I am not sure if the statement below is true. The statement is: Let $(M,d)$ be a connected metric space and $A, B$ be two nonempty subsets of $M.$ Assume the boundary $\partial A$ and $\partial B$ are nonempty. Suppose that $A\cap B=\emptyset.$ Then $$d(A,B)=d(\partial A, \partial B).$$ Here $d(E,F):=\inf\{d(x,y)\mid x\in E, y\in F\}, \forall E,F\subset M.$

If the statement is true, how to prove it? If it is not valid, give a counterexample, and is it valid if $(M, d)$ is complete?

  • 1
    $\begingroup$ As the answer below shows, you need to assume at least that $M$ is connected, otherwise any connected component has empty boundary, so $d(\partial A, \text{anything}) = \infty$. It is true if you have some rather strong regularity assumption, e.g. connected, something like locally compact to assure that bounded is totally bounded, (complete and) geodesic, so that any sequence of points in $A$ and $B$ subconverges to something in the closure, which cannot be an internal point (otherwise the geodesic between the limit points would give something closer). $\endgroup$ – Spi Jun 12 '14 at 6:25
  • $\begingroup$ I have changed the condition to that $M$ is connected. I am not sure if this modification is sufficient. $\endgroup$ – nuage Jun 12 '14 at 6:31

The answer is no. Consider $M=(-\infty,-1)\cup (1,\infty)$ with distance $d(x,y)=|y-x|.$ Let $A=(-2,-1)$ and $B=(1,2).$ It is clear that $d(A,B)=2.$ However $\partial A=\{-2\},$ $\partial B=\{2\},$ and $d(\partial A,\partial B)=d(-2,2)=4.$

  • $\begingroup$ What happens if $M$ is a normed vector space? $\endgroup$ – Omran Kouba Jun 12 '14 at 6:17
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    $\begingroup$ @mfl, What would happen if $M$ is connected? Your counterexample is wonderful, but that $M$ is disconnected! $\endgroup$ – nuage Jun 12 '14 at 6:29

Take two unit circles in $R^2$, one shifted up by $3$. Connect these circles with vertical lines on their left- and rightmost points. This a complete connected metric space by taking the induced distance of $R^2$ (completeness follows from closedness, connectedness from the lines left and right). If you now take $A$ the upper and $B$ the lower circle both without their right and left points, their distance is $1$, but the distance of their boundaries is 3.

Note however that the answer is yes for path metric spaces, i.e. spaces in which the distance of two points equals the length of the shortest continuous path connecting them (actually, equals the infimum of such paths). The proof is easy, just take any points $x,y$ in $A$ and $B$, respectively, take any connecting path, and take two points $x_0, y_0$ in the intersections of this path with the boundaries of $A$ and $B$. Then the same path but restricted only to connect $x_0$ and $y_0$ has shorter length, thus $d(x_0,y_0)< d(x,y)$ and since $x,y$ were arbitrary $d(\partial A, \partial B)\leq d(A,B)$


The answer is no even if $M$ is homeomorphic to the unit interval $[0,1]$ and thus as regular as can be.

$M=\{(x,y)\in\mathbb R^2\colon x^2+y^2=1\}\setminus(1/2,1]\times[-1,-1]$ with the euclidean metric

$A=\{(x,y)\in M\colon y<0,\ 0\le x\le 1/2$ and $B=\{(x,y)\in M\colon y>0,\ 0\le x\le 1/2\}$

$d(A,B)<2=d(\{(0,-1)\},\{(0,1)\})=d(\partial A,\partial B)$


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