Maximal Path-Connected Subsets

From Paths to Maximal Subsets

A path-connected space is one in which every pair of points can be joined by a continuous path. A general space need not be path-connected, but its points still fall into groups: two points belong to the same group exactly when some path in the space joins them. The purpose here is to show that these groups are precisely the maximal path-connected subsets. “Maximal” means that no strictly larger path-connected subset contains the given subset; it does not mean largest among all subsets by size.

Throughout, \((X,\mathcal{T})\) is a topological space, and \(I=[0,1]\) has its usual subspace topology from \(\mathbb{R}\). A path in \(X\) from \(x\) to \(y\) is a continuous map \(\gamma:I\to X\) such that \(\gamma(0)=x\) and \(\gamma(1)=y\). We use the convention that a path-connected subset is nonempty.

The requirement that the path stay in \(P\) matters: a path in \(X\) joining two points of \(P\) is not enough to make \(P\) path-connected. The maximal subsets we seek are defined by paths that remain within them.

The Path Relation and Its Equivalence Classes

Define a relation on \(X\) by \(x\sim y\) when there is a path in \(X\) from \(x\) to \(y\). For \(x\in X\), write \([x]=\{y\in X:y\sim x\}\). We first verify that the relation is an equivalence relation, including the constructions of paths needed below.

Proof. For each \(x\in X\), the constant map \(\gamma(t)=x\) is a path from \(x\) to itself, so \(x\sim x\). If \(\gamma\) is a path from \(x\) to \(y\), then \(t\mapsto\gamma(1-t)\) is continuous and is a path from \(y\) to \(x\), so \(x\sim y\) implies \(y\sim x\). If \(\gamma\) is a path from \(x\) to \(y\) and \(\eta\) is a path from \(y\) to \(z\), define \[ \delta(t)= \begin{cases} \gamma(2t),&0\le t\le \tfrac12,\\ \eta(2t-1),&\tfrac12\le t\le 1. \end{cases} \] At \(t=\tfrac12\), both formulas give \(y\). Each formula is continuous on its closed subinterval, and the formulas agree where the subintervals overlap. The finite closed-set pasting lemma therefore gives continuity of \(\delta\): for any closed target set, its preimage is the union of its closed preimages on the two closed subintervals, hence is closed in \(I\). Also \(\delta(0)=x\) and \(\delta(1)=z\), so \(x\sim z\). This proves transitivity. Finally, if two equivalence classes intersect, an element in their intersection is equivalent to representatives of both classes; symmetry and transitivity then show the classes contain exactly the same points. Thus distinct classes are disjoint. \(\square\)

Proof. First let \(u,v\in[x]\). By definition, \(u\sim x\) and \(v\sim x\). Symmetry gives a path from \(u\) to \(x\), and there is a path from \(x\) to \(v\). Concatenating these paths as in the proposition gives a path from \(u\) to \(v\). Every point of that path lies in \([x]\): if \(w\) is on the first path, its initial segment joins \(u\) to \(w\), so \(w\sim u\sim x\); if \(w\) is on the second path, its initial segment joins \(x\) to \(w\), so \(w\sim x\). Thus \([x]\) is path-connected.

To prove maximality, suppose \(Q\subseteq X\) is path-connected and \([x]\subseteq Q\). Choose any \(q\in Q\). Since \(x\in[x]\subseteq Q\), path-connectedness of \(Q\) gives a path in \(Q\), and therefore in \(X\), from \(x\) to \(q\). Hence \(q\in[x]\). This holds for every \(q\in Q\), so \(Q\subseteq[x]\). Combined with \([x]\subseteq Q\), it gives \(Q=[x]\), proving maximality.

For the converse, let \(P\) be a maximal path-connected subset and choose \(x\in P\), which is possible because path-connected subsets are required to be nonempty. For every \(p\in P\), path-connectedness gives a path in \(P\) from \(x\) to \(p\), so \(p\in[x]\). Thus \(P\subseteq[x]\). We have already proved that \([x]\) is path-connected. Maximality of \(P\) therefore forces \(P=[x]\). This proves both claims. \(\square\)

Why the Classes Form a Partition

Proof. For any \(x\in X\), the constant path shows \(x\in[x]\), so each class is nonempty and every point of \(X\) belongs to a class. Thus the classes cover \(X\). The proposition shows that any two classes are equal or disjoint, so distinct classes are disjoint. If \(X=\varnothing\), there are no points and hence no equivalence classes; the empty family covers the empty set and has no overlapping members. \(\square\)

The convention that path-connected subsets are nonempty handles the empty space without treating the empty set as an extra maximal subset. Under a convention that allows the empty set to count as path-connected, it must be excluded from the phrase “maximal path-connected subset” for this theorem to apply when \(X\) is empty.

A Common Mistake: Unions Without a Shared Point

It is tempting to reason that the union of path-connected subsets must be path-connected. That statement is false unless there is an additional condition connecting the subsets. For example, in the discrete space \(X=\{a,b\}\), each of \(\{a\}\) and \(\{b\}\) is path-connected: each has a constant path joining its only point to itself. But their union is not path-connected, because the preceding inverse-image argument shows there is no path from \(a\) to \(b\).

The missing condition is a shared point. If path-connected subsets \(P\) and \(Q\) have \(z\in P\cap Q\), then any \(p\in P\) can be joined to \(z\) inside \(P\), and \(z\) can be joined to any \(q\in Q\) inside \(Q\). Concatenating the two paths joins \(p\) to \(q\) inside \(P\cup Q\). Pairs lying both in \(P\) or both in \(Q\) are already joined there. Hence \(P\cup Q\) is path-connected when \(P\cap Q\ne\varnothing\). This agrees with the theorem: equivalence classes that intersect must be equal, not two distinct maximal pieces.

Keep the distinction between an arbitrary path-connected subset and a maximal one in view. A path-connected subset may sit inside a larger path-connected subset; only the maximal subsets are exactly the equivalence classes. The classes give a canonical partition because the relation “can be joined by a path in \(X\)” is defined on all of \(X\), whether or not \(X\) itself is path-connected.