Kreyszig 2.5 Compactness and Finite Dimension
Problem 1. Show that \(\mathbb{R}^n\) and \(\mathbb{C}^n\) are not compact.
Solution
To show that \(\mathbb{R}^n\) and \(\mathbb{C}^n\) are not compact, we can utilize the Heine-Borel theorem, which characterizes compact subsets of \(\mathbb{R}^n\). The theorem states that a subset of \(\mathbb{R}^n\) is compact if and only if it is closed and bounded.
For \(\mathbb{R}^n\):
Closedness: By definition, \(\mathbb{R}^n\) is closed because it contains all its limit points; every convergent sequence in \(\mathbb{R}^n\) has a limit that is also in \(\mathbb{R}^n\).
Boundedness: However, \(\mathbb{R}^n\) is not bounded. To see this, consider the sequence \(\{(x_k)\}_{k=1}^{\infty}\) where \(x_k = (k, 0, 0, \ldots, 0)\) in \(\mathbb{R}^n\). This sequence has no upper bound within \(\mathbb{R}^n\) because for any given \(M \in \mathbb{R}\), there exists an \(N\) such that for all \(n > N\), \(x_n > M\).
Thus, since \(\mathbb{R}^n\) is not bounded, it cannot be compact according to the Heine-Borel theorem.
For \(\mathbb{C}^n\):
Closedness: \(\mathbb{C}^n\) is also closed because it includes all its limit points; it is the entire space of \(n\)-tuples of complex numbers.
Boundedness: We can show that \(\mathbb{C}^n\) is not bounded in a similar manner to \(\mathbb{R}^n\). Consider the sequence of complex numbers \(\{(z_k)\}_{k=1}^{\infty}\) where \(z_k = (k, 0, \ldots, 0)\) in \(\mathbb{C}^n\), where \(k\) represents the complex number \(k + 0i\). This sequence, too, has no bound within \(\mathbb{C}^n\) for the same reasons as in \(\mathbb{R}^n\).
Thus, \(\mathbb{C}^n\) is not bounded and, hence, not compact.
Note that \(\mathbb{C}^n\) is isomorphic to \(\mathbb{R}^{2n}\) since each complex number corresponds to a pair of real numbers (the real and imaginary parts). Therefore, the non-compactness of \(\mathbb{C}^n\) follows from the non-compactness of \(\mathbb{R}^{2n}\).
In conclusion, neither \(\mathbb{R}^n\) nor \(\mathbb{C}^n\) is compact because, although both are closed, neither is bounded.
\(\blacksquare\)
The detailed explanation of why the sequence \(\{(x_k)\}_{k=1}^{\infty}\) with \(x_k = (k, 0, 0, \ldots, 0)\) in \(\mathbb{R}^n\) has no upper bound is as follows:
For any chosen real number \(M\), no matter how large, there exists a natural number \(N\) such that for all \(k > N\), the value of \(k\) is greater than \(M\). This means that the first component of the vector \(x_k\) is greater than \(M\). The formal expression of this statement is:
This expression states that for any real number \(M\) one might consider as a potential upper bound, there exists a point in the sequence, beyond the \(N\) th term, where the elements of the sequence exceed \(M\). Thus, for any \(M\) in \(\mathbb{R}\), we can find elements in the sequence \(\{(x_k)\}\) that are larger than \(M\), demonstrating that the sequence does not have an upper bound in \(\mathbb{R}^n\).
The absence of an upper bound implies that the sequence is unbounded. According to the Heine-Borel theorem, a necessary condition for a subset of \(\mathbb{R}^n\) to be compact is that it must be bounded. Since the sequence \(\{(x_k)\}\) is unbounded in \(\mathbb{R}^n\), it follows that \(\mathbb{R}^n\) itself is not compact as it fails to satisfy the boundedness condition required by the theorem.
Problem 2. Show that a discrete metric space \(X\) consisting of infinitely many points is not compact.
Solution
To prove that an infinite discrete metric space \(X\) is not compact, we use the definition of compactness in metric spaces. A metric space is compact if every open cover has a finite subcover.
In a discrete metric space, the metric is defined such that \(d(x, x) = 0\) for all \(x \in X\), and \(d(x, y) = 1\) for all \(x \neq y\). This implies that each point in \(X\) is isolated from every other point. We can then consider an open cover of \(X\) consisting of the open balls \(\{B(x, \frac{1}{2})\}\) for each \(x \in X\). Each of these balls is indeed an open set because it contains no points other than its center.
Since \(X\) contains infinitely many points, the collection \(\{B(x, \frac{1}{2})\}\) is an infinite cover for \(X\). If \(X\) were compact, there would exist a finite subcover that still covers \(X\). However, this is not possible because each open ball in our cover contains exactly one point of \(X\) and no two balls contain the same point. Thus, no finite collection of these balls can cover the entirety of \(X\).
Consequently, there is no finite subcover possible for the cover \(\{B(x, \frac{1}{2})\}\), which means that the discrete metric space \(X\) cannot be compact.
\(\blacksquare\)
Problem 3. Give examples of compact and noncompact curves in the plane \(\mathbb{R}^2\).
Solution
Compact Curves:
Unit Circle: The set of all points \((x, y)\) such that \(x^2 + y^2 = 1\). This curve is closed and bounded.
Square: The boundary defined by the set of all points \((x, y)\) with vertices at \((\pm1, \pm1)\). It is a closed and bounded shape.
Triangle: The boundary formed by connecting points \((0, 0)\), \((1, 0)\), and \((0, 1)\). Each edge is a closed line segment, making the whole triangle compact.
Closed Disk: The set of all points \((x, y)\) satisfying \(x^2 + y^2 \leq r^2\) for a fixed \(r\). This includes all points within and on the boundary, constituting a closed and bounded set.
Noncompact Curves:
Ray: The set of points \((x, y)\) forming a ray extending from the origin indefinitely, like \(\{(t, t) | t \geq 0\}\). This curve is unbounded.
Hyperbola: The set of points \((x, y)\) satisfying \(xy = 1\). This curve extends to infinity in all directions.
Infinite Line: A line like \(y = x\) that extends without bound in both directions.
Logarithmic Spiral: Defined by the polar equation \(r = e^{\theta}\), this curve winds away from the origin infinitely.
These examples illustrate the distinction in \(\mathbb{R}^2\) between compact sets, which are both closed and bounded, and noncompact sets, which are not closed, not bounded, or both.
Problem 4. Show that for an infinite subset \(M\) in the space \(s\) to be compact, it is necessary that there are numbers \(\gamma_1, \gamma_2, \ldots\) such that for all \(x = (\xi_k(x)) \in M\) we have \(|\xi_k(x)| \leq \gamma_k\). (It can be shown that the condition is also sufficient for the compactness of \(M\).)
Solution
To show the necessity of the condition for compactness in the space \(s\), as defined in the problem statement, we need to demonstrate that for any sequence in a compact subset \(M\) of \(s\), the elements of the sequence must be uniformly bounded by some sequence \(\{\gamma_k\}\).
Consider the space \(s\) as defined in the second image, where the metric \(d\) is given by:
with \(x = (\xi_k)\) and \(y = (\eta_k)\) being elements of \(s\).
The metric \(d\) is designed such that the "distance" it measures is the sum of a series of terms, each of which is a fraction of the absolute difference between the components of two elements \(x\) and \(y\), scaled by \(1/2^j\). This series converges because each term is less than or equal to \(1/2^j\), and \(\sum 1/2^j\) is a convergent geometric series.
Now let's consider the subset \(M \subset s\). If \(M\) is compact, then from the definition of compactness in metric spaces, every sequence in \(M\) has a convergent subsequence. For a sequence \(\{x^{(n)}\}\) with \(x^{(n)} = (\xi_k^{(n)})\) in \(M\), its convergence in \(s\) means that for every \(\epsilon > 0\), there exists an \(N\) such that for all \(m, n > N\), \(d(x^{(m)}, x^{(n)}) < \epsilon\).
For compactness, we require that this sequence has a convergent subsequence in \(s\). Due to the definition of the metric, this means that for each \(j\), the sequence \(\{\xi_j^{(n)}\}\) must be Cauchy, and hence bounded. Therefore, there must exist a bound \(\gamma_j\) for each \(j\) such that \(|\xi_j^{(n)}| \leq \gamma_j\) for all \(n\).
To see why the sequence \(\{\xi_j^{(n)}\}\) must be bounded, suppose it were not. If for some \(j\), \(\{\xi_j^{(n)}\}\) were unbounded, then we could choose \(\epsilon\) small enough (specifically \(\epsilon < 1/2^j\)) and a subsequence \(\{x^{(n_k)}\}\) such that \(|\xi_j^{(n_k)} - \xi_j^{(n_{k+1})}| > 1\) for all \(k\), which would imply \(d(x^{(n_k)}, x^{(n_{k+1})})\) would not converge to 0, contradicting the compactness of \(M\).
Therefore, for \(M\) to be compact, it is necessary that there exist numbers \(\gamma_1, \gamma_2, \ldots\) such that for all \(x = (\xi_k(x)) \in M\) we have \(|\xi_k(x)| \leq \gamma_k\). This condition is known to be sufficient as well for the compactness of \(M\) in the space \(s\), as a uniformly bounded and equicontinuous sequence in \(s\) will have a convergent subsequence by the Arzelà-Ascoli theorem.
\(\blacksquare\)
Problem 5. A metric space \(X\) is said to be locally compact if every point of \(X\) has a compact neighborhood. Show that \(\mathbb{R}\) and \(\mathbb{C}\), and more generally, \(\mathbb{R}^n\) and \(\mathbb{C}^n\) are locally compact.
Solution
To prove that \(\mathbb{R}\), \(\mathbb{C}\), and by extension \(\mathbb{R}^n\) and \(\mathbb{C}^n\), are locally compact, we utilize the following concepts:
A space is locally compact if each point has a compact neighborhood.
A set is compact if every open cover has a finite subcover, which, in a metric space, translates to the set being closed and bounded, as per the Heine-Borel theorem.
A neighborhood of a point includes an open set containing that point.
Detailed Proofs of Local Compactness
Detailed Proof for \(\mathbb{R}\):
For any point \(x \in \mathbb{R}\), we can identify a neighborhood around \(x\), such as the open interval \((x - \epsilon, x + \epsilon)\) for some \(\epsilon > 0\). The closure of this interval is the closed interval \([x - \epsilon, x + \epsilon]\), which encompasses its limit points and is delimited by the points \(x - \epsilon\) and \(x + \epsilon\). By the Heine-Borel theorem, as \([x - \epsilon, x + \epsilon]\) is both closed and bounded within \(\mathbb{R}\), it is compact. Therefore, every point \(x\) possesses a compact neighborhood in \(\mathbb{R}\), affirming its local compactness.
Detailed Proof for \(\mathbb{C}\):
Upon recognizing \(\mathbb{C}\) as topologically equivalent to \(\mathbb{R}^2\), for any \(z \in \mathbb{C}\), we consider the open disk centered at \(z\), denoted \(D(z, \epsilon)\) for some \(\epsilon > 0\). This disk serves as a neighborhood of \(z\). The closure of \(D(z, \epsilon)\), which consists of all points inside and on the boundary of the disk, constitutes a closed set. It is also bounded by the circumference of the disk. Thus, by the Heine-Borel theorem, the closure of \(D(z, \epsilon)\) is compact in \(\mathbb{C}\), corroborating its local compactness.
Detailed Proof for \(\mathbb{R}^n\):
For an arbitrary point \(x \in \mathbb{R}^n\), we select the open ball \(B(x, \epsilon)\) centered at \(x\) with a radius \(\epsilon > 0\). The closure of this ball, \(\overline{B(x, \epsilon)}\), which includes all points within and on the periphery of the sphere, is closed. Moreover, it is bounded as all points lie within a maximum distance \(\epsilon\) from \(x\). Consequently, \(\overline{B(x, \epsilon)}\) is compact as per the Heine-Borel theorem, demonstrating that \(\mathbb{R}^n\) is locally compact since \(x\) has a compact neighborhood.
Detailed Proof for \(\mathbb{C}^n\):
Given that \(\mathbb{C}^n\) aligns with \(\mathbb{R}^{2n}\) topologically, each complex coordinate having a real and imaginary part, for any point \(z \in \mathbb{C}^n\), an open ball in \(\mathbb{R}^{2n}\) can be centered at the point corresponding to \(z\) with a radius \(\epsilon > 0\). The closure of this ball is also a closed and bounded set in \(\mathbb{R}^{2n}\), and hence compact. This provides every point in \(\mathbb{C}^n\) with a compact neighborhood, certifying local compactness.
Each proof underlines the principle that local compactness is evidenced by the ability to encase any point within a closed and bounded (thus compact) subset, meeting the local compactness criterion.
Problem 6. Show that a compact metric space \(X\) is locally compact.
Proof
Let \(X\) be a compact metric space. We aim to prove that for every point \(x\) in \(X\), there exists a compact neighborhood around \(x\). In metric spaces, we have the luxury of using open balls as basic neighborhoods. For an arbitrary \(x \in X\) and for any positive real number \(\epsilon\), the open ball \(B(x, \epsilon)\) is an open set containing \(x\).
Due to the compactness of \(X\), any open cover has a finite subcover. Consider the collection of open balls \(\{B(x, \frac{1}{n})\}_{n \in \mathbb{N}}\), which is indeed an open cover of \(X\). By the compactness of \(X\), there exists a finite subcover of this collection, implying the existence of some \(N \in \mathbb{N}\) such that \(B(x, \frac{1}{N})\) is contained within an open set that is part of the finite subcover of \(X\).
The closure of \(B(x, \frac{1}{N})\), denoted by \(\overline{B(x, \frac{1}{N})}\), is a closed subset of the compact space \(X\). By the properties of compact spaces, closed subsets of compact spaces are also compact. Thus, \(\overline{B(x, \frac{1}{N})}\) is compact and contains the open ball \(B(x, \frac{1}{N})\), which is a neighborhood of \(x\). This establishes that \(x\) has a compact neighborhood.
Since the choice of \(x\) in \(X\) was arbitrary, and we have demonstrated that each point has a compact neighborhood, it follows that the metric space \(X\) is locally compact.
This detailed proof leverages the Heine-Borel theorem and the properties of open and closed sets in metric spaces to demonstrate the local compactness of a compact metric space.
\(\blacksquare\)
Problem 7. If \(\dim Y < \infty\) in Riesz's lemma 2.5-4, show that one can even choose \(\theta = 1\).
Proof Using Riesz's Lemma
Let us consider Riesz's lemma in the context where \(Y\) is a finite-dimensional subspace of \(Z\), a subspace of a normed space \(X\). Riesz's lemma asserts that given a closed subspace \(Y\) which is a proper subset of \(Z\), for every real number \(\theta\) in the interval (0,1), there exists a \(z \in Z\) such that \(\|z\| = 1\) and \(\|z - y\| \geq \theta\) for all \(y \in Y\).
Suppose \(v \in Z \setminus Y\) and denote the distance from \(v\) to \(Y\) by \(a\), where \(a = \inf\{\|v - y\| : y \in Y\}\). Since \(Y\) is closed and finite-dimensional, it is also a known fact that closed balls in \(Y\) are compact. Thus, the infimum \(a\) is actually achieved by some \(y_0 \in Y\). We have \(\|v - y_0\| = a\) and \(a > 0\) because \(v\) is not in \(Y\).
We proceed to define \(z\) as the normalization of \(v - y_0\), so \(z = c(v - y_0)\) where \(c = \frac{1}{\|v - y_0\|} = \frac{1}{a}\). This normalization ensures that \(\|z\| = 1\).
For any \(y \in Y\), we can express \(y\) as \(y_1 = y_0 + c^{-1}y\), with \(y_1\) also in \(Y\) due to the vector space properties of \(Y\). The norm \(\|z - y\|\) is then \(\|c(v - y_0) - y\| = c\|v - y_1\|\). Given that \(v\) is closest to \(y_0\) by the very definition of \(a\), it follows that \(c\|v - y_1\| \geq c\|v - y_0\| = c \cdot a = 1\). Consequently, \(\|z - y\| \geq 1\) for all \(y \in Y\).
Since the choice of \(y\) was arbitrary, this implies that \(\|z - y\| \geq \theta\) for any \(\theta \leq 1\). Thus, when \(Y\) is finite-dimensional, it is permissible to select \(\theta = 1\) in Riesz's lemma. The lemma is thereby applicable for \(\theta = 1\), which is due to the structure of the normed space and the finite-dimensionality of \(Y\), guaranteeing the existence of such a \(z\) with the specified characteristics.
\(\blacksquare\)
A little different approach
To show that \(\theta=1\) can be chosen in Riesz's lemma under the condition that the dimension of \(Y\) is finite, we will analyze the proof of Riesz's lemma and demonstrate that if \(Y\) has a finite dimension, then the distance from any \(v \in Z \setminus Y\) to \(Y\) can be made equal to 1, which implies that \(\theta\) can be taken as 1.
Riesz's Lemma states that for any two subspaces \(Y\) and \(Z\) of a normed space \(X\), with \(Y\) being closed and a proper subset of \(Z\), for every \(\theta\) in the interval (0,1), there exists a \(z \in Z\) such that \(\|z\| = 1\) and \(\|z - y\| \geq \theta\) for all \(y \in Y\).
Proof Using Riesz's Lemma
Suppose \(Y\) is a finite-dimensional subspace of \(Z\). By the properties of finite-dimensional normed spaces, we know that closed balls in \(Y\) are compact. Let \(v \in Z \setminus Y\) and denote its distance from \(Y\) by \(a\), that is, \(a = \inf\{\|v - y\| : y \in Y\}\). Since \(Y\) is closed and \(v\) is not in \(Y\), it follows that \(a > 0\).
In the finite-dimensional subspace \(Y\), due to compactness, the infimum \(a\) is actually attained for some \(y_0 \in Y\). That is, there exists a \(y_0 \in Y\) such that \(\|v - y_0\| = a\). Now, define \(z\) as a scaled vector of \(v - y_0\), specifically \(z = c(v - y_0)\), where \(c = \frac{1}{\|v - y_0\|} = \frac{1}{a}\). This scaling ensures that \(\|z\| = 1\).
Now, consider any \(y \in Y\). We examine the distance from \(z\) to \(y\). Note that any \(y\) can be written as \(y_1 = y_0 + c^{-1}y\), where \(y_1 \in Y\) due to \(Y\) being a vector space and thus closed under addition and scalar multiplication. We calculate:
Because \(v\) is closer to \(y_0\) than any other point in \(Y\) by the definition of \(y_0\), it follows that \(c\|v - y_1\| \geq c\|v - y_0\| = c \cdot a = 1\). Therefore, for all \(y \in Y\), \(\|z - y\| \geq 1\), which by the choice of our \(z\) implies \(\|z - y\| \geq \theta\) for any \(\theta \leq 1\). Hence, in the case where \(Y\) has finite dimension, we can choose \(\theta = 1\) in Riesz's lemma.
This shows that the lemma is not only true for any \(\theta\) in the open interval (0,1) but can be strengthened to include \(\theta = 1\) when the subspace \(Y\) is of finite dimension. The lemma holds trivially for \(\theta = 1\) because the normed space structure and finite dimensionality ensure the existence of such \(z\) with the required properties.
\(\blacksquare\)
Problem 8. In Problem 7, Section 2.4, show directly (without using 2.4-5) that there is an \(a > 0\) such that \(a\|x\|_2 \leq \|x\|\). (Use 2.5-7.)
Show directly that there is a constant \(a > 0\) such that \(a\|x\|_2 \leq \|x\|\) for a normed finite-dimensional vector space \(X\) without using the theorem on equivalent norms.
Proof
Let \(X\) be a finite-dimensional vector space equipped with two norms \(\|\cdot\|\) and \(\|\cdot\|_2\), where \(\|\cdot\|_2\) is the standard Euclidean norm. Consider the unit sphere \(S\) in \(X\) with respect to \(\|\cdot\|_2\), that is, \(S = \{x \in X : \|x\|_2 = 1\}\).
Since \(X\) is finite-dimensional, \(S\) is compact with respect to \(\|\cdot\|_2\). Now, define a mapping \(T: S \to \mathbb{R}\) by \(T(x) = \|x\|\) for all \(x \in S\). This mapping is continuous because the norms are continuous functions, and by the Corollary 2.5-7, since \(S\) is compact, \(T\) attains its maximum and minimum values on \(S\).
Let \(m = \min \{T(x) : x \in S\}\). Since all norms on a finite-dimensional space are positive definite, we have \(m > 0\) because if \(m = 0\), there would exist an \(x \in S\) such that \(\|x\| = 0\), which implies \(x = 0\), contradicting the fact that \(x\) is on the unit sphere \(S\).
Now, for any \(x \in X\) with \(x \neq 0\), we can write \(x\) as \(x = \|x\|_2 \cdot \left(\frac{x}{\|x\|_2}\right)\). Notice that \(\frac{x}{\|x\|_2} \in S\), hence \(\left\|\frac{x}{\|x\|_2}\right\| \geq m\). Multiplying both sides by \(\|x\|_2\), we get \(\|x\| = \|x\|_2 \cdot \left\|\frac{x}{\|x\|_2}\right\| \geq m \|x\|_2\).
Set \(a = m\), which is the positive minimum value of \(T\) on the compact set \(S\). We have established that \(a\|x\|_2 \leq \|x\|\) for all \(x \in X\), where \(a > 0\).
This completes the proof, establishing the existence of a positive constant \(a\) that provides a lower bound for the ratio of the norms \(\|\cdot\|\) and \(\|\cdot\|_2\) on a finite-dimensional vector space \(X\).
\(\blacksquare\)
Problem.9 If \(X\) is a compact metric space and \(M \subseteq X\) is closed, show that \(M\) is compact.
Proof
Consider \(X\), a metric space endowed with a metric \(d\), and let \(M \subseteq X\) be a closed subset. Our objective is to substantiate the compactness of \(M\) predicated on the compactness of the ambient space \(X\).
Compactness in a metric space is defined such that a subset \(M\) of \(X\) is compact if every open cover of \(M\) admits a finite subcover. Let us take an arbitrary open cover \(\mathcal{O}\) of \(M\), constituted by a family of open sets in \(X\) such that every point in \(M\) resides within some member of \(\mathcal{O}\).
Given the closure of \(M\) in \(X\), its complement \(X \setminus M\) is open in \(X\). Enhance the open cover \(\mathcal{O}\) of \(M\) by annexing the open set \(X \setminus M\), thereby generating a new open cover \(\mathcal{O}'\) that extends over the entirety of \(X\), for it encompasses every point in \(X\).
The compact nature of \(X\) necessitates that the open cover \(\mathcal{O}'\) of \(X\) must possess a finite subcover, designated as \(\mathcal{O}''\). This finite subcover aptly covers all points in \(X\), and by extension, all points in \(M\).
From the finite subcover \(\mathcal{O}''\), excise the set \(X \setminus M\) should it be included. The residual compendium of sets within \(\mathcal{O}''\) thus forms a finite subcollection originating from the initial cover \(\mathcal{O}\), which adequately covers \(M\). Consequently, \(M\) is furnished with a finite subcover from its open cover \(\mathcal{O}\).
Ergo, \(M\) aligns with the compactness criterion. We have thus rigorously delineated, utilizing the axioms of metric topology alongside the attributes of closed sets nestled within compact spaces, that a closed subset \(M\) of a compact metric space \(X\) is necessarily compact.
This consummates the proof, and we have methodically demonstrated, consistent with the tenets of metric space theory and the inherent properties of closed subsets within compact spaces, that a closed subset \(M\) of a compact metric space \(X\) must itself exhibit compactness.
\(\blacksquare\)