Uncountable set of irrational numbers such that no two elements sum to a rational I'm trying to find such a set as described in the title to prove that the following metric space is separable:
$([0,1],d)$
where
$ d(x,y) = \left\{
 \begin{array}{ll}
  |x-y|  & \mbox{if } x-y \in \mathbb{Q}\\
  5 & \mbox{if } x-y \not\in \mathbb{Q}
 \end{array}
\right.$
If there exists an easier approach then any suggestions are welcome. Also if I'm going the wrong way and it is separable I would like to ask how to do it the other way around.
 A: Note that $x\sim y\iff x-y\in\Bbb Q$ defines an equivalence relation on $[0,1]$. Moreover, this equivalence relation has uncountably many equivalence classes. Suppose that $V\subseteq[0,1]$ is a set which meets every equivalence class exactly on one point, and now consider $A=\{B(x,1/3)\mid x\in V\}$.
Show that $A$ is an uncountable set of pairwise disjoint open sets.
A: This metric space is not separable. To see this consider the partition on $[0,1]$ induced by the relation $x \sim y \iff x-y \in \mathbb  Q$. It can be shown each equivalence class is countable and is a translated copy of $\mathbb Q$. For example $\mathbb Q$ itself is the equivalence class of $0$ while $\mathbb  Q + \pi$ and $\mathbb Q + \pi + 1$ both represent the class of the irrational $\pi$. Since each equivalence class has measure zero, there must be uncountably many classes. Thus a countable subset $D$ of your space can only contain elements from countably many classes. Then if $z$ is in a different class the open ball $B(z,1)$ is disjoint from $D$.
A: Extend $\{1\}$ to a basis of $\mathbb R$ as a $\mathbb Q$-vector space, call it $B$.
Then no $\mathbb Q$-linear combination of elements of $B \setminus \{1\}$ is rational.
A: Zorn's Lemma yields a maximal subset $A\subset [0,1]$ with the property that for each $x,y\in A$ with $x\neq y$ we have $x-y\notin \mathbb Q$. Suppose $A$ was countable. Then for each $x\in A^c$ there is $a(x)\in A$ with $x-a(x)\in \mathbb Q$. Since $A^c$ is uncountable, there must be $z\in A$ such that $Z := a^{-1}(\{z\})$ is uncountable. Now for each $x,y\in Z$ we have $x-y = x-a(x)-(y-a(y))\in \mathbb Q$. But this yields an injection $Z\to \mathbb Q, y\mapsto x-y$ for fixed $x\in Z$. Contradiction. 
A: Here's an odd approach that relies heavily on the Axiom of Choice: for any irrational $r$, the set $\{s \mid r + s \in \mathbb{Q}\}$ is countable. So for any countable set $R$ of irrationals, the set $\{s \mid (\exists r \in R)s + r \in \mathbb{Q}\}$ is a countable union of countable sets, and hence countable.
Let $r_0$ be any irrational. For any countable ordinal $\alpha > 0$, let $r_{\alpha}$ be an irrational not equal to any previously selected $r_{\beta}$ and not contained in the set $\{s \mid (\exists \beta < \alpha)s + r_{\beta} \in \mathbb{Q}\}$.
The sequence $\langle r_{\alpha} \mid \alpha < \omega_1\rangle$ (recall that $\omega_1$ is the first uncountable ordinal) is an uncountable sequence of irrationals no two of which sum to a rational.
