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Let $(X,d)$ be a metric space, and let $A$ be a non-empty subset of $X$. Define the distance between a point and a set by the function $$d(x,A) = \inf_{z \in A}d(x,z).$$

Prove that for all $x,y \in X$, $$d(x,A) \leq d(x,y) + d(y,A).$$

Note that it is rather trivial to show that

  1. $d(x,A) \geq 0$ for all $x \in X$ and $A \subseteq X$ (follows directly from non-negativity of a metric);
  2. $d(x,A) = 0$ iff $x$ is in the closure of $A$; and
  3. $d(x,A) = d(A,x)$ (by abuse of notation).

By proving the final property of the triangle inequality, we can conclude that the modified distance $d: X \times 2^X \to R$ and its mirrored version $d: 2^X \times X \to R$ forms a sort of pseudo-metric where $d(x,A) = 0$ does not define an equivalence relation.

If we were to modify this function further by first defining $d(x,A)$ to be substituted with $d({x},A)$, we can form a more general pseudo-metric space $(2^X, d:2^X \times 2^X \to R)$ via defining $$d(A,B) = \inf\{d(a,b) : a \in A, b \in B\}.$$ It should be noted that in this particular metric, $A = B$ means that $A$ and $B$ are not separated. That is, if $A'$ denotes the closure of $A$, then $A \subseteq B'$ and $B \subseteq A'$. Then we know that (and can verify 4)

  1. $d(A,B) \geq 0$;
  2. $d(A,B) > 0$ if and only if $A$ and $B$ are separated (i.e., $A$ is not in the closure of $B$, and $B$ is not in the closure of $A$);
  3. $d(A,B) = d(B,A)$; and
  4. $d(A,B) \leq d(A,C) + d(C,B)$, for all subsets $C \subseteq X$.

Note that since (2) does not define an equivalence relation on $2^X$, this particular function does not define an actual metric space.

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    I don't think it makes sense to call what you're describing a metric. When you say "$A=B$ means that $A$ and $B$ are not seperated" that doesn't really count, since "not being seperated" is definitely not an equivalence relation. You can say that you're looking for an analogue of the triangle inequality for set distance, but that's about it2012-11-14

2 Answers 2