# Uncountable set of irrational numbers closed under addition and multiplication?

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?

• I think it would either have full measure or measure $0$ if it exists. Call the set in question $S$, and take some $r\in S$. Then $S\mod r = [0,r]\cap S$ is an uncountable set of irrationals contained in the unit interval (since $s\in S,n\in\mathbb{Z}\implies s+nr\in S$). The transformation given by adding $q\in S$ not a rational multiple of $r$ is ergodic but fixes $[0,r]\cap S$ so $[0,r]\cap S$ must have full measure or measure $0$, and this can be extended to $S$ using countable subadditivity of measures. – Alex Becker Dec 28 '11 at 18:47
• @Alex Becker: for any set $A$ of positive measure, $A+A$ contains an interval. So here the set would have to have measure 0. – Carl Mummert Dec 28 '11 at 19:03
• The property that a set $S$ contains only irrationals and is closed under multiplication and addition satisfies the ascending chain condition, so there must be a maximal such $S$ by Zorn's lemma. Can you prove that a maximal such $S$ must be uncountable? – Thomas Andrews Dec 28 '11 at 19:05
• Why is this set theory? Isn't that just group theory, or something else? – Asaf Karagila Dec 28 '11 at 19:23
• I'm sorry, I just don't see how [set-theory] is a "garbage collector" tag for anything related to sets. – Asaf Karagila Dec 28 '11 at 21:40

## 4 Answers

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.

• Nice answer! Is any construction of an uncountable set of algebraically independent reals known, or is existence all that has been shown? – Alex Becker Dec 28 '11 at 19:06
• @AlexBecker What do you mean by "construction?" The set of all computable real numbers is necessarily countable, since the set of all computer programs is countable, so, in a sense, no uncountable set is constructible. – Thomas Andrews Dec 28 '11 at 19:24
• @ThomasAndrews Sorry, construction was not the word I was thinking of. What I meant was, can the set be defined as a subset of the reals, in the sense that the reals can be defined as the subset of the complex numbers which have imaginary part 0? I'm not sure how to state this rigorously. – Alex Becker Dec 28 '11 at 19:28
• Very strong answer + 1 – Adam Dec 28 '11 at 19:30
• I suspect that the only way for this to be even true is with some axiom-of-choice argument - I assume the existence of a transcendental basis uses Zorn's lemma or some other Axiom-of-Choice argument. You probably don't need complete AofC. – Thomas Andrews Dec 28 '11 at 19:45

Let $A_0$ be a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$, which already has size of the continuum, and let $$B_0=\{\text{finite sums of elements of }A_0\},$$ $$A_1=\{\text{finite products of elements of }B_0\},$$ $$B_1=\{\text{finite sums of elements of }A_1\},$$ $$\cdots$$ and $S=\bigcup_{n=0}^\infty A_n$. Then $S$ is closed under addition and multiplication by definition, $S$ is uncountable, and every element of $S$ is irrational because no algebraic relation holds between the elements of $A_0$ over $\mathbb{Q}$.

EDIT: This is the same as Jim's suggestion above, just phrased slightly differently.

• Beautiful answer. – Carl Mummert Dec 28 '11 at 19:13
• Thank you for your kind words :) – Zev Chonoles Dec 28 '11 at 21:02

To extend the comment above.

The property: "$S\subset \mathbb R\setminus\mathbb Q$ is closed under addition and multiplication" satisfies Zorn's lemma, so there must be a maximal such set $S$.

Maximality means that for any irrational number $x\notin S$, there must be a polynomial of the form $p(z)-q$ where $q\in \mathbb Q$ and $p(z)$ is a non-zero non-constant polynomial with coefficients in $(S\cup\{0\})\oplus\mathbb N)$ such that $x$ is a root of the polynomial.

But if $S$ is countable, then the set of such polynomials is countable, and therefore the set of such roots is countable.

So any maximal $S$ must necessarily be uncountable.

This can be extended to a more general theorem: If $K\subset \mathbb R$ is uncountable and closed under addition and multiplication, then there is an uncountable $S\subseteq K\setminus\mathbb Q$ which is closed under addition and multiplication.

Proof: Since $K$ is uncountable, it contains a transcendental number, so the positive polynomials in that transcendental form a subset of $K\setminus\mathbb Q$ closed under addition and multiplication.

As before, by Zorn's lemma, there must be a maximal set $S\subset K\setminus\mathbb Q$ closed, and, as before, if $S$ is countable, then you get a contradiction - there are too many elements of $K\setminus (S\cup\mathbb Q)$ and not enough polynomials with coefficients in $(S\cup\{0\})\oplus\mathbb N$.

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.

• You are still using some limited form of the axiom of choice, if not full on choice - namely, you are using that there is a choice function from the subsets of $\mathbb R^n$ with positive measure. Still, a nice proof. – Thomas Andrews Dec 29 '11 at 13:45
• The choice can be made perfectly explicit, since the sets involved are actually dense open sets in a product of intervals: you can take the first $2^k$-tuple of rationals (in some convenient enumeration) that works. – Robert Israel Dec 29 '11 at 19:34