Prove o disprove that $X$ is connected?

Question

Given $ r> 0 $, let $ C_r $ be the circumference in the plane that has center at $ (0,0) $ and radius $ r $. Let $ X $ be a subset of $ \Bbb R ^ 2 $ that has the following properties,

1. For all $r\in\Bbb Q$, $C_r\subseteq X$ and,
2. for all $r\in\Bbb R\setminus \Bbb Q$, $X\cap C_r\neq \varnothing$.

Is $X$ connected?

I affirm that it is connected. And to demonstrate, I do it by contradiction and I assume that $ X $ is not connected, so there are two separate sets $ A $ and $ B $, such that $ X = A \cup B $.

I have a minimal idea, and that is that for example, $ C_1 $ being a connected set, being contained in $ X $, it must be completely contained in $ A $ or completely contained in $ B $, since it is $ 1 \in \Bbb Q $. Without loss of generality let's say that it is completely contained in $ A $.

So somehow show that all other $ C_r $ is contained in $ A $. And then use some density argument to show that the points that are in $ X \cap C_r $ with $ r \in \Bbb R \setminus \Bbb Q $, cannot be in $ B $. So $ B $ has to be empty and I would have my absurdity. Any help for the exercise or how I can develop my idea. Thanks a lot.

Answer 1

Suppose that $X$ is not connected. Then there exist disjoint open subsets $U$, $V$ of $\mathbb{R}^2$ that separate $X$.

Let $R_U = \{\|x\| : x \in U \cap X \}$ and $R_V = \{\|y\| : y\in V \cap X\}$. We verify the properties of $R_U$ and $R_V$.

• It is clear that $R_U \cup R_V = [0, \infty)$.

• Lemma 1. $R_U \cap R_V \cap \mathbb{Q}^+ = \varnothing$.

  Let $r \in \mathbb{Q}^+ = \mathbb{Q} \cap (0, \infty)$. Then $C_r \subseteq X \subseteq U \cup V$. Since $C_r$ is connected, $U$ and $V$ cannot separate $C_r$. This implies that either $C_r \subseteq U$ or $C_r \subseteq V$, but not both.

• Lemma 2. If $r \in R_U$, then there exists $\varepsilon > 0$ such that $B_{\mathbb{Q}^+}(r, \varepsilon) = \mathbb{Q}^+ \cap (r-\varepsilon, r+\varepsilon)$ lies in $R_U$, and a similar statement holds for $R_V$.

  Indeed, choose $x \in U \cap X$ such that $\|x\| = r$. Then there exists an open ball $B_{\mathbb{R}^2}(x, \varepsilon) $ contained in $U$. Then for each $s \in B_{\mathbb{Q}^+}(r, \varepsilon)$, we have $\varnothing \neq C_s \cap B_{\mathbb{R}^2}(x, \varepsilon) \subseteq C_s \cap U $ and hence $s \in R_U$.

• Lemma 3. Both $R_U$ and $R_V$ are closed.

  Suppose $r_n \in R_U$ and $r_n \to r$ in $\mathbb{R}$. If $r \notin R_U$, then $r \in R_V$, Then by the previous step, there exists $\varepsilon > 0$ such that $B_{\mathbb{Q}^+}(r, \varepsilon) \subseteq R_V$. However, $r_n \in (r-\varepsilon, r+\varepsilon)$ for any sufficiently large $n$, and so, by Lemma 2 there exists $\delta > 0$ such that $(r_n-\delta, r_n+\delta) \subseteq (r-\varepsilon, r+\varepsilon)$ and $B_{\mathbb{Q}^+}(r_n, \delta) \subseteq R_U$, a contradiction. Therefore $r \in R_U$.

• Lemma 4. Both $R_U$ and $R_V$ are open.

  Let $r \in R_U$. Then $B_{\mathbb{R}^+}(r, \varepsilon) \subseteq R_U$ for some $\varepsilon > 0$, hence by Lemma 3, $\overline{B_{\mathbb{Q}^+}(r, \varepsilon)}$ also lies in $R_U$. However, since $[0, \infty) \cap (r-\varepsilon, r+\varepsilon) \subseteq \overline{Q_{\mathbb{Q}^+}(r, \varepsilon)}$, the $\varepsilon$-neighborhood of $r$ lies in $R_U$ and hence $r$ is an interior point of $R_U$.

Now we are ready for the conclusion. Since $[0, \infty)$ is connected, the only non-empty clopen (both closed and open) subset of $[0, \infty)$ is $[0, \infty)$ itself. Then Lemma 3 and 4 together imply that $R_U = [0, \infty) = R_V$, a contradiction.