Open domain in general topology

Question

A subset of a topological space is an open domain if it is equal to the interior of its closure. Show that if U is the interior of a closed set, then it is an open domain. Futhermore, show that the intersection of two open domains is an open domain.

Any ideias?

Answer 1

Part 1: Let $Z$ be a closed set, and $U = \text{Int}(Z)$. We want to show $U$ is the interior of its closure.

Since $Z$ is a closed set containing $U$, $\overline{U} \subset Z$, and so $\text{Int}(\overline{U}) \subset \text{Int}(Z) = U$. On the other hand, since $U$ is an open set contained in $\overline{U}$, $U \subset \text{Int}(\overline{U}))$.

Part 2: Let $U$ and $V$ be open domains. We want to show $(U \cap V)$ is the interior of its closure. Since $U \cap V$ is an open set contained in $\overline{U \cap V}$, $U \cap V \subset \text{Int}(\overline{U \cap V})$. On the other hand, since $\overline{U \cap V} \subset \overline{U}$, $\text{Int}(\overline{U \cap V}) \subset \text{Int}(\overline{U}) = U$. For the same reason, $\text{Int}(\overline{U \cap V}) \subset V$. So $\text{Int}(\overline{U \cap V}) \subset U \cap V$.

Answer 2

$Int(S)=Int(\bar{S})$ for any closed set $S$.

So what do you know about closure of $U$?

Answer 3

A well known identity for interiors and closures is, for all $A$ a subset of a topological space:

$$\operatorname{int}\operatorname{cl}\operatorname{int}\operatorname{cl}(A) = \operatorname{int}\operatorname{cl}(A)$$

I proved it in my note here as formula (p).

This implies your statement immediately as we can write the interior of a closed set $S$ as $\operatorname{int}\operatorname{cl}(S)$.