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Let $(X, d)$ be metric space. Define $B_\epsilon = \{ x \in X : \exists b \in B \; d(x, b) \le \epsilon\} $. Let $F(X)$ be a family of all nonempty compact subsets of $X$ (so $\emptyset \notin F(X)$ ). We shall define Hausdorff metric by: \begin{equation} D(A, B) = \inf \{ \epsilon \in \mathbb{R}^+ : A \subset B_\epsilon \; \land B \subset A_\epsilon \}. \end{equation} Then $(F(X), D)$ is a metric space.

What I would like to know is what is the relation between connectedness of $(X, d)$ and connectedness of $(F(X), D)$.

So far I was able to prove that if $(X, d)$ is not connected, then $(F(X), D)$ also is not connected. Here the main idea was that if (A, B) is a pair of nonempty subsets of $X$ such that $A\cup B = X$ and $A \cap B = \emptyset$ and $A, B$ are open, then $(2^A \cap F(X), F(X) \setminus 2^A \cap F(X))$ is a pair of nonempty, open subsets of $F(X)$ which sum to $F(X)$ and have empty intersection. (Which by contraposition means that connectedness of $(F(X), D)$ implies connectedness of $(X, d)$).

Is the opposite implication true, that is, does connectedness of $(X, d)$ imply connectedness of $(F(X), D)$?


marked as duplicate by freakish, Community Mar 15 at 13:25

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$F(X)$ is connected when $X$ is.

Suppose $F(X)$ is disconnected by open subsets $U,V$. That is, they are disjoint collections of compact subsets of $X$ whose union is $F(X)$. Let $A:=\{x\in X:\{x\}\in U\}$, and similarly for $B$.

Let $a\in A$, $b\in B$ be any two points. Then the compact set $\{a,b\}$ is in either $U$ or $V$. Then $D(\{a,b\},\{b\})>\epsilon$ for some $\epsilon>0$, since $U$ and $V$ are open sets. This means that $d(a,b)>\epsilon$. Hence $B(\epsilon,a)\subseteq A$, and $A$ is open. Similarly for $B$, so $X$ is disconnected.

  • $\begingroup$ I'm afraid what you said is not true. Consider $X = \{0, 1\}$ and let $d$ be discrete metric, that is $d(x, y) = 1$. Now, $F(X) = \{ \{0 \}, \{ 1 \}, \{0, 1\} \}$. Distance $D$ between any two different sets in $F(X)$ is 1. We can choose $U = \{ \{0\} \}$ and $V = \{ \{1\}, \{0, 1 \} \}$. $U$ and $V$ are thus nonempty, they have empty intersection and they sum up to $F(X)$. They are also open, since ball $B(x, 1/2)$, where $x$ is any element of $F(X)$ is always equal to $\{ \{x\} \}$ and so it is a subset of the same subset of $F(X)$ that $x$ is an element of. $\endgroup$ – Kakuro Mar 15 at 10:11
  • $\begingroup$ I believe that the statement: "Let $x \in K$, then $D(\{x\}, K) = 0$" is incorrect, with exemplary counterexample being $K = \{0, 1\}$ and $x = 0$ from my comment. $\endgroup$ – Kakuro Mar 15 at 10:15
  • $\begingroup$ You're right. I misunderstood $D$, but I think the proof is still broadly correct. Do you take $\emptyset$ to be compact, in which case how is its distance defined? $\endgroup$ – Chrystomath Mar 15 at 13:04
  • 1
    $\begingroup$ @Chrystomath How do you derive that both $A$ and $B$ are nonempty? One of them surely is but why both? $\endgroup$ – freakish Mar 15 at 13:06
  • $\begingroup$ @Chrystomath while I think $\emptyset$ should be considered compact, it is true that here I didn't want for $\emptyset$ to be an element of $F(X)$ - I shall edit the question to avoid confusion. $\endgroup$ – Kakuro Mar 15 at 13:14

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