TITLE: Uncountable set of irrational numbers closed under addition and multiplication?
QUESTION [22 upvotes]: 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?

REPLY [8 votes]: Here's a more "consructive" answer, without using Axiom of Choice. Let ${\cal F}_n$ be the set of functions of $n$ variables corresponding to expressions using each variable once, with $+$ and $*$ (a convenient way to represent these is as trees where each leaf is one of the $n$ variables and each non-leaf node is labelled with either $+$ or $*$: in any case, there are finitely many for each $n$, and they can be enumerated).  Note that ${\cal F}_1$ consists of the one function $x \to x$. I will take $S = \bigcup_{n=1}^\infty \bigcup_{f \in {\cal F}_n} f(C, \ldots,C)$ for a suitable uncountable set $C$.  I need to ensure that for all $f \in {\cal F}_n$, $f(C,\ldots,C)$ does not contain any algebraic numbers.  
Enumerate the triples 
$(\alpha, n, f)$ where $\alpha$ is algebraic, $n$ is a positive integer and $f \in{\cal F}_n$, as $(\alpha_k, n_k, f_k)_{k=1}^\infty$.  $C$ will be obtained as a nested intersection $\bigcap_{k=0}^\infty C_k$, where $C_0 = [0,1]$,
each $C_k$ is the union of $2^k$ disjoint closed intervals, and each interval of $C_{k-1}$ contains two intervals of $C_{k}$.  These are to be chosen so that
$\alpha_k \notin f_k(C_{k},\ldots,C_k)$.  To do this, note that if we randomly choose  $x_i$ and $y_i$ in each interval of $C_{k-1}$, with probability $1$
none of the numbers $f(z_1,\ldots, z_{n_k})$ for $z_1, \ldots, z_{n_k} \in \{x_i: i=1..2^{k-1} \} \cup \{y_i: i =1..2^{k-1}\}$ is $\alpha_k$.  Taking $\epsilon > 0$  small enough, we can let $C_k$ be the union of the intervals $[x_i - \epsilon, x_i + \epsilon]$ and $[y_i - \epsilon, y_i + \epsilon]$ for $i=1 \ldots 2^{k-1}$, and this will have the required properties.