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Define the set:

$$X:=\{K \subset \mathbb{C}: K \text{ is bounded and closed } \}.$$

Define a function $d : X \times X \rightarrow \mathbb{R}$ via:

$$d(K_1,K_2) = \inf \{ \delta > 0 : K_1 \subset N_{\delta}(K_2) \text{ and } K_2 \subset N_{\delta}(K_1)\},$$

where

$$N_δ(K) = \{ x \in \mathbb{C} : \exists y \in K \text{ with } |x−y|< \delta \}.$$

Note that we could also write $N_{\delta}(K) = \bigcup \limits_{y \in K} N_{\delta}(y)$, where $N_{\delta}(y) = \{x \in \mathbb{C} : |x-y|<\delta\}$.

I'm having trouble with understanding $d$ and $N_δ(K)$. Can someone explain me why $d(K_1,K_2)$ is a metric?

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  • $\begingroup$ I have edited your post (it is currently being peer reviewed); please check that the edits correspond with your original intent. $\endgroup$ – Sambo Oct 13 '18 at 18:03
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Intuitively $N_{\delta}$ takes a set $K$ and 'fattens' it by radius of $\delta$. An alternative way of writing it would be $$N_\delta(K) = K + B_{\delta}(0) $$ $d$ then takes two sets $K_1$ and $K_2$, and measures how much we need to 'fatten' $K_1$ until it contains all of $K_2$ and vice versa. Another way of writing it is $$ d(K_1, K_2) = \inf \{\delta > 0 : (\forall x \in K_1)(\exists y \in K_2)\ \lvert x - y \rvert \leq \delta \ \text{and}\ (\forall y \in K_2)(\exists x \in K_1)\ \lvert x - y \rvert < \delta\} $$ $d$ is well defined since the sets $K_1$ and $K_2$ are bounded, thus we aren't taking the infinum over an empty set.

By $d$'s very definition, it is symmetric and positive. What remains is to show $d(K_1, K_2) = 0 \iff K_1 = K_2$ (this requires the compactness of $K_1$ and $K_2$) and the triangle inequality.

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    $\begingroup$ It's worth nothing that the sets $K_1$ and $K_2$ should also be non-empty for $d$ to be well-defined! $\endgroup$ – Sambo Oct 13 '18 at 18:15
  • $\begingroup$ @bitesizebo if $d(A,B) = 0$, then every point of A is within zero distance of B,and hence must belong to B since B is compact, so $A \subset B$ and similarly $B \subset A$. So $d(A, B) = 0$ implies that $A = B$. Is this it for definiteness? And how do I go with the triangle inequality? $\endgroup$ – dxdydz Oct 14 '18 at 14:30

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