[proofplan] The strategy is to show that in a locally path-connected space, every path component is open. Once this is established, all three assertions follow: (2) each connected component, being a disjoint union of path components that are open, must consist of a single path component (otherwise it would be disconnected by a nontrivial partition into open sets), giving (1); and (3) follows from (1) and (2) combined with the [Path Components Refine Connected Components](/theorems/1055) theorem. The key step — proving openness of path components — uses the local path-connectedness hypothesis to surround each point with a path-connected open neighborhood, which must be absorbed into the path component. [/proofplan] [step:Show that every path component of $X$ is open] Let $P$ be a path component of $X$ and let $x \in P$. Since $X$ is locally path-connected, there exists an open neighborhood $U$ of $x$ such that $U$ is path-connected. We claim $U \subset P$. Let $u \in U$. Since $U$ is path-connected, there exists a continuous map $\gamma_1: [0,1] \to U \subset X$ with $\gamma_1(0) = x$ and $\gamma_1(1) = u$. Since $x \in P$, there exists a point $p_0$ (the basepoint of $P$) and a path from $p_0$ to $x$ in $X$. Concatenating this path with $\gamma_1$ yields a path from $p_0$ to $u$ in $X$, so $u \in P$. Since $u \in U$ was arbitrary, $U \subset P$. Since $x \in P$ was arbitrary and $U$ is an open neighborhood of $x$ contained in $P$, the set $P$ is open. [guided] This is the step where local path-connectedness is consumed. The hypothesis says: every point of $X$ has arbitrarily small path-connected open neighborhoods. We use it once: given $x \in P$, find a path-connected open set $U$ with $x \in U$. Why does $U \subset P$ follow? Every point $u \in U$ can be reached from $x$ by a path within $U$ (since $U$ is path-connected). But $x$ already belongs to the path component $P$, meaning $x$ is path-connected to every other point of $P$. Composing: $u$ is path-connected to $x$ (via the path in $U$), and $x$ is path-connected to the basepoint of $P$ (via the defining path). By transitivity of the path-connectedness relation, $u$ belongs to $P$. The critical observation is that the path from $x$ to $u$ lies in $U$, but the path certifying $u \in P$ may leave $U$ — it only needs to be a path in $X$. We do not require the path component $P$ to be path-connected within any particular subset; all paths live in the ambient space $X$. Note what happens without local path-connectedness. Consider the topologist's sine curve $S = \{(x, \sin(1/x)) : x > 0\} \cup \{0\} \times [-1, 1]$. The path component of $(0,0)$ is $\{0\} \times [-1, 1]$, which is not open in $S$. Indeed, every open neighborhood of $(0,0)$ in $S$ contains points $(x, \sin(1/x))$ with $x > 0$ that are not path-connected to $(0,0)$. The space $S$ is not locally path-connected at $(0,0)$ — every neighborhood of that point contains points from the oscillating curve that cannot be reached by a path from the vertical segment. [/guided] [/step] [step:Deduce that path components are also closed, hence clopen] Since the path components of $X$ partition $X$ (by the equivalence relation argument in [Path Components Refine Connected Components](/theorems/1055)), each path component $P$ satisfies \begin{align*} X \setminus P = \bigcup_{\substack{Q \text{ path component} \\ Q \neq P}} Q. \end{align*} Each path component $Q \neq P$ is open (by the previous step). A union of open sets is open, so $X \setminus P$ is open. Therefore $P$ is closed. Since $P$ is both open and closed, $P$ is clopen. [guided] Every equivalence class in a partition has the property that its complement is the union of the remaining classes. When every equivalence class is open, every complement is a union of open sets, hence open. This makes every class both open and closed. This is a general phenomenon: if a topological space is partitioned into open sets, then each piece of the partition is also closed (and hence clopen). The partition into path components is the one we use here. [/guided] [/step] [step:Prove that the connected components and path components coincide] By [Path Components Refine Connected Components](/theorems/1055), every path component $P$ is contained in a connected component $C$, and $C = \bigsqcup_{i \in I} P_i$ is a disjoint union of path components. We must show $|I| = 1$, i.e., $C$ consists of a single path component. Suppose for contradiction that $|I| \geq 2$. Fix some $P_{i_0}$ in the partition. By the previous step, $P_{i_0}$ is open in $X$, and the union $\bigcup_{i \neq i_0} P_i = C \setminus P_{i_0}$ is a union of open sets, hence open in $X$. Both $P_{i_0}$ and $C \setminus P_{i_0}$ are non-empty (since $|I| \geq 2$), open in $X$ (hence open in the subspace topology on $C$), and their union is $C$. This is a disconnection of $C$, contradicting the connectedness of $C$. Therefore $|I| = 1$ and $C = P_{i_0}$. Every connected component is a path component, and every path component is a connected component. This establishes assertion (1). Assertion (2) — the components are open, hence clopen — follows immediately since path components are clopen by the previous step. [guided] The contradiction argument here is a standard application of the characterization of connectedness: a space is connected if and only if it admits no partition into two non-empty open sets. The key input is that path components are open (established in the first step via local path-connectedness). Without this openness, the argument fails — and indeed it must fail, since [Path Components Refine Connected Components](/theorems/1055) shows that in a general topological space, a connected component can contain strictly more than one path component. The topology of the subspace $C$ matters here. The sets $P_{i_0}$ and $C \setminus P_{i_0}$ are open in $X$, hence open in $C$ with the subspace topology (since an open subset of $X$ that is contained in $C$ is open in the subspace topology). Alternatively, a subset of $C$ is open in the subspace topology if and only if it equals $C \cap V$ for some open $V \subset X$; since $P_{i_0} \subset C$ and $P_{i_0}$ is open in $X$, we have $P_{i_0} = C \cap P_{i_0}$, which is open in $C$. [/guided] [/step] [step:Conclude that $X$ is connected if and only if it is path-connected] Assertion (3) follows from assertions (1) and (2). **If $X$ is path-connected**, then $X$ is connected by [Path Components Refine Connected Components](/theorems/1055), assertion (2) (or equivalently by [Path-Connected Implies Connected](/theorems/300)). **If $X$ is connected**, then $X$ has exactly one connected component, namely $X$ itself. By assertion (1), the connected components and path components coincide, so $X$ is also the unique path component. Therefore $X$ is path-connected. [guided] In a general topological space, "connected" is strictly weaker than "path-connected" — the topologist's sine curve provides the standard counterexample. The locally path-connected hypothesis closes the gap by ensuring that path components are open, which forces each connected component to be a single path component. This equivalence is frequently invoked in practice. Many of the spaces encountered in geometry and analysis — open subsets of $\mathbb{R}^n$, manifolds, CW complexes, topological groups — are locally path-connected. In these settings, one can freely interchange "connected" and "path-connected," which is the content of this theorem. The theorem [Connected Open Subsets of Euclidean Space are Path-Connected](/theorems/301) is a special case: open subsets of $\mathbb{R}^n$ are locally path-connected (every point has a path-connected open ball around it), so connectedness and path-connectedness coincide. [/guided] [/step]