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In a locally path-connected space, every path component is open: a path-connected open neighbourhood of a point lies within that point's path component. The complement of a path component is a union of the other open path components, hence is also open. In a nonempty connected space, a nonempty subset that is both open and closed must be the whole space. Thus a connected space that is locally path-connected is path-connected.

Every locally path-connected space is a locally connected space, but the converse can fail.

A space is locally path-connected when every point has a neighbourhood basis of path-connected open sets. Its path components are open.

Every point has arbitrarily small path-connected open neighbourhoods. Indeed, in a coordinate chart homeomorphic to ${\mathbb{R}}^n$, take the inverse image of a sufficiently small open Euclidean ball. The ball is convex, hence path-connected, and its inverse image is path-connected by the homeomorphism. If the supplied neighbourhood is not itself open, first take an open neighbourhood within it and a ball contained in its chart image. This proves that $X$ is a locally path-connected space.

Fix $x_0\in X$ and let $A$ consist of the points joined to $x_0$ by a path. For $x\in A$, a path-connected open neighbourhood $U_x$ is contained in $A$: concatenate a path from $x_0$ to $x$ with one from $x$ to any point of $U_x$. Hence $A$ is open. For $x\notin A$, the same kind of neighbourhood cannot meet $A$, since a meeting point would provide a path to $x$. Hence $X\setminus A$ is also open. Since $X$ is a connected space and $A$ is nonempty, $A=X$. This proves $X$ is path-connected, including the general principle that connected locally path-connected spaces are path-connected. The empty-space case, if allowed, is vacuous.

A topological space is connected when it is not the union of two disjoint nonempty open subsets. It is path-connected when every two points $x,y$ can be joined by a path, meaning a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.

To prove that $[0,1]$ is connected, suppose it had a separation $[0,1]=U\cup V$ with $0\in U$. LetWhichever of $U$ or $V$ contains $c$, its relative openness supplies an interval about $c$ that contradicts either the definition of the supremum or the fact that all points immediately below $c$ lie in $U$. Thus no separation exists.

If $X$ is path-connected and $X=U\cup V$ were a separation, choose $u\in U$ and $v\in V$. A path from $u$ to $v$ would make $[0,1]$ the union of the disjoint nonempty open sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$, contradicting the connectedness just proved. Hence

Now let $X\subseteq {\mathbb{R}}^n$ be open. Every point $x\in X$ lies in an open ball contained in $X$, and an open ball is path-connected by straight line segments. It follows that every path component of $X$ is open. If there were more than one path component, one component and the union of all the others would separate $X$. Consequently a connected open subset of Euclidean space is path-connected. The converse follows from the preceding implication.

The same argument answers the locally Euclidean question affirmatively. Every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$, hence a path-connected open neighbourhood after restricting to a sufficiently small ball. Thus $X$ is locally path-connected, its path components are open, and connectedness forces there to be only one.

For the final example, putThe set $Y$ is path-connected: each vertical segment $C_n$ meets the horizontal segment $A$. The segment $B$ is also path-connected and lies in the closure of $Y$, because $(1/n,y)\to (0,y)$ for every $y\in [1/2,1]$. A connected set together with a connected subset of its closure is connected, so

It is not path-connected. The image of the first coordinate of any path in $X$ lies inbut, while the path has positive height, its first coordinate lies in the totally disconnected set $\{0\}\cup \{1/n:n\ge 1\}$. A path beginning on $B$ cannot leave $x=0$ before reaching height zero, and $X$ contains no point $(0,y)$ with $0\le y<1/2$. Hence no path joins $B$ to $Y$, and

A subset is connected when it has no separation into two disjoint nonempty relatively open sets. It is path-connected when every pair of its points can be joined by a continuous path from $[0,1]$ lying in the subset.

Path-connectedness implies connectedness; the reverse fails in general, for example for the topologist's sine curve. They are equivalent for open subsets of every Euclidean space ${\mathbb{R}}^n$. More generally this holds for open subsets of locally path-connected spaces.

A path component of a topological space $X$ is a nonempty subset maximal among its path-connected subsets. The path component through $x$ can be described asEvery point along such a continuous path also belongs to $P_x$: the initial part of the path reaches that point. Thus the paths witnessing membership stay inside $P_x$. This makes path components useful for deciding which allowed positions can be reached from one another by continuous motion.

Every path component lies inside one connected component, because every path-connected space is connected. Thus each connected component is a union of path components. In a locally path-connected space the two classifications agree, by connected locally path-connected spaces are path-connected. For a subset of the real line with its usual subspace topology, the classifications also agree: every connected component is a real interval, and interpolation supplies a continuous path within it.

The topologist's sine curve shows why the distinction matters in general. Use specifically the closed versionwith the subspace topology from the usual Euclidean plane. It has one connected component, $T$, but two path components, $G$ and $L$. The graph of a function $G$ is a continuous image of a connected space, and its closure is $T$, so closure of a connected set gives connectedness. Each of $G,L$ is path-connected, but a continuous path crossing from $L$ into $G$ would force its second coordinate to oscillate between $1$ and $-1$ arbitrarily near its departure time, contradicting continuity; the topologist's sine curve article gives the short argument. Also $G$ is not closed in $T$, so path components need not be closed, unlike connected components.^[ref]

Being joined by a continuous path is an equivalence relation. Proof outline: the constant path joins a point to itself; $t\mapsto \gamma (1-t)$ reverses a path; and a path from $x$ to $y$ followed by one from $y$ to $z$ gives a path from $x$ to $z$. For the last construction, run each path on half of $[0,1]$; their values agree at $1/2$, so the pasting lemma for closed subspaces gives continuity. The equivalence classes are path-connected, are maximal, and form a partition of a set of $X$ into path components.^[ref]