SectionPath Connectedness as an Equivalence Relation
We saw that we could define an equivalence relation using connected subsets of a topological space, which partitions the space into a disjoint union of connected components. We might expect to be able to do something similar with path connectedness. The main difficulty will be transitivity. As illustrated in Figure 19.5, if we have a path \(p\) from \(a\) to \(b\) and a path \(q\) from \(b\) to \(c\text{,}\) it appears that we can just follow the path \(p\) from \(a\) to \(b\text{,}\) then path \(q\) from \(b\) to \(c\) to have a path from \(a\) to \(c\text{.}\) But there are two problems to consider: how do we define this path as a function from \([0,1]\) into our space, and how do we know the resulting function is continuous. The next lemma will help.
Figure19.5.A path from \(a\) to \(c\text{.}\)
Lemma19.6.The Gluing Lemma.
Let \(A\) and \(B\) be closed subsets of a space \(X = A \cup B\text{,}\) and let \(f:A \to Y\) and \(g: B \to Y\) be continuous functions into a space \(Y\) such that \(f(x)=g(x)\) for all \(x \in (A \cap B)\text{.}\) Then the function \(h:X \to Y\) defined by
\begin{equation*}
h(x) = \begin{cases}f(x) \amp \text{ if } x \in A \\ g(x) \amp \text{ if } x \in B \end{cases}
\end{equation*}
is continuous.
Figure19.7.The Gluing Lemma.
Proof.
Let \(A\) and \(B\) be closed subsets of a space \(X = A \cup B\text{,}\) and let \(f:A \to Y\) and \(g: B \to Y\) be continuous functions into a space \(Y\) such that \(f(x)=g(x)\) for all \(x \in (A \cap B)\) as illustrated in Figure 19.7. Define \(h: X \to Y\) by
\begin{equation*}
h(x) = \begin{cases}f(x) \amp \text{ if } x \in A \\ g(x) \amp \text{ if } x \in B. \end{cases}
\end{equation*}
To show that \(h\) is continuous, let \(C\) be a closed subset of \(Y\text{.}\) Then
Since \(f\) is continuous, \(f^{-1}(C)\) is closed in the subspace topology on \(A\) and since \(g\) is continuous \(g^{-1}(C)\) is closed in the subspace topology on \(B\text{.}\) So \(f^{-1}(C) = A \cap D\) and \(g^{-1}(C) = B \cap E\) for some closed sets \(D\) and \(E\) of \(X\text{.}\) The fact that \(A\) is closed in \(X\) implies that \(A \cap D\) is closed in \(X\text{.}\) Similarly, the fact that \(B\) is closed in \(X\) implies that \(B \cap E\) is closed in \(X\text{.}\) Thus,
is a finite union of closed sets in \(X\) and so is closed in \(X\text{.}\) Since \(h^{-1}(C)\) is closed for every closed set in \(Y\text{,}\) it follows that \(h\) is continuous.
We can use the Gluing Lemma to create a path from \(a\) to \(c\) given a path from \(a\) to \(b\) and a path from \(b\) to \(c\text{.}\)
Activity19.3.
Use the Gluing Lemma to explain why the path product given in the following definition is actually a path from \(p(0)\) to \(q(1)\text{.}\)
Definition19.8.
Let \(p\) be a path from \(a\) to \(b\) and \(q\) a path from \(b\) to \(c\) in a space \(X\text{.}\) The path product \(q*p\) is the path in \(X\) defined by
\begin{equation*}
(q*p)(x) = \begin{cases}p(2x) \amp \text{ for } 0 \leq x \leq \frac{1}{2} \\ q(2x-1) \amp \text{ for } \frac{1}{2} \leq x \leq 1. \end{cases}
\end{equation*}
Now we can show that path connectedness defines an equivalence relation on a topological space.
Activity19.4.
Let \((X,\tau)\) be a topological space. Define a relation on \(X\) as follows:
\begin{equation}
a \sim b \text{ if there is a path in } X \text{ from } a \text{ to } b\text{.}\tag{19.1}
\end{equation}
(a)
Explain why \(\sim\) is a reflexive relation.
(b)
Explain why \(\sim\) is a symmetric relation.
(c)
Explain why \(\sim\) is a transitive relation.
Since \(\sim\) as defined in (19.1) is an equivalence relation, the relation partitions \(X\) into a union of disjoint equivalence classes. The equivalence class of an element \([a]\) is called a path component of \(X\text{,}\) and is the largest path connected subset of \(X\) that contains \(a\text{.}\)
Definition19.9.
The path component of an element \(a\) in a topological space \((X, \tau)\) is the largest path connected subset of \(X\) that contains \(a\text{.}\)
