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

  • Thread starter Uncanny
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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?

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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##.
 
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The author is assuming that ## q > 0 ## which as ## \frac1{-2} = \frac{-1}2 ## is fine.
 
Thank you, friends!
 

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