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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?

Andrés E. Caicedo
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Dana
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  • I don't understand the third last sentence, about $(0,\sqrt2)$. Can you clarify? – arkeet Oct 11 '16 at 00:13
  • @arkeet So, let's say that for every $q \in \mathbb{Q}^2$, we always pick some radius lower than $d(q,(0,\sqrt{2}))$ to draw our open ball around $q$. Then wouldn't the resulting countable union of balls around rationals not include $(0,\sqrt{2})$? – Dana Oct 11 '16 at 00:17
  • What do you do if $x = (0,\sqrt2)$? – arkeet Oct 11 '16 at 00:19
  • @arkeet: there is nothing to do if $x \not\in \Bbb{Q}^{2}$. – Rob Arthan Oct 11 '16 at 00:22
  • OK, misread something. I guess the concern is how we cover the point $a=(0,\sqrt2)$? Well you pick a point $q$ close enough to $a$, but then you can't pick the ball around $q$ to be too small, or else it won't contain $a$. – arkeet Oct 11 '16 at 00:25
  • Yes, you understand my concern. I do not understand how reducing from uncountably many balls to countably many balls centered on rationals will necessarily still cover the set. – Dana Oct 11 '16 at 00:26
  • In the proof, we choose the radius first, and then the choice of $q$ depends on the radius. Let me write out a proof in an answer to hopefully clarify it. – arkeet Oct 11 '16 at 00:33
  • See Any open subset of R is a at most countable union of disjoint open intervals. Very loosely speaking, you first construct a union of maximal open subsets, and only then you apply the rational argument. The maximality established in the first step ensures that coverage is not dropped. – dxiv Oct 11 '16 at 00:34
  • @dxiv: I don't see how you can get the maximality trick to work in the plane. – Rob Arthan Oct 11 '16 at 00:42
  • @RobArthan Fair point. Maybe I read too fast, but I thought that was the main idea behind this answer which could extend to $\mathbb{R}^2$. – dxiv Oct 11 '16 at 00:47
  • If $S$ is open and $p=(0,\sqrt 2)\in S$ then we certainly can find a countable family $F$ of open balls with centers in $\Bbb Q^2$ and rational radii, such that $\cup F=S \setminus {p},$ because $S$ \ ${p}$ is also open. The point of the theorem is that there does exist at least one family $G$ of open balls with centers in $\Bbb Q^2$ and rational radii, such that $\cup G=S.$ – DanielWainfleet Mar 13 '20 at 22:04

2 Answers2

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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 $B(x, \delta_x)$.

Rob Arthan
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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.

arkeet
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