Cardinal of C
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From: William Elliot on 5 Jun 2010 08:02 On Sat, 5 Jun 2010, hagman 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! > > Your next task is to find a bijection that is continuous in one of the > two directions (it can't be i nboth directions) :) > Between R^2 and R, between C and R or between R^2 and C?
From: Pollux on 5 Jun 2010 08:00 I was just trying to find a proof that |C| = |R^2| = |R| = c = 2^aleph0. I thought one way would be to find a bijection. That might be hard to construct. In that case, are there other ways to find the cardinal of |C|? Pollux
From: Pollux on 5 Jun 2010 08:39 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
From: William Elliot on 6 Jun 2010 05:35 On Sat, 5 Jun 2010, Pollux 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? Yes, that's the Cantor-Bernstein theory: if f:A -> B and g:B -> A are injections, then A and B are equinumerous, ie there's a bijection between A and B. > 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.... f:(0,1) -> R, r -> tan pi(r - 1/2) is a continuous bijection. You can now construct a bijection from R^2 to (0,1)^2 to (0,1) to R. If you want an injection that's not surjective, then at the last step use h:(0,1) -> R, r -> tan pi.r/2. An injection f, from R into [0,oo) is to place the diget one before r if 0 <= r and the digit two if r < 0. Thus (x,y) -> (f(x),f(y)) is an injection from R^2 into (R+)^2 which, if composed with the interweaving injection, is an injection
From: Pollux on 6 Jun 2010 06:36
Thanks! It doesn't look so hard after all... :-) Pollux |