Let $C$ be a compact set in $\Bbb{R}^n$ with the following property:
for any $\varepsilon> 0$ there exists a finite number $M(=M(\varepsilon))$ of open balls $B_i$ with radius $r_i$, $i\in\{1,\ldots,M\}$ s. t.
$\bigcup\limits_{i=1}^{M}B_i \supset C $
and
$\sum\limits_{i=1}^{M} r_i \leqslant \varepsilon $
Prove the following :
(1) for $n \geqslant 2 \; \Bbb{R}^n \setminus C$ is connected
(2) for $n \geqslant 3 \;\Bbb{R}^n\setminus C$ is simply connected (I mean by that it has trivial fundamental group)
The only thing I was able to try were proof by contradiction or trying to understand intuitively how this space is made but I did not conclude . In particular , I tried this:
for $\varepsilon > 0$ call $S_{\varepsilon }$ the finite cover of $C$ made by a finite number of balls whose sum of radii is less or equal than $\varepsilon$,say it $\{B_{i}^{\varepsilon}\}_{i=1}^{M(\varepsilon)},$which exists by hypotesis. I noticed that, since $C$ is closed , if a point $x$ does not belong to $C$ , then $L=\operatorname{dist}(x,C) > 0$ . Choosing a cover of $C$ of the form $S_{\varepsilon } $ for $\varepsilon < L$ and assuming that every element of $S_{\varepsilon }$ intersects nontrivially $C$, I can argue that $x \notin B_i^\varepsilon $ for any $i$. Otherwise, there would be a point $y$ in $C$ s. t. $\operatorname{dist}(x,y)< L$ which is not possible.
Define $A_{\varepsilon}= \bigcup\limits _{i=1}^{M(\varepsilon)} B_i^\varepsilon$.From what said, if I consider a small enough $\varepsilon$,then $x \notin A_{\varepsilon }$. Choosing $\varepsilon = \frac{1}{i}$ for $i \in \Bbb{N}$ from what said I could argue that
$\mathbb{R}^n \setminus C = \bigcup\limits_{i=1}^{\infty}(\Bbb{R}^n\setminus A_{\frac{1}{i}})$
Manipulating a bit this equality (not straightforward ) ,if, for example I may prove that $\exists j \geqslant 1 $ s. t. $\mathbb{R}^n\setminus A_{\frac{1}{k}}$ is connected for any $k \geqslant j$, I may easily get point (1). Anyway I do not see how to prove it. I do not even see clearly if my previous observations are correct.
REMARK: Someone suggested this set is discrete , I think that this not set is necessarily discrete: If I take $C=\{(\frac{1}{n},0) \cup {(0,0)}\}$ ,$n \in \mathbb{N}$,and choose $\epsilon > 0$ I can find a cover of $C$ according to the hypotesis:
Choose a ball centered in $(0,0)$ of radius less than $\frac{\epsilon}{2}$. You are left to cover a finite number $M$ of points. For any point,choose a open ball centered in itself with radius less than $\frac{\epsilon}{2M}$. Formally you can do it because a point has a zero lebesgue measure and you can cover it with a ball of any radius you want. This cover respects the hypotesis.
Since I found in a standard topology exercise set ,its solution should not require big machinery, so I would really appreciate a solution of this kind. In particular, it is a exercise from an admission test to a fourth year of universiy, so I assume it is possible to use standard general topology(from a university course), fundamental groups, universal convering and homology.Please help. Thanks in advance.