4.12 Connectedness
Definition 4.12.1 (Connected). Let $X$ be a topological space, then the following are equivalent:
For any $\emptyset \ne U, V \subset X$ open with $U \cup V = X$, $U \cap V \ne \emptyset$.
There exists no surjective $f \in C(X; \bracs{0, 1})$.
For any $U \subset X$ open and closed, either $U = \emptyset$ or $U = X$.
If the above holds, then $X$ is connected.
Proof. $(1) \Rightarrow (2)$: Let $f \in C(X; \bracs{0, 1})$, then $X = f^{-1}(0) \cup f^{-1}(1)$. If $f$ is surjective, then $f^{-1}(0) \cap f^{-1}(1) \ne \emptyset$, which is impossible.
$\neg (1) \Rightarrow \neg (2)$: If there exists $\emptyset \ne U, V \subset X$ open with $X = U \sqcup V$, then $f = \one_{U} \in C(X; \bracs{0, 1})$ is surjective.
$\neg (3) \Rightarrow \neg (1)$: Let $\emptyset \ne U \subsetneq X$ be open and closed, then $X = U \sqcup U^{c}$ is a disjoint union of two non-empty open sets.
$\neg (1) \Rightarrow \neg (3)$: Let $\emptyset \ne U, V \subset X$. If $X = U \sqcup V$, then $\emptyset \ne U \subsetneq X$ is open and closed.$\square$
Proposition 4.12.2. Let $X$ be a topological space, $\seqi{A}\subset 2^{X}$ be connected with $\bigcap_{i \in I}A_{i} \ne \emptyset$, then $\bigcup_{i \in I}A_{i}$ is connected.
Proof. Let $x \in \bigcap_{i \in I}A_{i} = A$ and $U, V \subset X$ be open with $U \cap A, V \cap A \ne \emptyset$. Assume without loss of generality that $x \in U$. Let $i \in I$ such that $V \cap A_{i} \ne \emptyset$, then since $U \cap A_{i} \ne \emptyset$, $U \cap V \ne \emptyset$ by connectedness of $A_{i}$.$\square$
Proposition 4.12.3. Let $X$ be a connected space, $Y$ be a topological space, and $f \in C(X; Y)$, then $f(X)$ is connected.
Proof. Let $U, V \subset Y$ be open with $U \cap f(X), V \cap f(X) \ne \emptyset$. By continuity of $f$ and connectedness of $X$, $f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V) \ne \emptyset$. Hence $U \cap V \ne \emptyset$.$\square$
Definition 4.12.4 (Connected Component). Let $X$ be a topological space and $A \subset X$ be connected, then there exists a unique connected set $C \supset A$ such that for any $C' \supset A$ connected, $C \supset C'$. The set $C$ is the connected component of $A$.
Proof. Let $C$ be the union of all connected sets that contain $A$, then $C$ is connected by Proposition 4.12.2, and is the maximum connected set containing $A$ by definition.$\square$