Let $n\geq 2$ and denote by $B\subset \mathbb{R}^n$ the closed unit ball.


Does there exist a closed subset $A\subset B$ containing $0\in \mathbb{R}^n$ with the following properties i,ii,iii?


i) $\{0\}$ is a connected component of $A$.


ii) $0\in \overline{A\setminus\{0\}}$ and


iii) each connected component $C\neq \{0\}$ of $A$ intersects $\partial B$: $C\cap \partial B\neq \emptyset$.


A comment: $A$ needs to have infinitely many connected components by i) and ii). The first two conditions i) and ii) are easily satisfied for a set $A$ consisting of the elements of a sequence in $B\setminus \{0\}$ converging to $0$ together with its limit point. However I do not know how to construct an example of a closed set $A$ where in addition iii) holds.



Closed subset of unit ball with peculiar connected components
No.


Note that for any integer $n>0$ there is a component $C\_n\subset A$ that contains two points $x\_n$ and $y\_n$ such that $|x\_n|=\tfrac1n$ and $|y\_n|=1$.
Recall that $C\_n$ is a closed set.


Pass to a subsequence of $C\_n$ that converges in the sense of Hausdorff; denote its limit by $C\_\infty$.
We may assume that $y\_n\to y\_\infty$.


Observe that $C\_\infty$ is a closed connected subset of $A$ that contains $0$ and $y\_\infty$ --- a contradiction.


