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.

  • $\begingroup$ 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). $\endgroup$ – ItsJustVennDiagramsBro Mar 10 at 20:56
  • $\begingroup$ What's a 'deleted' neighborhood? Notice that in any neighborhood of $p$ there's $p$ itself that is in $A^c$ $\endgroup$ – Exodd Mar 10 at 21:01
  • $\begingroup$ 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 $\endgroup$ – Dylan Mehrer Mar 10 at 21:25
  • $\begingroup$ Deleted neighborhood is that "other than p itself" part $\endgroup$ – 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 $A$.

  • $\begingroup$ Ok but the question before this one asked to show that p is a limit point, but not a boundary point. $\endgroup$ – Dylan Mehrer Mar 10 at 21:24

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