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A perfect metric space is defined as one where $M=M'$, where $M$ is the space itself and $M'$ is the set of all cluster points of $M$, that is, points whose $\epsilon$-neighbourhoods contain infinitely many elements of $M$.

I think that it is sufficient for a space to be perfect for the closure of each open ball to equal the corresponding closed ball.

Is it necessary though? Looking at this proof, it seems like the accepted answer has a weaker necessary condition than perfectness, but I'm not certain.

Student
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    On $\mathbb{R}$, consider $d(x,y) = \min { 1, \lvert x-y\rvert}$. – Daniel Fischer Sep 04 '20 at 18:01
  • This metric space is perfect, isn't it? Each $\epsilon$-ball will contain infinitely many points. But I don't think the closure of an open $1$-ball is the closed $1$-ball with same centre......so sufficiency may not hold either. – Student Sep 04 '20 at 18:08
  • Now I wonder whether I parsed your second paragraph as you intended it. Did you mean "closure property implies perfect" or "perfect implies closure property"? – Daniel Fischer Sep 04 '20 at 18:13
  • @DanielFischer I meant to ask whether closure property implies perfect. But now looking at your first comment, I don't think sufficiency holds, whereas I have written the opposite in the question. Which in turn could mean that the condition in the other answer is in fact stronger, not weaker. – Student Sep 04 '20 at 18:14
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    It does, except for spaces containing only one point. If there are at least two points, "For any two distinct points $x,y$ in the space, and any positive $\epsilon$, there is a point $z$ within $\epsilon$ of $y$, and [something that implies $z \neq y$]" tells you that $y$ is a cluster point. Since $y$ is arbitrary, the space is perfect. The empty space is perfect too. But singleton spaces aren't perfect, yet they have the closure property (trivially). – Daniel Fischer Sep 04 '20 at 18:22
  • Okay so fundamentally, I seem to have things the other way around. Closure property implies perfect (for non-singleton spaces), and perfection is weaker than the property mentioned in the linked proof. Thank you! (maybe you could put this in an answer?) @DanielFischer – Student Sep 04 '20 at 18:36

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A space being perfect or not depends only on the topology of the space, while the closure property (the closure of the open ball $B_r(x) = \{ y : d(x,y) < r\}$ is the closed ball $K_r(x) = \{ y : d(x,y) \leqslant r\}$ for all $r \in (0,+\infty)$ and all $x \in X$) clearly depends on the metric. On every metric space $(X,d)$ with at least two points we can find an equivalent metric (i.e. a metric $d'$ inducing the same topology as $d$) such that there is at least one open ball $B_r(x)$ whose closure is a proper subset of $K_r(x)$. A uniform construction works. There always is a point $x_0 \in X$ and a $c > 0$ such that $B_c(x_0)$ is not dense in $X$, and there is a $y \in X$ with $d(x_0,y) = c$. Then taking $d'(x,y) = \min \{ c, d(x,y)\}$ works.

Hence a perfect metric space need not have the closure property.

Conversely, the closure property implies that the space is perfect if the space is not a singleton space.

This is vacuously true for the empty space. If the space contains more than one point, choose an arbitrary $x \in X$, and $y \in X \setminus \{x\}$. Let $r = d(x,y)$. Then by the closure property $x \in \overline{B_r(y)}$, hence for all $\varepsilon > 0$ we have $B_r(y) \cap B_{\varepsilon}(x) \neq \varnothing$. Since $x \notin B_r(y)$ it follows that $B_{\varepsilon}(x) \setminus \{x\} \neq \varnothing$. Hence $x \in X'$. Since $x$ was arbitrary, $X$ is perfect.

Singleton spaces aren't perfect, but they trivially have the closure property.

Daniel Fischer
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