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.

  • $\begingroup$ You mean a non-empty $B$, right? Also, the purported counterexample does not work: any finite $B$ closed under taking negatives sums to $0$ $\endgroup$
    – ac15
    Commented Jan 1 at 18:29
  • $\begingroup$ @ac15 True. I amended the question accordingly. $\endgroup$ Commented Jan 1 at 18:50
  • $\begingroup$ If $Q$ is any countable subset of $\mathbb R$, there is a nonempty perfect set $A$ such that $\sum_{x\in B}x\notin Q$ for every nonempty finite set $B\subset A$. $\endgroup$
    – bof
    Commented Jan 2 at 4:19

3 Answers 3


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 $$ \sum_{B} x \not\in \mathbb{Q}. $$

  • 2
    $\begingroup$ The Axiom of Choice is not needed, though, since it is enough to find an uncountable set where no two points are equivalent. And we can do that without the Axiom of Choice. $\endgroup$
    – Asaf Karagila
    Commented Jan 1 at 18:32
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    $\begingroup$ @AsafKaragila You know these things better than I---the AoC may not be needed, but I don't know the details of the argument. $\endgroup$
    – Xander Henderson
    Commented Jan 1 at 18:33
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    $\begingroup$ math.stackexchange.com/questions/1696284/… $\endgroup$
    – Asaf Karagila
    Commented Jan 1 at 18:34
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    $\begingroup$ Ulam wrote on Page 14: "von Neumann shows the existence of a set $M$ of real numbers, of the power of the continuum, such that any finite number of the elements of $M$ are algebraically independent. The proof is given effectively without the axiom of choice" $\endgroup$ Commented Jan 4 at 8:35

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.


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.)

  • $\begingroup$ +1, I like this counter-example. Instead of every $4n^2$ digit, every $2^n$ digit would also work, right? I wonder if these set is perfect or almost perfect ? $\endgroup$ Commented Jan 5 at 12:31

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.)


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