Section Connected Subsets of \(\R\)
With
Theorem 18.5 in hand, we are just about ready to show that any interval in
\(\R\) is connected. Let us return for a moment to our example of
\(A = (-1,0) \cup (4,5)\) in
\(\R\text{.}\) It is not difficult to see that if
\(U\) and
\(V\) are a separation of
\(A\text{,}\) then the subset
\((-1,0)\) must be entirely contained in either
\(U\) or in
\(V\text{.}\) The reason for this is that
\((-1,0)\) is a connected subset of
\(A\text{.}\) This result is true in general.
Activity 18.5.
Let \(X\) be a topological space, and let \(A\) be a subset of \(X\text{.}\) Assume that \(U\) and \(V\) form a separation of \(A\text{.}\) Let \(C\) be a connected subset of \(A\text{.}\) In this activity we want to prove that \(C \subseteq U\) or \(C \subseteq V\text{.}\)
(a)
Use the fact that \(U\) and \(V\) form a separation to \(A\) to wxplain why \(C \subseteq U \cup V\) and \(C \cap U \cap V = \emptyset\text{.}\)
(b)
Given that \(C\) is connected, what conclusion can we draw about the sets \(U' = U \cap C\) and \(V' = V \cap C\text{?}\)
(c)
Complete the proof of the following lemma.
Lemma 18.7.
Let \(X\) be a topological space, and let \(A\) be a subset of \(X\text{.}\) Assume that \(U\) and \(V\) form a separation of \(A\text{.}\) If \(C\) is a connected subset of \(A\text{,}\) then \(C \subseteq U\) or \(C \subseteq V\text{.}\)
Now we can prove that any interval in \(\R\) is connected. Since \([a,b]\text{,}\) \([a,b)\text{,}\) and \((a,b]\) are all sets that lie between \((a,b)\) and \(\overline{(a,b)}\text{,}\) we can address their connectedness all at once with the next result.
Theorem 18.8.
Let \(X\) be a topological space and \(C\) a connected subset of \(X\text{.}\) If \(A\) is a subset of \(X\) and \(C \subseteq A \subseteq \overline{C}\text{,}\) then \(A\) is connected in \(X\text{.}\)
Proof.
Let
\(X\) be a topological space and
\(C\) a connected subset of
\(X\text{.}\) Let
\(A\) be a subset of
\(X\) such that
\(C \subseteq A \subseteq \overline{C}\text{.}\) To show that
\(A\) is connected, assume to the contrary that
\(A\) is disconnected. Then there are nonempty open subsets
\(U\) and
\(V\) of
\(X\) that form a separation of
\(A\text{.}\) Lemma 18.7 shows that
\(C \subseteq U\) or
\(C \subseteq V\text{.}\) Without loss of generality we assume that
\(C \subseteq U\text{.}\) Since
\(U \cap V \cap A = \emptyset\text{,}\) it follows that
\begin{equation*}
C \cap V = (C \cap A) \cap V = C \cap (A \cap V) \subseteq U \cap A \cap V = \emptyset\text{.}
\end{equation*}
Since \(A \cap V \neq \emptyset\text{,}\) there is an element \(x \in A \cap V\text{.}\) Since \(x \notin C\) and \(x \in A \subseteq \overline{C}\text{,}\) it must be the case that \(x\) is a limit point of \(C\text{.}\) Since \(V\) is an open neighborhood of \(x\text{,}\) it follows that \(V \cap C \neq \emptyset\text{.}\) This contradiction allows us to conclude that \(A\) is connected.
One consequence of
Theorem 18.8 is that any interval of the form
\([a,b)\text{,}\) \((a,b]\text{,}\) \([a,b]\text{,}\) \((-\infty,
b]\text{,}\) or
\([a, \infty)\) in
\(\R\) is connected. This prompts the question, are there any other subsets of
\(\R\) that are connected?
Activity 18.6.
Let \(A\) be a subset of \(\R\text{.}\)
(a)
Let \(A = \{a\}\) be a single point subset of \(\R\text{.}\) Is \(A\) connected? Explain.
(b)
Now suppose that \(A\) is a subset of \(\R\) that contains two or more points. Assume that \(A\) is not an interval. Then there must exist points \(a\) and \(b\) in \(A\) and a point \(c\) in \(\R \setminus A\) between \(a\) and \(b\text{.}\) Use this idea to find a separation of \(A\text{.}\) What can we conclude about \(A\text{?}\)
(c)
Explicitly describe the connected subsets of \((\R, d_E)\text{.}\)
