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I have to show that for a complete metric space $Z$, if $Z\ne\overline{\operatorname{iso}\left(Z\right)}$, then $\operatorname{iso}\left(Z\setminus\overline{\operatorname{iso}\left(Z\right)}\right)=\emptyset$. Here $\operatorname{iso}(Z)$ is the set of isolated points of $Z$.

I'm having trouble proving this because I thought that if we remove the isolation points of a metric space $Z$, then the set of isolation points of $Z\setminus\overline{\operatorname{iso}\left(Z\right)}$ would automatically be empty, regardless of if $Z$ is complete and that $Z\ne\overline{\operatorname{iso}\left(Z\right)}$.

Is there a metric space Z such that when we remove the requirement of completeness or that $Z\ne\overline{\operatorname{iso}\left(Z\right)}$, then $\operatorname{iso}\left(Z\setminus\overline{\operatorname{iso}\left(Z\right)}\right)\ne\emptyset$?

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  • $\begingroup$ What is $\operatorname{iso}(Z)$? $\endgroup$ – MereMortal47 Oct 24 '17 at 21:44
  • $\begingroup$ @PotatoHead47 it is the set of isolated points of $Z$. $\endgroup$ – Jake Oct 24 '17 at 21:46
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    $\begingroup$ To see what's going on, consider the space $\{0\}\cup \{1/n:n\in\mathbb N\}$. If you remove all isolated points, you are left with $\{0\}$, which becomes an isolated point. This is why in this problem one removes the closure. The assumption $Z\ne\overline{\operatorname{iso}\left(Z\right)}$ is not important; if it doesn't hold, then the complement is just empty space, which doesn't have isolated points either. $\endgroup$ – user357151 Oct 24 '17 at 22:09
  • $\begingroup$ I found an example of a topological space which satisfies the condition you want, I couldn't come up with a metric space that does though. Consider the $X$={a,b,c} equipped with the topology T={$\phi$, {a},{b}, {a,b}, {b,c}, {a,b,c}} $\operatorname{iso}(X)$ ={a,b}=$ \overline{\operatorname{iso}\left(X\right)}$ then $c \in$ $\operatorname{iso}\left(X\setminus\overline{\operatorname{iso}\left(X\right)}\right)$ $\endgroup$ – MereMortal47 Oct 24 '17 at 22:36

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