# Hausdorff metric and connectedness [duplicate]

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: $$$$D(A, B) = \inf \{ \epsilon \in \mathbb{R}^+ : A \subset B_\epsilon \; \land B \subset A_\epsilon \}.$$$$ Then $$(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

$$F(X)$$ 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.
• 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. – Kakuro Mar 15 at 10:11
• 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. – Kakuro Mar 15 at 10:15
• 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? – Chrystomath Mar 15 at 13:04
• @Chrystomath How do you derive that both $A$ and $B$ are nonempty? One of them surely is but why both? – freakish Mar 15 at 13:06
• @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. – Kakuro Mar 15 at 13:14