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

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

• What is $\operatorname{iso}(Z)$? Commented Oct 24, 2017 at 21:44
• @PotatoHead47 it is the set of isolated points of $Z$.
– user281997
Commented Oct 24, 2017 at 21:46
• 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.
– user357151
Commented Oct 24, 2017 at 22:09
• 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 $c \in$ $\operatorname{iso}\left(X\setminus\overline{\operatorname{iso}\left(X\right)}\right)$ Commented Oct 24, 2017 at 22:36

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!)