Isolate points of a metric space with some properties?

Question

Suppose that all dense subspace of a metric space $(X,d)$ is open.

Prove that the set of the isolate points of $X$ is dense in $X$.

My Idea: all isolate points of $X$ are in any dense subspace, so the set of isolate points is open, that implies $X$ is finite set.

Answer 1

HINT: The easiest approach is to prove the contrapositive: suppose that the set of isolated points of $X$ is not dense in $X$, and show that $X$ must have a dense subset that is not open in $X$. Let $I$ be the set of isolated points of $X$, and let $U=X\setminus\operatorname{cl}I$; if $I$ is not dense in $X$, then $U\ne\varnothing$. Fix a point $p\in U$; $p$ is not isolated, so there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X\setminus\{p\}$ that converges to $p$. Let $A=\{x_n:n\in\Bbb N\}$.

• Show that without loss of generality we may assume that $A\subseteq U$.
• Show that $A$ is infinite.
• Show that $X\setminus A$ is a dense set that is not open.