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 $\mathbb{R}^n$:

1. 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 $\mathbb{R}^n$.

2. 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 $\mathbb{C}^n$:

1. 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.

2. 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:

1. Unit Circle: The set of all points $(x, y)$ such that $x^2 + y^2 = 1$. This curve is closed and bounded.

2. Square: The boundary defined by the set of all points $(x, y)$ with vertices at $(\pm1, \pm1)$. It is a closed and bounded shape.

3. 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.

4. 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:

1. 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.

2. Hyperbola: The set of points $(x, y)$ satisfying $xy = 1$. This curve extends to infinity in all directions.

3. Infinite Line: A line like $y = x$ that extends without bound in both directions.

4. 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:

1. A space is locally compact if each point has a compact neighborhood.

2. 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.

3. 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.

$\blacksquare$