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Consider a bounded subset $A$ of a metric space $(X,d)$. Define the function $d_{A}:X \rightarrow \mathbb{R}$ as follows: $d_{A}(x)= \inf \{ d(x,a)| a \in A \}$. This function thus gives the distance from the set $A$. Now consider bounded sets $A$ and $B$. Show that for every $x \in X$: $d_{A \cup B}(x)= \min(d_{A}(x),d_{B}(x))$, $d_{A \cap B}(x) \geq max(d_{A}(x),d_{B}(x))$.

I find the first equality easy to prove, but I'm a little stuck on the second one. Can someone give me a hint? Thanks!

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    $\begingroup$ I think you need to assume $A$ and $B$ are not disjoint, else you will be measuring the distance to the empty set, which is not defined. For that last inequality, you just need to show that $d_{A\cap B}(x) \geq d_A(x)$, then show (similarly) that $d_{A \cap B}(x) \geq d_B(x)$. Note that $A\cap B \subseteq A$. $\endgroup$ – Michael Jun 12 '17 at 20:32
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For $d_{A \cap B}(x)$ there is a sequence $x_n \in A \cap B$, such that $\lim |x_n-x|=d_{A \cap B}(x)$. But such a sequence is necessarily in $A , B$ and hence $d_A(x), d_B(x) \leq d_{A \cap B}(x)$, and hence our result.

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This follows from the following fact:

If $\emptyset\neq C\subset A$ then $d_C\ge d_A$.

Proof: Since $C\subset A$ one has $\{d(x,c):c\in C\} \subset \{d(x,a):a\in A\}$. It follows that $d_C(x)\ge d_A(x)$ (since we are taking infimum over larger set, we have more choices for values of $d(x,a)$ and hence infimum can only be smaller or equal.

Apply this to $A\cap B \subset A,B$

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