If isolated points are not dense, then removing their closure leaves a space without isolated points

Question

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 $\operatorname{iso}\left(Z\setminus\overline{\operatorname{iso}\left(Z\right)}\right)\ne\emptyset$?

Answer 1

(Getting another question off the unanswered list.)

The result actually holds for all topological spaces.

Let $\langle Z,\tau\rangle$ be a topological space, let $I$ be the set of isolated point of $Z$, and let $Y=Z\setminus\operatorname{cl}I$. Let $\tau_Y$ be the subspace topology on $Y$, and let $y\in Y$ be arbitrary. Suppose, to get a contradiction, that $y$ is an isolated point in the subspace $\langle Y,\tau_Y\rangle$. Then $\{y\}$ is open in $Y$, so there is a $U\in\tau$ such that $\{y\}=U\cap Y$. But $Y$ is open in $X$, so $Y\in\tau$, and hence $\{y\}=U\cap Y\in\tau$, and $y$ is an isolated point of $Z$, i.e., $y\in I$. But $y\in Y$, so $y\in Y\cap I=\varnothing$, which is absurd. This contradiction shows that $y$ cannot be an isolated point of $\langle Y,\tau_Y\rangle$ and hence that $Y$ has no isolated points. (Of course $Y$ is empty if $I$ is dense in $Z$, but the empty space certainly has no isolated points!)