0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.