Cardinal of C
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From: Pollux on 4 Jun 2010 18:16 Is |C| = 2^aleph0 or |C|= 2^2^aleph0? where C is the usual set of all complex numbers on the usual field R. I couldn't put a bijection between R and C. But of course, I'm not much of a mathematician. Is there one? Pollux
From: Daniel Giaimo on 4 Jun 2010 22:38 On 6/4/2010 10:16 PM, Pollux wrote: > Is |C| = 2^aleph0 or |C|= 2^2^aleph0? where C is the usual set of all complex numbers on the usual field R. I couldn't put a bijection between R and C. But of course, I'm not much of a mathematician. Is there one? > Yes there is one. One simple way to get one is to map a complex number a + bi to the real number you get by interleaving the decimal representations of a and b. -- Dan Giaimo
From: Daniel Giaimo on 4 Jun 2010 22:45 On 6/4/2010 10:38 PM, Daniel Giaimo wrote: > On 6/4/2010 10:16 PM, Pollux wrote: >> Is |C| = 2^aleph0 or |C|= 2^2^aleph0? where C is the usual set of all >> complex numbers on the usual field R. I couldn't put a bijection >> between R and C. But of course, I'm not much of a mathematician. Is >> there one? >> > > Yes there is one. One simple way to get one is to map a complex number a > + bi to the real number you get by interleaving the decimal > representations of a and b. Actually, I posted too soon. This doesn't quite work due to the non-uniqueness of decimal representations. It does, however give you an injection from C to R, so, at least assuming the axiom of choice, it proves that |C| <= |R|. |R| <= |C| is clear, so |R| = |C|.
From: Pollux on 4 Jun 2010 19:11 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
From: W. Dale Hall on 4 Jun 2010 23:40
Daniel Giaimo wrote: > On 6/4/2010 10:38 PM, Daniel Giaimo wrote: >> On 6/4/2010 10:16 PM, Pollux wrote: >>> Is |C| = 2^aleph0 or |C|= 2^2^aleph0? where C is the usual set of all >>> complex numbers on the usual field R. I couldn't put a bijection >>> between R and C. But of course, I'm not much of a mathematician. Is >>> there one? >>> >> >> Yes there is one. One simple way to get one is to map a complex number a >> + bi to the real number you get by interleaving the decimal >> representations of a and b. > > Actually, I posted too soon. This doesn't quite work due to the > non-uniqueness of decimal representations. It does, however give you an > injection from C to R, so, at least assuming the axiom of choice, it > proves that |C| <= |R|. |R| <= |C| is clear, so |R| = |C|. > Google "Cantor Schroeder Bernstein Theorem". Choice is unnecessary. |