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Topology: An Inquiry-Based Approach

Steven Schlicker

Section Connected Sets

As we learned in our preview activity, connected sets are those sets that cannot be separated into a union of disjoint open sets. Another characterization of connectedness is established in the next activity.

Activity 18.2.

Let \((X, \tau)\) be a topological space.

(a)

Assume that \(X\) is a connected space, and let \(A\) be a subset of \(X\) that is both open and closed. What happens if we combine \(A\) and \(X \setminus A\text{?}\) What does the fact that \(X\) is connected tell us about \(A\text{?}\)

(b)

Now assume that the only subsets of \(X\) that are both open and closed are \(\emptyset\) and \(X\text{.}\) Must it follow that \(X\) is connected? Prove your assertion.

(c)

Summarize the result of this activity into a theorem of the form “A topological space \((X, \tau)\) is connected if and only if ...”.
A standard example of a connected topological space is the metric space \((\R, d_E)\text{.}\)

Proof.

We proceed by contradiction and assume that there are nonempty open sets \(U\) and \(V\) such that \(\R = U \cup V\) and \(U \cap V = \emptyset\text{.}\) Let \(a \in U\) and \(b \in V\text{.}\) Since \(U \cap V = \emptyset\text{,}\) we know that \(a \neq b\text{.}\) Without loss of generality we can assume \(a \lt b\text{.}\) Let \(U' = U \cap [a,b]\) and let \(V' = V \cap [a,b]\text{.}\) The set \(V'\) is bounded below by \(a\text{,}\) so \(x = \inf\{v \mid v \in V'\}\) exists. Since \(\R = U \cup V\) it must be the case that \(x \in U\) or \(x \in V\text{.}\)
Suppose \(x \in U\text{.}\) The fact that \(U\) is an open set implies that there exists \(\epsilon \gt 0\) such that \(B(x, \epsilon) \subseteq U\text{.}\) But then \(B(x, \epsilon) \cap V = \emptyset\) and so \(d(x,v) \geq \epsilon\) for every \(v \in V\text{.}\) This means that \(x+\epsilon \lt v\) for every \(v \in V'\text{,}\) contradicting the fact that \(x\) is the greatest lower bound. We conclude that \(x \notin U\text{.}\)
It follows that \(x \in V\text{.}\) Since \(a \in U\text{,}\) we know that \(x \neq a\text{.}\) The fact that \(V\) is an open set tells us that there exists \(\delta \gt 0\) such that \(B(x, \delta) \subseteq V\text{.}\) We can choose \(\delta\) to ensure that \(\delta \lt x-a\text{.}\) Since \(x > a\text{,}\) the interval \((x-\delta, x)\) is a subset of \(V'\text{,}\) and so \(x\) is not a lower bound for \(V\text{.}\)
Each possibility leads to a contradiction, so we conclude that the sets \(U\) and \(V\) cannot exist. Therefore, \((\R, d_E)\) is a connected topological space.
As you might expect, connectedness is a topological property.

Activity 18.3.

Let \((X, \tau_X)\) and \((Y, \tau_Y)\) be topological spaces, and let \(f : X \to Y\) be a continuous function. Assume that \(X\) is a connected subset of \(X\text{.}\) Our goal is to prove that \(f(X)\) is a connected subspace of \(Y\text{.}\)
Let \(Z = f(X)\) and define \(g: X \to Z\) by \(g(x) = f(x)\text{.}\) Then \(g\) is a continuous function that maps \(X\) onto \(Z\text{.}\) So we consider \(g\) instead of \(f\text{.}\)

(a)

Assume to the contrary that \(Z\) is not connected. What do we then assume about \(Z\text{?}\)

(b)

Suppose that \(U\) and \(V\) are disjoint nonempty open sets in \(Z\) such that \(U \cup V = Z\text{.}\) Let \(R = g^{-1}(U)\) and \(S = g^{-1}(V)\text{.}\)
(i)
Explain why \(R\) and \(S\) are open sets in \(X\text{.}\)
(ii)
Show that \(R \cup S = X\text{.}\) (Hint: \(X = g^{-1}(Z)\text{.}\))
(iii)
Show \(R\) and \(S\) are nonempty sets.
Hint.
Use the fact that \(g\) is a surjection.
(iv)
Now show that \(R \cap S = \emptyset\text{.}\) (Hint: \(R \cap S = g^{-1}(U) \cap g^{-1}(V)\text{.}\))

(c)

Explain how we have proved the following.
The fact that connectedness is preserved by continuous functions means that connectedness is a property that is shared by any homeomorphic topological spaces, as the next corollary indicates.

Proof.

Let \((X, \tau_X)\) and \((Y, \tau_Y)\) be topological spaces and let \(f: X \to Y\) be a homeomorphism. Assume that \(X\) is connected. Since \(f\) is continuous, Theorem 18.3 shows that \(f(X) = Y\) is connected. The reverse implication follows from the fact that \(f^{-1}\) is a homeomorphism.
Recall that \((\R,d_E)\) is homeomorphic to the topological subspaces \((a,b)\text{,}\) \((-\infty, b)\text{,}\) and \((a,\infty)\) for any \(a, b \in \R\text{.}\) The fact that \((\R, d_E)\) is connected (Theorem 18.2) allows us to conclude that all open intervals are connected. It would seem natural that all closed (or half-closed) intervals should also be connected. We address this question next. Before we get to this result, we consider an alternate formulation of connected subsets.
Consider the set \(A = (-1,0) \cup (4,5)\) in \(\R\text{.}\) Let \(U = (-2,3)\) and \(V=(2,6)\) in \(\R\text{.}\) Note that \(U' = U \cap A = (-1,0)\) and \(V' = V \cap A = (4,5)\text{,}\) and so \(U\) and \(V\) are open sets in \(\R\) that separate the set \(A\) into two disjoint pieces. We know that \(U'\) and \(V'\) are open in \(A\) and \(A = U' \cup V'\) with \(U' \cap V' = \emptyset\text{.}\) So to show that a subset of a topological space \(X\) is not connected, this example suggests that it suffices to find nonempty open sets \(U\) and \(V\) in \(X\) with \(U \cap V \cap A = \emptyset\) and \(A \subseteq (U \cup V)\text{.}\) Note that it is not necessary to have \(U \cap V = \emptyset\text{.}\) That this works in general is the result of the next theorem.

Proof.

Let \(X\) be a topological space, and let \(A\) be a subset of \(X\text{.}\) We first assume that \(A\) is disconnected and show that there are open sets \(U\) and \(V\) in \(X\) that satisfy the given conditions. Since \(A\) is disconnected, there are nonempty open sets \(U'\) and \(V'\) in \(A\) such that \(U' \cup V' = A\) and \(U' \cap V' = \emptyset\text{.}\) Since \(U'\) and \(V'\) are open in \(A\text{,}\) there exist open sets \(U\) and \(V\) in \(X\) so that \(U' = U \cap A\) and \(V' = V \cap A\text{.}\) Now
\begin{equation*} A = U' \cup V' = (U \cap A) \cup (V \cap A) = (U \cup V) \cap A\text{,} \end{equation*}
and so \(A \subseteq U \cup V\text{.}\) By construction, \(U \cap A = U'\) and \(V \cap A = V'\) are not empty. Finally,
\begin{equation*} U \cap V \cap A = (U \cap A) \cap (V \cap A) = U' \cap V' = \emptyset\text{.} \end{equation*}
So we have found sets \(U\) and \(V\) that satisfy the conditions of our theorem.
The proof of the reverse implication is left to the next activity.

Activity 18.4.

Let \(X\) be a topological space, and let \(A\) be a subset of \(X\text{.}\) Assume that there exist open sets \(U\) and \(V\) in \(X\) with \(A \subseteq U \cup V\text{,}\) \(U \cap A \neq \emptyset\text{,}\) \(V \cap A \neq \emptyset\text{,}\) and \(U \cap V \cap A = \emptyset\text{.}\) Prove that \(A\) is disconnected.
The conditions in Theorem 18.5 provide a convenient way to show that a set is disconnected, and so any pair of sets \(U\) and \(V\) that satisfy the conditions of Theorem 18.5 is given a special name.

Definition 18.6.

Let \(X\) be a topological space, and let \(A\) be a subset of \(X\text{.}\) A separation of \(A\) is a pair of nonempty open subsets \(U\) and \(V\) of \(X\) such that
  • \(A \subseteq (U \cup V)\text{,}\)
  • \(U \cap A \neq \emptyset\text{,}\)
  • \(V \cap A \neq \emptyset\text{,}\) and
  • \(U \cap V \cap A = \emptyset\text{.}\)
If \(X\) is a disconnected topological space, then a separation of \(X\) is a pair \(U\text{,}\) \(V\) of disjoint nonempty open sets such that \(U \cup V = X\text{.}\)
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