Let the set A be a punctured disk around p = (0,0) of radius 1 in R^2. Is the point p part of the interior, exterior, or boundary of A?
Certainly p is a limit point of A. As a limit point of A, it is not in the exterior of A.
Now an open disk around p of radius > 0 is not contained in A because p is not contained in A.
But also a deleted neighborhood around p of radius > 0 does not contain points in both A and A^c.
So the exterior is out of the question, but I can't seem to place p in the interior or in the boundary. Interior is defined with a neighborhood and boundary is defined with a deleted neighborhood and that is what seems to be the issue. Surely p must be in either the interior, exterior, or boundary.
Best Answer
$\mathrm{Int}(A) \subset A$ so $p \notin \mathrm{Int}(A)$.
But $p$ is in the closure $\overline{A}$ of $A$, so $p \in \overline{A} \setminus \mathrm{Int}(A)$, i.e. $p$ is in the boundary of $A$.