# Is C bigger than R?

1. Jan 4, 2009

### samkolb

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?

2. Jan 4, 2009

### marcus

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: Jan 4, 2009
3. Jan 4, 2009

### MathematicalPhysicist

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: Jan 4, 2009
4. Jan 4, 2009

### Big-T

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 $$\mathbb{R}^n$$.

5. Jan 4, 2009

### MathematicalPhysicist

Correct Big-T, but at least it's onto.
(-:

6. Jan 7, 2009

|C| = |R2| = |R|.