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This question popped up the other day while browsing about perfect spaces.

Suppose $(X,\rho)$ is a perfect metric space, and I take any ball $B_r(x)$, the ball centered at $x$ of radius $r$. Out of curiosity, for any $b>0$ sufficiently small, can I then find distinct points $x,y\in B_r(x)$ such that the closures $Y:=\text{cl}{B_b(y)}$ and $Z:=\text{cl}{B_b(z)}$ are disjoint, both contained in $\text{cl}{B_r(x)}$, and both perfect in the induced topology?


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up vote 2 down vote accepted

Yes. You can do better than that. Let $y$ and $z$ be distinct points in $B_r(x)$. Choose a positive number $b < \min\{r-\rho(x,y)$, $r-\rho(x,z),\rho(y,z)/2\}$. Then $$\begin{align*}&\operatorname{cl}B_b(y) \subseteq B_r(x), \text{ since }b<1-\rho(x,y);\\ &\operatorname{cl}B_b(z) \subseteq B_r(x), \text{ since }b<1-\rho(x,z);\text{ and}\\ &\operatorname{cl}B_b(y) \cap \operatorname{cl}B_b(z), \text{ since }b<\rho(y,z)/2. \end{align*}$$

Suppose that $p \in \operatorname{cl}B_b(y)$. If $p \in B_b(y)$, $p$ is clearly not isolated in $\operatorname{cl}B_b(y)$ (since it’s not isolated in $X$, and $B_b(y)$ is open in $X$). If $p \in \operatorname{cl}B_b(y) \setminus B_b(y)$, let $V$ be any relatively open nbhd of $p$ in $\operatorname{cl}B_b(y)$. Then there is an open $U$ in $X$ such that $V = U \cap \operatorname{cl}B_b(y)$, and since $p \in \operatorname{cl}B_b(y)$, $U \cap B_b(y) \ne \varnothing$. But $U\cap B_b(y) \subseteq V$, so $V\cap B_b(y) \ne \varnothing$. Since $p\notin B_b(y)$, this implies that $V \ne \{p\}$, and since $V$ was an arbitrary relatively open nbhd of $p$ in $\operatorname{cl}B_b(y)$, $p$ is not isolated in $\operatorname{cl}B_b(y)$. Thus, $\operatorname{cl}B_b(y)$ is perfect. The argument that $\operatorname{cl}B_b(z)$ is perfect is mutatis mutandis identical.

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