Is such a thing even possible?
There's not much to say really. Obviously if there was a set it would be full of transcendental numbers. This led me to think of a function generating transcendental numbers (given a transcendental number) closed under addition and multiplication – an easy one is algebraic function but these are clearly countable however. (Even worse the numbers would have to be algebraically independent which means you have to cut back on the algebraic functions.).
Any insights, guys?
Best Answer
This is possible.
First, consider the set of all numbers of the form $$ a_1\pi + a_2\pi^2 + \cdots + a_n\pi^n $$ where $n \geq 1$, the coefficients $a_1,\ldots,a_n$ are non-negative integers, and at least one $a_i$ is positive. This set is clearly closed under both addition and multiplication. However, it is not uncountable.
We can make this set larger by adding another number. For example, we can consider two-variable polynomials involving $\pi$ and $e$ with the same restrictions: there is no constant term, all of the coefficients are non-negative integers, and at least one of the coefficients is positive. Assuming that $\pi$ and $e$ are algebraically independent (which is not known), all of these polynomials are distinct and nonzero, so we get a larger set of transcendental numbers which is closed under addition and multiplication. However, this set is still not uncountable.
To make an uncountable set that is closed under addition and multiplication, we must start with an uncountable set $S$ of algebraically independent transcendental real numbers. It is known that such a set exists, since the transcendence degree of the real numbers over the rationals is uncountable. If we now take all polynomials over the elements of $S$ satisfying the same conditions (no constant term, non-negative integer coefficients, at least one positive coefficient), the result will be an uncountable set of transcendental numbers that is closed under addition and multiplication.