General Topology – Equivalent Metrics and Induced Hausdorff Metrics

general-topologymetric-spaces

Let $(E, d)$ be a metric space and $\mathcal C$ the set of all compact subsets of $E$. We define $d_H: \mathcal C \times \mathcal C \to \mathbb R$ by
$$
d_H (A, B) := \max \left \{ \max_{x \in B} d(x, A), \max_{y \in A} d (y, B) \right \}\quad \forall A,B \in \mathcal C.
$$
with $d(x, A) := \min_{y \in A} d(x, y)$. I proved that $d_H$ is indeed a metric. It is mentioned in this Wikipedia page that the topology of $\mathcal C$ depends on that of $E$, not on $d$.

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I formalize it as follows.

Theorem: Let $d'$ be another metric on $E$ that is equivalent to $d$, i.e., $d, d'$ give rise to the same topology. Let $d'_H$ be the Hausdorff metric on $\mathcal C$ that is induced from $d'$. Then $d_H, d'_H$ give rise to the same topology on $\mathcal C$.

Could you elaborate on how to prove this claim?

Best Answer

I'm following the idea from the @Alessandro Codenotti's comment.

Define:

  • $$K_r:=\bigcup_{k\in K}B(k,r) = \{x\in X:d(x,K)<r\}.$$

  • $\mathcal T$ -- topology on $E$ generated by $d$.

  • for $U_1,U_2,\ldots,U_k\in\mathcal T$ $$\mathcal U(U_1,U_2,\ldots,U_k)=\left\{C\in\mathcal C:C\subset \bigcup_{i=1}^kU_i\text{ and }\forall_i\,C\cap U_i\neq\emptyset\right\}.$$

  • $\mathscr U=\{\mathcal U(U_1,U_2,\ldots,U_k):k\in\Bbb N,\ U_1,U_2,\ldots,U_k\in\mathcal T\}$

We claim that $\mathscr U$ is a basis for the topology generated by $d_H$. Observe that this means that the topology generated by $d_H$ can be expressed by the topology $\mathcal T$.

Proof.

  • Take any $U_1,U_2,\ldots,U_k\in\mathcal T$ and $C\in \mathcal U:=\mathcal U(U_1,U_2,\ldots,U_k)$. Let $c_i\in C\cap U_i$ for all $i$. We can find $r>0$ such that $B(c_i,r)\subset U_i$ and $C_r\subset \bigcup_i U_i$ (from Lemma 1.). We claim that $B_{d_H}(C,r)\subset \mathcal U$. To show it consider any $K\in \mathcal C$ such that $d_H(C,K)<r$. Then $K\subset C_r\subset \bigcup_i U_i$. Moreover for all $i$ we have $c_i\in C\subset K_r$, so $$K\cap U_i\supset K\cap B(c_i,r)\neq\emptyset.$$
  • Take any $C\in \mathcal C$ and $r>0$. Since $C$ is is totally bounded, there exist $c_1,c_2\ldots,c_k\in C$ such that $$C\subset \bigcup_i U_i,\quad \text{ where } U_i=B(c_i,r/3).\tag{1}$$ We claim that $\mathcal U:=\mathcal U(U_1,\ldots,U_k)\in B_{d_H}(C,r)$. To show it consider any $K\in \mathcal U$. Then $$K\subset \bigcup_iB(c_i,r/3)\subset C_{r/3}.\tag{2}$$ Since $K\cap B(c_i,r/3)\neq \emptyset$, we have $c_i\in K_{r/3}$ and therefore $U_i=B(c_i,r/3)\subset K_{2r/3}$. From (1) we thus get $$C\subset K_{2r/3}.\tag{3}$$ From (2) and (3) we know that $d_H(C,K)\leq 2r/3<r$.

We used the following two lemmata.

Lemma. If $K$ is compact, $U$ is open and $K\subset U$ then $K_r\subset U$ for some $r>0$.

Sketch of the proof. The function $d(\cdot,E\setminus U)\colon K\to (0,\infty)$ attains it's minimum $r>0$. $\ \square$

Lemma. We have $d_H(C,K)=\inf\{r>0:C\subset K_r,\ K\subset C_r\}$. $\ \square$

Corollary.

  • if $d_H(C,K)<r$ then $C\subset K_r$ and $K\subset C_r$
  • if $C\subset K_r$ and $K\subset C_r$ then $d_H(C,K)\leq r$.
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