Distance $\Psi(A,B)=\sup_{x\in E}\vert d_A(x)-d_B(x)\vert$ where $d_A(x)=\inf_{y\in A}d(x,y)$.

Question

Let $(X,d)$ be a metric space, assume that $d$ is bounded. Denote $F$ the set of all closed set of $X$.

Define $$\Psi(A,B)=\sup_{x\in X}\vert d_A(x)-d_B(x)\vert$$ where $d_A(x)=\inf_{y\in A}d(x,y)$.

One can prove that $\Psi(A,B)$ is a distance and $d(A,B)\le\Psi(A,B)$, Indeed $\inf_{b\in B}d(a,b)=d_B(a)-d_A(a)\le\Psi(A,B)$.

I was wondering if we can express $d(a,b)=\Psi(?,?)$. Any ideas please?

@LeeMosher Perhaps! I fail to see any geometrical point of view.. If you can be extend a little more it will (very) helpful. — , Sep 29 '14 at 18:59
Looking more closely, I believe your question has some errors. For instance, in your equation for $\Psi(A,B)$, the symbol "$E$" has not been defined. — Lee Mosher, Sep 29 '14 at 19:42

score 0 · Accepted Answer · answered Sep 29 '14 at 19:42

The function $\Psi$ is the Hausdorff distance. Its geometric meaning is: $\Psi(A,B)$ is the infimum of numbers $\delta$ such that no matter where you are in one set, you can jump into the other by traveling at most $\delta$.

It's not immediately obvious that my description agrees with your definition of $\Psi$. But it does. For any $\delta$ as above, $\Psi(A,B)\le \delta$ by the triangle inequality: the distances $d_A(x)$ and $d_B(x)$ cannot differ by more than $\delta$, since once you can get from $x$ to one of these sets, the other one is at most $\delta$ away. Conversely, $\Psi(A,B)\ge \delta$ because when the supremum in the definition of $\Psi$ is restricted to $x\in A\cup B$, it describes what is written in my first paragraph.

Given the above, the formula $d(a,b) = \Psi(\{a\},\{b\})$, stated by Lee Mosher, should be geometrically clear.

Distance $\Psi(A,B)=\sup_{x\in E}\vert d_A(x)-d_B(x)\vert$ where $d_A(x)=\inf_{y\in A}d(x,y)$.

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