Connected Sets
By Sébastien Boisgérault, Mines ParisTech, under CC BY-NC-SA 4.0
November 28, 2017
Contents
Exercises
Image of Path-Connected/Connected Sets
Let \(f: A \subset \mathbb{C} \to \mathbb{C}\) be a continuous function.
Question
Show that if \(A\) is path-connected/connected, its image \(f(A)\) is path-connected/connected.Answer
Suppose that \(A\) is path-connected. Let \(a, b \in f(A)\); there are some \(c, d \in A\) such that \(f(c) = a\) and \(f(d) = b\). As \(A\) is path-connected, there is a path \(\gamma\) that joins \(c\) and \(d\) in \(A\). By continuity of \(f\), it is plain that its image \(f \circ \gamma\) is a path of \(f(A)\) that joins \(a\) and \(b\). Consequently, \(f(A)\) is path-connected.Now suppose that \(A\) is connected. Let \(g\) be a locally constant function defined on \(f(A)\). The function \(g \circ f\) is locally constant on \(A\): if \(a \in A\), there is a radius \(r>0\) such that \(g\) is constant on \(D(f(a), r) \cap f(A)\); by continuity of \(f\), there is a \(\epsilon>0\) such that if \(b \in D(a, \epsilon) \cap A\), \(f(b) \in D(f(a),\epsilon) \cap f(A)\), thus \(g \circ f\) is constant on \(D(a, \epsilon) \cap A\) and finally, \(g \circ f\) is locally constant. Since \(A\) is connected, \(g \circ f\) is actually constant and \(g\) itself is constant: \(f(A)\) is connected.
Complement of a Compact Set
Question
Prove that the complement \(\mathbb{C} \setminus K\) of a compact subset \(K\) of the complex plane has a single unbounded component.Answer
The compact set \(K\) is closed, hence its complement is open. Therefore, the connected and path-connected components of \(\mathbb{C} \setminus K\) are the same. The compact set \(K\) is also bounded, hence there is a \(r>0\) such that the annulus \[ A = \{z \in \mathbb{C} \; | \; |z|> r\} \] is included in \(\mathbb{C} \setminus K\). The annulus \(A\) is path-connected: if \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\) are in \(A\), the path \(\gamma= [r_1 \to r_2] e^{i[\theta_1 \to \theta_2]}\), which is defined by \[ \gamma(t) = ((1-t) r_1 + t r_2) e^{i((1-t)\theta_1 + t\theta_2)} \] belongs to \(A\) and joins \(z_1\) and \(z_2\). Hence, \(A\) is included in some path-connected component of \(\mathbb{C} \setminus K\). The collection of these path-connected components are a partition of \(\mathbb{C} \setminus K\), hence every other component \(C\) is a subset of \(\mathbb{C} \setminus A = \overline{D}(0,r)\): it is bounded.Union of Separated Sets
Source: “Sur les ensembles connexes” (Knaster and Kuratowski 1921)
If \(A \cap B = \varnothing\), is \(A \cup B\) always disconnected ?
Assume that \(d(A, B) > 0\); show that \(A \cup B\) is not connected.
Assume that \(\overline{A} \cap B = \varnothing\) and \(A \cap \overline{B} = \varnothing\); show that \(A \cup B\) is not connected.
Answers
No. For example, the sets \(A = \{z \in \mathbb{C} \; | \; \Re (z) < 0\}\) and \(B = \mathbb{C} \setminus A\) are disjoints, but their union is \(\mathbb{C}\), which is connected.
Let \(r = d(A, B) / 2\). Under the assumption, the sets \[ A' = \cup_{a \in A} D(a, r), \; B' = \cup_{b \in B} D(b, r), \] which are both open sets, are disjoints, hence their union is not path-connected. However \(A' \cup B'\) is a dilation of \(A \cup B\), hence \(A \cup B\) is not connected.
Alternatively, consider the function \(f\) equal to \(1\) on \(A\) and \(0\) on \(B\). It is locally constant – if \(z \in A \cup B\), f is constant on \((A\cup B)\cap D(z, r)\) with \(r=d(A, B)\) for example – but not constant, hence \(A \cup B\) is not connected.
Consider again the function \(f\) introduced in the previous answer. The assumption yields \(A \cap B = \varnothing\); as \(A\) and \(B\) are non-empty, \(f\) is not constant. If this function was not locally constant around some \(a \in A\), we could find a sequence of \(b_n \in (A \cup B) \setminus A = B\) such that \(b_n \to a\). But that would imply that \(a \in A \cap \overline{B}\) and would lead to a contradiction. Similarly, if it was not constant around some \(b \in B\), that would lead to \(b \in \overline{A} \cap B\), another contradiction. Hence, it is locally constant and \(A \cup B\) is not connected.
Anchor Set
Questions
Prove that if \(\mathcal{A}\) is a collection of path-connected/connected sets and there is a set \(A^* \in \mathcal{A}\) such that \(\forall \, A \in \mathcal{A}\), \(A \cap A^*\neq \varnothing\), then the union \(\cup \mathcal{A}\) is path-connected/connected.
A deformation retraction of a subset \(A\) of the complex plane onto a subset \(B\) of \(A\) is a “continuous shrinking process” of \(A\) into \(B\); formally, it is a collection of paths \(\gamma_a\) of \(A\), indexed by \(a \in A\), such that:
\(\forall \, a \in A, \; \gamma_a(0) = a \, \mbox{ and } \, \gamma_a(1) \in B\),
\(\forall \, a \in B, \; \forall \, t \in [0,1], \; \gamma_a(t) = a\),
the function \((t, a) \in [0,1] \times A \mapsto \gamma_a(t)\) is continuous.
(see e.g. (Hatcher 2002)). Show that if there is a deformation retraction of \(A\) onto \(B\) and \(B\) is path-connected/connected, then \(A\) is also path-connected/connected.
Answers
Let \(\mathcal{A}'\) be the collection of all the sets \(A^* \cup A\) for \(A \in \mathcal{A}\). For any \(A \in \mathcal{A}\), the collection \(\{A, A^{*}\}\) is composed of two path-connected/connected sets with a non-empty intersection; hence all the sets of \(\mathcal{A}'\) in path-connected/connected. Moreover, the unions \(\cup \mathcal{A}\) and \(\cup \mathcal{A}'\) are identical. By assumption, unless \(\mathcal{A}\) is empty, \(A^{*}\) is not empty; hence the intersection \(\cap \mathcal{A}'\) that contains \(A^*\) is not empty. Therefore, \(\cup \mathcal{A} = \cup \mathcal {A}'\) is path-connected/connected.
For any \(a \in A\), \(\gamma_a(0) = a\) and \(\gamma_a([0,1]) \subset A\), hence \[ A = \bigcup_{a \in A} \gamma_a([0,1]). \] For any \(a \in A\), the set \(\gamma_a([0,1])\) is path-connected (as the image of a path-connected set by a continuous function), and \(\gamma_a([0,1]) \cap B\) is non-empty (it contains \(\gamma_a(1)\)). Consequently, the collection \[ \mathcal{A} = \{B\} \cup \{\gamma_a([0,1]) \, | \, a \in A\} \] satisfies the assumption of the previous question with \(A^* = B\). Consequently, \(A = \cup \mathcal{A}\) is path-connected/connected.