Let $F$ be a nonempty closed set of reals and denote by $I$ the set of its isolated points.
By definition, $F\setminus I$ is dense-in-itself.
Is it true that $F\setminus I$ is closed?
[Math] A closed set without isolated points
general-topology
general-topology
Let $F$ be a nonempty closed set of reals and denote by $I$ the set of its isolated points.
By definition, $F\setminus I$ is dense-in-itself.
Is it true that $F\setminus I$ is closed?
Best Answer
Yes. If $x\in I$, then $\{x\}$ is an open subset of $F$. Since $I=\bigcup_{x\in I}\{x\}$, $I$ is an open subset of $F$ and therefore $F\setminus I$ is closed.