Disconnected and connected sets
Definition (Disconnected and connected space)
A space \((X, d)\) is disconnected if there exist non-empty open sets \(A, B\) such that
\[X = A \cup B, \quad A \cap B = \emptyset.\]A space is said to be connected if it is not disconnected. Equivalently, if \(X = A \cup B\) with \(A, B\) disjoint open sets, then \(A = X\) or \(B = X\).
Definition (Disconnectedness of a subset)
\(E \subseteq X\) is disconnected if there exist sets \(A, B\) open in \((X, d)\) such that
\[E = (A \cap E) \cup (B \cap E), \quad (A \cap B) \cap E = \emptyset, \quad A \cap E \neq \emptyset \neq B \cap E.\]That is, \(E\) splits into two non-empty disjoint relatively open sets.
Exercise (Characterisation of the open sets in subspaces)
Given \(E \subseteq X\), define \(d_{E} : E \times E \to \mathbb{R}\) by \((x, y) \mapsto d(x, y)\). Then \((E, d_{E})\) is a metric space. Prove that \(D \subseteq E\) is open in \((E, d_{E})\) if and only if there exists \(O \subseteq X\) open in \((X, d)\) such that \(D = E \cap O\).
Connected sets in R and properties
Lemma (Open intervals are connected)
For \(a < b\), the interval \(I = (a, b)\) is connected.
Proof: Suppose that \(I = (I \cap A) \cup (I \cap B)\) with \(A, B\) disjoint open sets and both pieces non-empty. Let \(s \in I \cap A\), \(t \in I \cap B\) with \(s < t\); then \([s, t] \subseteq I\). Since \(s \in A\) there exists \(\delta_{1} > 0\) with \((s - \delta_{1}, s + \delta_{1}) \subseteq A\), so \([s, s + \delta_{1}/2] \subseteq [s, t] \cap A\).
Define \(u = \sup\{ x \in [s, t] : [s, x] \subseteq A\}\). By construction \(s < u \leq t\). If \(u \in B\), there exists \(\delta_{2} > 0\) with \((u - \delta_{2}, u + \delta_{2}) \subseteq B \cap [s,t]\); but by properties of the supremum there exists \(w \in [s, t] \cap A\) with \(u - \delta_{2} < w \leq u\), which would imply \(w \in A \cap B\) (since \(w \in [s, t] \cap A\) by the definition of the supremum and \(w \in (u - \delta_{2}, u + \delta_{2}) \subseteq B\)), contradicting \(A \cap B = \emptyset\). If \(u \in A\), there exists \(\delta_{3} > 0\) with \([u, u + \delta_{3}] \subseteq [s, t] \cap A\), which contradicts the definition of the supremum when \(u < t\), or contradicts \(t \in B\) when \(u = t\).
Lemma (Continuous functions preserve connectedness)
Let \(f : X \to Y\) be continuous. If \(E \subseteq X\) is connected, then \(f(E)\) is connected.
Proof: Suppose that there exist open sets \(B, C \subseteq Y\) such that \(f(E) = (f(E) \cap C) \cup (f(E) \cap B)\) with both pieces non-empty and disjoint. Then
\[\emptyset \neq f^{-1}(f(E) \cap C) = E \cap f^{-1}(C), \quad \emptyset \neq f^{-1}(f(E) \cap B) = E \cap f^{-1}(B),\]and \(E = (E \cap f^{-1}(C)) \cup (E \cap f^{-1}(B))\) with \(f^{-1}(C), f^{-1}(B)\) disjoint open sets in \(X\), contradicting that \(E\) is connected.
Corollary (\(\mathbb{R}^{d}\) is connected)
\(\mathbb{R}^{d}\) is connected for every \(d \geq 1\).
Proof: Suppose \(\mathbb{R}^{d} = A \cup B\) with \(A, B\) non-empty disjoint open sets. Take \(x \in A\), \(y \in B\) and \(f : [0, 1] \to \mathbb{R}^{d}\), \(t \mapsto (1-t)x + ty\), which is continuous. Since \([0, 1]\) is connected, \([x, y] = f([0,1])\) is connected. But \([x, y] = ([x,y] \cap A) \cup ([x,y] \cap B)\) with both pieces non-empty, a contradiction.
Lemma (Union of connected sets with a common point)
Let \(\{ E_{\alpha} : \alpha \in A\}\) be a family of connected sets such that \(\bigcap_{\alpha \in A} E_{\alpha} \neq \emptyset\). Then \(E = \bigcup_{\alpha \in A} E_{\alpha}\) is connected.
Proof: Let \(x \in \bigcap_{\alpha \in A} E_{\alpha}\) and suppose, for a contradiction, that there exist open sets \(A', B'\) with \(E = (A' \cap E) \cup (B' \cap E)\), \(A' \cap E \neq \emptyset\), \(B' \cap E \neq \emptyset\) and \(A' \cap B' \cap E = \emptyset\). Without loss of generality \(x \in B'\).
Since \(A' \cap E \neq \emptyset\), take \(y \in A' \cap E\); since \(E = \bigcup_{\alpha} E_{\alpha}\), there exists \(\alpha_{0}\) with \(y \in E_{\alpha_{0}}\), hence \(A' \cap E_{\alpha_{0}} \neq \emptyset\). On the other hand \(x \in E_{\alpha_{0}}\) (since \(x\) lies in every \(E_{\alpha}\)) and \(x \in B'\), so \(B' \cap E_{\alpha_{0}} \neq \emptyset\).
Let us see that this splits \(E_{\alpha_{0}}\) into two non-empty disjoint relatively open sets:
-
Cover: every \(z \in E_{\alpha_{0}}\) satisfies \(z \in E\), hence (by the global decomposition) \(z \in A'\) or \(z \in B'\). Therefore
\[E_{\alpha_{0}} \;\subseteq\; (E_{\alpha_{0}} \cap A') \cup (E_{\alpha_{0}} \cap B');\]the reverse inclusion is immediate. We conclude \(E_{\alpha_{0}} = (E_{\alpha_{0}} \cap A') \cup (E_{\alpha_{0}} \cap B')\).
- Disjointness: \((E_{\alpha_{0}} \cap A') \cap (E_{\alpha_{0}} \cap B') = E_{\alpha_{0}} \cap (A' \cap B') \subseteq E \cap (A' \cap B') = \emptyset\).
- Both non-empty: already proved above.
This contradicts that \(E_{\alpha_{0}}\) is connected.
Connected components
Definition (Connected component of a point)
Let \((X, d)\) be a metric space and \(x \in X\). The connected component of \(x\) is
\[C(x) = \bigcup\{ E \subseteq X : E \text{ connected}, \; x \in E\}.\]Note (The connected component is connected)
By the lemma on unions of connected sets with a common point, \(C(x)\) is connected for every \(x \in X\).
Exercise (The components are equivalence classes)
Define the relation \(\mathcal{R}\) on \(X\) by \(x \mathcal{R} y \iff \exists\, C \text{ connected}\,(x, y \in C)\). Prove that \(\mathcal{R}\) is an equivalence relation and that \([x] = C(x)\).
Characterisation of the connected sets in R
Lemma (A connected subset of \(\mathbb{R}\) contains the segments between its points)
Let \(E \subseteq \mathbb{R}\) be connected and \(a, b \in E\) with \(a \leq b\). Then \([a, b] \subseteq E\).
Proof: Suppose that there exists \(x \in [a, b] \setminus E\). Then
\[E = (E \cap (-\infty, x)) \cup (E \cap (x, \infty)),\]a decomposition of \(E\) into two non-empty disjoint relatively open sets, contradicting that \(E\) is connected.
Theorem (The connected subsets of \(\mathbb{R}\) are the intervals)
\(E \subseteq \mathbb{R}\) is connected if and only if \(E\) is an interval (bounded or not, open, closed or half-open).
Proof: \((\impliedby)\): Open intervals are connected by a previous lemma, and the increasing union \(\bigcup_{n} (a + 1/n, b - 1/n)\), \(\bigcup_{n} [a, b - 1/n]\), etc., with non-empty intersection yields all the remaining types by applying the union lemma.
\((\implies)\): If \(E\) is connected and bounded, define \(\alpha = \inf E\), \(\beta = \sup E\). Take \(\{a_{n}\}, \{b_{n}\} \subseteq E\) with \(a_{n} \downarrow \alpha\) and \(b_{n} \uparrow \beta\). Then, by the previous lemma, \([a_{n}, b_{n}] \subseteq E\), and therefore \((\alpha, \beta) = \bigcup_{n} [a_{n}, b_{n}] \subseteq E \subseteq [\alpha, \beta]\). Thus \(E\) is one of \([\alpha, \beta], (\alpha, \beta], [\alpha, \beta), (\alpha, \beta)\), according to whether \(\alpha, \beta \in E\). If \(E\) is not bounded above, define \(\beta = +\infty\) and take \(b_{n} \in E\) with \(b_{n} \to +\infty\); then \((\alpha, +\infty) = \bigcup_{n} [a_{n}, b_{n}] \subseteq E\), so \(E \in \{ (\alpha, +\infty),\; [\alpha, +\infty) \}\). Analogously for \(E\) not bounded below, and the case \(E = \mathbb{R}\) is obtained by combining the two.
Exercise (Structure of the open sets in \(\mathbb{R}\))
Let \(G \subseteq \mathbb{R}\) be open. Prove that there exist disjoint open intervals \(\{ (a_{i}, b_{i})\}_{i=1}^{\infty}\) such that \(G = \bigcup_{i=1}^{\infty} (a_{i}, b_{i})\).
Path-connectedness
Definition (Path-connected set in \(\mathbb{R}^{d}\))
\(E \subseteq \mathbb{R}^{d}\) is path-connected if for all \(x_{0}, x_{1} \in E\) there exists a curve \(\gamma : [0, 1] \to \mathbb{R}^{d}\) such that
- \(\gamma(0) = x_{0}\) and \(\gamma(1) = x_{1}\);
- \(\gamma(t) \in E\) for every \(t \in [0, 1]\).
Example (The topologist’s sine curve: connected but not path-connected)
Consider
\[E = \{(0, 0)\} \cup \left\{ (x, y) : 0 < x \leq 1,\; y = \sin\!\left(\tfrac{1}{x}\right) \right\}.\]\(E\) is connected. Let \(A, B\) be open with \(A \cap E \neq \emptyset \neq B \cap E\), \(E \subseteq A \cup B\) and \(A \cap B = \emptyset\), and suppose \((0, 0) \in A\). For \(n \in \mathbb{N}\) with \(n\) large, \(x_{n} = (\tfrac{1}{\pi n}, \sin(\pi n)) = (\tfrac{1}{\pi n}, 0) \in A\) (by convergence to \((0,0)\) and \(A\) open). For each \(n\),
\[E_{n} = \left\{ (x, y) : \tfrac{1}{n\pi} \leq x \leq 1,\; y = \sin\!\left(\tfrac{1}{x}\right)\right\}\]is the continuous image of the interval \([\tfrac{1}{\pi n}, 1]\) and therefore is connected. Since \(E_{n} \cap A \neq \emptyset\), \(E_{n} \cap B = \emptyset\), so \(E_{n} \subseteq A\), and taking unions \(\{(x, \sin(1/x)) : 0 < x \leq 1\} \subseteq A\). We conclude \(E \subseteq A\), contradicting \(B \cap E \neq \emptyset\).
\(E\) is not path-connected. Suppose that there exists a continuous \(\tilde{\gamma} : [0, 1] \to E\) with \(\tilde{\gamma}(0) = (0, 0)\) and \(\tilde{\gamma}(1) = (1/(n_{0}\pi), 0)\). Let \(t_{0} = \sup\{ t > 0 : \tilde{\gamma}(t) = (0, 0)\}\). Taking \(\{t_{n}\}\) decreasing with \(t_{n} \to t_{0}\), \(\tilde{\gamma}(t_{n}) \to (0, 0)\), so \(\tilde{\gamma}_{1}(t_{n}) \to 0\) and \(\tilde{\gamma}_{2}(t_{n}) = \sin(1/\tilde{\gamma}_{1}(t_{n}))\) does not converge (it oscillates between \(-1\) and \(1\) along suitable \(t_{n}\)), contradicting the continuity of \(\tilde{\gamma}_{2}\) at \(t_{0}\).
Exercise (Details of the path-connectedness counterexample)
In the previous example, prove:
- If \(x_{n} < x_{n+1}\) then, for \(x_{n} \leq x \leq x_{n+1}\), \((x, \sin(1/x)) \in \tilde{\gamma}([0,1])\).
- \(B = \{ (x, \sin(1/x)) : 0 < x < 1/(n_{0}\pi)\} \subseteq \tilde{\gamma}([0,1])\).
- \(\tilde{\gamma}([t_{0}, 1]) \setminus B = \{ (0, 0)\}\).
- \(\tilde{\gamma}\) is not continuous at \(t_{0}\).
Lemma (On open subsets of \(\mathbb{R}^{d}\), connected is equivalent to path-connected)
Let \(E \subseteq \mathbb{R}^{d}\) be open. Then \(E\) is connected if and only if \(E\) is path-connected.
Proof: \((\impliedby)\): By the general theorem, every path-connected set is connected.
\((\implies)\): Fix \(x_{0} \in E\) and consider \(A = \{ x \in E : x \text{ is joined to } x_{0} \text{ by a curve in } E\}\). Since \(E\) is open and curves may be concatenated with balls, \(A\) and \(E \setminus A\) are both open in \(E\) (around any \(y \in A\) a ball in \(E\) is joined to \(y\) by a straight segment and, by concatenation, to \(x_{0}\)). By connectedness and \(A \neq \emptyset\), \(A = E\). (Details: exercise.)