Differential Geometry and Its Application - Exercise 2.1.2

Problem:

This exercise is for those with some knowledge of topology. It will be used in Theorem 6.7.7. Suppose $M$ is connected. Show that a subset $Z \subseteq M$ which is both open and closed must be either $M$ or $\emptyset$.

Solution:

Suppose the contrary that $\emptyset \neq A \subsetneq M$ is both open and closed, now we can write $A \cup (M - A) = M$ where $A \neq \emptyset$ is both open and closed. By definition, $M - A \neq \emptyset$ is also open because $A$ is closed.

But $M$ is connected, a connected set cannot be written as a disjoint union of two non-empty open sets, so we have reached a contradiction that proved the required proposition.