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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.

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1 Answer 1

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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.

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