Prove that the set of transcendental numbers has cardinality c

1. Oct 28, 2007

andytoh

Prove that the set T of transcendental numbers (numbers that do not satisfy some polynomial equation of positive degree with rational coefficients) has the power of the continuum, i.e. has cardinality c.

Here's what I have: Since T is uncountable, then |T|>alephnull . Also, since T is a subset of R , then |T| not> c . Thus, by the Continuum Hypothesis, we must have |T|=c .

But is there a way to get a proof without using the Continuum Hypothesis, by showing directly a bijection between T and R?

2. Oct 28, 2007

andytoh

Oh, perhaps I can use the Schroder-Bernstein theorem, i.e. find an injection from T to R and an injection from R to T, thus showing that T and R have the same cardinality.

The identity map : T -> R is clearly injective, but an injection from R to T??? Inserting the decimal places of pi between all the decimal places of a real number will make the number irrational, but will it be transcendental???

Last edited: Oct 28, 2007
3. Oct 29, 2007

matt grime

How do you know T is uncountable? Is it because the real numbers are the union of the transcendentals and the algebraics and the latter set is countable? Well, that is how you might start to prove what you want to prove.

4. Oct 29, 2007

andytoh

Ok, let A be the set of algebraic numbers. Then since A intersect T = empty, then

|A|+|T| = |A U T| = |R| = c
alephnull + |T| = c
|T| = c,

but that again uses the Continuum Hypothesis.

Is there no explicit bijection between the transendentals and the reals? If not, is it because the Continuum Hypothesis cannot be proven?

Last edited: Oct 29, 2007
5. Oct 29, 2007

matt grime

It is not the continuum hypothesis. It is straightforward to write down a bijection from any infinite cardinal to itself omitting countably many elements. It is no harder than showing that the naturals are in bijection with the even natural numbers.

I also think you misunderstand the concept of proof - the Continuum Hypothesis can be proven to be true from some starting point, just not from the axioms of ZF. That does not mean it 'can't be proven' - with that logic nothing can be proven.

6. Oct 29, 2007

andytoh

Ok, here's my bijection construction. Let A be the set of algebraic numbers. Let S be any countable subset of the transcendentals T (constructable by the axiom of choice). Then R = A U (T-S) U S. Define f:R -> T by

A -> S (with odd indices) (e.g. a_i -> s_(2i-1) )
T-S -> T-S (identity map)
S -> S (with even indices) (e.g. s_i -> s_(2i)

I think this will be a bijection, proving |T| = |R| = c.

7. Oct 29, 2007

matt grime

You don't need the axiom of choice to find an countable subset of T. The numbers

n+e, n in N

are a countable subset of T.

8. Oct 29, 2007

andytoh

Ok S= {pi + n | n is a natural number}

A -> S: a_n -> pi + (2n-1)
T-S -> T-S: r -> r
S -> S: pi + n -> pi + 2n

is a bijection from R to T. Thanks Matt!