[proofplan] We use the path-connectability relation on points of $X$. The result [citetheorem:9162] says that this relation is an [equivalence relation](/page/Equivalence%20Relation), so its equivalence classes are pairwise disjoint and cover $X$. It remains to check that each class is path-connected as a subspace, not merely that its points can be joined by paths in $X$; this is done by showing that every point on such a path stays in the same equivalence class. Finally, a nonempty path-connected subset lies in the class of any one of its points. [/proofplan] [step:Identify path components with equivalence classes of path-connectability] Define a relation $\sim_p$ on $X$ by declaring that, for $x,y \in X$, \begin{align*} x \sim_p y \end{align*} if and only if there exists a continuous map $\gamma : [0,1] \to X$ such that $\gamma(0)=x$ and $\gamma(1)=y$. By [citetheorem:9162], applied to the [topological space](/page/Topological%20Space) $(X,\tau)$, the relation $\sim_p$ is reflexive, symmetric, and transitive. For each $x \in X$, define \begin{align*} P_x := \{y \in X : x \sim_p y\}. \end{align*} Thus $P_x$ is the equivalence class of $x$ under $\sim_p$, and by the standard definition of path component, the path components of $X$ are precisely the sets $P_x$ with $x \in X$. [/step] [step:Use the equivalence relation to obtain disjointness and covering] Since $\sim_p$ is an equivalence relation on $X$, its equivalence classes are pairwise disjoint and their union is $X$. Therefore, if $x,y \in X$, then either $P_x=P_y$ or $P_x \cap P_y=\varnothing$, and \begin{align*} X = \bigcup_{x \in X} P_x. \end{align*} Hence the path components cover $X$ and are pairwise disjoint. [/step] [step:Show each equivalence class is path-connected as a subspace] Fix $x \in X$. We prove that $P_x$, equipped with the [subspace topology](/page/Subspace%20Topology) inherited from $X$, is path-connected. Let $u,v \in P_x$. Since $u \in P_x$ and $v \in P_x$, we have $x \sim_p u$ and $x \sim_p v$. By symmetry and transitivity of $\sim_p$, it follows that $u \sim_p v$. Therefore there exists a continuous map $\gamma : [0,1] \to X$ such that $\gamma(0)=u$ and $\gamma(1)=v$. We claim that $\gamma([0,1]) \subset P_x$. Let $t \in [0,1]$. Define \begin{align*} \gamma_t : [0,1] &\to X \end{align*} \begin{align*} s &\mapsto \gamma(ts). \end{align*} The map $s \mapsto ts$ is continuous from $[0,1]$ to $[0,1]$, and $\gamma$ is continuous, so $\gamma_t$ is continuous as a composition. Also $\gamma_t(0)=\gamma(0)=u$ and $\gamma_t(1)=\gamma(t)$. Hence $u \sim_p \gamma(t)$. Since $x \sim_p u$ and $\sim_p$ is transitive, $x \sim_p \gamma(t)$, so $\gamma(t) \in P_x$. Since $t$ was arbitrary, $\gamma([0,1]) \subset P_x$. Now regard $\gamma$ as a map \begin{align*} \widetilde{\gamma} : [0,1] &\to P_x \end{align*} \begin{align*} t &\mapsto \gamma(t), \end{align*} where $P_x$ has the subspace topology. Because $\gamma : [0,1] \to X$ is continuous and its image is contained in $P_x$, the corestriction $\widetilde{\gamma}$ is continuous for the subspace topology. Thus $\widetilde{\gamma}$ is a path in $P_x$ from $u$ to $v$. Since $u,v \in P_x$ were arbitrary, $P_x$ is path-connected. [guided] Fix $x \in X$. We want to prove that $P_x$ is path-connected as a subspace. The subtle point is that the definition of $P_x$ only tells us that two points in $P_x$ can be joined by a path whose codomain is $X$. For path-connectedness of the subspace $P_x$, the path must actually have codomain $P_x$. Let $u,v \in P_x$. By definition of $P_x$, we have $x \sim_p u$ and $x \sim_p v$. Since $\sim_p$ is symmetric, $u \sim_p x$; since $\sim_p$ is transitive, $u \sim_p v$. Therefore there exists a continuous map $\gamma : [0,1] \to X$ satisfying $\gamma(0)=u$ and $\gamma(1)=v$. We now prove that this path never leaves $P_x$. Let $t \in [0,1]$. Define \begin{align*} \gamma_t : [0,1] &\to X \end{align*} \begin{align*} s &\mapsto \gamma(ts). \end{align*} The map $s \mapsto ts$ is continuous from $[0,1]$ to $[0,1]$, and composing it with the continuous map $\gamma$ gives a continuous map $\gamma_t : [0,1] \to X$. Its endpoints are \begin{align*} \gamma_t(0)=\gamma(0)=u \end{align*} and \begin{align*} \gamma_t(1)=\gamma(t). \end{align*} Thus $\gamma_t$ is a path in $X$ from $u$ to $\gamma(t)$, so $u \sim_p \gamma(t)$. Since $x \sim_p u$ and $\sim_p$ is transitive, we get $x \sim_p \gamma(t)$. Hence $\gamma(t) \in P_x$. Because this holds for every $t \in [0,1]$, we have $\gamma([0,1]) \subset P_x$. We may therefore corestrict $\gamma$ to a map \begin{align*} \widetilde{\gamma} : [0,1] &\to P_x \end{align*} \begin{align*} t &\mapsto \gamma(t). \end{align*} The subspace topology on $P_x$ is exactly what makes this corestriction continuous: a map into a subspace is continuous precisely when its composition with the inclusion into the ambient space is continuous. The composition of $\widetilde{\gamma}$ with the inclusion $P_x \hookrightarrow X$ is the original continuous map $\gamma$. Hence $\widetilde{\gamma}$ is a continuous path in $P_x$ from $u$ to $v$. Since every pair of points $u,v \in P_x$ can be joined by a path lying in $P_x$, the subspace $P_x$ is path-connected. [/guided] [/step] [step:Contain every nonempty path-connected subset in one path component] Let $A \subset X$ be nonempty and path-connected with the subspace topology. Choose $a \in A$. We prove that $A \subset P_a$. Let $b \in A$. Since $A$ is path-connected, there exists a continuous map $\alpha : [0,1] \to A$ such that $\alpha(0)=a$ and $\alpha(1)=b$. Let $\iota : A \to X$ denote the inclusion map. The inclusion $\iota$ is continuous by the definition of the subspace topology, so the composite \begin{align*} \iota \circ \alpha : [0,1] &\to X \end{align*} is continuous and satisfies $(\iota \circ \alpha)(0)=a$ and $(\iota \circ \alpha)(1)=b$. Hence $a \sim_p b$, so $b \in P_a$. Since $b \in A$ was arbitrary, $A \subset P_a$. Thus every nonempty path-connected subset of $X$ is contained in a single path component of $X$. Combining this with the pairwise disjoint covering and path-connectedness proved above completes the proof. [/step]