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A topological space $X$ is connected when it has no separation of a topological space: no partition of a set into two nonempty open subsets $U,V$ satisfyingHere the union covers all of $X$, and the intersection is the empty set, so no point belongs to both pieces. For a subset $X$ of a larger topological space, "open" means open in the subspace topology: each piece is the intersection of $X$ with an ambient open set. Both pieces must be open; merely dividing a drawing into two colours does not establish a separation of a topological space. ^[ref]

For example, the closed interval $I=[0,1]$ with its usual subspace topology from the real line is connected. The proposed split $[0,1/2)\cup [1/2,1]$ fails: the second subset is not open in $I$, since every neighbourhood of $1/2$ in $I$ contains points less than $1/2$. More generally, a connected subset of the real line is precisely a real interval. In contrast, for the unionthe open intervals $(-3,0)$ and $(0,3)$ exhibit $U,V$ as nonempty open subsets in the subspace topology of $D$. Their intersection is empty and their union is $D$, so $D$ is not connected. Although $U,V$ are closed intervals in the real line, they are also open within $D$.

Why is connectedness useful? A continuous image of a connected space is connected: a proposed separation of a topological space in the image would pull back through the continuous map to a forbidden separation of a topological space in its source. Thus a continuous map from a real interval to the real line cannot skip values between two attained values. This is the intermediate value theorem. For a computed example, the continuous function $f(x)=x^2$ on $[0,1]$ has image $[0,1]$; the value $1/2$ occurs at $x=1/\sqrt{2}\approx 0.707107$. Connectedness guarantees an input attaining that value even when we have no formula for it.

Path connectedness is a stronger condition: it supplies a continuous path between every pair of points. The topologist's sine curve below shows that a connected space need not be a path-connected space.

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]

A topological space $X$ is path-connected if, for every $a,b\in X$, there is a continuous mapThis continuous path must stay within $X$ for its entire duration. Its parameter need not represent physical time, and continuity does not require a derivative or finite arc length. ^[ref]

In the closed interval $[0,1]$, any $a,b$ are joined by $\gamma (t)=(1-t)a+tb$. For $a=1/4$ and $b=3/4$, this becomes $\gamma (t)=1/4+t/2$, so $\gamma (0)=1/4$, $\gamma (1/2)=1/2$, and $\gamma (1)=3/4$. All its values stay in $[0,1]$. In the union $D=[-2,-1]\cup [1,2]$, however, a continuous path from $-1$ to $1$ would attain $0$ by the intermediate value theorem, although $0\notin D$. Thus $D$ is not path-connected.

Every path-connected space is a connected space. Proof outline: if $X=U\cup V$ admitted a separation of a topological space, choose $a\in U$ and $b\in V$. A continuous path between them would have nonempty open inverse images ${\gamma }^{-1}(U),{\gamma }^{-1}(V)$ with empty intersection and union $[0,1]$. This contradicts the connectedness of that closed interval.

The converse fails: the topologist's sine curve is connected but has two path components. Path connectedness is useful when we need an actual continuous path of allowed positions, rather than only the absence of a separation of a topological space. For instance, when modelling allowed positions as a topological space, membership in the same path component answers whether one can move continuously between two positions while remaining in that topological space.

Consider the graph of a function $x\mapsto \sin (1/x)$, using the sine of the multiplicative inverse of $x$, together with a vertical closed interval:Give $T$ the subspace topology from the Euclidean plane. The symbol $\times$ denotes a Cartesian product. The right endpoint $(1,\sin 1)\approx (1,0.841471)$ belongs to $G$. The continuous map $x\mapsto (x,\sin (1/x))$ sends the connected real interval $(0,1]$ onto $G$. Hence $G$ is connected, and its closure $\overline{G}=T$ is connected by closure of a connected set. Every height in $[-1,1]$ is approached as $x$ tends to zero; for example,For $n=1$, these inputs are $p_1=2/(5\pi )\approx 0.127324$ and $q_1=2/(7\pi )\approx 0.090946$. For any other height $y\in [-1,1]$, choose $\theta \ge 1$ with $\sin \theta =y$; then the sequence $(1/(\theta +2\pi n),y)$ lies in $G$ and has limit of a sequence $(0,y)$.

Nevertheless, $T$ has exactly two path components, $G$ and $L$. Within $G$, join inputs $a,b\in (0,1]$ by the continuous path $t\mapsto ((1-t)a+tb,\sin (1/((1-t)a+tb)))$. Within $L$, use $t\mapsto (0,(1-t)c+td)$ for heights $c,d\in [-1,1]$.

The obstruction to a continuous path between $L$ and $G$ is persistent oscillation, not a visible gap. Proof outline: suppose such a path exists, directed from $L$ to $G$, and take its last time in $L$. Immediately afterwards its first coordinate $x(t)>0$ approaches zero and its second coordinate is $\sin (1/x(t))$. The intermediate value theorem forces the first coordinate to pass through inputs of both displayed types arbitrarily close to that departure time. The second coordinate therefore takes both values $1$ and $-1$ arbitrarily close to that time, contradicting continuity. ^[ref]

The sampled graph of a function illustrates the formulas; it does not prove connectedness or the failure of path connectedness. Taking a closure preserves connectedness, but this example shows that it can destroy path connectedness.