Cardinal of C
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From: hagman on 6 Jun 2010 12:57 On 5 Jun., 13:56, A N Niel <ann...(a)nym.alias.net.invalid> wrote: > In article > <cfd67c23-7633-4968-8e3e-25395b72e...(a)x21g2000yqa.googlegroups.com>, > > > > hagman <goo...(a)von-eitzen.de> wrote: > > On 5 Jun., 05:11, Pollux <frank.ast...(a)gmail.com> wrote: > > > I like that bijection. I had started thinking about multiplying a and b in > > > a + bi to get a single real (that would be one half of the bijection, I > > > needed something eld for the other half), but of course, this naive > > > multiplication wouldn't work (z1 = a1 + b1i and z2 = b1 + a1i would map to > > > the same real :-( ). Interleaving is going to work in both directions > > > though. Great! > > > > Thanks for your help! > > > > Pollux > > > Your next task is to find a bijection that is continuous in one of the > > two directions (it can't be i nboth directions) :) > > > hagman > > Why suggest impossible tasks to a beginner? Oops, of course I should rather have asked for a continuous surjection R -> C
From: cbrown on 6 Jun 2010 15:11 On Jun 6, 2:35 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Sat, 5 Jun 2010, Pollux wrote: <snip> > > I'm still curious about seeing a bijection, or an injection in each > > direction, if somebody could show that. > > f:R -> R^2, r -> (r,0) is a continuous injection from R into R^2. > > A bijection g, from (0,1)^2 onto (0,1) is the interweaving of > digits as describe earlier. Note, no real ends with 999.... > g is an injection, not a bijection: e.g., for what pair (x,y) has g(x,y) = 0.90909090909...? Cheers - Chas
From: cbrown on 6 Jun 2010 16:04
On Jun 5, 9:39 am, Pollux <po....(a)gmail.com> wrote: > Looks like not, two infinite sets having the same cardinal is equivalent to the existence of a bijection between those two sets. So, by definition, there is no other way to find the cardinal of C. Is that correct? > > I guess it's enough to prove that there exists a bijection, even if you can't show it? And for that, it's enough to show that there is an injection in each direction, right? I'm still curious about seeing a bijection, or an injection in each direction, if somebody could show that. > > Thanks! > > Pollux It's a bit overwrought, but just for fun... There is an explicit bijection between the half open interval [0,1) and B*, the set of /all/ binary sequences (including those with infinite trailing 1's), which can be found at the end of http://groups.google.com/group/sci.math/msg/15cb222f8e6c844f An explicit bijection between (0,1) and [0,1) is given at http://groups.google.com/group/sci.math/msg/82d6ac5aec38028d And using the tricks that William Elliot described to get a bijection from R -> (0,1), we can compound these bijections to construct a bijection between R and B*. Then there's an obvious bijection between B* and (B*)^2 via interleaving of sequences; and so that's enough to get us an explicit bijection between R and R^2. Cheers - Chas |