Properties Of Complex N-Space $C^n$ Complete Normed And Compact

In the realm of mathematics, understanding the properties of different spaces is crucial for various applications. One such space is the complex n-space, denoted as $C^n$. This space, which consists of ordered n-tuples of complex numbers, plays a significant role in fields like linear algebra, functional analysis, and quantum mechanics. This article delves into the fundamental properties of $C^n$, focusing on whether it is a complete space, a compact space, and a normed space. We will explore each of these concepts in detail, providing definitions, explanations, and examples to clarify the characteristics of $C^n$.

Before diving into the properties, it's essential to define what $C^n$ actually is. Complex n-space, denoted as $C^n$, is the set of all ordered n-tuples of complex numbers. Mathematically, it can be represented as:

$C^n = \{ (z_1, z_2, ..., z_n) \mid z_i \in C \text{ for } i = 1, 2, ..., n \}$

Each $z_i$ is a complex number, which can be written in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit ($i^2 = -1$). For example, in $C^2$, an element might look like $(2 + 3i, 1 - i)$, and in $C^3$, it could be $(1, i, 2 - i)$.

$C^n$ is more than just a set; it's a vector space over the field of complex numbers. This means we can perform vector addition and scalar multiplication within $C^n$. Vector addition is defined component-wise:

$(z_1, z_2, ..., z_n) + (w_1, w_2, ..., w_n) = (z_1 + w_1, z_2 + w_2, ..., z_n + w_n)$

Scalar multiplication is also defined component-wise:

$c(z_1, z_2, ..., z_n) = (cz_1, cz_2, ..., cz_n)$

where $c$ is a complex scalar. These operations make $C^n$ a vector space, which is a crucial foundation for understanding its properties.

A normed space is a vector space on which a norm is defined. A norm is a function that assigns a non-negative real number to each vector, representing its length or magnitude. To determine if $C^n$ is a normed space, we need to define a norm on it and verify that it satisfies the properties of a norm. The most common norm used in $C^n$ is the Euclidean norm (also known as the 2-norm), denoted as $||.||_2$, which is defined as follows:

$||(z_1, z_2, ..., z_n)||_2 = \sqrt{|z_1|^2 + |z_2|^2 + ... + |z_n|^2}$

Here, $|z_i|$ represents the modulus (or absolute value) of the complex number $z_i$. The modulus of a complex number $z = a + bi$ is given by $|z| = \sqrt{a^2 + b^2}$. The Euclidean norm essentially calculates the length of the vector in $C^n$ using the Pythagorean theorem in n dimensions.

To confirm that this is indeed a norm, we must verify the following properties for all vectors $x, y \in C^n$ and scalar $c \in C$:

1. Non-negativity: $||x|| \geq 0$, and $||x|| = 0$ if and only if $x = 0$.
2. Homogeneity: $||cx|| = |c| \cdot ||x||$.
3. Triangle inequality: $||x + y|| \leq ||x|| + ||y||$.

Let's verify these properties for the Euclidean norm:

1. Non-negativity: Since $|z_i|^2$ is non-negative for all $i$, their sum is also non-negative, and the square root of a non-negative number is non-negative. Thus, $||x||_2 \geq 0$. If $||x||_2 = 0$, then $\sqrt{|z_1|^2 + |z_2|^2 + ... + |z_n|^2} = 0$, which implies $|z_i| = 0$ for all $i$, meaning $z_i = 0$ for all $i$, and hence $x = 0$.
2. Homogeneity: $||cx||_2 = \sqrt{|cz_1|^2 + |cz_2|^2 + ... + |cz_n|^2} = \sqrt{|c|^2|z_1|^2 + |c|^2|z_2|^2 + ... + |c|^2|z_n|^2} = |c|\sqrt{|z_1|^2 + |z_2|^2 + ... + |z_n|^2} = |c| \cdot ||x||_2$.
3. Triangle inequality: This property is a bit more involved and requires the Cauchy-Schwarz inequality. For complex numbers, the Cauchy-Schwarz inequality states that for any two vectors $x, y \in C^n$:

$|\sum_{i=1}^{n} x_i \overline{y_i}| \leq ||x||_2 \cdot ||y||_2$

where $\overline{y_i}$ is the complex conjugate of $y_i$. Using this, we can show that the triangle inequality holds.

Therefore, $C^n$ equipped with the Euclidean norm is indeed a normed space. This is a fundamental property that allows us to measure distances and define concepts like convergence and continuity in $C^n$.

A complete space (also known as a Cauchy space) is a metric space in which every Cauchy sequence converges to a limit within the space. To determine if $C^n$ is complete, we need to understand what a Cauchy sequence is and whether all such sequences in $C^n$ converge within $C^n$.

A sequence $(x_k)$ in $C^n$ is called a Cauchy sequence if for every $\epsilon > 0$, there exists an integer $N$ such that for all $m, k > N$, we have $||x_m - x_k|| < \epsilon$. In simpler terms, the terms of the sequence become arbitrarily close to each other as the sequence progresses.

To show that $C^n$ is complete, we need to demonstrate that every Cauchy sequence in $C^n$ converges to a limit that is also in $C^n$. Let $(x_k)$ be a Cauchy sequence in $C^n$, where $x_k = (z_{k1}, z_{k2}, ..., z_{kn})$. For each component $j$ (where $1 \leq j \leq n$), the sequence of complex numbers $(z_{kj})$ is a Cauchy sequence in the complex plane $C$. Since the complex plane $C$ is complete (a well-known result), each of these component sequences converges to a limit, say $z_j \in C$.

Let $x = (z_1, z_2, ..., z_n)$. We claim that the Cauchy sequence $(x_k)$ converges to $x$ in $C^n$. To show this, we need to prove that for every $\epsilon > 0$, there exists an integer $N$ such that for all $k > N$, $||x_k - x|| < \epsilon$.

Since each component sequence $(z_{kj})$ converges to $z_j$, for every $\epsilon > 0$, there exists an integer $N_j$ such that for all $k > N_j$, $|z_{kj} - z_j| < \frac{\epsilon}{\sqrt{n}}$. Let $N = \max(N_1, N_2, ..., N_n)$. Then, for all $k > N$, we have:

$||x_k - x|| = \sqrt{|z_{k1} - z_1|^2 + |z_{k2} - z_2|^2 + ... + |z_{kn} - z_n|^2} < \sqrt{\frac{\epsilon^2}{n} + \frac{\epsilon^2}{n} + ... + \frac{\epsilon^2}{n}} = \sqrt{\epsilon^2} = \epsilon$

Thus, the Cauchy sequence $(x_k)$ converges to $x$ in $C^n$, and since $x$ is an element of $C^n$, we conclude that $C^n$ is a complete space. This property is crucial for many analytical results, as it ensures that certain types of sequences have limits within the space.

A compact space is a topological space in which every open cover has a finite subcover. In the context of metric spaces (like $C^n$ with the Euclidean norm), compactness is equivalent to being both complete and totally bounded. We already know that $C^n$ is complete. Now, we need to examine whether $C^n$ is totally bounded.

A subset $S$ of a metric space is totally bounded if for every $\epsilon > 0$, there exists a finite set of points $x_1, x_2, ..., x_m$ in the space such that $S$ is covered by the union of open balls of radius $\epsilon$ centered at these points. In other words, we can cover $S$ with finitely many small balls.

However, $C^n$ itself is not totally bounded. To see why, consider the closed unit ball in $C^n$, which is the set of all vectors with a norm less than or equal to 1:

$B = \{ x \in C^n \mid ||x|| \leq 1 \}$

If $C^n$ were totally bounded, then this closed unit ball would also be totally bounded. However, it is not. For any $\epsilon$ with $0 < \epsilon < 1$, we cannot cover the unit ball with finitely many balls of radius $\epsilon$. This can be shown by constructing an infinite sequence of points in the unit ball that are pairwise separated by a distance greater than $\epsilon$.

Since $C^n$ is not totally bounded, it is not compact. However, closed and bounded subsets of $C^n$ are compact. This is a consequence of the Heine-Borel theorem, which states that a subset of $R^n$ (and analogously, $C^n$) is compact if and only if it is closed and bounded.

Therefore, while $C^n$ itself is not a compact space, its closed and bounded subsets are. This distinction is important in many applications, especially in analysis and topology.

In summary, we have explored the properties of complex n-space, $C^n$, and determined that it is a normed space and a complete space. The Euclidean norm provides a way to measure distances, and the completeness property ensures that Cauchy sequences converge within the space. However, $C^n$ itself is not a compact space, although its closed and bounded subsets are compact. These properties collectively define the nature of $C^n$ and make it a fundamental space in various areas of mathematics and physics. Understanding these characteristics is essential for further exploration and applications in these fields.

The correct answers are:

A. Complete space C. Normed space