Topology: Reducing Open Set as Union of Open Balls to Countable Union I have seen some proofs on the topic of how to prove that an open set $A$ in $\mathbb{R}^2$ is a countable union of open balls by selecting one such ball for every $x \in A \cap \mathbb{Q}^2$. The justification is that since Q2 is dense, any selection of a distance and a point in R2 will yield some rational number within that range of that point, which I understand. However, can we not select balls so as to exclude this point? Why couldn't we select, for every rational $r \in \mathbb{Q}^2$, some ball of radius just slightly less than every than $\|r|-|(0, \sqrt{2})\|$? This would not contradict the theorem itself, but it would make the step of reducing to every non-Q2 ball potentially lose complete coverage of the set. What am I missing?
 A: To prove that any open subset $A$ of $\Bbb{R}^2$ is a countable union of open disks, you need to take the union over all points $x \in A \cap \Bbb{Q}^2$ of all open disks $B(x, \delta)$ such that $\delta \in \Bbb{Q}$ and $B(x, \delta) \subseteq  A$. If you just picked one $\delta$ for each $x$, then you might indeed omit some points of $A$.
For example,, take $A = \Bbb{R}^2$, and, for each $x \in A \cap \Bbb{Q}^2 = \Bbb{Q}^2$, pick a rational number $\delta_x > 0$ such that $d(x, (\sqrt{2}, 0)) < \delta_x$, then $(\sqrt{2}, 0)$ is not in any of the disks $B(x, \delta_x)$.
A: Let me write out a proof and try to address your concern in the proof.
Given an open set $A \subseteq \mathbb{R}^2$, suppose we have a point $x \in A$. Then there is some $\epsilon_x > 0$ such that the open ball $B(x,2\epsilon_x)$ is contained in $A$. We may assume $\epsilon_x$ is rational by making it smaller.
Because $\mathbb{Q}^2$ is dense, we can pick a point $q_x \in \mathbb{Q}^2$ with $\|x-q_x\| < \epsilon_x$. Here we have already fixed the radius $\epsilon_x$; we are then choosing $q_x$ dependent on $\epsilon_x$. Note that $q$ is in $A \cap \mathbb{Q}^2$. Then $x \in B(q_x,\epsilon_x)$. If $y \in B(q_x,\epsilon_x)$, then $\|y - x\| \le \|y - q_x\| + \|q_x - x\| < 2\epsilon_x$, so $B(q_x,\epsilon_x) \subseteq B(x,2\epsilon_x) \subseteq A$.
So I have shown two things:


*

*For all $x \in A$, $x \in B(q_x, \epsilon_x)$. Therefore, $A \subseteq \bigcup_{x \in A} B(q_x, \epsilon_x)$.

*For all $x \in A$, $B(q_x, \epsilon_x) \subseteq A$. Therefore, $\bigcup_{x \in A} B(q_x, \epsilon_x) \subseteq A$.


Then $A = \bigcup_{x \in A} B(q_x, \epsilon_x)$. In fact this is a countable union, because there are only countably many balls with rational center and rational radius.
