# Interior, exterior, and boundary of deleted neighborhood

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.

• What is your definition of the boundary of a set? Mine is that the boundary of $S$, denoted $\partial S$, is $\bar{S}\setminus S^{\circ}$ (that is, the closure minus the interior). – ItsJustVennDiagramsBro Mar 10 at 20:56
• What's a 'deleted' neighborhood? Notice that in any neighborhood of $p$ there's $p$ itself that is in $A^c$ – Exodd Mar 10 at 21:01
• No definition of boundary was given other than the following theorem: If X has non-trivial metric d and A is a subset of X then p is a boundary point of A iff for any r > 0 the ball around p of radius r contains at least one point in A and at least one point in A^c other than p itself – Dylan Mehrer Mar 10 at 21:25
• Deleted neighborhood is that "other than p itself" part – Dylan Mehrer Mar 10 at 21:28

$$\mathrm{Int}(A) \subset A$$ 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 $$A$$.