# A definition of metric space

Can you please help me solve the question below? I have no idea how to prove this one.

Define the set $$X:=\{K\subset\mathbb C:K\text{ is bounded and closed}\}$$ Define a function $d\colon X \times X \to \mathbb{R}$ via $$d(K_1,K_2)=\inf\{\delta>0:K_1\subset N_\delta(K_2)\text{ and }K_2\subset N_\delta(K_1)\}$$ where $$N_\delta(K):=\bigcup_{y\in K}N_\delta(y)=\{x\in\mathbb C:\exists y\in K\text{ with }|x-y|<\delta\}.$$

i) Show that $d$ defines a metric on $X$.

ii) Is $d$ still a metric if $X$ contains all bounded sets in $\mathbb C$? All closed sets?

• $d:X \to X$ or $d:X\times X \to \mathbb{R}$ – Kunnysan Jul 14 '13 at 21:16
• Where are you stuck? Have you tried to show any of the metric axioms? Most of them should be fairly easy (non-negativity, non-degeneracy, and symmetry). – icurays1 Jul 14 '13 at 21:17
• Oh! My bad. I solved i). Part ii is the problem! – InfimumMaximum Jul 14 '13 at 21:21

This is called Hausdorff distance, by the way.

i) The problem statement is wrong. You must exclude the empty set from $X$., ie. I shall work with $X:=\{K\subset\mathbb C:K\text{ is nonempty and compact}\}$. Also, $d$ is not a map $X\to X$ but rather $X\times X\to\mathbb R$ as required for a metric.

You need to show the properties of a metric:

• $d(K_1,K_2)\ge 0$: Since both sets are bounded and nonempty, the set we take the infimum of is nonempty, hence the infimum is $<\infty$. As all $\delta$ in the set are $>0$, the infimum is also $\ge 0$.
• $d(K_1,K_2)=0\Rightarrow K_1=K_2$: Assume $K_1\ne K_2$ and wlog $x\in K_2\setminus K_2$. Since $K_2$ is closed, there is an $\epsilon$ neighbourhood of $x$ such that $B_\epsilon(x)\cap K_2=\emptyset$. Then $x\notin N_\delta(K_2)$ if $\delta<\epsilon$ and hence $d(K_1,K_2)\ge \epsilon$.

• $d(K_1,K_3)\le d(K_1,K_2)+d(K_2,K_3)$: Note that $N_{\delta_1+\delta_2}(K)\subseteq N_{\delta_1}(N_{\delta_2}(K))$. Use this and write down the exlicit condition we want to show and you are done.

ii) Let $A=\{z\in\mathbb C:|z|<1\}$ and $B=\{z\in\mathbb C:|z|\le 1\}$. These are bounded sets with $d(A,B)=0$ and $A\ne B$. Hence this is no longer a metric.

Let $A=\mathbb R$ and $B=i\mathbb R$. These are closed sets with $d(A,B)=\infty$, hence this is no longer a metric.

• I'm not sure about how you proved d(K1, K2) >= 0. Can you explain it one more time please? – InfimumMaximum Jul 17 '13 at 5:12
• Maybe if K1 is not equal to K2, then K2-K1 is nonempty. Then for x in K2-K1, we can say d(K1,K2) >= d(x,K1) > 0. Is this correct? (though I'm not sure if I can just say d(x,K1) > 0...) – InfimumMaximum Jul 17 '13 at 5:14