1

Let $X$ be a topological space and $A\subseteq X$. We say that $A$ is a regular closed if $A=\text{cl}(\text{int}(A))$

Is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set?

Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists?

Thanks in advance.

Adrina
  • 127
  • 1

1 Answers1

2

$\DeclareMathOperator\Cl{cl}\DeclareMathOperator\Int{int}\let\bez\smallsetminus$If every closed set is regular, then every closed (or open) set is clopen: given a closed set $A$, $B=A\bez\Int(A)$ is a closed set with $\Int(B)=\varnothing$, thus $B=\Cl(\Int(B))=\varnothing$, and $A$ is open.

It follows that any such space $X$ is a disjoint union of indiscrete spaces: for any point $x$, $\Cl(\{x\})$ is a minimal clopen set, and $\{\Cl(\{x\}):x\in X\}$ forms a partition of $X$ into indiscrete subspaces. This gives a full characterization (clearly, all closed sets are clopen and thus regular in any disjoint union of indiscrete spaces).

Emil Jeřábek
  • 1,995