> [!definition] > > Let $(X, \topo_X)$ be a [[Topological Space|topological space]] and let $A \subseteq X$. $A$ is **path connected** if for any $x, y \in A$, there exists a [[Continuity|continuous]] function $f: [0, 1] \to X$ with $f(0) = x$ and $f(1) = y$. > [!theorem] > > Let $(X, \topo_X)$ and $A \subseteq X$ be a **path connected** set. Then $A$ is [[Connected|connected]]. > > *Proof*. Suppose that $A$ is path connected but not connected. Then > $ > \exists U, V \in \topo_X: U, V \cap A \ne \emptyset, U \cap V \ne \emptyset > $ > Let $x \in U \cap A$ and $y \in V \cap A$. Then there exists a continuous function $f: [0, 1] \to A$ with $f(0) = x$ and $f(1) = y$. Then $f([0, 1])$ is connected but covered by two disjoint open sets. Therefore $A$ must also be connected.