Lets define: $$F(X) :=\{A \subseteq X \mid A \neq \emptyset , A = \overline{A}\}.$$

For $A, B \in F(X)$ and $p \in X$ define $$d_p(A,B) = \sup_{x \in X} \{ | \operatorname{dist}(x,A) - \operatorname{dist}(x,B) | e^{- \rho(p,x)} \}.$$ This function is called Busemann metric.

Now we can also define: $$ B(X) := \{ A \in F(X) \mid A \text{ bounded} \}.$$ For $A, B \in B(X)$ lay: $$d(A,B) = \max \{ \sup_{x \in X} \operatorname{dist} (x,A), \sup_{x \in X} \operatorname{dist} (x,B) \}. $$ This one is called Hausdorff metric. $$ $$

I am trying to prove the following theorem:

If $(X, \rho )$ is a metric space, then $d_p$ and $d$ are equivalent over $B(X)$.

To attain that, I need to show, that there exists (for $A \in B(X) $) $r>0$ s.th. $ \bigcup K_p (A,r) $ is bounded in $X$.

Does anyone know what can I do to achieve that?

  • 2
    $\begingroup$ I changed \underset{x\in X}{\sup} to \sup_{x\in X}. In a displayed, as opposed to inline, setting, that does the same thing and is simpler, and is standard usage. Also, \mid results in proper spacing where | does not and I changed it. \operatorname{dist} also provides proper spacing where \text{dist} does not. I edited accordingly. $\endgroup$ – Michael Hardy Jun 25 '15 at 15:09
  • $\begingroup$ You haven't defined $K_p(A,r)$. $\endgroup$ – user147263 Jun 26 '15 at 4:18
  • $\begingroup$ Oh, sorry, this is a ball in $d_p$ metric $\endgroup$ – Kuba Jun 26 '15 at 20:15

It helps to observe that the Hausdorff metric $d$ can also be given by the formula $$d (A,B) = \sup_{x \in X} \{ | \operatorname{dist}(x,A) - \operatorname{dist}(x,B) | \}$$ Indeed, the original definition of $d$ amounts to taking this supremum over $A\cup B$ only; but you can check with the triangle inequality that that for other values of $x$, the difference $| \operatorname{dist}(x,A) - \operatorname{dist}(x,B) | $ does not exceed $d(A,B)$.

Thus, the only difference between the definitions is the factor of $e^{-\rho(x,p)}$. In particular, $d_p\le d$.

Next, I would focus on considering the set $$B_R(X) := \{ A \in F(X) \mid A\subset B(p,R) \}$$ where $B(p,R)=\{x\mid \rho(x,p)\le R\}$. For such sets, $d_p\ge e^{-R}d$ because restricting the supremum in the definition of $d_p$ to $x\in A\cup B$ already gives at least $e^{-R} d(A,B)$.

Thus, we see that the metrics are equivalent on each $B_R(X)$. Thus, the identity map from one metric to the other is continuous both ways, as required.

  • $\begingroup$ Are you sure that if there is a Cauchy sequence in one metric which isnt Cauchy in the second, then these metric are not equivalent? I don't think so. Thank you for answering!:) $\endgroup$ – Kuba Jun 26 '15 at 20:41
  • $\begingroup$ I was thinking of another notion of equivalence. Corrected now. $\endgroup$ – user147263 Jun 26 '15 at 20:46
  • $\begingroup$ one more question before I accept your answer - how do you know that the identity map is continuous on $B(X)$? Is it because that's the limit of continuous mappings on $B_R (X)$ ? $\endgroup$ – Kuba Jun 27 '15 at 13:16
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    $\begingroup$ Continuity is a local property. It suffices to check if at every point. So, you want to prove that it's continuous at point $a$, and choose $R$ large enough so that $a$ is in there. $\endgroup$ – user147263 Jun 27 '15 at 16:29
  • $\begingroup$ but we check the continuity at point $A$ which is a set, and we do not know if $d_p(A,{p})$ is finite. $\endgroup$ – Kuba Jun 29 '15 at 6:54

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