Cardinality of transcendental numbers Hello everyone I'm struggling to prove that $|T|=|R|$, where $T$ is the set of all transcendental numbers.
I've seen a few proofs that are based on the statement $$(1)\ \ \mathbb{R}=T\cup \{algebraic\ numbers\}$$
which is simple enough to finish. HOWEVER, on the wikipedia page of algebraic/transcendental numbers it is strictly stated that both transcendentals/algebraics can be complex or real.
I was hoping somone with more knowledge can shed some light on me.
Edit for clairty: I see that (1) would imply that there are $|T|\geq|R|$ many trascendental numbers, but why would that inequality be strictly an equality?

So, there are the real algebraic numbers (those that satisfy an even polynomial of odd degree over $\mathbb{Q}$) and the complex algebraic numbers (those that satisfy any polynomial over $\mathbb{Q}$). Recall the fact that ever polynomial has degree many roots. Now, how many polynomials can we write? (Answer: Only countably many  -- you should be able to show this). Since there are only countably many polynomials (over $\mathbb{Q}$), each corresponding to only finitely many real/complex numbers, we have that the real algebraic numbers and the complex algebraic numbers is the countable union of finitely sets, which you should know is also countable. Since transcendental numbers are those defined as "not" algebraic, we have 2 cases: 
The real case is trivial (since $|\mathbb{R}|=2^{\aleph_0} > |\mathbb{N}|$ and $2^{\aleph_0}$ minus a countable set is still $2^{\aleph_0}$)
The complex case is realizing that $|\mathbb{C}|=|\mathbb{R}|$, which then boils down to the argument above. 
