4 ms·

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 f

• 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.