One thinking problem of uncountable set in real analysis Assume: $E \subseteq R^2$ and $E$ is a uncountable set, try to prove that there exists $x_0 \in E$ s.t.$\forall B(x_0)$($B(x_0)$ is an open circular neighborhood, $x_0 \in B(x_0)$ and the center is unnecessarily $x_0$), $B(x_0) \cap E$ is a uncountable set.
The answer proves it reversely: assume that: $ \forall x \in E$, there exists one circular neighborhood $B(x)$($x \in B(x)$) s.t. $ E \cap B(x)$ is a countable set. Then $\forall x\in E$, choose a $r_x  \in \mathbb Q$, s.t.$x \in B(x,r_x)$. Therefore, $E=E \cap \bigcup _{x}{B(x,r_x)}=\bigcup_{r_x}{ E \cap B(x,r_x)}$,thus $E$ is a countable set, which leads to contradiction.
Q:For two different $x$, there may be one same $r_x$, which means it is not a single-shot mapping from the set of $B(x,r_x)$ to $ \mathbb Q$. So it can’t prove that the set of all $B(x,r_x)$ is countable? The answer is wrong?
 A: I think the answer is wrong. Your idea is right. For some $r_{x_0}$, there are uncountable many $x$ s.t. $r_x=r_{x_0}$. (Otherwise $E$ would be countable.) Therefore, the map $x\mapsto r_x$ is not injective. So $E\cap\bigcup_xB(x,r_x)=\bigcup_xE\cap B(x,r_x)\neq \bigcup_{r_x}E\cap B(x,r_x)$.
The correct solution can be found in other answers.
A: I thought it for a long time and I got another answer, please help me verify the answer, thanks!
Prove it by contradiction: $ \forall x \in E$, there exists one circular neighborhood $B(x)$($x \in B(x)$) s.t. $ E \cap B(x)$ is a countable set. Then $\forall x\in E$, choose a rational dot $q=(r_x,r_y)$ as the center and a rational number $ \delta$ as the radius, s.t.$x \in B(q,\delta) \subseteq B(x)$. Set $\Gamma=\{B(q,\delta)|\exists x\in E$,s.t.$x\in B(q,\delta) \subseteq B(x)$}. We claim that $ \Gamma$ is countable cause there is an injective mapping from $\Gamma$ to the subset of $\mathbb Q^3$. Thus, $E \subseteq \bigcup_{S \in \Gamma}{S} \subseteq \bigcup_{S\in \Gamma}{B(x)}$. Therefore, $E=E\cap\bigcup_{S\in\Gamma}{S}=E \cap \bigcup_{S \in  \Gamma}{B(x)}=\bigcup_{S \in \Gamma}{E \cap B(x)}$, which leads to the contradiction.
A: I believe your concern regarding the supplied proof is valid. As a matter of fact, I think the chosen proof strategy isn't the optimal one - it is often said that a direct proof is better than a proof by contradiction. So I will attempt to produce one (although an intermediate result I'll use will be by contradiction).
Let $E\subseteq \mathbb{R}$, and $E$ is uncountable. We claim that $E$ must contain an open interval, say $(a,b)$. Suppose $E$ does not contain any open intervals. Then $E$ is a discrete set (i.e. consists entirely of isolated points). This means that for any $x\in E$ we can find an open neighbourhood $B_{\varepsilon(x)}(x)$ such that $B_{\varepsilon(x)}(x)\cap E = \{x\}$. But each $B_{\varepsilon(x)}(x)$ contains a rational number, say $q_x$ (due to the density of $\mathbb{Q}$ in $\mathbb{R}$). Because the open neighbourhoods do not intersect, we can 'index' all elements of $E$ by a subset of the rationals, and so $E$ itself must be countable, a contradiction. Therefore $\exists a,b\in \mathbb{R}$ such that $a<b$ and $(a,b)\subseteq E$.
The rest of the proof follows quite smoothly simply by taking any $x\in E$ such that $a<x<b$. Let $B(x)$ be an open neighbourhood containing $x$. Pick some $y\in B(x)\cap (a,b)$ such that $y> x$ (it exists because $B(x)$ and $(a,b)$ are both open intervals containing the same point). Then set $\varepsilon = \min\{y-x, x-a\}$. Clearly, the resulting (uncountable) open neighbourhood  $B_{\varepsilon}(x)\subseteq (a,b)$ and $B_{\varepsilon}(x)\subseteq B(x)$, hence $$B_{\varepsilon}(x)\subseteq (a,b)\cap B(x)\subseteq E \cap B(x).$$ Since $E\cap B(x)$ has an uncountable subset, it must be uncountable too.
$\underline{\text{P.S.}}$ I've just realised $E$ was supposed to be a subset of $\mathbb{R}^2$, not $\mathbb{R}$. Sorry for not spotting it earlier. But I believe my argument generalises to $\mathbb{R}^2$ without any issues (just replace 'open interval' by 'open disc', and modify the proof accordingly).
