accumulation point of infinite compact subset of $\mathbb{R}^2$

Question

Let $A$ be an infinite compact subset of $\mathbb{R}^2$ and acc$(A)=\{0\}$. Then show that

$0 \in \mbox{bd}(A)$.

Here bd($A$) denotes the boundary of $A$ and acc($A$) is the set of accumulation points of $A$.

(A point $x$ is an accumulation point of $A$ if every neighborhood of $x$ x contains infinitely many points of $A$. The boundary of $A$ is defined as bd$(A)=\bar{A} \cap \bar{A^c}.$)

Answer 1

Weird problem IMHO.

$0$ being an accumulation point of $A$ implies $0 \in \overline{A}$. If it were not in the boundary of $A$, it must be in the interior of $A$ (as $A$ is closed (from compact), $A = \operatorname{int}(A) \cup \operatorname{Bd}(A)$), so for some $r>0$, $B(0,r) \subseteq A$. But every point of $B(0,r)$ is an accumulation point of $B(0,r)$ and hence of $A$. But this contradicts $\operatorname{acc}(A)=\{0\}$. So $0 \in \operatorname{Bd}(A)$.

We use very little of the compactness of $A$ or properties of $\mathbb{R}^2$. We could have sufficed with: let $A$ be an infinite subset of $(X,d)$, a metric space where every open ball is infinite, with a unique accumulation point $p$. Then $p$ is a boundary point of $A$.