Is this an interesting generalization of the notion of an open set? Let $X$ denote a topological space. Some subsets $A \subseteq X$ might have the property that $\partial A = \partial(\mathrm{int}\,A).$ This is certainly true if $A$ is open (since open implies $\mathrm{int}\,A = A$).
Thus, the property that $\partial A = \partial(\mathrm{int}\,A)$ can be viewed as a generalization of "open." Edit: Lets call this property "weakly open."
Other $A$'s might have the property that $\partial A = \partial(\mathrm{cl}\,A)$. Lets call this "weakly closed."
So any given $A \subseteq X$ may be weakly open only, weakly closed only, it may be have both properties, or neither.
Is there anything interesting that can be said about such $A$'s? Reference, anyone?
Edit: As Arthur points out in his answer, the arbitrary union of weakly open sets is weakly open. Thus, we can take the "weak interior" of a subset $B$, defined as the union of the set of all weakly open $A$ such that $A \subseteq B$. The weak interior will always be a superset of the usual interior.
 A: For any $\newcommand{\Int}{\mathrm{Int}}A \subseteq X$ we have $\partial A = \overline{A} \setminus \Int (A)$, so in particular $$\partial ( \Int ( A ) ) = \overline{ \Int (A) } \setminus \Int ( \Int ( A ) ) = \overline{ \Int (A) } \setminus \Int ( A ).$$ As $\Int ( A ) \subseteq \overline{ \Int (A) } \subseteq \overline{A}$ your condition is equivalent to $\overline{ \Int ( A ) } = \overline{A}$.
A couple of notes:

*

*Regular closed sets ($A = \overline{ \Int ( A ) }$) are weakly open.


*Weak openness is not necessarily closed under finite intersections (e.g., in $\mathbb R$ both $[-1,0]$ and $[0,1]$ are regular closed sets).


*Weak openness is closed under finite unions (if $A$ and $B$ satisfy this condition, then $\overline{ A \cup B } = \overline{A} \cup \overline{B} = \overline{ \Int (A) } \cup \overline{ \Int(B) } = \overline{ \Int (A) \cup \Int (B) } \subseteq \overline{ \Int ( A \cup B ) } \subseteq \overline{ A \cup B }$.)


*Weak openness is even closed under arbitrary unions
Given a family $\{ A_i \}_{i \in I}$ of weakly open sets, let $x \in \overline{ \bigcup_{i \in I} A_i }$ and let $U$ be an arbitrary open neighbourhood of $x$.  As $U \cap \bigcup_{i \in I} A_i \neq \emptyset$ there is an $i \in I$ such that $U \cap A_i \neq \emptyset$.  Clearly $A_i \subseteq \overline{ A_ i} = \overline{ \Int ( A_i ) }$ and so $\emptyset \neq U \cap \Int ( A_i ) \subseteq U \cap \Int ( \bigcup_{i \in I} A_i )$.  Thus $x \in \overline{ \Int ( \bigcup_{i \in I} A_i ) }$.


*Note that if we define $A \subseteq X$ to be weakly closed iff $X \setminus A$ is weakly open it follows that $A$ is weakly closed iff $\Int ( A ) = \Int ( \overline{A} )$, which is the natural dual to weak openness.

In my opinion, due to (2) this is a fairly poor generalisation of openness (we generally like our set-structures to be closed under finite unions and intersections).  I was actually looking for counterexample to (4) for a while, and was a bit surprised to see that this holds.  At least with this you can unproblematically define the "weak interior" of a set to be the largest weakly open subset of that set.
I suspect (though have not yet been able to prove) that you can have two non-homeomorphic topologies on a set for which the families of weakly open sets coincide.  This would almost seem to put a nail in the coffin of this notion, but that would be a rash judgement: note that you can have non-homeomorphic topologies on a set for which the families of Borel sets coincide, and no-one questions the importance of Borel sets.
