Paths, path-connected spaces and path components

Throughout, $I := [0,1] = \{\, t \in \mathbb{R} : 0 \le t \le 1 \,\}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) carries the subspace topology inherited from $\mathbb{R}$ with its usual topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). It is called the unit interval.

Let $X$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $x, y \in X$.

• A path in $X$ from $x$ to $y$ is a continuous map $\gamma : I \to X$ (Continuity of a map of topological spaces at a point and globally) with $\gamma(0) = x$ and $\gamma(1) = y$. Its image is $\gamma[I]$.
• $X$ is path-connected when for every pair $x, y \in X$ there is a path in $X$ from $x$ to $y$. A subset $A \subseteq X$ is a path-connected subset when the space $A$ with its subspace topology is path-connected; equivalently, when any two of its points are joined by a path whose image lies in $A$, by the characteristic property of a map into a subspace (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).
• Write $x \sim y$ when a path in $X$ from $x$ to $y$ exists. The path component of $x$ is its equivalence class $P(x) \;:=\; \{\, y \in X : x \sim y \,\} .$
• The empty space is path-connected, the defining condition quantifying over no pair of points, and so is every one-point space, the constant path joining its point to itself.

$\sim$ is an equivalence relation on $X$, and the obligation is discharged here, so that "equivalence class" above denotes.

Reflexive. The constant map $\gamma(t) = x$ is continuous, every preimage being $\varnothing$ or $I$ (Continuity of a map of topological spaces at a point and globally), and joins $x$ to $x$.

Symmetric. If $\gamma$ joins $x$ to $y$, put $\bar\gamma(t) := \gamma(1-t)$. The map $r : I \to I$, $r(t) = 1 - t$, is continuous: for $s, t \in I$ one has $|r(s) - r(t)| = |s - t|$, so a ball of radius $\varepsilon$ around $r(t)$ pulls back to contain the ball of radius $\varepsilon$ around $t$ (Open ball, closed ball and sphere in a metric space, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Hence $\bar\gamma = \gamma \circ r$ is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, claim 1) and joins $y$ to $x$.

Transitive. Let $\gamma_1$ join $x$ to $y$ and $\gamma_2$ join $y$ to $z$. Define $\delta : I \to X$ by

$\delta(t) \;:=\; \begin{cases} \gamma_1(2t), & 0 \le t \le 1/2, \\ \gamma_2(2t - 1), & 1/2 \le t \le 1. \end{cases}$

The two clauses agree at $t = 1/2$, both giving $\gamma_1(1) = y = \gamma_2(0)$, so $\delta$ is a well-defined function. The sets $[0,1/2]$ and $[1/2,1]$ are closed in $I$ and cover it, and there are two of them, so the finite closed form of the pasting lemma applies (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, claim 3). On $[0,1/2]$ the map $\delta$ is $\gamma_1 \circ a_1$ with $a_1(t) = 2t$, and on $[1/2,1]$ it is $\gamma_2 \circ a_2$ with $a_2(t) = 2t - 1$; each $a_k$ is continuous into $I$, since $|a_k(s) - a_k(t)| = 2|s-t|$, so the ball of radius $\varepsilon/2$ around $t$ maps into the ball of radius $\varepsilon$ around $a_k(t)$ (Open ball, closed ball and sphere in a metric space, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). So both restrictions are continuous by Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous claim 1, hence $\delta$ is continuous, and it joins $x$ to $z$.

The path components partition $X$, being the classes of an equivalence relation, and each is a path-connected subset of $X$: two points of $P(x)$ are joined to $x$, hence to each other by the transitivity construction above, and the resulting path has image inside $P(x)$: if $\delta$ is a path from $x$ and $s \in I$, then $t \mapsto \delta(st)$ is a path from $x$ to $\delta(s)$, continuous because $t \mapsto st$ satisfies $|st_1 - st_2| \le |t_1 - t_2|$ and is therefore continuous into $I$ by the ball criterion used above, so every point of the image is itself joined to $x$.

• Why the unit interval and not an arbitrary closed bounded interval. Any $[a,b]$ with $a < b$ would give the same relation, since $t \mapsto a + t(b-a)$ carries $[0,1]$ onto $[a,b]$ and is continuous with continuous inverse. Fixing $[0,1]$ removes a parameter from every statement below and costs nothing.

• A path is a map, not a subset. The image $\gamma[I]$ is a subset of $X$, but the path is the map: two different paths may have the same image, and the concatenation above depends on the maps rather than on their images. Nothing in this library identifies a path with its image.

• Path components are not asserted to be closed, or open, or to coincide with components. Each of those is false in general, and each is taken up separately on this page. What is proved here is only that they partition $X$ and that each is path-connected.

• The finiteness in the pasting lemma is what makes concatenation legal. The cover $\{[0,1/2], [1/2,1]\}$ has two members. An infinite closed cover would not do, and the standing warning is $\mathbb{R}$ covered by its closed singletons: every restriction of the indicator of $\{0\}$ is continuous and the map is not, so the closed pasting lemma needs finiteness; this is worth naming here because the temptation to concatenate infinitely many paths is exactly what fails for the zigzag curve later on this page.