Section Path Connectedness and Connectedness in Finite Topological Spaces
In this section we will demonstrate that connectedness and path connectedness are equivalent concepts in finite topological spaces. In the following section, we prove that path connectedness and connectedness are not equivalent in infinite topological spaces. Throughout this section, we assume that \(X\) is a finite topological space. We begin with an example to motivate the main ideas.
Activity 19.5.
Let \(X = \{a,b,c,d\}\) and \(\tau = \{\emptyset, \{b\}, \{c\}, \{a,b\}, \{b,c\}, \{a,b,c\}, \{b,c,d\}, X\}\text{.}\) Assume that \(\tau\) is a topology on \(X\text{.}\)
(a)
Is \(X\) connected? Explain.
(b)
For each \(x \in X\text{,}\) let \(U_x\) be the intersection of all open sets that contain \(x\) (we call \(U_x\) a minimal neighborhood of \(x\)).
Definition 19.11.
For \(x \in X\text{,}\) the minimal neighborhood \(U_x\) of \(x\) is the intersection of all open sets that contain \(x\text{.}\)
Find \(U_x\) for each \(x \in X\text{.}\)
(c)
We will see that the minimal neighborhoods of \(X\) are path connected. Here we will illustrate with \(U_d\text{.}\)
(i)
Let \(p : [0,1] \to X\) be defined by
\begin{equation*}
p(t) = \begin{cases}b \amp \text{ if } 0 \leq t \lt \frac{1}{2} \\ d \amp \text{ if } \frac{1}{2} \leq t \leq 1. \end{cases}
\end{equation*}
Show that \(p\) is a path in \(U_d\) from \(b\) to \(d\text{.}\)
(ii)
Let \(p : [0,1] \to X\) be defined by
\begin{equation*}
p(t) = \begin{cases}c \amp \text{ if } 0 \leq t \lt \frac{1}{2} \\ d \amp \text{ if } \frac{1}{2} \leq t \leq 1. \end{cases}
\end{equation*}
Show that \(p\) is a path in \(U_d\) from \(c\) to \(d\text{.}\)
(iii)
Explain why \(U_d\) is path connected.
The terminology in
Definition 19.11 is apt. Since every neighborhood
\(N\) of a point
\(x \in X\) must contain an open set
\(O\) with
\(x \in O\text{,}\) it follows that
\(U_x \subseteq O \subseteq N\text{.}\) So every neighborhood of
\(x \in X\) has
\(U_x\) as a subset. In addition, when
\(X\) is finite, the set
\(U_x\) is a finite intersection of open sets, so the sets
\(U_x\) are open sets (this is not true in general in infinite topological spaces — you are asked to find an example in
Exercise 1). In
Activity 19.5 we saw that
\(U_x\) was path connected for a particular
\(x\) in one example. The next activity shows that this result is true in general in finite topological spaces.
Activity 19.6.
Let \(X\) be a finite topological space, and let \(x \in X\text{.}\) In this activity we demonstrate that \(U_x\) is path connected. Let \(y \in U_x\) and define \(p : [0,1] \to X\) by
\begin{equation*}
p(t) = \begin{cases}y \amp \text{ if } 0 \leq t \lt \frac{1}{2} \\ x \amp \text{ if } \frac{1}{2} \leq t \leq 1. \end{cases}
\end{equation*}
To prove that \(p\) is continuous, let \(O\) be an open set in \(X\text{.}\) We either have \(x \in O\) or \(x \notin O\text{.}\)
(a)
Suppose \(x \in O\text{.}\) Why must \(y\) also be in \(O\text{?}\) What, then, is \(p^{-1}(O)\text{?}\)
(b)
Now suppose \(x \notin O\text{.}\) There are two cases to consider.
(i)
What is \(p^{-1}(O)\) if \(y \in O\text{?}\)
(ii)
What is \(p^{-1}(O)\) if \(y \notin O\text{?}\)
(c)
Explain why \(p\) is a path from \(y\) to \(x\text{.}\)
(d)
Show that we can find a path between any two points in \(U_x\text{.}\) Conclude that \(U_x\) is path connected.
The sets \(U_x\) collectively form the space \(X\text{,}\) and each of the \(U_x\) is a path connected subspace. So every point in \(X\) is contained in some neighborhood with a path connected subset containing \(x\text{.}\) Spaces with this property are called locally path connected.
Definition 19.12.
A topological space \((X, \tau)\) is locally path connected at \(x\) if every neighborhood of \(x\) contains a path connected open neighborhood with \(x\) as an element. The space \((X, \tau)\) is locally path connected if \(X\) is locally path connected at every point.
If
\(X\) is a finite topological space, for any
\(x \in X\) the set
\(U_x\) is the smallest open set containing
\(x\text{.}\) This means that any neighborhood of
\(N\) of
\(x\) will contain
\(U_x\) as a subset. Thus, a finite topological space is locally path connected (this is not true in general of infinite topological spaces, see
Exercise 4 for example). One consequence of a locally path connected space is the following.
Lemma 19.13.
A space \(X\) is locally path connected if and only if for every open set \(O\) of \(X\text{,}\) each path component of \(O\) is open in \(X\text{.}\)
Proof.
Let \(X\) be a locally path connected topological space. We first show that for every open set \(O\) in \(X\text{,}\) every path component of \(O\) is open in \(X\text{.}\) Let \(O\) be an open set in \(X\) and let \(P\) be a path component of \(O\text{.}\) Let \(p \in P\text{.}\) Since \(X\) is locally path connected, the neighborhood \(O\) of \(x\) contains an open path connected neighborhood \(Q\) of \(p\text{.}\) The fact that \(p \in Q\) and \(P\) is a path component of \(O\) implies that \(Q \subseteq P\text{.}\) Thus, \(P\) contains a neighborhood of \(p\) and \(P\) is open.
Now we show that if for every open set \(O\) in \(X\) the path components of \(O\) are open in \(X\text{,}\) then \(X\) is locally path connected. Let \(x \in X\) and let \(N\) be a neighborhood of \(x\text{.}\) Then \(N\) contains an open set \(U\) with \(x \in U\text{.}\) Let \(P\) be the path component in \(U\) that contains \(x\text{.}\) Now \(P\) is path connected and, by hypothesis, \(P\) is open in \(X\) and so is an open path connected neighborhood of \(x\text{.}\) Thus, \(N\) contains a path connected neighborhood of \(x\) and \(X\) is locally path connected at every point.
Since
\(X\) is open in
\(X\) whenever
\(X\) is a topological space, a natural corollary of
Lemma 19.13 is the following.
Corollary 19.14.
Let \(X\) be a locally path connected topological space. Then every path component of \(X\) is open in \(X\text{.}\)
Since there are only finitely many open sets in the finite space \(X\text{,}\) any arbitrary intersection of open sets in \(X\) just reduces to a finite intersection. So the intersection of any collection of open sets in \(X\) is again an open set in \(X\text{.}\) We will show that \(X\) is a union of path connected components, which will ultimately allow us to prove that if \(X\) is connected, then \(X\) is also path connected.
Activity 19.7.
Let \(X\) be a locally path connected topological space. In this activity we will prove that the components and path components of \(X\) are the same.
(a)
Let \(x \in X\text{,}\) and let \(C\) be the component of \(X\) containing \(x\) and \(P\) be the path component of \(X\) containing \(x\text{.}\) Show that \(P \subseteq C\text{.}\)
(b)
To complete the proof that \(P = C\text{,}\) proceed by contradiction and assume that \(C \neq P\text{.}\) Let \(Q\) be the union of all path components of \(X\) that are different from \(P\) and that intersect \(C\text{.}\) Each such path component is connected, and is therefore a subset of \(C\text{.}\) So \(C = P \cup Q\text{.}\) Explain why \(P\) and \(Q\) form a separation of \(C\text{.}\)
Hint.How do we use the fact that \(X\) is locally path connected?
We can now complete our main result of this section.
Theorem 19.15.
Let \(X\) be a finite topological space. Then \(X\) is connected if and only if \(X\) is path connected.
Proof.
Let
\(X\) be a finite topological space.
Theorem 19.10 demonstrates that if
\(X\) is path connected, then
\(X\) is connected. For the reverse implication, assume that
\(X\) is path connected. Then
\(X\) is composed of a single path component,
\(P=X\text{.}\) Since the path components and components of
\(X\) are the same, we conclude that
\(P= X\) is a component of
\(X\) and that
\(X\) is connected.
