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.
> 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.
If you assume the axiom of choice it's enough to have surjections both way.
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.
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
Sorry, didn't even look at the bounds - assumed things went to infinity.
Aleph_0^2 = Aleph_0
Aleph_1^2 = Aleph_1
I can see space filling curves through I^I, but I can't see them covering R^I.
How does a space-filling curve cover a plane without being filled with infinitesimal discontinuities?
(d being the dimensionality of the curve.)
As I think about this there's an unfilled infinitesimal R ^ (d - 1) space orthogonal to every point on the curve.
I don't see how it is that that does not manifest as discontinuities in the filled space.
The problem that I'm having is in getting from R to R^d, which I still don't think is possible - look at the comment where I mention "surjections" and "bijections".
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)
> 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.
I think his iterator is generating infinitesimals of rational and not continuous nature.
If there are holes in the map does it really cover the whole space?
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.
There's a one to one mapping between the integers and the cartesian product of the integers (I^I == I), but (I believe) that's not true for the reals (someone correct me if I'm wrong about that).
However, you can take some subset of R^n and stretch it to infinity across all dimensions, giving you back R^n.
So:
There are the same number of integers as odd integers,
There are more points on a continuous plane than on a continuous line.
The number of points on a line is the same as the number of points on the edge of a circle.
The number of points in a plane is the same as the number of points in a semiplane.
There are the same number of rationals as integers.
There are more reals than integers.