Visualizing $\partial A$ I have read the Wikipedia page on boundary sets and I'm just trying to piece things together, make sure I understand them entirely (I'm still a beginner in this realm of mathematics). The book I am reading tells me it can be defined as
\begin{align}
\bar A -A^0
\end{align}
which (I think) can be rewritten as
\begin{align}
\underbrace{\bigcap\left\{B\big| A\subset B\:\text{and}\:X-B\in T\right\}}_{a}-\underbrace{\bigcup\left\{B\big| B\subset A\:\text{and}\:B\in T\right\}}_{b}=\partial A.
\end{align}
Could I visualize this by considering $a$ to be the entire closed set, including the set's boundary, and subtracting from that the entire interior of the set, $b$?
 A: Your second displayed formula looks horrible, and nobody trying to understand $\partial A$ would understand $\partial A$ better after studying it. The notion of "boundary of a set" $A\subset X$ is present in any topological space $X$; but for the understanding of this idea it is sufficient to think of $X={\mathbb R}^n$.
Given any set $A\subset X$, be it beautyful or ugly, full or with holes, like a sponge, with or without boundary (whatever that means), this set $A$ induces a partition of all of space into three disjoint subsets:


*

*The interior of $A$, consisting of all points $x\in X$ possessing a neighborhood $U(x)$ lying completely in $A$;

*The exterior of $A$, consisting of all points $x\in X$ possessing a neighborhood $U(x)$ that does not intersect $A$;

*The boundary of $A$, denoted by $\partial A$, consisting of all remaining points $x\in X$. These points neither have an $U(x)\subset A$ nor an $U(x)$ not meeting $A$. This means: Each and every neighborhood of an $x\in\partial A$ intersects as well $A$ as the complement of $A$.
A: Yes, this is indeed true. Another way to look at it: $\partial A$ is the set of all $x \in X$, such that for any open $O$ that contains $x$ (so any $x \in O\in T$) we have that $O \cap A \neq \emptyset$, and also $O \cap (X - A) \neq \emptyset$.
So all open neighbourhoods of a boundary point intersect both $A$ and its complement.  
