When I read the anecdote about how Peano was inspired to give an explicit construction because Cantor’s original proof didn’t, I decided to not bother looking up info on Cantor’s proof.
ETA: Re-reading my own link, I find that I misremembered what inspired Peano to search for space-filling curves—he was looking specifically for continuous mappings, and my link was more specific than necessary, just as you pointed out in your first reply to me in this thread.
Huh, did Cantor do it by well-ordering or something? I wouldn’t know. In any case it’s pretty easy to explicitly put 2^N x 2^N in bijection with 2^N, because the former is just 2^(2N), and 2N is in bijection with N. What this cashes out to is, if you have two elements of 2^N, and want to make one that encodes them both, you just interleave them. Note also this works for any 2^S when you have an explicit bijection between S and 2S. If you want it for R all you need is an explicit bijection between R and 2^N. It’s a more general consequence of well-ordering that S x S is of the same cardinality as S for any infinite S, and that is necessarily nonconstructive, but for “practical” infinite sets (by which I basically mean ℶ_n for n finite) many bijections can be made explicitly that would in general require choice.
When I read the anecdote about how Peano was inspired to give an explicit construction because Cantor’s original proof didn’t, I decided to not bother looking up info on Cantor’s proof.
ETA: Re-reading my own link, I find that I misremembered what inspired Peano to search for space-filling curves—he was looking specifically for continuous mappings, and my link was more specific than necessary, just as you pointed out in your first reply to me in this thread.
Huh, did Cantor do it by well-ordering or something? I wouldn’t know. In any case it’s pretty easy to explicitly put 2^N x 2^N in bijection with 2^N, because the former is just 2^(2N), and 2N is in bijection with N. What this cashes out to is, if you have two elements of 2^N, and want to make one that encodes them both, you just interleave them. Note also this works for any 2^S when you have an explicit bijection between S and 2S. If you want it for R all you need is an explicit bijection between R and 2^N. It’s a more general consequence of well-ordering that S x S is of the same cardinality as S for any infinite S, and that is necessarily nonconstructive, but for “practical” infinite sets (by which I basically mean ℶ_n for n finite) many bijections can be made explicitly that would in general require choice.