Tranfinite induction, 0.9...=1

Question

Can transfinite induction be used to demonstrate that 0.9...=1?

More generally, can it be used to prove limits of sequences?

Answer 1

No, transfinite induction is not designed for this job.

The job that transfinite induction is designed for is to prove universally quantified statements of the form

For all $x \in \omega$, $P(x)$ is true

where $\omega$ is a well-ordered set. In ordinary induction $\omega$ is the natural numbers, in transfinite induction $\omega$ can be a more general well-ordered set.

But the equation $.999... = 1$, which translates into the equation $$\lim_{n \to \infty} \sum_{i=1}^n \frac{9}{10^i} = 1 $$ is not a universally quantifed statement of the required form for transfinite induction. It is, of course, a universally quantified statement of a different sort:

For all $\epsilon > 0$ there exists $N \in \mathbb N$ such that if $n \in \mathbb N$ and if $n \ge N$ then $\left| 1 - \sum_{i=1}^n \frac{9}{10^i} \right| = 0$.

but there's not any way (that I can see) to twist this into an equivalent form that matches the form of transfinite induction.

Answer 2

Sequences and series such as the one we use for $0.999\ldots=1$ all have the same infinite length, so there isn't anywhere to "transfinitely induct" to.

Moreover, we already use ordinary finite induction to prove limits of sequences.

So I really don't see how transfinite induction applies here.