As part of my algebra homework a few weeks ago, I was asked to prove some things about the relation $R$, defined by $(x,y) \in R$ if $x - y \in \mathbb{Q}$. The homework problem itself wasn't particularly out of the ordinary, but it did lead me to a more challenging question. Is there a set containing exactly $1$ element from each equivalence class? The feel the answer should be yes, but finding such a set has eluded me so far.

To be more specific:

Does there exist a set $S \subseteq \mathbb{R}$ such that for any $x_1,x_2 \in S$, that $x_1 - x_2 \notin \mathbb{Q}$, and for any $y \in \mathbb{R} - S$ there exists an $x \in S$, such that $x - y \in \mathbb{Q}$. If so, what would be an example of such a set?

This is what I have managed to determine:

The set $S$ must have an uncountable number of elements.

Suppose $S$ has a countable number of points, then $T$ defined by $$T = \bigcup_{a \in \mathbb{Q}} \bigcup_{b \in S} a+b$$ must also be countable as the countable union of countable sets is countable. Since for every $y \in \mathbb{R}$ there is an $x \in S$ such that $y - x \in \mathbb{Q}$, then $y \in T$. Thus $ \mathbb{R} \subseteq T$, but $\mathbb{R}$ is uncountable, thus $T$ is uncountable. This is a contradiction so $S$ must be uncountable.

The set $S$ can not contain any intervals or intervals of irrationals.

If the set $S$ contained an interval of irrationals, $(a,b) \cap (\mathbb{R} - \mathbb{Q}) \subseteq S$, and for some $x \in (a,b) \cap (\mathbb{R} - \mathbb{Q})$.

If $x \leq \frac{a+b}{2}$, then $x + \frac{b-a}{3} < b$ so $x + \frac{1}{\frac{1}{\lceil 3(b-a) \rceil}} \in (a,b) \cap (\mathbb{R} - \mathbb{Q})$

If $x \geq \frac{a+b}{2}$, then $x - \frac{b-a}{3} > a$ so $x - \frac{1}{\frac{1}{\lceil 3(b-a) \rceil}} \in (a,b) \cap (\mathbb{R} - \mathbb{Q})$. As these points have a rational difference with $x$ this is a contradiction. As $(a,b) \cap (\mathbb{R} - \mathbb{Q}) \subseteq (a,b)$, this also applies to regular intervals.

I have also tried a number of methods of generating $S$, but none of them were able to produce a result. So How would I go about finding an $S$ or proving one doesn't exist?

  • $\begingroup$ Have you heard of the Axiom of Choice? $\endgroup$ – Joey Zou Sep 30 '15 at 21:04
  • $\begingroup$ I have, but I have never seen it introduced formally. It does make quick work of this problem now that I've looked at it. $\endgroup$ – ixsetf Sep 30 '15 at 21:38
  • $\begingroup$ Well, allow me, then. The axiom of choice, this is ixsetf; ixsetf this is the axiom of choice. And you may have already met her brother Zorn's lemma, this is ixsetf. :-P $\endgroup$ – Asaf Karagila Sep 30 '15 at 22:07

What you're describing is what's called a Vitali set. It can be created by taking one element from each equivalence class generated from the equivalence relation '$ \sim $' defined by $x \sim y$ if $x-y \in \mathbb{Q}$. It's an example of a set which is not Lebesgue measurable. For further info have a look at the wikipedia page: https://en.wikipedia.org/wiki/Vitali_set


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