5
$\begingroup$

I tried to prove the following: There exists a set $A\subseteq \mathbb{R}$ such that $|A|>\aleph_0$, and, for every $x_1,\ldots,x_n\in A$, $$x_1+\ldots+x_n \notin \mathbb{Q}$$

I found a solution using Zorn's lemma. I'm interested to know whether there exists a solution which doesn't apply the axiom of choice.

My solution: Apply Zorn's lemma to find a maximal element among subsets $A\subseteq \mathbb{R}$ such that any finite sum of elements from $A$ doesn't belong to $\mathbb{Q}$. Now, if the maximal $A$ was countable, then we could extend it to $A\cup \{r\}$, where $r\in \mathbb{R}$ is picked as follows: take $r\in \mathbb{R}$ which doesn't belong to the countable set of elements of the form $q-S$, where $q\in \mathbb{Q}$ and $S$ is a sum of finitely many elements of $A$.

$\endgroup$
  • $\begingroup$ My guess is you need some form of choice, like dependent choice. I don't think any statement concerning some (reasonable) construction on the reals needs the entirety of AC. $\endgroup$ – Arthur Aug 25 '18 at 15:45
5
$\begingroup$

Yes: You can take $$ A=\biggl\{ \sum_{n=0}^\infty 10^{-n^2} f(n) \biggm| f \in \mathbb \{1,2\}^{\mathbb N} \biggr\}$$ This clearly has $2^{\aleph_0}$ elements. And the nonzero digit in the decimal expansions are ever further from each other, so for every finite sum of elements from the set, there will be increasingly long runs of $0$s in the decimal expansion of the sum ...

If we change the exponents in the scaling factor $10^{-n^2}$ to grow fast enough (something like $10^{-n!}$ should suffice), we can even make sure that all of the finite sums will be Liouville numbers and therefore transcendental.

$\endgroup$
  • $\begingroup$ Beautiful! Thank you very much. $\endgroup$ – user333618 Aug 26 '18 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.