3 ms·

First of all the first link explicitely constructs (in a very strong sense) an explicit bijection R->R^2 so I don't know how why you seem to think otherwise. I

First of all the first link explicitely constructs (in a very strong sense) an explicit bijection R->R^2 so I don't know how why you seem to think otherwise. It is true that there is no continuous bijection R -> R^2. It is also true that this has nothing to with cardinality, which is only about the existence of bijections (and indeed is defined for arbitrary sets with no topological structure for which continuity makes no sense)

yarg 4y ago
I don't think that it's actually formulating a bijection across the reals - the generator function looks to me to be generating the complete set of rationals. > Represent each real number x∈[0,1) as a sequence of DIGITS, where each DIGIT is either in {0,…,8} or is of the form 10∗(10k−1)+i with k≥1 and i∈{0,…,8} (i.e., in {90,…,98;990,…,998;9990,…,9998;…}. He's iterating to infinity, that sure as shit sounds countable to me. Anyway, the conclusion that I've come to is that you can map R onto a subset of R^n such that for all values {r[1] ... r[n]} in R^n there is a value in the R -> R^n mapping within all finite bounds. So as long as you stay away from infinitesimals (not all of them), the domains are equivalent. https://en.wikipedia.org/wiki/Infinitesimal https://en.wikipedia.org/wiki/Infinitesimal > In mathematics, an infinitesimal number is a quantity that is closer to zero than any standard real number, but that is not zero. But as far as I can tell, the reverse mapping has an R^(n - 1) orthogonal that is completely unaccounted for, within those finite bounds. (And breaking infinitesimals might cause some trouble for calculus.)
- ginnungagap 4y ago
  He's associating to each x in [0,1) an infinite sequence of natural numbers (just like decimal representations do), and there's uncountably many such sequences
  - yarg 4y ago
    But he's defining the numbers with a property that makes them countably infinite? prefix = '0.'; queue = new Queue(prefix); for(;;){ s = queue.poll(); for(c:['0'...'9']){ queue.add(s + c); } } When the constraint is "run forever" I don't think that the generator he defined produces results any different from the snippet above.
  - yarg 4y ago
    I may have misread that one, but I still think the reverse mapping only gets as far as within arbitrary bounds.
  - yarg 4y ago
    The part that seems problematic is "each real number" the reals exist across bounds, I'm not sure there can be an "each". I think his iterator is generating infinitesimals of rational and not continuous nature.

yarg 4y ago

I don't think that it's actually formulating a bijection across the reals - the generator function looks to me to be generating the complete set of rationals. > Represent each real number x∈[0,1) as a sequence of DIGITS, where each DIGIT is either in {0,…,8} or is of the form 10∗(10k−1)+i with k≥1 and i∈{0,…,8} (i.e., in {90,…,98;990,…,998;9990,…,9998;…}. He's iterating to infinity, that sure as shit sounds countable to me. Anyway, the conclusion that I've come to is that you can map R onto a subset of R^n such that for all values {r[1] ... r[n]} in R^n there is a value in the R -> R^n mapping within all finite bounds. So as long as you stay away from infinitesimals (not all of them), the domains are equivalent. https://en.wikipedia.org/wiki/Infinitesimal https://en.wikipedia.org/wiki/Infinitesimal > In mathematics, an infinitesimal number is a quantity that is closer to zero than any standard real number, but that is not zero. But as far as I can tell, the reverse mapping has an R^(n - 1) orthogonal that is completely unaccounted for, within those finite bounds. (And breaking infinitesimals might cause some trouble for calculus.)

ginnungagap 4y ago
He's associating to each x in [0,1) an infinite sequence of natural numbers (just like decimal representations do), and there's uncountably many such sequences
- yarg 4y ago
  But he's defining the numbers with a property that makes them countably infinite? prefix = '0.'; queue = new Queue(prefix); for(;;){ s = queue.poll(); for(c:['0'...'9']){ queue.add(s + c); } } When the constraint is "run forever" I don't think that the generator he defined produces results any different from the snippet above.
- yarg 4y ago
  I may have misread that one, but I still think the reverse mapping only gets as far as within arbitrary bounds.
- yarg 4y ago
  The part that seems problematic is "each real number" the reals exist across bounds, I'm not sure there can be an "each". I think his iterator is generating infinitesimals of rational and not continuous nature.