consider the field of all numbers. meaning, of course, the complex numbers. we say that a number is representable if we can describe it by a text (for example, by a binary coded string of 0′s and 1′s of finite length). the set of all representable numbers is countable, as there’s a surjection of the set of binary strings (which is countable) onto it. moreover, it is a field, as if a and b are representable numbers, we have that a + b, a − b, a ⋅ b and a / b are represented by strings as “sum of (description of a) and (description of b)”. obviously, every algebraic number is representable, so our field of representable numbers contains the algebrically closed field of the algebraic numbers. but then, our field also contains euler’s number e and archimedes’ constant π, so it’s strictly larger. this opens the question: how does it’s algebraic closure looks like? not too surprisingly, it turns out to be already algebraically closed: every element in its closure can be represented by “root of polynomial with coefficients (description of coefficients)”, as all coefficients are representable. hence, our field, being countable, is strictly larger than the smallest algebraically closed subfield of the complex numbers, but still countable. and it contains lots of transcendental numbers. isn’t that cool?