Proposition

Prove that a topological space $X$ is disconnected if and only if there exists a non-constant function from $X$ to the discrete space $\{ 0, 1 \}$.

Solution

Suppose $X$ is disconnected. Then there exist a pair of disjoint, open sets $U, V$ such that $U \cup V = X$. Let $f: X \rightarrow \{ 0, 1 \}$ be defined such that

• $f$ is non-constant because $U$ and $V$ are nonempty.
• Is $f$ continuous?
  • There are exactly $4$ open subsets of $\{ 0, 1 \}$.
    • $f^{-1}(\emptyset) = \emptyset$.
    • $f^{-1}(\{ 0 \}) = U$.
    • $f^{-1}(\{ 1 \}) = V$.
    • $f^{-1}(\{ 0, 1 \}) = X$.
  • Therefore, $f^{-1}(S)$ is open for any open $S \subset \{ 0, 1 \}$.

On the contrary, suppose that there exists a non-constant function from $X$ to the discrete space $\{ 0, 1 \}$ Let $f$ be such a function. Let $U = f^{-1}(\{ 0 \})$ and $V = f^{-1}(\{ 1 \})$. Then $\{ 0 \}$ and $\{ 1 \}$ are both open in $\{ 0, 1 \}$, and $f$ is continuous, so $U$ and $V$ are both open. Since $f^{-1}(\{ 0, 1 \}) = X$ and $f^{-1}(\{ 0, 1 \}) = f^{-1}(\{ 0 \}) \cup f^{-1}(\{ 1 \}) = U \cup V$, $X = U \cup V$. $U \cap V = \emptyset$ because there exists no element $x \in X$ such that $f(x) = 0$ and $f(x) = 1$. Therefore, $U, V$ are disjoint open sets whose union equals $X$.