Connectedness is transitive

Let $𝑋$ be a topological space and $𝑥, 𝑦 \in 𝑋$ . If $𝑥$ is (path-)connected to $𝑦$ and $𝑦$ is (path-)connected to $𝑧$ , then $𝑥$ is (path-)connected to $𝑧$ . #m/thm/topology Hence both forms of connectedness form an equivalence relation.

For plain connectedness

Let $𝑥 \sim' 𝑦$ and $𝑦 \sim' 𝑧$ , i.e. there exists connected subspaces $𝑈 1 ∋ 𝑥, 𝑦$ and $𝑈 2 ∋ 𝑦, 𝑧$ . We claim that $𝑉 = 𝑈 1 \cup 𝑈 2$ is a connected subspace of $𝑋$ . Let $𝜄 𝑘 : 𝑈 𝑘 \to 𝑉$ denote that natural inclusions of $𝑈 𝑘$ in $𝑉$ , and $𝑓 : 𝑉 \to {0, 1}$ be a continuous function. Since $𝑓$ and $𝜄 𝑘$ are continuous, so too are $𝑓 𝜄 𝑘 : 𝑈 𝑘 \to {0, 1}$ . Thus $𝑓 𝑤 = 𝑓 𝑦$ for all $𝑤 \in 𝑈 1 \cup 𝑈 2 = 𝑉$ . Hence $𝑓$ is constant for $𝑉$ . Therefore $𝑥 \sim' 𝑧$ .

For path connectedness

Let $𝑓$ be a continuous path from $𝑥$ to $𝑦$ and $𝑔$ be a continuous path from $𝑦$ to $𝑧$ . Then the product $𝑓 \cdot 𝑔$ is a continuous path from $𝑥$ to $𝑧$ . Hence $𝑥 \sim 𝑦 \land 𝑦 \sim 𝑧 ⟹ 𝑥 \sim 𝑧$ .

#state/tidy | #lang/en | #SemBr