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https://mathoverflow.net/questions/126069/bijection-from-mathbbr-to-mathbbr2 https://mathoverflow.net/questions/126069/bijection-from-mat... https://mathoverfl

by yarg 3y ago

https://mathoverflow.net/questions/126069/bijection-from-mathbbr-to-mathbbr2 https://mathoverflow.net/questions/126069/bijection-from-mat... https://mathoverflow.net/questions/240249/continuous-map-from-mathbb-r2-to-mathbb-r https://mathoverflow.net/questions/240249/continuous-map-fro... https://mathoverflow.net/questions/34232/injective-maps-mathbbrn-to-mathbbrm https://mathoverflow.net/questions/34232/injective-maps-math... The second link says that there's a continuous surjection from R -> R^2, but no such bijection. (Which, after some googling looks in line with what I was imagining.) If there's no bijection, how can the cardinalities be the same?

ginnungagap 3y ago
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 3y 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 3y 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 3y 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 3y ago
      I may have misread that one, but I still think the reverse mapping only gets as far as within arbitrary bounds.
    - yarg 3y 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.
Skeime 3y ago
Because continuity is not required here. There are non-continuous bijections between R and R^2.
- yarg 3y ago
  How is continuity not required? If there are holes in the map does it really cover the whole space?
  - Skeime 3y ago
    I think you have misconceptions about what continuity means and I’m not even sure what you mean with holes. As a simpler example: The map that multiplies each rational number by -1 and keeps irrational numbers as they are is bijective (it’s its own inverse), however it is nowhere continuous.

by yarg 3y ago

ginnungagap 3y ago

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 3y 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 3y 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 3y 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 3y ago
    I may have misread that one, but I still think the reverse mapping only gets as far as within arbitrary bounds.
  - yarg 3y 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.

Skeime 3y ago

Because continuity is not required here. There are non-continuous bijections between R and R^2.

yarg 3y ago
How is continuity not required? If there are holes in the map does it really cover the whole space?
- Skeime 3y ago
  I think you have misconceptions about what continuity means and I’m not even sure what you mean with holes. As a simpler example: The map that multiplies each rational number by -1 and keeps irrational numbers as they are is bijective (it’s its own inverse), however it is nowhere continuous.