452 karma · joined November 21, 2017
"He said he didn’t need a translator since, “I feel very comfortable now,” " from https://chess24.com/en/read/news/ding-liren-passes-acid-test...
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...
If you assume the axiom of choice it's enough to have surjections both way.
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)
In general even presenting a topological space to a computer is not easy. This can be done for very good spaces, for example those admitting (finite) triangulations, and there are some algorithms there, but my understanding is that you run into undecidability issues very quickly. But for the kind of spaces I work with (which are usually compact metrizable connected, so way better than arbitrary topological spaces) there seems to be no hope to get a computer to work with them.
lea rax, [eiz+rbx*4]
instead of lea rax, [rbx*4]
because the latter will produce an heavier binary by being interpreted as lea rax, [00000000h+rbx*4]