Dubious proof in Halmos's book: Two similar ordinal numbers are always equal

Question

I'm studying ordinal numbers using Naive set theory of Halmos.

I think there was a small mistake in his proof of the statement "if two ordinal numbers are similar, then they are equal"

The proof goes as follow:

To prove this, suppose that $\alpha$ and $\beta$ are ordinal numbers and that $f$ is a similarity from $\alpha$ onto $\beta$; we shall show that $f(\theta) = \theta$ for each $\theta \in \alpha$. The proof is a straightforward transfinite induction.

Write $S$ = {$\theta \in \alpha: f(\theta) = \theta $}. For each $\theta \in \alpha$, the least element of $\alpha$ that does not belong to $s(\theta)$ is $\theta$ itself (here s$(\theta)$ is the initial segment of $\theta$ in $\alpha$). Since $f$ is a similarity, it follows that the least element of $\beta$ that does not belong to the image of $s(\theta)$ under $f$ is $f(\theta)$.

These assertions imply that if $s(\theta) \subset S$, then $f(\theta)$ and $\theta$ are ordinal numbers with the same initial segments, and hence $f(\theta) = \theta$. We have proved thus that $\theta \in S$ whenever $s(\theta) \subset S$. The principle of transfinite induction implies that $S = \alpha$, and from this it follows that $\alpha = \beta$

I think that there are 2 suspicious points in the proof of the author:

1. He does not prove the base case for $\theta = 0$. In fact, to prove the base case, we must prove that $f(0) = 0$, and I don't see any reason why should $f(0) = 0$. Therefore, the transfinite induction may be broken even for the base case.

2. There maybe a problem in the definition of the set $S$ by the author. Here he denotes $S$ = {$\theta \in \alpha: f(\theta) = \theta $}. I think that this definition implies implicitly that the range of $f$ is the set $\alpha$ which may not be true (because $f$ maps from $\alpha$ to $\beta$ which can be 2 completely different sets). So therefore, it is not sure that the set $S$ is not empty.

Could you please tell me if there is a small mistake in the proof or it's me who says stupid things ? Thank you very much for your help!

Answer 1

Q.1 :Halmos himself gives an explanation as to why base case is not needed in transfinite induction, on p.67 of Naive Set Theory:

The second difference between ordinary induction and transfinite induction (no starting element required for the latter) is more linguistic than conceptual. If $x_0$ is the smallest element of $X$, then $s(x_0)$ is empty, and, consequently, $s(x_0) \subset S$; the hypothesis of the principle of transfinite induction requires therefore that $x_0$ belong to $S$ [my emphasis].

(Halmos is pointing out that since $s(x_0)=\emptyset$ and we always have $\emptyset \subset S$, having proved the hypothesis we will automatically obtain $x_0 \in S$.)

Q.2:

(Actually given the above remark I don't think there's anything left for further explanation, but anyway:) that $f(\theta)=\theta$ is just what we want to assert for all $\theta \in \alpha$, hence we assume that it holds for some initial segment of $\alpha$, which is simply the way any transfinite induction proceeds.

Or, in particular: we know that for any non-trivial case (i.e. when $\alpha \ne 0$ and $\beta \ne 0$, otherwise the equality holds trivially), $0 \in \alpha$ and $0 \in \beta$, since $f$ is a similarity, least element must be mapped to least element, so indeed $0 \in S$.

But what's said above is actually a paraphrase (or application) of Halmos'

These assertions imply that if $s(\theta) \subset S$, then $f(\theta)$ and $\theta$ are ordinal numbers with the same initial segments, and hence $f(\theta) = \theta$.

on $x_0=0$.