[proofplan] We establish each of the three assertions in turn. For (1), we show that every path component is connected — by observing that any two of its points are joined by the continuous image of an interval — and appeal to the fact that connected subsets are contained in connected components. For (2), a single-component consequence of (1). For (3), we verify the partition properties: the path components cover $C$ (every point belongs to its own path component) and are pairwise disjoint (path-connectedness is an equivalence relation), so $C$ decomposes as a disjoint union of path components. [/proofplan]

[step:Show that every path component is path-connected, hence connected]Let $P$ be a path component of $X$. By definition, $P$ is the equivalence class of some point $x_0 \in X$ under the relation "$x \sim y$ if and only if there exists a continuous map $\gamma: [0,1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$." In particular, for any two points $x, y \in P$, there exist continuous maps $\gamma_1: [0,1] \to X$ with $\gamma_1(0) = x_0$, $\gamma_1(1) = x$ and $\gamma_2: [0,1] \to X$ with $\gamma_2(0) = x_0$, $\gamma_2(1) = y$. Define the concatenation \begin{align*} \gamma: [0,1] &\to X \\ t &\mapsto \begin{cases} \gamma_1(1 - 2t) & \text{if } t \in [0, \tfrac{1}{2}], \\ \gamma_2(2t - 1) & \text{if } t \in [\tfrac{1}{2}, 1]. \end{cases} \end{align*} The map $\gamma$ is continuous by the pasting lemma (the two restrictions agree at $t = \tfrac{1}{2}$ since $\gamma_1(0) = x_0 = \gamma_2(0)$), and satisfies $\gamma(0) = \gamma_1(1) = x$ and $\gamma(1) = \gamma_2(1) = y$. Therefore $P$ is path-connected. Since $[0,1]$ is connected and paths are continuous maps from $[0,1]$ into $X$, each path image $\gamma([0,1])$ is connected by the [Continuous Image of a Connected Space is Connected](/theorems/296) theorem. For any $x \in P$, choose a path $\gamma_x: [0,1] \to X$ from $x_0$ to $x$. Then \begin{align*} P = \bigcup_{x \in P} \gamma_x([0,1]). \end{align*} Each set $\gamma_x([0,1])$ is connected and contains $x_0$, so their union is connected (a union of connected sets with a common point is connected).[/step]

[guided]Why must a path component be connected? The argument has two layers. First, we verify that $P$ is path-connected: any two points $x, y \in P$ can be joined by a path in $X$ lying entirely in $P$. Both $x$ and $y$ are path-connected to the basepoint $x_0$ (that is what membership in the equivalence class $P$ means). We concatenate the reverse of the path from $x_0$ to $x$ with the path from $x_0$ to $y$, obtaining a path from $x$ to $y$. Every point on this path is itself path-connected to $x_0$ (by restricting the path), so the entire image lies in $P$. Therefore $P$ is path-connected. Second, we deduce that $P$ is connected. Write $P = \bigcup_{x \in P} \gamma_x([0,1])$ where $\gamma_x: [0,1] \to X$ is a path from $x_0$ to $x$. By the [Continuous Image of a Connected Space is Connected](/theorems/296) theorem, each $\gamma_x([0,1])$ is a connected subset of $X$ (since $[0,1]$ is connected and $\gamma_x$ is continuous). All these connected sets share the common point $x_0$. A union of connected sets with a common point is connected: if $\{C_i\}_{i \in I}$ are connected subsets of $X$ with $p \in \bigcap_{i \in I} C_i$, and $f: \bigcup_i C_i \to \{0, 1\}$ is continuous, then $f$ is constant on each $C_i$ (by connectedness), and since all $C_i$ share $p$, the constant value must equal $f(p)$ on every $C_i$. Hence $f$ is constant on $\bigcup_i C_i$. This gives us: $P$ is a connected subset of $X$.[/guided]

[step:Conclude that every path component is contained in a connected component] Since $P$ is a connected subset of $X$ (established in the previous step), and connected components are the maximal connected subsets of $X$, the set $P$ must be contained in the unique connected component $C$ that contains any (hence every) point of $P$. This proves assertion (1). [/step]

[step:Deduce that path-connected spaces are connected]Suppose $X$ is path-connected. Then $X$ has exactly one path component, namely $X$ itself (every pair of points is joined by a path). By assertion (1), $X$ is contained in a connected component of $X$. Since connected components are subsets of $X$, this connected component equals $X$. Therefore $X$ is connected, which is assertion (2).[/step]

[guided]Assertion (2) is a global consequence of (1). If the entire space $X$ is a single path component, then $X$ is connected as a subset of itself. This recovers the classical result that [path-connected implies connected](/theorems/300). The point of stating it as part of this theorem is to emphasize that it is an immediate corollary of the containment relationship between path components and connected components — not a separate argument.[/guided]

[step:Verify that the path components within a connected component form a partition]Let $C$ be a connected component of $X$. We verify the two properties of a partition. **Covering.** Every point $x \in C$ belongs to a path component of $X$ — namely, the equivalence class of $x$ under the path-connectedness relation. Since $C$ is a connected component containing $x$, assertion (1) guarantees that this path component is contained in $C$. Therefore $C = \bigcup_{i \in I} P_i$ where $\{P_i\}_{i \in I}$ is the collection of all path components meeting $C$. **Disjointness.** Path components are equivalence classes of the relation "$x \sim y$ if and only if there exists a path from $x$ to $y$." This relation is reflexive (the constant path), symmetric (traverse the path in reverse: if $\gamma: [0,1] \to X$ is a path from $x$ to $y$, then $t \mapsto \gamma(1-t)$ is a path from $y$ to $x$), and transitive (concatenate paths, as in the first step). Equivalence classes of an equivalence relation are pairwise disjoint. Therefore distinct path components $P_i, P_j$ with $i \neq j$ satisfy $P_i \cap P_j = \varnothing$. Combining both properties: $C = \bigsqcup_{i \in I} P_i$, which is assertion (3).[/step]

[guided]The partition claim rests entirely on the fact that path-connectedness defines an equivalence relation on $X$. Let us verify each axiom explicitly. **Reflexivity:** For any $x \in X$, the constant map $\gamma: [0,1] \to X$, $t \mapsto x$ is continuous and satisfies $\gamma(0) = \gamma(1) = x$. So $x \sim x$. **Symmetry:** If $\gamma: [0,1] \to X$ is a path from $x$ to $y$, define $\overline{\gamma}: [0,1] \to X$ by $\overline{\gamma}(t) = \gamma(1-t)$. The map $t \mapsto 1-t$ is continuous from $[0,1]$ to $[0,1]$, so $\overline{\gamma}$ is continuous as a composition of continuous maps. Moreover $\overline{\gamma}(0) = \gamma(1) = y$ and $\overline{\gamma}(1) = \gamma(0) = x$, so $\overline{\gamma}$ is a path from $y$ to $x$. **Transitivity:** If $\gamma_1: [0,1] \to X$ is a path from $x$ to $y$ and $\gamma_2: [0,1] \to X$ is a path from $y$ to $z$, define \begin{align*} \gamma: [0,1] &\to X \\ t &\mapsto \begin{cases} \gamma_1(2t) & \text{if } t \in [0, \tfrac{1}{2}], \\ \gamma_2(2t-1) & \text{if } t \in [\tfrac{1}{2}, 1]. \end{cases} \end{align*} The restrictions to the closed sets $[0, \tfrac{1}{2}]$ and $[\tfrac{1}{2}, 1]$ are continuous (each is a composition of a continuous map with an affine reparametrisation), and they agree at $t = \tfrac{1}{2}$ since $\gamma_1(1) = y = \gamma_2(0)$. By the pasting lemma, $\gamma$ is continuous. We have $\gamma(0) = \gamma_1(0) = x$ and $\gamma(1) = \gamma_2(1) = z$, so $x \sim z$. Since path-connectedness is an equivalence relation, the equivalence classes (= path components) partition $X$. Each path component meeting $C$ is contained in $C$ by assertion (1). Conversely, every point of $C$ belongs to some path component. Therefore the path components meeting $C$ partition $C$.[/guided]