Suppose our carrier set S is a metric space.
Say that set B is arbitrarily close to set A iff, for any point x in A, for all positive $\epsilon$, there are points of B within $\epsilon$ distance of x. And say that the boundary of $A\subseteq S$ is the set of points in S arbitrarily close to A and to the complement of A in S.
My question is: Can we then define $A\subseteq S$ to be open iff the intersection of A and the boundary of A is empty, and define $A\subseteq S$ to be closed iff the boundary of A is a subset of A?
Best Answer
Yes; in fact, the result holds for general topological spaces. $\partial A$ is the set of $p \in S$ such that every open neighbourhood $p \in O$ intersects $A$ and its complement $S \setminus A$.
If $A$ is open, then we simply take our open neighbourhood to be $O = A$, which does not intersect $S \setminus A$, therefore every point in $A$ is not in the boundary.
Suppose $A \cap \partial A$ is empty, and let $p \in A$. In particular, $p \not\in \partial A$, so there exists an open neighbourhood $p \in O_p$ that does not intersect $S \setminus A$, or, equivalently, $O_p \subseteq A$. We can write $$A = \bigcup_{p \in A} O_p$$ so by closure under arbitrary unions, $A$ is open.
An analogous proof holds for metric spaces, replacing neighbourhoods with balls and $\epsilon$'s.