Let $(X,d)$ be a metric space, and $comp(X)$ denote the collection of compact sets of $X$. For every $K_1,K_2\in com(X)$, we define $$ d_H(K_1, K_2) = \inf\{r>0\ |\ d(x,K_2), d(y,K_1) < r,\ \text{for all }x\in K_1, y\in K_2\} $$ I need to show that this defines a metric over $comp(X)$.
I have two questions:
- For the triangle inequality, I have proven that for all $x\in K_1$, $d(x,K_2) \leq d_H(K_1,K_2) + d_H(K_2,K_3)$ and symmetrically for all $y\in K_2$. This would have been enough if the inequality was strict, because then $d_H(K_1,K_2) + d_H(K_2,K_3) \in \{r>0\ |\ d(x,K_2), d(y,K_1) < r,\ \text{for all }x\in K_1, y\in K_2\}$, which yields the needed inequality. But, the inequality is not strict, hence I cannot infer this. How can I proceed?
- Why is compactness needed? I have proven all these properties of metric using only the closeness of compact sets. What fails if the sets are not compact?
Note: I use only notions from metric space theory, no knowledge in topology exists.