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I'm trying to verify that $d_H$ is indeed a bona fide metric. Could you have a check on the triangle inequality part?

Theorem: Let $(E, d)$ be a metric space and $\mathcal C$ the set of all compact subsets of $E$. We define $d_H: \mathcal C \times \mathcal C \to \mathbb R \cup \{+\infty\}$ by $$ d_H (A, B) := \max \{ \sup \{ d(x, A) \mid x \in B\} , \sup \{ d(y, B) \mid y \in A\} \} \quad \forall A,B \in \mathcal C. $$ with $d(x, A) := \inf_{y \in A} d(x, y)$. Then $d_H$ is a metric.

My attempt: First of all, $\mathcal C$ contains compact sets, so $$ d_H (A, B) := \max \{ \max \{ d(x, A) \mid x \in B\} , \max \{ d(y, B) \mid y \in A\} \}. $$

Hence $d_H$ is $\mathbb R_{\ge 0}$-valued and symmetric. Also, $$ \begin{align} d_H (A, B) = 0 &\iff \max \{ d(x, A) \mid x \in B\} = \max \{ d(y, B) \mid y \in A\} = 0 \\ &\iff d(x,A) = d(y, B) = 0 \quad \forall (x,y) \in B \times A \\ &\iff (x, y) \in A \times B \quad \forall (x,y) \in B \times A \\ &\iff A = B. \end{align} $$

We want to prove $$ d_H (A, B) + d_H (B, C) \ge d_H (A, C) \quad \forall A, B, C \in \mathcal C. $$

This is equivalent to $$ \begin{align} & \max \{ \max \{ d(b, A) \mid b \in B\} , \max \{ d(a, B) \mid a \in A\} \} \\ +& \max \{ \max \{ d(c, B) \mid c \in C\} , \max \{ d(b, C) \mid b \in B\} \} \\ \ge& \max \{ \max \{ d(c, A) \mid c \in C\} , \max \{ d(a, C) \mid a \in A\} \}. \end{align} $$

It suffices to $$ \begin{align} \max \{ d(b, A) \mid b \in B\} + \max \{ d(c, B) \mid c \in C\} &\ge \max \{ d(c, A) \mid c \in C\} \quad (1) \\ \max \{ d(a, B) \mid a \in A\} + \max \{ d(b, C) \mid b \in B\} &\ge \max \{ d(a, C) \mid a \in A\} . \end{align} $$

By symmetry, it is enough to prove (1) by showing that $$ \max \{ d(b, A) \mid b \in B\} + d(c, B) \ge d(c, A) \quad \forall c \in C \quad (2). $$

Fix $c \in C$. There is $b_c \in B$ such that $d(c, b_c) = d(c, B)$. Then $$ \max \{ d(b, A) \mid b \in B\} + d(c, B) \ge d(b_c, A) + d(c, b_c) \ge d(c, A). $$

This completes the proof.

Theo Bendit
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Analyst
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    I've added a new tag: [tag:hausdorff-distance], as this metric is the Hausdorff distance/Hausdorff metric. I'll also point out that $\mathcal{C}$ should not contain $\emptyset$, otherwise $d_H$ is not well-defined. – Theo Bendit Jul 28 '22 at 04:19
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    Other than that, the proof looks fine. I did note that you're using the fact that the distance function $x \mapsto d(x, A)$ is non-expansive in the last step, which may be good to mention. The symmetry used to prove $(2)$ would also be good to elaborate on; I think you mean the symmetric roles of $A$ and $C$, but you also took away the maximum over $c \in C$ in the same step. Also, consider using the notation $\max\limits_{c \in C} d(c, A)$, instead of $\max{d(c, A) \mid c \in C}$, as it makes the proof more difficult to read, IMHO. – Theo Bendit Jul 28 '22 at 04:43
  • @TheoBendit Could you transfer your comment to an answer so that I can accept it? – Analyst Jul 28 '22 at 04:46
  • @TheoBendit Could you elaborate on the problem "...took away the maximum over $c \in C$ in the same step"? I could not get it. – Analyst Jul 28 '22 at 07:28
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    Ignore that. It was never a big deal; just something I thought could be slightly more clearly written. But, upon re-reading it, I think it's absolutely fine, and my suggestion would not even necessarily improve your proof. So, just ignore it. – Theo Bendit Jul 28 '22 at 12:05

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