Since by definition, it remains to prove that if is countable then is countable.Show is countable.
Let for .
Thus, .
Also, is clear by defining a bijection.
Since each ( ) is countable.
We have that is a countable union of countable sets.
Thus, it is countable.
.Show there are uncountably many transcendental numbers
Given an algebraic number it needs to satisfy a non-zero polynomial over . There are countably many non-zero polynomials over and each one has finitely many zeros. Since every algebraic number is among those polynomials and conversely a zero of one of those polynomials is algebraic it follows there are countably many algebraic numbers. Thus, there have to be uncountably many transcendental numbers. Because .