Is C > R? Complex vs Real Set Size

  • Thread starter Thread starter samkolb
  • Start date Start date
AI Thread Summary
The discussion centers on whether the set of complex numbers is larger than the set of real numbers. Participants argue that the cardinality of the complex numbers, represented as C, is equivalent to the cardinality of the Cartesian product of real numbers, R x R, suggesting that C = R x R. One participant proposes a one-to-one mapping between R x R and R, while another counters that this mapping is not valid due to issues with decimal expansions. Ultimately, the consensus leans towards the idea that the cardinalities of C and R are indeed the same, with some nuances regarding the definitions of bijections. The conversation highlights the complexities of set theory and cardinality in mathematics.
samkolb
Messages
37
Reaction score
0
Is it true that the set of complex number is bigger than the set of real numbers?

I know that card C = card (R x R) and I think that card (R x R) > card R. Is this true, and if so, why?
 
Physics news on Phys.org
samkolb said:
Is it true that the set of complex number is bigger than the set of real numbers?

I know that card C = card (R x R) and I think that card (R x R) > card R. Is this true, and if so, why?

I think card (RxR) = card R

I would show this by setting up a one-to-one map between RxR and R

I will just show you a one-to-one between the unit square [0,1]x[0,1] and the unit interval [0,1]
You just look at the two decimal expansions and merge

(0.abcdefg..., 0.mnopqrs...) -> 0.ambncodpeq...
 
Last edited:
C is with cardinality c, or aleph if you want, the same as R.

The simple bijection is a+ib |-> (a,b) into RxR.

If you want a bijection from C to R, then z=x+iy|->Im(z)/Re(z) it's a bijection to [-infinity,infinity] which is RU{infininity,-infinity} this cardinality is aleph+2=aleph.

QED
 
Last edited:
loop quantum gravity said:
If you want a bijection from C to R, then z=x+iy|->Im(z)/Re(z) it's a bijection to [-infinity,infinity] which is RU{infininity,-infinity} this cardinality is aleph+2=aleph.
How could that possibly be a bijection? Obviously, z_1=a+ib is mapped to the same point as z_2=a z_1, so it is not an injection.

Marcus has already provided a valid bijection, his "decimal merging" is the classical example of this. Notice how it is also valid in \mathbb{R}^n.
 
Correct Big-T, but at least it's onto.
(-:
 
|C| = |R2| = |R|.

There's some discussion about that in this thread.

Minor point: marcus's function isn't even well-defined; consider decimal expansions with infinite trailing "9"s. (For example, 0.0999... = 0.1000..., but (0.0999..., 0.0000...) maps to 0.00909090..., and (0.1000..., 0.0000) maps to 0.10000000... .) However, the mapping from 0.abcdefgh... to (0.acef..., 0.bdfh...) is a well-defined surjection from [0, 1) to [0, 1)2, and that's all you need.
 
Marcus' function would be well defined if we agreed to use trailing nines wherever the decimal expansion is terminating, this should of course have been specified.
 

Similar threads

Replies
10
Views
2K
Replies
1
Views
1K
Replies
5
Views
2K
Replies
5
Views
2K
Replies
3
Views
2K
Replies
13
Views
2K
Replies
1
Views
1K
Back
Top