Proposition

Suppose $X$ is a connected topological space, and $\sim$ is an equivalence relation on $X$ such that every equivalence class is open. Show that there is exactly one equivalence class, namely $X$ itself.

Solution

For each $x \in X$, let $U_x = \{ y \in X \mid x \sim y \}$. Let $x_0 \in X$ be given. If $U_{x_0} = X$, we are done. Suppose otherwise. Let \(V = \bigcup_{x \in U_{x_0}^c} U_x\). Then $U_{x_0} \cup V = X$ and $U_{x_0} \cap V = \emptyset$. $U_{x_0}$ is open since it is an equivalence class. $V$ is open since it is a union of equivalence classes, each of which is open.

Therefore, $X$ is disconnected. This is a contradiction, so $U_{x_0} = X$.