# 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.

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.

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)

### Questions

Let $$A$$ and $$B$$ be two non-empty subsets of the complex plane.
1. If $$A \cap B = \varnothing$$, is $$A \cup B$$ always disconnected ?

2. Assume that $$d(A, B) > 0$$; show that $$A \cup B$$ is not connected.

3. Assume that $$\overline{A} \cap B = \varnothing$$ and $$A \cap \overline{B} = \varnothing$$; show that $$A \cup B$$ is not connected.

1. 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.

2. 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.

3. 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

1. 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.

2. 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.

1. 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.

2. 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.

# References

1. Algebraic Topology
Allen Hatcher, 2002.
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2. Sur les Ensembles Connexes
B. Knaster and C. Kuratowski, 1921.
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