Let $A\subset\mathbb{R}$ be an uncountable set of irrational numbers. Does there exist a finite $B\subset A$ such that $\sum_{x\in B} x\in\mathbb{Q}?$

Question

Let $A\subset\mathbb{R}$ be an uncountable set of irrational numbers.

Does there exist a nonempty finite subset $B\subset A$ such that $\displaystyle\sum_{x\in B}x \in \mathbb{Q}\ ?$

If we change "uncountable" to "countable" then the answer is trivially no, as $\ A=\{ q\pi: q\in\mathbb{Q}_{>0} \}\ $ is a counter-example. I believe there is no analogue to this counter-example to my question above.

I am unsure how to answer the question, although I sense that maybe the Baire Category Theorem could be helpful, but I have poor familiarity with this theorem and it's applications.

Answer 1

Let $\mathscr{B}$ be any basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Note that $\mathscr{B}$ is uncountable. Since $\mathbb{Q}$ is a subset of $\mathbb{R}$, there must be some finite set $Q = \{x_1, \dotsc, x_m\} \subseteq \mathscr{B}$ and a collection of nonzero rational numbers $q_1, \dotsc, q_m$ such that $$ 1 = \sum_{j=1}^{m} q_j x_j. $$ By linear independence, this set must be unique (if not, then zero can be written as $$0 = 1-1 = \sum q_j x_j + \sum r_k y_k, $$ with $x_j, y_k \in \mathscr{B}$ and $q_j, r_k \in \mathbb{Q}\setminus \{0\}$, which is a contradicts the linear independence of the elements of $\mathscr{B}$). Let $$ A = \mathscr{B} \setminus Q. $$ By construction, any finite subset $B$ of $A$ has the property that $$ \sum_{B} x \not\in \mathbb{Q}. $$

Answer 2

Here's a simple example without the axiom of choice: let $A$ be the set of real numbers in $[0,1)$ whose binary representations have a $1$ at every index-$4n^2$ digit, a zero at all non-square indices, and whatever we like at index $(2n+1)^2$ digits.

$$0.?0010000?000000100000000?000000000010\ldots$$

Given any finite sum of such values, if we go far enough out into the binary expansion we'll avoid any overflow from the $n^2$ place to the $(n-1)^2$ place, so we'll preserve the property that we have ever-longer strings of zeros between the occasional 1 in the binary expansion, thereby ruling out periodicity and hence rationality.

(If we replace $n^2$ with $n!$ in this definition, we can additionally guarantee the sums aren't algebraic.)

Answer 3

More generally, let $R$ be an infinite set, let $S$ be any set of less than $|R|$ finitary operations on $R$ of positive arity that are injective in each of their inputs when the other inputs are fixed, and let $Q\subset R$ be a subset of cardinality less than $R$. Then there exists $A\subseteq R$ of cardinality $|R|$ such that $f(x_1,\dots,x_n)\not\in Q$ for all $f\in S$ and all distinct $x_1,\dots,x_n\in A$. (In your case, $R=\mathbb{R}$, $S$ is the set containing the operations $f_n(x_1,\dots,x_n)=\sum_i x_i$ for each $n>0$, and $Q=\mathbb{Q}$.)

The proof is a simple transfinite recursion. Having chosen elements $(a_\beta)_{\beta<\alpha}$ for $\alpha<|R|$ which satisfy the required property for elements of $A$, we can always choose one more element $a_\alpha$ since there are at most $|S|\cdot|Q|\cdot|\alpha|<|R|$ elements of $R$ that would cause an element of $S$ to give an output in $Q$ when combined with some of the $a_{\beta}$ we have already chosen (here we use injectivity of elements of $S$ in each input). So we can iterate this and obtain a sequence $(a_\beta)_{\beta<|R|}$ which gives the desired set $A$ of cardinality $|R|$.

(Strictly speaking, the bound $|S|\cdot|Q|\cdot|\alpha|$ is only correct when one of these cardinals is infinite. When they are all finite, though, there are still only finitely many elements of $R$ that we can't choose as $a_\alpha$ so we are fine since $R$ is infinite.)