Just thinking if this proof works, and i have some "not so explicit" ending so i would love if you suggest a way for refining this.
Theorem : $S \subset \mathbb R^n$ is a set. Then the set of isolated points of $S$ is countable.
Proof: Let $S_{\lambda}$ denote the set of isolated points of $S$.
Thus $\forall ~ \bf{x}$ $ \in S_{\lambda}$, $\exists ~ \varepsilon_{\bf{x}} \in \mathbb R^+$ such that $B(\bf{x}$ ,$~\varepsilon_{\bf{x}})\cap S =\phi$
We know that the set of open balls of rational coordinates is countable.
Say $\bf{x}$ $=(x_1,x_2,\cdots,x_n)$ and chooae rationals $r_j$ such that $||x_j -r_j||<\frac{\varepsilon_{\bf{x}}}{4n}$.
Set $r_{\phi_x}=(r_1,r_2,\dots, r_n)$
Then $||~ \bf{x}$ $-r_{\phi_x}~||< \frac{\varepsilon_{\bf{x}}}{4}$
Thus $B(r_{\phi_x,\frac{\varepsilon_{\bf{x}}}{4}}) \subseteq B(\bf{x}$ $,\varepsilon_{\bf{x}})$.
Like this we can generate a ball with rational coordinate which is a subset of the ball with center at $\bf{x}$ with radius $\varepsilon_{\bf{x}}$ for all $x \in S_{\lambda}$.
To end, we define the map $\Gamma:B_{\mathbb Q}\to B_{S_{\lambda}}$ (where $B_{\mathbb Q}$ is the set kf balls with rational coordinates (and is countable) and $B_{S_{\lambda}}$ is the set of balls $B(\bf{y}$ $,\varepsilon_{\bf{y}})$ for all $\bf{y}$ in $S_{\lambda}$) such that $\Gamma (B(r_{\phi_x},\frac{\varepsilon_{\bf{x}}}{4}))=B(\bf{x}$ $,\varepsilon)$.
This map is injective when the $r_{\phi_y}$'s are kept fixed and hence $B_{S_{\lambda}}$ is countable which gives (please suggest a good way, i mean intuitively it makes me think its correct but, here i am in need of much help) $S_{\lambda}$
Much thanks!
Best Answer
Around each isolated point $x ∈ S$ is an open $n-ball$ $B(x)$ such that $B(x)\cap S=\phi$.Then there is an $n-ball$ $A_x$ with rational radius and rational center coordinates such that $x ∈ A_x ⊂ B(x)$. The map $x\longmapsto A_x$ is a one-to-one correspondence between the isolated points of $S$ and a subset of the countable set of all open $n-balls$ with rational center and radius.