Multidimensional Spaces
Multidimensional Spaces
In this chapter, we introduce multidimensional spaces, laying the foundation for the core themes of Analysis II.
Basics of Metric Spaces
The Euclidean Structure
In this section, we discuss spaces endowed with a Euclidean structure. We begin by defining the standard vector space \(\mathbb{R}^n\) as the set of ordered \(n\)-tuples of real numbers: \[\begin{equation} \mathbb{R}^n = \{x = (x_1, \dots, x_n) \mid x_i \in \mathbb{R}\} \end{equation}\] where \(n \in \mathbb{N}\) represents the dimension. The space \(\mathbb{R}^n\) is a linear space equipped with the following operations:
Vector Addition: For all \(x, y \in \mathbb{R}^n\), \[\begin{equation} x + y = (x_1+y_1, \dots, x_n + y_n) \end{equation}\]
Scalar Multiplication: For all \(\lambda \in \mathbb{R}\) and \(x \in \mathbb{R}^n\), \[\begin{equation} \lambda x = (\lambda x_1, \dots, \lambda x_n) \end{equation}\]
Inner Product, Norm, and Distance
The geometric structure of \(\mathbb{R}^n\) arises from the introduction of the standard scalar product (or dot product), defined as:
\[\begin{equation} x \cdot y = \langle x, y \rangle := \sum_{i=1}^n x_i y_i \end{equation}\]
Based on the scalar product, we induce the Euclidean norm (the length of a vector):
\[\begin{equation} \lVert x \rVert := \sqrt{\langle x, x \rangle} = \sqrt{\sum_{i=1}^n x_i^2} \end{equation}\]
Finally, we define the Euclidean distance between two points \(x\) and \(y\):
\[\begin{equation} d(x,y) := \lVert y-x \rVert \end{equation}\]
For all \(x, y \in \mathbb{R}^n\), the following inequality holds: \[\begin{equation} |\langle x, y \rangle| \leq \lVert x \rVert \cdot \lVert y \rVert \end{equation}\]
Proof. If either \(x = 0\) or \(y = 0\), the inequality holds trivially (\(0 \leq 0\)). Assume without loss of generality that \(x, y \neq 0\). We start from the elementary inequality valid for all real numbers \(a, b \in \mathbb{R}\): \[\begin{equation} 2ab \leq a^2 + b^2 \end{equation}\] Let \(\lambda > 0\) be an arbitrary scalar. By applying the inequality above to each component, we have: \[\begin{equation} \frac{2 x_i y_i}{\lambda} = 2 (\lambda x_i) \left( \frac{y_i}{\lambda} \right) \leq \lambda^2 x_i^2 + \frac{y_i^2}{\lambda^2} \end{equation}\] Summing over \(i = 1, \dots, n\): \[\begin{equation} 2 \sum_{i=1}^n x_i y_i \leq \lambda^2 \sum_{i=1}^n x_i^2 + \frac{1}{\lambda^2} \sum_{i=1}^n y_i^2 = \lambda^2 \lVert x \rVert^2 + \frac{1}{\lambda^2} \lVert y \rVert^2 \end{equation}\] To obtain the tightest bound, we choose \(\lambda^2 = \frac{\lVert y \rVert}{\lVert x \rVert}\). Substituting this back: \[\begin{equation} 2 \langle x, y \rangle \leq \left( \frac{\lVert y \rVert}{\lVert x \rVert} \right) \lVert x \rVert^2 + \left( \frac{\lVert x \rVert}{\lVert y \rVert} \right) \lVert y \rVert^2 = \lVert y \rVert \lVert x \rVert + \lVert x \rVert \lVert y \rVert = 2 \lVert x \rVert \lVert y \rVert \end{equation}\] Dividing by 2 gives \(\langle x, y \rangle \leq \lVert x \rVert \lVert y \rVert\). The same argument applies to \(-x\) and \(y\), proving the absolute value. ◻
For all \(x, y \in \mathbb{R}^n\): \[\begin{equation} \lVert x + y \rVert \leq \lVert x \rVert + \lVert y \rVert \end{equation}\] This implies the metric triangle inequality: \(d(x, z) \leq d(x, y) + d(y, z)\).
Proof. We proceed by squaring the norm: \[\begin{align} \lVert x+y \rVert^2 &= \langle x+y, x+y \rangle \\ &= \langle x, x \rangle + 2\langle x, y \rangle + \langle y, y \rangle \\ &= \lVert x \rVert^2 + 2\langle x, y \rangle + \lVert y \rVert^2 \end{align}\] By the Cauchy-Schwarz inequality, we know that \(\langle x, y \rangle \leq \lVert x \rVert \lVert y \rVert\). Therefore: \[\begin{align} \lVert x+y \rVert^2 &\leq \lVert x \rVert^2 + 2\lVert x \rVert \lVert y \rVert + \lVert y \rVert^2 \\ &= (\lVert x \rVert + \lVert y \rVert)^2 \end{align}\] Taking the square root of both sides yields the result. ◻
A metric space is defined as a set endowed with a distance function: \[\begin{equation} (X, d) \end{equation}\] is a metric space where \(X\) is a non-empty set and \(d: X \times X \rightarrow [0, \infty)\) is a distance function.
A metric space must satisfy three axioms:
\(\forall x, y \in X, \quad d(x,y) = 0 \iff x = y\)
\(\forall x,y \in X, \quad d(x,y) = d(y,x)\)
\(\forall x,y,z \in X, \quad d(x,z) \leq d(x,y) + d(y,z)\)
Some examples of metric spaces are:
\((\mathbb{R}^n, d_{\text{Euclidean}})\)
\((\mathbb{R}^2, d_{\text{NY}})\) where \(d_{\text{NY}}(x,y) = |x_1-y_1| + |x_2-y_2|\) is called the New York (or Manhattan) distance.
Note that for a metric space \((X, d)\), if \(Y \subset X\), then \((Y, d|_{Y \times Y})\) is also a metric space.
The space of continuous functions on a real interval \([a,b]\) with \(a < b\), defined as \(X = \{ f: [a,b] \rightarrow \mathbb{R} \mid f \text{ is continuous} \}\). We can endow this set with several metrics, for instance: \[\begin{equation} d_{\infty}(f,g) = \max_{x \in [a,b]} |f(x)-g(x)| \end{equation}\] and the integral metric: \[\begin{equation} d_2(f,g) = \left( \int_a^b (f(x)-g(x))^2 \, dx \right)^{\frac{1}{2}} \end{equation}\]
We can define sequences in multidimensional spaces similarly to how we have defined sequences in \(\mathbb{R}\).
Let \(X\) be a set. We call a sequence in \(X\) a map \(x: \mathbb{N} \rightarrow X\), and write \(x_n\) to indicate the \(n\)-th element of the sequence.
There are many notations; the most common are: \((x_n)_{n\geq 0}\), \((x_n)_{n \in \mathbb{N}}\), and \((x_n)_{n = 0}^{\infty}\).
Let \((X,d)\) be a metric space. We say that a sequence \((x_n)_{n\geq 0}\) has a limit \(x \in X\) if and only if \(d(x_n, x) \rightarrow 0\) as a sequence of real numbers.
Equivalently, \(\forall \varepsilon > 0\), \(\exists N > 0\) such that \(d(x_n, x) < \varepsilon\) for all \(n \geq N\).
We will use this notation: \[\begin{equation} \lim_{n\rightarrow \infty} x_n = x \quad \text{or} \quad x_n \rightarrow x \end{equation}\]
Let \((X, d)\) be a metric space, and \((x_n)_{n \geq 0}\) a sequence in \(X\). Assume \(x_n \rightarrow x\) and \(x_n \rightarrow y\) with \(x,y \in X\). Then, \(x = y\).
Proof. Assume by contradiction that \(x \neq y\). Then \(d(x,y) > 0\). Let \(\varepsilon = \frac{d(x,y)}{3} > 0\).
By the definition of a limit, since \(x_n \rightarrow x\), we have: \[\begin{equation} \exists N_x > 0 \quad \text{s.t.} \quad d(x_n, x) < \varepsilon \quad \forall n \geq N_x \end{equation}\] And since \(x_n \rightarrow y\): \[\begin{equation} \exists N_y > 0 \quad \text{s.t.} \quad d(x_n, y) < \varepsilon \quad \forall n \geq N_y \end{equation}\]
Now take \(N = \max(N_x, N_y)\). Then for all \(n \geq N\), applying the triangle inequality yields: \[\begin{equation} 3\varepsilon = d(x,y) \leq d(x, x_n) + d(x_n, y) = d(x_n, x) + d(x_n, y) < \varepsilon + \varepsilon = 2\varepsilon \end{equation}\] This implies \(3\varepsilon < 2\varepsilon\), which is a contradiction since \(\varepsilon > 0\). Therefore, \(x = y\). ◻
Let \((x_n)_{n \geq 0}\) be a sequence in \(X\). We define a subsequence as any sequence of the form \((x_{f(k)})_{k \geq 0}\), where \(f: \mathbb{N} \rightarrow \mathbb{N}\) is a strictly increasing function.
Let \((X,d)\) be a metric space.
Given a subset \(Y \subset X\), we say that \(y \in X\) is an accumulation point (or limit point) of \(Y\) if and only if there exists a sequence \((y_n)_{n \geq 0} \subset Y \setminus \{y\}\) such that \(y_n \rightarrow y\).
Equivalently, given a sequence \((x_n)_{n \geq 0}\) in \(X\), we say that \(x\) is an accumulation point of the sequence if and only if there exists a strictly increasing function \(f: \mathbb{N} \rightarrow \mathbb{N}\) such that the subsequence \(x_{f(k)} \rightarrow x\).
Let \((X,d)\) be a metric space and \((x_n)_{n \geq 0} \subset X\) a sequence. Then \((x_n)_{n \geq 0}\) converges to some \(x \in X \iff \forall\) subsequence \((x_{n_k})_{k \geq 0}\) we have \(x_{n_k} \rightarrow x\).
Proof. Recall that if \(x_n \rightarrow x\) and \(x_{n_k} \rightarrow y\), then \(x = y\) (a subsequence limit must match the sequence limit if the sequence converges). We will prove both implications:
(\(\Leftarrow\)) Take as a subsequence the sequence itself (where \(n_k = k\)). Then \((x_{n_k})_{k \geq 0} = (x_n)_{n \geq 0}\), which by assumption implies \(x_n \rightarrow x\).
(\(\Rightarrow\)) Let \(f: \mathbb{N} \rightarrow \mathbb{N}\) be a strictly increasing function, so \(n_k = f(k)\). The goal is to show that \(x_{f(k)} \rightarrow x\).
By definition: \[\begin{equation} x_n \rightarrow x \iff \forall \varepsilon > 0, \exists N > 0 \quad \text{s.t.} \quad d(x_n, x) < \varepsilon \quad \forall n \geq N \end{equation}\] Since \(f\) is strictly increasing, we have \(f(k) \geq k\) for all \(k \in \mathbb{N}\). Therefore, for all \(k \geq N\), it follows that \(f(k) \geq N\). This implies: \[\begin{equation} d(x_{f(k)}, x) < \varepsilon \quad \forall k \geq N \end{equation}\] which means \(x_{f(k)} \rightarrow x\).
◻
A sequence \((x_n)_{n \geq 0}\) is a Cauchy sequence if and only if \(\forall \varepsilon > 0, \exists N > 0\) such that \(d(x_n, x_m) < \varepsilon\) for all \(n,m \geq N\).
A metric space \((X,d)\) is complete if and only if every Cauchy sequence in \(X\) converges to some limit in \(X\).
Let \((x_m)_{m \geq 0} \subset \mathbb{R}^n\). Then \(x_m \rightarrow x \in \mathbb{R}^n \iff x_{m,i} \rightarrow x_i \quad \forall i \in \{1, \dots , n\}\).
An element of a sequence in a multidimensional space is defined by its components: \[\begin{equation} x_m = (x_{m,1}, \dots , x_{m,n}) \end{equation}\] And its limit is: \[\begin{equation} x = (x_1, \dots , x_n) \end{equation}\]
Proof.
(\(\Rightarrow\)) Assume \(x_m \rightarrow x\). By definition, \(\forall \varepsilon > 0, \exists N > 0\) such that \(\lVert x_m - x \rVert < \varepsilon\) for all \(m \geq N\). For any component \(i\), this implies: \[\begin{equation} |x_{m,i} - x_i| = \sqrt{|x_{m,i} - x_i|^2} \leq \sqrt{\sum_{j = 1}^n |x_{m,j} - x_j|^2} = \lVert x_m - x \rVert < \varepsilon \end{equation}\] Thus, \(x_{m,i} \rightarrow x_i\).
(\(\Leftarrow\)) Assume \(x_{m,i} \rightarrow x_i\) for all \(i = 1, \dots, n\). Let \(\varepsilon > 0\). For each \(i\), there exists \(N_i > 0\) such that \(|x_{m,i} - x_i| < \frac{\varepsilon}{\sqrt{n}}\) for all \(m \geq N_i\). Let \(N = \max(N_1, \dots, N_n)\). Then for all \(m \geq N\), we have: \[\begin{equation} \lVert x_m - x \rVert = \sqrt{\sum_{i = 1}^n |x_{m,i} - x_i|^2} < \sqrt{\sum_{i=1}^n \left(\frac{\varepsilon}{\sqrt{n}}\right)^2} = \sqrt{n \frac{\varepsilon^2}{n}} = \varepsilon \end{equation}\] Thus, \(x_m \rightarrow x\).
◻
The metric space \((\mathbb{R}^n, d_{\text{Euclidean}})\) is complete.
Proof. Given a Cauchy sequence \((x_m)_{m \geq 0}\) in \(\mathbb{R}^n\), we must show that there exists an \(x \in \mathbb{R}^n\) such that \(x_m \rightarrow x\).
Since \((x_m)_{m \geq 0}\) is a Cauchy sequence with respect to the Euclidean norm, each component sequence \((x_{m,i})_{m \geq 0}\) is a Cauchy sequence in \(\mathbb{R}\) for every \(i \in \{1, \dots, n\}\).
Since the field of real numbers \(\mathbb{R}\) is complete, every Cauchy sequence in \(\mathbb{R}\) converges. Therefore, for each \(i\), there exists a real number \(x_i \in \mathbb{R}\) such that \(x_{m,i} \rightarrow x_i\) as \(m \rightarrow \infty\).
Let \(x = (x_1, \dots, x_n) \in \mathbb{R}^n\). By the previous lemma, convergence component by component implies convergence in the Euclidean metric. Thus, \(x_m \rightarrow x\). This proves that every Cauchy sequence in \(\mathbb{R}^n\) converges to a limit in \(\mathbb{R}^n\), meaning the space is complete. ◻
Topology of Metric Spaces
Open and Closed Sets
Let \((X, d)\) be a metric space, \(x\in X\), and \(r > 0\) a real number. In this context, we write \[\begin{equation} B(x,r) := \{y \in X \mid d(x,y) < r \} \end{equation}\] and refer to the set \(B(x,r)\) as the open ball with centre \(x\) and radius \(r\).
Let \((X,d)\) be a metric space. Then:
A subset \(E \subset X\) is called open if for every \(x \in E\) there exists \(r > 0\) such that \(B(x,r) \subset E\).
The collection of all open sets, \(\mathcal{T}_d = \{ U \subset X \mid U \text{ is open}\}\), is called the topology generated by \(d\).
A subset \(E \subset X\) is called closed if its complement \(X \setminus E\) is open.
The empty set \(\varnothing\) and the whole space \(X\) are always both open and closed. A set that is both open and closed is referred to as "clopen". In general, a subset \(U\) need not be either open or closed.
Let \((U_i)_{i \in I}\) be any family of open subsets of \(X\). Then the arbitrary union \(\bigcup_{i \in I}U_i\) is also open. Moreover, if \(I\) is a finite set, the finite intersection \(\bigcap_{i\in I} U_i\) is open as well.
Proof. We first prove that an arbitrary union of open sets is open.
To this aim, set \[\begin{equation} \mathcal{U} = \bigcup_{i \in I}U_i \end{equation}\] and let \(x \in \mathcal{U}\). Then there exists an index \(i \in I\) such that \(x \in U_i\). Since \(U_i\) is open, there exists an \(\varepsilon > 0\) such that \(B(x, \varepsilon) \subset U_i\), which implies \(B(x, \varepsilon) \subset \mathcal{U}\). Thus, \(\mathcal{U}\) is open.
Now, to prove that a finite intersection of open sets is open, let \((U_i)_{i \in I}\) be a finite family of open subsets of \(X\). Set \[\begin{equation} \mathcal{V} = \bigcap_{i \in I}U_i \end{equation}\] and let \(x \in \mathcal{V}\). Then \(x\in U_i\) for all \(i \in I\). Since each \(U_i\) is open, for every \(i \in I\) there exists \(\varepsilon_i > 0\) such that \(B(x, \varepsilon_i) \subset U_i\). Now, since \(I\) is finite, we can choose the minimum radius \(\varepsilon := \min \{ \varepsilon_i \mid i \in I \}\). We have \(\varepsilon > 0\) and \(B(x, \varepsilon) \subset U_i\) for all \(i \in I\). Thus, \(B(x, \varepsilon) \subset \mathcal{V}\), completing the proof. ◻
Let \((A_i)_{i \in I}\) be any family of closed subsets of \(X\). Then \(\bigcap_{i \in I}A_i\) is also closed. Moreover, if \(I\) is a finite set, \(\bigcup_{i \in I}A_i\) is closed as well.
Proof. To prove this lemma, simply apply the previous lemma to the (open) complements of the closed sets, using De Morgan’s laws (i.e., the complement of an intersection is the union of complements, and the complement of a union is the intersection of complements). ◻
Let \((X, d)\) be a metric space, and \(E \subset X\) a subset. We define:
The interior \(E^\circ := \bigcup_{U \subset E, \, U \text{ open}} U\), which is the largest open set contained in \(E\). We will also use the notation \(\text{int}(E)\).
The closure \(\overline{E} := \bigcap_{A \supset E, \, A \text{ closed}} A\), which is the smallest closed set that contains \(E\).
The (topological) boundary \(\partial E := \overline{E}\setminus E^\circ\).
Let \((X, d)\) be a metric space. Then:
A subset \(U \subset X\) is open if and only if, for every convergent sequence in \(X\) with a limit in \(U\), the sequence eventually lies in \(U\).
A subset \(A \subset X\) is closed if and only if, for every convergent sequence \((x_n)_{n \geq 0} \subset A\), the limit also lies in \(A\). In other words, \(A\) is closed if and only if it contains all its accumulation points.
Proof. We will first prove the statement for open sets.
Let \(U \subset X\) be an open subset, and let \((x_n)_{n \geq 0}\) be a sequence in \(X\) with a limit \(x_0 \in U\). Then, there exists an \(\varepsilon > 0\) such that \(B(x_0, \varepsilon) \subset U\). Since \((x_n)_{n \geq 0}\) converges to \(x_0\), there exists an \(N \in \mathbb{N}\) such that \(x_n \in B(x_0, \varepsilon)\) for all \(n \geq N\). Therefore, \(x_n \in U\) for all \(n \geq N\).
Conversely, let \(V \subset X\) be a non-open subset. Then there exists a point \(x_0 \in V\) such that for all \(\varepsilon > 0\), \(B(x_0, \varepsilon) \setminus V \neq \varnothing\). For each \(n \geq 1\), we can choose a point \(x_n \in B(x_0, 2^{-n}) \setminus V\). The sequence \((x_n)_{n \geq 1}\) lies entirely in \(X \setminus V\) and converges to \(x_0 \in V\). Thus, we have found a sequence that converges to a point in \(V\) but never enters \(V\), completing the proof of the first statement.
Now we proceed to the statement for closed sets.
Let \(A \subset X\) be a closed subset, and let \((x_n)_{n \geq 0}\) be a convergent sequence in \(X\) with \(x_n \in A\) for all \(n \geq 0\). Let \(x_0\) be the limit of this sequence. The set \(U = X \setminus A\) is open. It cannot contain the limit point \(x_0\), because otherwise (by the first part of this lemma) the sequence \((x_n)_{n \geq 0}\) would eventually have to lie in \(U\), which is impossible since \(x_n \in A\). Therefore, the limit \(x_0\) belongs to \(A\).
For the converse, suppose \(A \subset X\) is not closed. Then \(U = X \setminus A\) is not open. According to the previous argument, there exists a sequence \((x_n)_{n \geq 0} \subset X \setminus U = A\) converging to a point \(x_0 \in U = X \setminus A\). Thus, \(A\) does not contain all its sequential limits. ◻
Continuity
The aim of this section is to generalize the notion of continuity to functions defined between metric spaces.
Let \((X, d_X)\) and \((Y, d_Y)\) be two metric spaces, and let \(f : X \rightarrow Y\) be a function. We say that \(f\) is continuous if one of the following equivalent conditions holds:
\(\varepsilon-\delta\) continuity: For a given \(x_0 \in X\) and every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for all \(x \in X\), \(d_X(x, x_0) < \delta \implies d_Y(f(x), f(x_0)) < \varepsilon\). In other words, \(f(B(x_0, \delta)) \subset B(f(x_0), \varepsilon)\).
Sequential continuity: For every convergent sequence \((x_n)_{n \geq 0}\) in \(X\) converging to some \(x_0 \in X\), the sequence \((f(x_n))_{n \geq 0}\) converges in \(Y\), and \(\lim_{n \rightarrow \infty} f(x_n) = f(x_0)\).
Topological continuity: For every open subset \(U \subset Y\), its preimage \(f^{-1}(U) = \{x \in X \mid f(x) \in U \}\) is an open set in \(X\).
The three statements of continuity:
\(f\) is \(\varepsilon-\delta\) continuous.
\(f\) is sequentially continuous.
\(f\) is topologically continuous.
are equivalent.
Proof. (1) \(\implies\) (2): Let \((x_n)_{n \geq 0}\) be a converging sequence in \(X\) with limit \(x_0 \in X\), and let \(\varepsilon > 0\). Then there exists a \(\delta > 0\) such that \(f(x) \in B(f(x_0), \varepsilon)\) for all \(x \in B(x_0, \delta)\). Since \((x_n)_{n \geq 0}\) converges to \(x_0\), there exists an \(N \in \mathbb{N}\) such that \(x_n \in B(x_0, \delta)\) for all \(n \geq N\). In particular, for all \(n \geq N\), \(f(x_n) \in B(f(x_0), \varepsilon)\). Since \(\varepsilon > 0\) is arbitrary, it follows that \(f(x_n) \rightarrow f(x_0)\).
\(\neg\) (3) \(\implies \neg\) (2): Assume \(f\) is not topologically continuous. Then there exists an open subset \(U \subset Y\) such that \(f^{-1}(U)\) is not open in \(X\). Therefore, there is a point \(x_0 \in f^{-1}(U)\) and a sequence \((x_n)_{n \geq 0} \subset X \setminus f^{-1}(U)\) such that \(x_n \rightarrow x_0\). This means \(f(x_0) \in U\), but \(f(x_n) \in Y \setminus U\) for all \(n \geq 0\). Since \(U\) is open, \(f(x_n)\) cannot converge to \(f(x_0)\), breaking sequential continuity.
(3) \(\implies\) (1): Let \(x_0 \in X\) and \(\varepsilon > 0\). The ball \(B(f(x_0), \varepsilon)\) is open in \(Y\). By assumption, its preimage \(f^{-1}(B(f(x_0), \varepsilon))\) is open in \(X\) and clearly contains the point \(x_0\). Because this preimage is open, there exists a \(\delta > 0\) such that \(B(x_0, \delta) \subset f^{-1}(B(f(x_0), \varepsilon))\). This implies that if \(x \in B(x_0, \delta)\), then \(f(x) \in B(f(x_0), \varepsilon)\). Thus, \(f\) is \(\varepsilon-\delta\) continuous. ◻
Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces. We say that \(f: X \rightarrow Y\) is:
Uniformly continuous if for every \(\varepsilon > 0\) there exists a \(\delta > 0\), such that \[\begin{equation} d_X(x, y) < \delta \implies d_Y(f(x), f(y)) < \varepsilon \quad \forall \ x,y \ \in X. \end{equation}\]
Lipschitz continuous if there exists a constant \(L > 0\) such that \[\begin{equation} d_Y(f(x), f(y)) \leq Ld_X(x,y) \quad \forall \ x,y \ \in X. \end{equation}\] The constant \(L\) is called Lipschitz constant of \(f\).
Every Lipschitz continuous function is also uniformly continuous. Indeed, let \(f: X \rightarrow Y\) be Lipschitz continuous. Then, given \(\varepsilon > 0\), let \(\delta \coloneqq \frac{\varepsilon}{L}\). Since \(f\) is Lipschitz continuous we know that \(d_Y(f(x), f(y)) \leq Ld_X(x,y) \quad \forall \ x,y \ \in X\), so if \(d_X(x,y) < \delta\) this implies that \(d_Y(f(x), f(y)) < L\delta = L \frac{\varepsilon}{L} = \varepsilon\).
Banach’s Fixed-Point Theorem
Fixed-point theorems, as the name suggests, investigate when a mapping \(f: X \rightarrow Y\) possesses a fixed point, i.e. a point \(x \in X\) such that \(f(x) = x\). They can be a rather powerful tool to prove existence theorems.
In this section we will prove Banach’s fixed-point theorem. The idea behind this result is flexible and useful: for example, it is possible to use this result to prove the Implicit Function Theorem and the existence and uniqueness of solutions to ODEs via Picard’s iteration.
Let \((X,d)\) be a complete metric space, and let \(T: X \rightarrow Y\) be a Lipschitz map with Lipschitz constant \(L \in (0,1)\). Then, there exists a unique element \(a \in X\) such that \(T(a) = a\).
The function \(T\) is a Lipschitz contraction. A point \(x\in X\) with \(T(x) = x\) is called a fixed point of the mapping \(T\), hence the theorem states that a Lipschitz contraction has a unique fixed point, provided the ambient space is complete.
Proof. First, we show uniqueness of a putative fixed point. Let \(a \in X\) and \(a' \in X\) be fixed points of \(T\). Then, \[\begin{equation} d(a,a') = d(T(a), T(a')) \leq Ld(a,a') \end{equation}\] which since \(L < 1\) implies \(d(a,a') = 0\) and thus \(a = a'\).
Now, to prove the existence of a fixed point, choose any \(x_0 \in X\) and define a sequence \((x_n)_{n\geq 0}\) recursively by \(x_{n+1} = T(x_n)\), for \(n \geq 0\). We claim that the sequence \((x_n)_{n\geq 0}\) is a Cauchy sequence. Iterating the contractivity assumption, we find that, for all integers \(p \geq 0\), \[\begin{equation} d(x_{n+1}, x_n) = d(T(x_n), T(x_{n-1})) \leq Ld(x_n, x_{n-1}) \leq \dots \leq L^nd(x_1, x_0) \end{equation}\] Pick now any integers \(m \geq n \geq N\), then using this observation and the triangle inequality we find: \[\begin{equation} d(x_m, x_n) \leq \sum_{p = n}^{m-1}d(x_{p+1}, x_p) \leq \sum_{p = n}^{m-1} L^pd(x_1, x_0) \leq d(x_1, x_0) \sum_{p \geq N}^{\infty}L^p \end{equation}\] which is equal to: \[\begin{equation} \frac{L^N}{1-L}d(x_1, x_0) \end{equation}\] where we crucially used \(L < 1\) to sum the whole geometric series. Now given any \(\varepsilon >0\) we can find some \(N\) so large that \(\frac{L^N}{1-L}d(x_1, x_0) < \varepsilon\), thus proving that \((x_n)_{n\geq 0}\) is a Cauchy sequence.
Now we use the completeness assumption to infer that \(x_n \rightarrow a\) for some \(a \in X\). Since \(T\) is continuous, we have: \[\begin{equation} T(a) = \lim_{n \rightarrow \infty}T(x_n) = \lim_{n \rightarrow \infty} x_{n+1} = \lim_{n \rightarrow \infty} x_n = a \end{equation}\] Which shows that \(a\) is a fixed point of \(T\). ◻
Compactness
A closed and bounded interval on the real line is called compact interval. Some fundamental properties of continuous functions on compact intervals are boundedness, existence of maxima and minima, and uniform continuity. We intend to investigate these and other properties in the broader context of metric spaces.
In a general metric space, it is not necessarily true that a closed and bounded set is compact.
Nevertheless when considering \(\mathbb{R}^n\) with its Euclidean structure, which is the main focus of Analysis II, it will turn out that a closed and bounded set is indeed compact, and viceversa.
Let \(E \subset X\) and let \(\mathcal{U} = \{ U_i \}_{i \in I}\) be a family of subsets of \(X\), where \(I\) is some set of indexes. We say that \(\mathcal{U}\) covers \(E\) if: \[\begin{equation} E \subset \bigcup \mathcal{U} = \bigcup_{i \in I} U_i. \end{equation}\] Any family \(\mathcal{V}\subset \mathcal{U}\) that still covers \(E\) is called a subcover of \(\mathcal{U}\).
Let \((X,d)\) be a metric space. A subset \(K \subset X\) is called compact if one of the following equivalent conditions hold:
\(K\) is sequentially compact: every sequence \((x_n)_{n\geq 0}\) in \(K\) has a subsequence that is convergent in \(K\).
\(K\) is topologically compact: every family of open sets \(\{U_i \}_{i \in I}\) that covers \(K\) has a finite subcover.
Let \((X,d)\) be a metric space, and \(K \subset X\) a subset. Then the following statements are equivalent:
\(K\) is sequentially compact.
\(K\) is topologically compact.
Proof. The proof of this equivalence is quite lengthy and beyond the scope of this introductory chapter. The interested reader is encouraged to prove it as an exercise or consult a dedicated text on topology. ◻
Let \(X\) be a metric space, \(A \subset X\) a closed subset and \(K \subset X\) a compact subset. Then \(A \cap K\) is compact.
Proof. We will prove that \(A \cap K\) is sequentially compact.
Take any sequence \((x_n)_{n \geq 0} \subset A \cap K\), by compactness of \(K\), it has a converging subsequence \(x_{n_k} \rightarrow x\), for some \(x \in K\). On the other hand, since \(A\) is closed, it contains its accumulation points, so we have \(x \in A\) as well. ◻
In a metric space \(X\), every compact subset \(K \subset X\) is closed.
Proof. Take any sequence in \(K\) which converges to some \(x \in X\). By compactness a suitable subsequence is converging to some \(x' \in K\), and by uniqueness of the limit we have \(x = x'\). It follows \(x \in A\). ◻
From this result, we can also derive quite easily the following version of the Heine-Borel theorem in the Euclidean space \(\mathbb{R}^n\).
A subset \(K \subset \mathbb{R}^n\) is compact if and only if it is closed and bounded.
In the statement of the Heine-Borel Theorem [thm:Heine-Borel], a set \(E \subset \mathbb{R}^n\) is called bounded if there exists an \(N \in \mathbb{N}\) such that \(E \subset B_N(0)\). i.e. if \(\sup \{ d(x, x') \mid (x, x') \in E \times E \} < \infty\).
Proof. If \(K\) is compact, by Corollary [cor:compact_closed], then it is closed.
Moreover, \(K\) is bounded. If it were not bounded, for each \(m \in \mathbb{N}\) one can pick an \(x_m \in K\) with \(\lVert x_m \rVert \geq m\), which does not admit a convergent subsequence, since every convergent sequence in \(\mathbb{R}^n\) is bounded, contradicting compactness.
For the converse, assume that \(K \subset \mathbb{R}^n\) is closed and bounded. Then \(K \subset B_N(0) \subset[-N, N]^n\) for some \(N > 0\). By Corollary [cor:compact_closed], it is sufficient to show that \([-N, N]^n\) is compact.
To this aim, let \((x_k)_{k \geq 0}\) be a sequence in \([-N, N]^n\), and write \(x_k = (x_{k,1}, \dots , x_{k,n})\). For each fixed \(i \in \{1, \dots , n\}\) the real sequence \((x_{k,i})_{k \geq 0}\) is bounded, hence it admits a converging subsequence.
We then build a converging subsequence of \((x_k)_{k \geq 0}\) by successive extraction.
Choose increasing sequence indices \(k_1^{(1)}\) such that \((x_{k_j^{(1)},1})_j\) converges to some \(l_1\). From \((k_j^{(1)})_{j \geq 0}\) choose a subsequence of indices \(k_j^{(2)}\) such that \((x_{k_j^{(2)},2})_j\) converges to some \(l_2\); then also \((x_{k_j^{(2)},1})_j \rightarrow l_1\). Continuing in this way, after \(n\) steps we obtain indices \(k_j^{(n)}\) and numbers \(l_1, l_2, \dots , l_n\) such that for every \(i = 1, \dots , n\) we have \[\begin{equation} x_{k_j^{(n)},i} \rightarrow l_i \quad (j \rightarrow \infty). \end{equation}\] Now let \(x \coloneqq (l_1, l_2, \dots , l_n) \in \mathbb{R}^n\) (recall that convergence in \(\mathbb{R}^n\) is equivalent to component-wise convergence).
Therefore, every sequence in \([-N, N]^n\) has a convergent subsequence, so \([-N, N]^n\) is compact. ◻
Let \(X\) and \(Y\) be metric spaces, let \(f: X \rightarrow Y\) be a continuous function, and let \(K \subset X\) be a compact subset. Then, \(f(K)\) is a compact subset of \(Y\).
Proof. We will prove this theorem using sequential compactness.
Suppose that \((y_n)_{n \geq 0}\) is a sequence of points in \(f(K)\). Then \(y_n = f(x_n)\) for some \((x_n)_{n \geq 0}\). Since \(K\) is compact, there exists a convergent subsequence \(x_{n_k} \rightarrow x \in K\). But then, since \(f\) is continuous \(y_{n_k} \rightarrow y = f(x) \in f(K)\). ◻
Let \(X\) be a metric space, \(f:X \rightarrow \mathbb{R}\) be a continuous function, and \(K \subset X\) a compact subset. Then, \(f\) admits a maximum point, i.e. there exists \(\overline{x} \in X\) such that \(f(\overline{x}) = \sup_K f = \sup \{f(x) \mid x \in K \}\). An analogous statement holds for the minimum. In particular, \(f\) must be bounded.
Proof. We know by Theorem [thm:compact_continuous_image] that \(f(K) \subset \mathbb{R}\) is compact and non-empty, so \(s \coloneqq \sup f(K) < \infty\) (as \(f(K) \subset \mathbb{R}\) is bounded) and \(s \in f(K)\) (as \(f(K) \subset \mathbb{R}\) is closed). Then, any element \(\overline{x} \in f^{-1}(s) = f^{-1}(\sup f(K))\) attains the supremum. ◻
A continuous function in a compact set is uniformly continuous Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces, and let \(f: X \rightarrow Y\) be a continuous function. If \(X\) is compact, then \(f\) is uniformly continuous.
Proof. Let \(\varepsilon >0\). By definition of continuity, for each \(x \in X\), there exists a \(\delta_x > 0\) such that \[\begin{equation} f(B(x, \delta_x)) \subset B(f(x), \frac{\varepsilon}{2}). \end{equation}\] The collection \(\{ B(x, \frac{\delta_x}{2}) \mid x \in X \}\) forms an open cover of \(X\). Since \(X\) is compact by assumption, there exists a finite subcover of this collection. This implies the existence of \(x_1, \dots , x_n \in X\) such that: \[\begin{equation} X = B(x_1, \frac{\delta_{x_1}}{2}) \cup \dots \cup B(x_n, \frac{\delta_{x_n}}{2}) \end{equation}\] Now, let \(\delta = \min \{\delta_{x_1}, \dots , \delta_{x_n} \}\), and consider \(x, x' \in X\) with \(d_X(x, x') < \delta\). By the above, there exists \(k\) such that \(x \in B(x_k, \frac{\delta_{x_k}}{2})\); this implies \(x' \in B(x_k, \delta _{x_k})\), leading to \[\begin{equation} d_Y(f(x), f(x')) \leq d_Y(f(x), f(x_k)) + d_Y(f(x_k), f(x')) \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{equation}\] which concludes the proof. ◻