In a metric space $(M,d)$, I have compact nonempty sets $A, B$ and $C$ with $A\subset \operatorname{int} B$ and $B \subset \operatorname{int} C$, where $\operatorname{int}$ denotes the interior of a set and $\partial$ its boundary. I'm trying to find out if $d(A,C - \operatorname{int} B) = d(A, \partial B).$ If this were to be true it would solve a problem I'm working on, but can't prove it nor find a counterexample. It is always true that $d(A,C - \operatorname{int} B) \leq d(A, \partial B),$ since $\partial B \subseteq C- \operatorname{int} B, $ and the reverse inequality seems true because taking a point in $C - \operatorname{int} B$ which is not in the boundary of $B$, it would be "further away" from $A$ than a point in the boundary of $B,$ but I'm thinking geometrically on the plane, and I can't work on the triangle inequality to give me this result.

Is what I'm trying prove to even true? Any hints are appreciated, thanks.

(OBS: the metric space is also a smooth manifold in my problem, if any extra structure helps).

(OBS2: in the original question, $M$ could be a topological manifold, since I thought this problem could be solved with topology alone. User @theo-bendit showed this is false in general, but more research made me see this may be true for a smooth manifold, as in this other question When is distance to the boundary always less than that to the exterior? . In not used to work with riemann metrics so I can't be sure).

  • $\begingroup$ Do we have a proof for $d \left( A, C - \text{int } B \right) \leq d \left( A, \partial B \right)$? $\endgroup$ – Aniruddha Deshmukh Jan 4 at 3:21
  • $\begingroup$ Sure. Since $B \subset C$ and $B$ is compact, so is closed in the metric topology, and $B = int \ B \cup \partial B,$ where the union is disjoint, therefore $\partial B \subseteq C - int \ B.$ Now the result follows since the distance is the infimum of $d$ on $A \times C - int \ B$ and $A \times \partial B \subseteq A \times C - int \ B.$ $\endgroup$ – Vic Jan 4 at 3:31

I can't comment on the topological manifold part here, but in general, no, this is not the case.

Let $K$ be the Cantor Middle Third set, and \begin{align*} M &= C = K \cup [100, 103] \\ B &= \left(K \cap \left[0, \frac13\right]\right) \cup [101,102] \\ A &= K \cap \left[0, \frac19\right]. \end{align*} Then, \begin{align*} C \setminus \operatorname{int} B &= \left(K \cap \left[\frac23, 1\right]\right) \cup [100, 101] \cup [102, 103] \\ \partial B &= \{ 101, 102\} \\ d(A, C \setminus \operatorname{int} B) &= \frac59 \\ d(A, \partial B) &= \frac{908}{9}. \end{align*}

  • $\begingroup$ I'll check these computations, but this should solve the problem, since the real line is the best manifold there is, thanks! $\endgroup$ – Vic Jan 4 at 3:46
  • $\begingroup$ Bear in mind that $M$ is not the real line; in fact it has an open, totally disconnected compact subspace $K$. $\endgroup$ – Theo Bendit Jan 4 at 3:52
  • $\begingroup$ Right, I was thinking of it as a subset of the real line. $\endgroup$ – Vic Jan 4 at 3:56
  • $\begingroup$ @Vic The example doesn't work if you take $M = \mathbb{R}$, as $C$ will only have the interior $(100, 103)$. $\endgroup$ – Theo Bendit Jan 4 at 3:57
  • $\begingroup$ @Vic In fact, the $[100, 103]$ bit was only included to prevent $\partial B = \emptyset$. $\endgroup$ – Theo Bendit Jan 4 at 3:58

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