Let $(X,d)$ be a metric space. Let us define a relation $x \sim y$ on $X$ by declaring $x \sim y$ iff there exists a connected subset of $X$ which contains both $x$ and $y$. Show that this is an equivalence relation (i.e., it obeys the reflexive, symmetric, and transitive axioms). Also, show that the equivalence classes of this relation (i.e., the sets of the form $\{y \in X : y\sim x\}$ for some $x \in X$) are all closed and connected.
Let $x \in X$. Then $\{x\}$ is a connected subset of $X$. This proves the reflexive axiom. The symmetric axiom is obvious. Suppose that there exists a connected subset $U$ such that $x,y \in U$ and another connected subset $V$ such that $y,z \in V$. For sake of contradiction, suppose that every set containing $x,z$ is disconnected. How can I proceed from here?
For the second part, I know that if we have a collection of connected sets $\{E_\alpha\}_{\alpha \in I}$, and $\bigcap_{\alpha \in I} E_\alpha$ is non-empty, then $\bigcup_{\alpha \in I}E_\alpha$ is connected. $\{y \in X : y\sim x\}$ is the union of all connected subsets of $X$ containing $x$, so it is connected. I also know that if $E$ is connected, then $\overline{E}$ is also connected. I guess that I need to use this information to prove closedness, but I don't know how to do this.
I appreciate if you give some help.
Best Answer
The lemma you quoted about connected sets with non-empty intersection also applies to prove transitivity: $x \sim y$ means we have connected set $C_1$ with $x,y \in C_1$, and $y \sim z$ means we have a connected set $C_2$ with $y,z \in C_2$. So $y \in C_1 \cap C_2$ and $C_1 \cup C_2$ is connected with $x,z$ in it, so $x \sim z$ by definition.
The class of $C_x=\{y: x \sim y\}$ is also connected by the same argument: we have $C_y$ with $x,y \in C(y)$ for a connected set $C(y)$ every $y \in C_x$ by definition. All $C(y), y \in C_x$ contain $x$ (so their intersection is non-empty) and
$$C_x = \bigcup \{C(y): y \in C_x\}$$ is thus connected by the same lemma.
(that this equality holds is not so hard: $z \in C_x$ implies $z \in C(z)$ which is a subset of the union; if $z \in C(y)$ for some $y \in C_x$, this $C(y)$ witnesses that $x \sim z$ so that $z \in C_x$ again).
If $z \in \overline{C_x}$, then that latter set is connected as the closure of the connected $C_x$ and $x \in C_x \subseteq \overline{C_x}$ and so it witnesses $z \sim x$ and so $z \in C_x$ and we see that $\overline{C_x} \subseteq C_x$ and $C_x$ is closed.