K. Jänich, Topology. Chapter I: Fundamental Concepts. 6: Connectedness

Definition 1: Connected A topological space \(X\) is connected iff there do not exist disjoint nonempty open sets \(U\) and \(V\) such that \(X = U \cup V\).

Example 2 Any interval in \(\mathbb{R}\) is connected.

Proof Let \(I\) be an interval in \(\mathbb{R}\). Assume for a contradiction \(U\) and \(V\) are open in \(I\), nonempty, disjoint, and \(I = U \cup V\). Pick \(a \in U\) and \(b \in V\). Assume w.l.o.g. \(a < b\). Let \(s = \inf \{x \in V : a < x \}\). We have \(a \leq s \leq b\) and so \(s \in I\). We must have \(s \in U\) or \(s \in V\); we will derive a contradiction from both of these possibilities.

Assume first \(s \in U\). Pick \(\epsilon > 0\) such that \((s - \epsilon, s + \epsilon) \cap I \subseteq U\). We have \(s < b\); pick \(t\) such that \(s < t < \min(s + \epsilon,b)\). By the minimality of \(s\), we have \(t\) is not a lower bound for \(\{x \in V : a < x \}\), so pick \(x \in V\) such that \(a < x < t\). We have \( x < b \) and so \(x \in I\), and \(x < s + \epsilon\) so \(x \in U\). This contradicts the fact that \(U\) and \(V\) are disjoint.

Now assume \(s \in V\). Pick \(\epsilon > 0\) such that \((s - \epsilon, s + \epsilon) \cap I \subseteq V\). We have \(a < s\); pick \(t\) such that \(\max(s - \epsilon,a) < t < s\). Then we have \(t \in V\). But, since \(s\) is a lower bound for \(\{x \in V : a < x \}), we have \(t \leq a\). This is a contradiction. \(\Box\)

Definition 3: Path Connected A topological space \(X\) is path connected iff, for all \(a,b \in X\), there exists a continuous function \(\alpha : [0,1] \rightarrow X\) such that \(\alpha(0) = a\) and \(\alpha(1)=b\). We call \(\alpha\) a path from \(a\) to \(b\).

Proposition 4 Every path connected space is connected.

Proof Let \(X\) be a path connected space. Assume for a contradiction \(X = U \cup V\) where \(U\) and \(V\) are nonempty, disjoint, and open. Pick \(a \in U\) and \(b \in V\). Pick a path \(\alpha : [0,1] \rightarrow X\) from \(a\) to \(b\). Then \([0,1] = \alpha^{-1}(U) \cup \alpha^{-1}(V)\), contradicting the fact that \([0,1]\) is connected (Example 2).

Example 5 An example of a connected space that is not path connected is the topologist's sine curve: \(\{(x, \sin 1/x) : x > 0 \} \cup \{(0,y) : -1 \leq y \leq 1 \} \).

Proposition 6 The continuous image of a path connected space is path connected.

Proof Let \(X\) be a path connected space and \(f : X \rightarrow Y\) a surjective continuous function. Let \(a,b \in Y\). Pick \(c,d \in X\) such that \(f(c) = a\) and \(f(d) = b\). Pick a path \(\alpha : [0,1] \rightarrow X\) from \(c\) to \(d\). Then \(f \circ \alpha\) is a path from \(a\) to \(b\) in \(Y\). \(\Box\)

Proposition 7 The continuous image of a connected space is connected.

Proof Let \(X\) be a connected space and \(f : X \rightarrow Y\) a surjective continuous function. Assume for a contradiction \(Y = U \cup V\) where \(U\) and \(V\) are nonempty, disjoint and open. Then \(X = f^{-1}(U) \cup f^{-1}(V)\), contradicting the fact that \(X\) is connected. \(\Box\)

Proposition 8 Let \(X\) be a topological space. Assume \(X = A \cup B\) where \(A\) and \(B\) are not disjoint. If \(A\) and \(B\) are connected then \(X\) is connected.

Proof Pick \(p \in A \cap B\).

Assume for a contradiction \(X = U \cup V\) where \(U\) and \(V\) are nonempty, disjoint and open. Then \(A = (U \cap A) \cup (V \cap A) \) and so one of \(U \cap A\), \(V \cap A\) must be empty. Assume w.l.o.g. \(U \cap A = \emptyset\). Then \(p \in U\).

We also have \(B = (U \cap B) \cup (V \cap B)\) and \(p \in U \cap B\) so \(U \cap B\) is nonempty. Therefore \(V \cap B\) is empty.

This means \(V = \emptyset\) which is a contradiction. \(\Box\)

Proposition 9 Let \(X\) be a topological space. Assume \(X = A \cup B\) where \(A\) and \(B\) are not disjoint. If \(A\) and \(B\) are path connected then \(X\) is path connected.

Proof Pick \(p \in A \cap B\).

Let \(x,y \in X\). If \(x,y \in A\) or \(x,y \in B\) then there is a path from \(x\) to \(y\) by assumption.

If \(x \in A\) and \(y \in B\), pick paths \(\alpha : [0,1] \rightarrow A\) from \(x\) to \(p\) and \(\beta : [0,1] \rightarrow B\) from \(p\) to \(y\). Define \(\gamma : [0,1] \rightarrow X\) by \(\gamma(x) = \alpha(2x)\) if \(x \leq 1/2\), and \(\gamma(x) = \beta(2 x-1)\) if \(x \geq 1/2\). Then \(\gamma\) is a path from \(x\) to \(y\). \(\Box\)

Proposition 10 The product of two connected spaces is connected.

Proof Let \(X\) and \(Y\) be connected spaces. Assume w.l.o.g. \(X\) and \(Y\) are nonempty. Assume for a contradiction \(X = U \cup V\) where \(U\) and \(V\) are nonempty, disjoint, and open.

For all \(a \in X\), the function \(f_a : Y \rightarrow X \times Y\) that maps \(y\) to \((a,y)\) is continuous. We have \(Y = f_a^{-1}(U) \cup f_a^{-1}(V)\), and so one of \(f_a^{-1}(U)\), \(f_a^{-1}(V)\) is empty. Let \(U' = \{a \in X : f_a^{-1}(V) = \emptyset \}\), and \(V' = \{a \in X : f_a^{-1}(U) = \emptyset \}\).

We have \(X = U' \cup V'\) and \(U'\) and \(V'\) are disjoint.

We now prove \(U'\) is open. Let \(a \in U'\). Pick \(b \in Y\). We have \((a,b)\in U\). Pick neighbourhoods \(W\) of \(a\) and \(W'\) of \(b\) such that \(W \times W' \subseteq U\). Then \(a \in W \subseteq U'\). Thus \(U'\) is open. Similarly \(V'\) is open.

Since \(X\) is connected, it must therefore be that one of \(U'\) and \(V'\) is empty. Assume w.l.o.g. \(U' = \emptyset\). Then \(U = \emptyset\), which is a contradiction. \(\Box\)

Proposition 11 The product of two path connected spaces is path connected.

Proof Let \(X\) and \(Y\) be path connected. Let \((x_1,y_1),(x_2,y_2) \in X \times Y\). Pick paths \(\alpha : [0,1] \rightarrow X\) and \(\beta : [0,1] \rightarrow Y\) be paths from \(x_1\) to \(x_2\) and from \(y_1\) to \(y_2\) respectively. Then \(\langle \alpha, \beta \rangle : [0,1] \rightarrow X \times Y\) is a path from \((x_1,y_1)\) to \((x_2,y_2)\). \(\Box\)

Proposition 12 The disjoint union of two nonempty spaces is not connected.

Proof We have \(X + Y = (X + \emptyset) \cup (\emptyset + Y) \). \(\Box\)

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