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I'm looking for an uncountable set $X \subset \mathbb{R}_{\geq 0}$ such that for all $a, b \in X$, $a \neq b \implies a + b \not\in X$.

Two points about this question: first of all, I'm not sure which area of mathematics this falls under so I used the set-theory tag; feel free to correct me on this. Secondly, note immediately that if $X$ is allowed to be countable, then e.g. $X = \{2^k \mid k \in \mathbb{Z}\}$ works.

JMP
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  • @AdityaAgarwal true, but ${7-\pi, \pi}$ is not even infinite, let alone uncountable. –  Feb 14 '21 at 13:34
  • There are uncountable positive irrationals. All of them are in $X$. Let one of them be $p$. Then for any rational $q>p$, $q-p$ would be in $X$. – Aditya Agarwal Feb 14 '21 at 13:36
  • @AdityaAgarwal Perhaps I'm missing something, but since $\sqrt{2} + \sqrt{3}$ is irrational, how can all positive irrationals be in $X$? –  Feb 14 '21 at 13:39
  • Oh right, I misinterpreted it. My bad. – Aditya Agarwal Feb 14 '21 at 13:41
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    I don't think ${2^k | k \in \mathbb{Z}}$ works, because the wanted property doesn't mention $a \neq b$. For example, $2 \in X$ and $2+2 \in X$. For a countably infinite $X$ we could just take all odd integers. – aschepler Feb 14 '21 at 13:48
  • @aschepler you're right! My intention was for ${2^k \mid k \in \mathbb{Z}}$ to work - I've edited in the condition $a \neq b$. Thanks :) –  Feb 14 '21 at 14:02

2 Answers2

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Let $B$ be a Hamel basis of $\mathbb R$ over $\mathbb Q$. Then let $X = \{|x| \colon x\in B\}$. This clearly works.

If you want an example avoiding axiom of choice, consider the set of all numbers whose all digits in the decimal expansion are either $1$ or $3$.

timon92
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Based on your countable example what about $X=\bigcup\limits_{k=1}^\infty[3^k,3^k+1]$

We have $x=3^a+\{x\}$ and $y=3^b+\{y\}$ then

$$x+y=(3^a+3^b)+(\underbrace{\{x\}+\{y\}}_{\in[0,2]})\notin [3^c,3^c+1]$$

Indeed, wlog let assume $1\le a\le b$.

  • $3^a+3^b>3^b+1$
  • $3^{b+1}-(3^a+3^b+2)=3^b(3-\overbrace{\frac 1{3^{b-a}}}^{\le 1}-1-\overbrace{\frac 2{3^b}}^{\le \frac 23})>3^b\times \frac 13>0$

And we don't even need the $x\neq y$ condition.

zwim
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