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> (I believe) that's not true for the reals (someone correct me if I'm wrong about that). It is also true for the reals. R^2 can be trivially mapped to R. Her

> (I believe) that's not true for the reals (someone correct me if I'm wrong about that). It is also true for the reals. R^2 can be trivially mapped to R. Here's one mapping from R to R^2: Let b_{n} b_{n-1} .. b_0 . b_{-1} b_{-2} ... Be the binary representation of the real 'x'. For example, if x = 2.5, then: b_1 = 1 b_0 = 0 b_{-1} = 1 Have x map to the following value in R^2 (y, z) where y = b_{n} b_{n-2} ... b_{0} . b_{-2} b_{-4} .. z = b_{n-1} b_{n-3} .. b_{1} . b_{-1} b_{-3} .. So 2.5 -> (0, 1.5) 1 -> (1, 0) 1.75 -> (1.5, 0.5) etc.

yarg 3y ago
> It is also true for the reals. R^2 can be trivially mapped to R. > Here's one mapping from R to R^2: That's the easy part. It's very easy to map R to R^2, the other direction not so much. From my searches, based upon what other people said, the mapping is a surjection and not a bijection. If that's the case, then either they don't have the same cardinality or I'm confused as to what cardinality actually is.
- ginnungagap 3y ago
  There is no need to exhibit bijections, if you have an injection X -> Y and an injection Y -> X, then X and Y have the same cardinality. This is the statement of the Cantor-Schroeder-Bernstein theorem. If you assume the axiom of choice it's enough to have surjections both way.
  - yarg 3y ago
    I get how you can map every value in R -> R^2, but I have yet to see any proof that there's a complete mapping in reverse. For the most reasonable attempt at flooding a plane with R, I still end up with something where I can trivially define values {r0, r1} that do not map back to R.
    - ginnungagap 3y ago
      What are those trivially defined {r0,r1} for the "interleaving digits" map f:(0,1) -> (0,1)^2 defined by taking a real x in (0,1), writing its decimal expansion as x=0.a1a2a3a4a5a6... (always pick the nonrepeating decimal expansion in case x has two) and sending it to the pair (y,z) where y=0.a1a3a5... z=0.a2a4a6... (Using (0,1) is no issue since we can compose later with bijections (0,1) -> R) Edit: I'm not sure why I cannot reply to your reply, but {π,π} is not even in (0,1)^2 We can also do a map f:R -> R^2 directly, it's just slightly more annoying to write. Take a real number x, written in decimal expansion as x=bn...b2b1b0.a0a1a2a3a4... where again we choose the representation which is not repeating in case x has two. We also choose our first index n so that bn is zero but b(n-1) is not. We now map x to the pair (y,z), where y=...b5b3b1.a1a3a5... z=...b4b2b0.a0a2a4... So that, for example, (π,π) is the image of x=33.114411559922...
      - yarg 3y ago
        {π, π} will generally do. Sorry, didn't even look at the bounds - assumed things went to infinity.
yarg 3y ago
As an example, try using a Hilbert curve to map {π, π} back onto R (and I mean π exactly, not π ± an infinitesimal). (There are, however, infinitely many mappings for {π, π ± ε} -> R and {π ± ε, π} -> R, such that ε is infinitesimal.)

yarg 3y ago

> It is also true for the reals. R^2 can be trivially mapped to R. > Here's one mapping from R to R^2: That's the easy part. It's very easy to map R to R^2, the other direction not so much. From my searches, based upon what other people said, the mapping is a surjection and not a bijection. If that's the case, then either they don't have the same cardinality or I'm confused as to what cardinality actually is.

ginnungagap 3y ago
There is no need to exhibit bijections, if you have an injection X -> Y and an injection Y -> X, then X and Y have the same cardinality. This is the statement of the Cantor-Schroeder-Bernstein theorem. If you assume the axiom of choice it's enough to have surjections both way.
- yarg 3y ago
  I get how you can map every value in R -> R^2, but I have yet to see any proof that there's a complete mapping in reverse. For the most reasonable attempt at flooding a plane with R, I still end up with something where I can trivially define values {r0, r1} that do not map back to R.
  - ginnungagap 3y ago
    What are those trivially defined {r0,r1} for the "interleaving digits" map f:(0,1) -> (0,1)^2 defined by taking a real x in (0,1), writing its decimal expansion as x=0.a1a2a3a4a5a6... (always pick the nonrepeating decimal expansion in case x has two) and sending it to the pair (y,z) where y=0.a1a3a5... z=0.a2a4a6... (Using (0,1) is no issue since we can compose later with bijections (0,1) -> R) Edit: I'm not sure why I cannot reply to your reply, but {π,π} is not even in (0,1)^2 We can also do a map f:R -> R^2 directly, it's just slightly more annoying to write. Take a real number x, written in decimal expansion as x=bn...b2b1b0.a0a1a2a3a4... where again we choose the representation which is not repeating in case x has two. We also choose our first index n so that bn is zero but b(n-1) is not. We now map x to the pair (y,z), where y=...b5b3b1.a1a3a5... z=...b4b2b0.a0a2a4... So that, for example, (π,π) is the image of x=33.114411559922...
    - yarg 3y ago
      {π, π} will generally do. Sorry, didn't even look at the bounds - assumed things went to infinity.

yarg 3y ago

As an example, try using a Hilbert curve to map {π, π} back onto R (and I mean π exactly, not π ± an infinitesimal). (There are, however, infinitely many mappings for {π, π ± ε} -> R and {π ± ε, π} -> R, such that ε is infinitesimal.)