# How to show that Hausdorff distance is a metric on the set of all compact non-empty subsets of a Polish space?

For each perfect Polish space $$X$$, let $$H[X]$$ be the set of all compact non-empty subsets of $$X$$. If $$x ∈ X$$ and $$A ∈ H[X]$$, put $$d(x,A) = \inf \{d(x, y) : y ∈ A\}$$ where on the right $$d$$ is the distance function on $$X$$. The Hausdorff distance between two compact sets is defined by $$d_H(A,B) = \max \{ \sup \{d(x,B) : x ∈ A\}, \sup\{d(y,A) : y ∈ B\}\}$$ Prove that $$d_H$$ is a metric on $$H[X]$$.

This is an exercise on page 13, Descriptive Set Theory, Yiannis N. moschovakis(2009). I got stuck on how to show $$d_H(A,B)+d_H(B,C) \ge d_H(A,C)$$.

• How about distinguishing two distance functions (default distancing function and Hausdorff distance on H(X)) by different notations? Sep 23, 2021 at 13:17

Let $$d\colon X\times X\to E^1$$ be the distance in $$X$$ and $$d_H\colon H[X]\times H[X]\to E^1$$ be the Hausdorff metric on $$H[X]$$.
We have $$d(a,C)\le d(a,b)+d(b,C)\le d(a,b)+d_H(B,C)$$ for all $$b\in B$$, so, taking $$\inf_{b\in B}$$, we have $$d(a,C)\le d(a,B)+d_H(B,C)$$ which is $$\le d_H(A,B)+d_H(B,C)$$ then take $$\sup_{a\in A}$$, and similarly, we can show that $$d(A,c)\le d_H(A,B)+d_H(B,C)$$ as well.
• (For the first inequality: $d(a,C)\le d(a,c)\le d(a,b)+d(b,c)$, and take $\inf_{c\in C}$ on the right hand side.) Mar 28, 2013 at 22:27