. Clearly $X^{(\alpha)}$ is countably infinite.

Next, we analyze $X^{(\alpha)}$ (there we will find our singly-periodic configurations). It is known that a $\mathbb{Z}^2$-subshift is finite if and only if every configuration in it has finite-index stabilizer, so some configuration $X^{(\alpha)}$ has infinite-index stabilizer. Let $x$ be such a configuration, and let $V \leq \mathbb{Z}^2$ be its stabilizer.

It now suffices to show that we cannot have $|V| = 1$: Suppose for a contradiction that we had. Observe that all $\sigma^{\vec v}(x)$ are distinct, so any limit point of such shifts is in $X^{(\beta)}$. Then for every $\epsilon > 0$, every shift $\sigma^{\vec v}(x)$ with $|\vec v|$ large enough is at distance at most $\epsilon$ from the SFT $X^{(\beta)}$.

Since $X^{(\beta)}$ is finite, it has finite-index pointwise stabilizer, generated by some non-collinear $\vec u_1, \vec u_2 \in \mathbb{Z}^2$. Since all limit points of $x$ have to be in $X^{(\beta)}$, we must have $x_{\vec v} = x_{\vec v + \vec u_1} = x_{\vec v + \vec u_2}$ for all $|\vec v|$ large enough (again otherwise we have infinitely many distinct shifts $\sigma^{\vec v}(x)$ and we can extract a limit point which is not fixed by $\sigma^{\vec u_1}$ and $\sigma^{\vec u_2}$, thus is not in $X^{\beta}$).

We conclude that if $x$ does not have finite-index stabilizer, then $x$ is periodic apart from a finite "period-breaker" area, and since $X$ is an SFT, it is easy to see that it is uncountable, since we can glue these period-breakers all around $x$ and we will not see a problem with the defining clopen set $C$, as such a set will only look at the local picture. (I'm being very quick here, here you perhaps need to work a bit and draw a picture, or read the paper.)

So $x$ has infinite stabilizer which is not of finite index, i.e. $x$ is singly periodic. Square.

(There are some characterizations also for the other cardinalities. A $\mathbb{Z}^2$-subshift is finite if and only if every configuration has finite-index stabilizer. In another paper called "Structuring multi-dimensional subshifts", Ballier and Jeandel also give a characterization of uncountable SFTs, but I'll skip that.)

usein his work on convergence of trigonometric series? Maybe he didn't reallyneedordinals, maybe he could have gotten by without them if he had been more clever (and if he'd been aware that it was somehow desirable to avoid them), but that's not the same thing as saying they'reuseless, is it? And what's the sense of objecting specifically tocountableordinals? Since ordinals are made of smaller ordinals, how can you have uncountable ordinals if you don't have countable ones?? $\endgroup$ – bof Apr 5 at 3:57