A Topological preliminaries
A.1 Metrics and related concepts
A metric on a set \(X\) is a function \(\mathsf{d}\colon X \times X \to [0,\infty )\) such that for all \(p_1, p_2, p_3 \in X\) we have
The set \(X\) equipped with the metric \(\mathsf{d}\) on it is called a metric space.
The formula
defines a metric on the complex plane \(\mathbb {C}\).
(Thus \(\mathbb {C}\) becomes a metric space. Also any subset of \(\mathbb {C}\), in particular \(\mathbb {R}\subset \mathbb {C}\), becomes a metric space when equipped with the metric given by the above formula restricted to the subset.)
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Let \(X\) be a metric space with metric \(\mathsf{d}\colon X \times X \to [0,\infty )\). Let \(p_0 \in X\) be a point and let \(r{\gt}0\).
The set
is called an open ball in \(X\), centered at \(p_0\), and with radius \(r\).
The set
is called a closed ball in \(X\), centered at \(p_0\), and with radius \(r\).
(In the case of the complex plane \(\mathbb {C}\), the term disk is often used instead of the general metric space theory term ball.)
Let \(X\) be a metric space, and \(A \subset X\) a subset. A point \(p \in A\) is said to be an interior point of \(A\) if for some \(r {\gt} 0\) we have \(\mathcal{B}(p; r) \subset A\).
Let \(X\) be a metric space, and \(A \subset X\) a subset. A point \(p \in X \setminus A\) is said to be an exterior point of \(A\) if for some \(r {\gt} 0\) we have \(\mathcal{B}(p; r) \subset X \setminus A\).
(It is easy to see that the exterior points of \(A\) are exactly the interior points)
Let \(X\) be a metric space, and \(A \subset X\) a subset. A point \(p \in X\) is said to be a boundary point of \(A\) if for all \(r {\gt} 0\) we have that \(\mathcal{B}(p; r)\) contains points of \(A\) and \(X \setminus A\) (i.e. \(\mathcal{B}(p; r) \cap A \ne \emptyset \) and \(\mathcal{B}(p; r) \setminus A \ne \emptyset \)).
The set of all boundary points of \(A\) is denoted \(\partial A\) and called the boundary of \(A\).
(It is easy to see that the boundary \(\partial A \subset X\) is exactly the set of points of \(X\) which are neither interior nor exterior points of \(A\).)
Let \(X\) be a metric space. A subset \(U \subset X\) is said to be an open set if each point \(p \in U\) is an interior point of \(U\).
Let \(X\) be a metric space. A subset \(F \subset X\) is said to be a closed set if the complement \(X \setminus F \subset X\) is an open set.
(Equivalently, each point \(p \in X \setminus F\) in the complement of \(F\) is an exterior point of \(F\).)
Let \(X\) be a metric space. with metric \(\mathsf{d}\colon X \times X \to [0,\infty )\).
A subset \(A \subset X\) is bounded if there exists a number \(M{\gt}0\) such that \(\mathsf{d}(p,q) \le M\) for all \(p,q \in A\). (If \(X\) is nonempty, an equivalent definition would be that \(A\) is bounded if it is a subset of some ball in \(X\).)
A function \(f \colon Z \to X\) with values in a metric space \(X\) is bounded if the set \(f[Z] \subset X\) of its values is a bounded subset of \(X\).
(In the case \(X = \mathbb {C}\) we have the following further characterizations: A subset \(A \subset \mathbb {C}\) is bounded if and only if there exists an \(R{\gt}0\) such that \(|z| \le R\) for all \(z \in A\). A function \(f \colon Z \to \mathbb {C}\) is bounded if and only if there exists an \(R{\gt}0\) such that \(|f(z)| \le R\) for all \(z \in Z\).)
A.2 Limits
Let \(X\) be a metric space and let \((x_n)_{n \in \mathbb {N}}\) be a sequence of points in \(X\). We say that the sequence \((x_n)_{n \in \mathbb {N}}\) converges to a limit \(x \in X\) if for any \(\varepsilon {\gt} 0\) there exists an \(N \in \mathbb {N}\) such that for all \(n \ge N\) we have \(x_n \in \mathcal{B}(x; \varepsilon )\) (i.e., \(\mathsf{d}(x_n,x) {\lt} \varepsilon \)). We then denote
(It is straightforward to check that the limit is unique if it exists.)
Let then \(X\) and \(Y\) be metric spaces, with respective metrics \(\mathsf{d}_X\) and \(\mathsf{d}_Y\), and let \(f \colon X \to Y\) be a function. We say that the function \(f\) has a limit \(y \in Y\) at a point \(p_0 \in X\) if for any \(\varepsilon {\gt} 0\) there exists a \(\delta {\gt} 0\) such that for all \(p \in \mathcal{B}(p_0; \delta ) \setminus \left\{ p_0 \right\} \) we have \(f(p) \in \mathcal{B}(y; \varepsilon )\). We then denote
(It is straightforward to check that the limit is unique if it exists.)
(Equivalently, written in terms of distances, \(\lim _{p \to p_0} f(p) = y\) means that for any \(\varepsilon {\gt} 0\) there exists a \(\delta {\gt} 0\) such that we have \(\mathsf{d}_Y \big( f(p), y \big) {\lt} \varepsilon \) whenever \(0 {\lt} \mathsf{d}_X (p,p_0) {\lt} \delta \).)
For a sequence \((z_n)_{n \in \mathbb {N}}\) of complex numbers we have
if and only if
Let \(X\) be a metric space, let \(f \colon X \to \mathbb {C}\) a complex-valued function on \(X\), and let \(p_0 \in X\) be a point. Then we have
if and only if
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Let \((z_n)_{n \in \mathbb {N}}\) and \((w_n)_{n \in \mathbb {N}}\) be complex number sequences converging to limits
Then we have
Let \(X\) be a metric space, let \(p_0 \in X\) be a point, and let \(f,g \colon X \to \mathbb {C}\) be two complex-valued functions on \(X\) such that
Then we have
The arguments are similar to the proofs given in MS-C1541 Metric Spaces for the real-valued cases.
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If a real number sequence \((x_n)_{n \in \mathbb {N}}\) is Cauchy, then it converges to a limit \(\lim _{n \to \infty } x_n \in \mathbb {R}\).
(This property is known as completeness of the metric space \(\mathbb {R}\).)
See MS-C1541 Metric Spaces.
If a complex number sequence \((z_n)_{n \in \mathbb {N}}\) is Cauchy, then it converges to a limit \(\lim _{n \to \infty } z_n \in \mathbb {C}\).
(This property is known as completeness of the metric space \(\mathbb {C}\).)
See MS-C1541 Metric Spaces.
(Idea: This follows from Lemma A.14 by considering real and imaginary parts separately and picking a subsequence of a subsequence.)
A.3 Continuity
Let \(X\) and \(Y\) be metric spaces. A function \(f \colon X \to Y\) is said to be continuous at a point \(p_0 \in X\) if \(\lim _{p \to p_0} f(p) = f(p_0)\).
(Equivalently, for every \(\varepsilon {\gt} 0\) there exists a \(\delta {\gt} 0\) such that for any \(p \in \mathcal{B}(p_0; \delta )\) we have \(f(p) \in \mathcal{B}(f(p_0); \varepsilon )\).)
A function \(f \colon X \to Y\) is said to be continuous if it is continuous at every point \(p_0 \in X\).
Let \(X\) be a metric space, and let \(f \colon X \to \mathbb {C}\) be a complex-valued function on \(X\). Then \(f\) is continuous at \(p_0 \in X\) if and only if its real and imaginary parts \(p \mapsto \Re \mathfrak {e}\big( f(p) \big)\) and \(p \mapsto \Im \mathfrak {m}\big( f(p) \big)\) are continuous at \(p_0\).
…
The coordinate projections
are continuous functions.
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Let \(X\) be a metric space, let \(p_0 \in X\) be a point, and let \(f,g \colon X \to \mathbb {C}\) be two complex-valued functions on \(X\) which are continuous at \(p_0\). Then also the functions
are continuous at \(p_0\).
If moreover \(g(p_0) \ne 0\), then also the function \(p \mapsto \frac{f(p)}{g(p)}\) is continuous at \(p_0\).
…
Let \(X\) and \(Y\) be metric spaces, and let \(f \colon X \to Y\) be a function. Then the following are equivalent:
\(f\) is a continuous function;
for every open set \(V \subset Y\), the preimage \(f^{-1}[V] = \big\{ x \in X \, \big| \, f(x) \in V \big\} \) is an open set in \(X\);
for every closed set \(A \subset Y\), the preimage \(f^{-1}[A] = \big\{ x \in X \, \big| \, f(x) \in A \big\} \) is a closed set in \(X\).
See MS-C1541 Metric Spaces.
Let \(X\), \(Y\), and \(Z\) be metric spaces, and let \(f \colon X \to Y\) and \(g \colon Y \to Z\) be functions. If \(f\) is continuous at \(x_0 \in X\) and \(g\) is continuous at \(f(x_0) \in Y\), then the composition \(g \circ f \colon X \to Z\) is continuous at \(x_0\).
(The composition \(g \circ f\) is defined by the formula \((g \circ f)(x) = g \big( f(x) \big)\).)
…
Let \(N \in \mathbb {N}\) be a natural number, and let \(c_{n,m} \in \mathbb {R}\) be real numbers for \(n,m \in \{ 0,1,\ldots ,N\} \). Then the function \(p \colon \mathbb {C}\to \mathbb {R}\) defined by
is continuous.
See MS-C1541 Metric Spaces.
Let \(X\) and \(Y\) be metric spaces. A function \(f \colon X \to Y\) is uniformly continuous if for every \(\varepsilon {\gt} 0\) there exists a \(\delta {\gt} 0\) such that for any \(p_0 \in X\) and \(p \in \mathcal{B}(p_0; \delta )\) we have \(f(p) \in \mathcal{B}(f(p_0); \varepsilon )\).
If a function \(f \colon X \to Y\) is uniformly continuous, then it is continuous.
See MS-C1541 Metric Spaces.
(The easy proof is also a good exercise.)
A.4 Connectedness and path-connectedness
A set \(A \subset X\) in a metric space \(X\) is disconnected if there exists a continuous surjective function \(f \colon A \to \left\{ 0,1 \right\} \) onto the two-element discrete set \(\left\{ 0,1 \right\} \). Otherwise \(A\) is connected; then every continuous function \(A \to \left\{ 0,1 \right\} \) must be either constant \(0\) or constant \(1\).
(The usual definition in topology textbooks reads slightly differently, but it is equivalent to the one we chose here by Lemma A.20.)
A set \(A \subset X\) in a metric space \(X\) is path connected if for any two points \(p, q \in X\) there exists a continuous function \(\gamma \colon [0,1] \to X\) such that \(\gamma (0) = p\) and \(\gamma (1) = q\) (a parametrized path in \(X\) starting from \(p\) and ending at \(q\)).
If a metric space \(X\) is path-connected, then it is connected.
See MS-C1541 Metric Spaces.
Suppose that \(U \subset \mathbb {C}\) is an open subset of the complex plane. Then \(U\) is connected if and only if it is path-connected.
See MS-C1541 Metric Spaces.
A.5 Compactness
Let \(X\) be a metric space. A subset \(K \subset X\) is compact if every sequence \((x_n)_{n \in \mathbb {N}}\) of points \(x_n \in K\) has a subsequence \((x_{n_k})_{k \in \mathbb {N}}\) which converges to a limit \(\lim _{k \to \infty } x_{n_k} \in K\) in the set \(K\).
A subset \(B \subset \mathbb {R}\) of the real line is compact if an only if it is closed and bounded.
A subset \(A \subset \mathbb {C}\) of the complex plane is compact if an only if it is closed and bounded.
See MS-C1541 Metric Spaces.
Suppose that \(X\) is compact. Then every continuous function \(f \colon X \to \mathbb {R}\) is bounded.
…
If \(X\) is compact and a function \(f \colon X \to Y\) is continuous, then it is uniformly continuous.
See MS-C1541 Metric Spaces.
Let \(X\) and \(Y\) be metric spaces and assume that \(X\) is compact. Then for any continuous bijection \(f \colon X \to Y\), also the inverse \(f^{-1} \colon Y \to X\) is continuous.
See MS-C1541 Metric Spaces.
Let \(X\) be a metric space. Suppose that \(K_1, K_2, K_3, \ldots \) are nonempty compact subsets of \(X\) nested so that \(K_1 \supset K_2 \supset K_3 \supset \cdots \). Then the intersection \(\bigcap _{n=1}^\infty K_n\) is nonempty.
See MS-C1541 Metric Spaces.
A.6 Simple connectedness
Let \(X\) be a metric space and \(\gamma _0 \colon [a,b] \to X\) and \(\gamma _0 \colon [a,b] \to X\) two closed paths in \(X\). If there exists a continuous function (called a homotopy)
such that
and
then we say that the closed paths \(\gamma _0\) and \(\gamma _1\) are homotopic.
Let \(X\) be a metric space. A closed path \(\gamma \colon [a,b] \to X\) is called contractible if it is homotopic to a constant path.
A metric space is said to be simply connected if every closed path \(\gamma \colon [a,b] \to X\) in \(X\) is contractible.