How do max and union commute in Hausdorff measure?

Question

Recall that $d_H(A,B) = \max\{\max_{a \in A} \min_{b \in B} d(a,b),\max_{b \in B} \min_{a \in A} d(a,b)\}$

Theorem: Let $A,B,C \in H(X)$ (where $H(X)$ is the set of non-empty compact subsets of $X$). Then $d_H(A \cup B,C) = \max \{d_H(A,C),d_H(B,C)\}$

This claim appears in the book on fractals and splines by Peter Massopust. However, I'm not able to connect the hint below with the statement of the theorem:

Proof: Note that \begin{align} d(A\cup B, C) &= \max_{\alpha\in A\cup B} d(\alpha, C) = \max\left\{ \max_{\alpha\in A} d(\alpha,C), \max_{\alpha\in B} d(\alpha,C)\right\} \\ &= \max\{ d(A,C), d(B,C)\} \end{align} which implies that claim. (The reader is encouraged to verify this).

It seems that the definition of $d$ is (see the book Fractals everywhere) taken to be $d(A,B) = max\{d(a,B):a \in A\}$. So the above hint makes sense.

So how can I use this hint to prove my theorem?

My try:

$d_H(A \cup B,C) = \max \{d(A \cup B,C) , d(C, A \cup B) \}$

by definition. Then, by the hint:

$d_H(A \cup B,C) = \max \{\max\{d(A \cup B,C), d(C,A \cup B)\} , d(C, A \cup B) \}$

then, I take the other member:

$\max\{d_H(A,C),d_H(B,C)\} = \max \{\max\{d(A,C), d(C,A)\} , \{\max\{d(B,C), d(C,B)\} \}$

again by definition. Two terms are in the two sides, but I would need an equality of the sort $d(C,A \cup B) = \max \{d(C,A),d(C,B)\}$. This reduces to show:

$\max_{c \in C}\{\min_{a \in A \cup B} d(c,a)\} = \max\{\max_{c \in C}\{min_{a \in A} d(c,a),\max_{c \in C}\{min_{a \in B} d(c,a)\}\}$

Answer 1

The keypoint is the following inequality:

$d(C, A \cup B) = \max\limits_{c \in C} \min\limits_{x \in A \cup B} d(c,x) = \max\limits_{c \in C} \min \{ \min\limits_{a \in A} d(c,a), \min\limits_{b \in B} d(c,b)\} \le \min\{ \max\limits_{c \in C} \min\limits_{a \in A} d(c,a), \max\limits_{c \in C} \min\limits_{b \in B} d(c,b) \} = \min \{ d(C,A),d(C,B) \}$

This inequality gives us that:

$d(C,A \cup B) \le d(C,A),d(C,B) \implies d(C,A \cup B) \le \max\{d(C,A),d(C,B)\}$.

On the other hand, we also know that:

$d(B \cup C, A) = \max\limits_{b \in B \cup C} \min\limits_{a \in A} d(b,a) = \max\{\max\limits_{b \in B} \min\limits_{a \in A} d(b,a),\max\limits_{b \in C} \min\limits_{a \in A} d(b,a)\} = \max\{d(B,A),d(C,A)\}$.

Finally,

$d_H(A \cup B,C) = \max \{d(C,A \cup B),d(A \cup B,C) \} \le \max \{ d(C,A),d(A,C),d(C,B),d(B,C)\} = \max\{d_H(A,C),d_H(B,C)\}$

The other inequality is immediate, since:

$d_H(A,C),d_H(B,C) \le d_H(A \cup B,C)$

so that $\max\{d_H(A,C),d_H(B,C)\} \le d_H(A \cup B,C)$.

References: Barnsley's "Superfractals", 2006.