I Error(?) in proof that the rational numbers are denumerable

[Uncanny]

TL;DR Summary

I am working through J. H. Wiliamson’s Book on Lebesgue Integration on my own and have come across a proof I find rather “sketchy.”

If someone can straighten out my logic or concur with the presence of a mistake in the proof (even though the conclusion is correct, of course), I would be much obliged.

I’m looking at the proof of the corollary near the middle of the page (image of page attached below). I simply don’t find that the set, for instance, A_1 is finite, for if n=1, then wouldn’t it contain the infinite sequence of elements (writing only one memeber of each equivalence class of the rationals): 0/1, 1/-1, 1/-2, 1/-3,...,2/-3,...?

I understand the structure of the proof- it uses the theorem presented above it, which proves that the union of countably infinite sets is countably infinite. I just don’t find how the particular portion of the statement of the proof mentioned above is accurate. Did the author, perhaps, mean to write “positive rationals, R_0?” But, if so, then why the inclusion of the absolute value in the equation governing the property of inclusion for the indexed sets?

 

[PeterDonis]

if n=1, then wouldn’t it contain the infinite sequence of elements (writing only one memeber of each equivalence class of the rationals): 0/1, 1/-1, 1/-2, 1/-3,...,2/-3,...?

I suspect that the author is using a convention where the numerator $p$ of rationals carries the sign, so $q$ is always assumed to be positive. That would explain why he writes the definition for the set $A_n$ as $|p| + q \le n$, putting the absolute value only on $p$. If $q$ is always positive, then it should be obvious that the set $A_n$ is finite for any $n$.

 

[pbuk]

The author is assuming that $q > 0$ which as $\frac1{-2} = \frac{-1}2$ is fine.

 

[Uncanny]

Thank you, friends!