Existence of a bounded ball 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.
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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? 
 A: 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. 
