I have attempted to prove that every sequence in $\mathbb{C}^k$, equipped with the standard inner product, is convergent if and only if each component sequence is convergent. I am using this result as a stepping stone to proving that $\mathbb{C}^k$ is a Hilbert space with respect to the standard inner product.
The reason I am posting my attempt here is that I haven't had any proof writing feedback in a while, so I have an itching feeling that I might be overlooking something. It is summer holidays where I attend university and I am currently overseas, so I can't ask anyone there. Any help is greatly appreciated!
My questions are:
- Is the proof valid?
- Does anyone have suggestions as to how to improve the presentation of the proof?
A detail of the presentation I am wondering about is in the first part of the proof, where the convergence of $(\mathbf{x}_n)$ is invoked:
Therefore, since $(\mathbf{x}_n)$ is convergent to $\mathbf{x}$, there is a natural number $N$ such that for all natural numbers $n \geq N \dots$
A little later on the following is written:
However, for all $i \in \{1,\dots,k\}$, we have $$|x_n^i - x^i|^2 \leq \sum_{j=1}^{k} |x_n^j - x^j|^2 \dots$$
I feel like it might not be clear here that $n$ is a natural number which is greater than or equal to $N$, since I never defined $n$ in a sentence of the form "let $n \dots$" Does anyone have any suggestions regarding how to fix this?
Something similar happens in the second part of the proof, after I write
Secondly, suppose that for all $i \in \{1,\dots, k\}$, that the sequence $(x_n^i)$ converges with limit $z^i \in \mathbb{C}.$
And later,
Furthermore, since $(x_n^i)$ converges to $z^i \dots$
I feel like i'm referring to an object $i$ without having defined it, relying on the context of the previous sentence.
I have seen slight variations of the definitions of convergence, some using $\leq$ where others use $<$, so for the sake of clarity, here are the ones I used:
Definition. A sequence $(z_n)_{n \in \mathbb{N}}$ in $\mathbb{C}$ is said to converge with limit $z \in \mathbb{C}$ if for all $\epsilon >0$, there is a natural number $N \in \mathbb{N}$, such that for all $n \in \mathbb{N}$ satisfying $n \geq N$, we have $|z_n - z | < \epsilon.$
Definition. A sequence $(\mathbf{x}_n)_{n \in \mathbb{N}}$ in an inner product space $V$ over $\mathbb{R}$ or $\mathbb{C}$ is said to converge with limit $\mathbf{x} \in V$ if for all $\epsilon>0$, there is a natural number $N$ such that for all natural numbers $n \geq N$, we have $||\mathbf{x}_n - \mathbf{x}|| < \epsilon,$ where $||\mathbf{v}|| := \sqrt{\langle v | v \rangle}$, for all $\mathbf{v} \in V$.
Theorem. Consider the inner product space $\mathbb{C}^k$ under the standard inner product and the associated induced norm. Then a sequence $(\mathbf{x}_n)_{n \in \mathbb{N}}$ in $ \mathbb{C}^k$ is convergent if and only if each coordinate sequence $(x_n^i)_{n \in \mathbb{N}}$ is convergent, where for all $n \in \mathbb{N}$, $x_n := (x_n^1,\dots, x_n^k)$ .
Proof. Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{C}^k$ under the standard inner product, where for every $n \in \mathbb{N}$, $x_n := (x_n^1,\cdots,x_n^k).$
Firstly, suppose that $(\mathbf{x}_n)$ is convergent, with limit $\mathbf{x} := (x^1,\dots, x^k) \in \mathbb{C}^k$. Let $\epsilon > 0$, noting that $\epsilon^2 >0$. Therefore, since $(\mathbf{x}_n)$ is convergent to $\mathbf{x}$, there is a natural number $N$ such that for all natural numbers $n \geq N$, we have \begin{align} ||\mathbf{x}_n - \mathbf{x}|| &< \epsilon^2 \\ \iff \sum_{j=1}^{k} |x_n^j - x^j|^2 &< \epsilon^2. \end{align} However, for all $i \in \{1,\dots,k\}$, we have $$|x_n^i - x^i|^2 \leq \sum_{j=1}^{k} |x_n^j - x^j|^2,$$ since each term in the sum is non-negative, and $|x_n^i - x^i|^2$ is a term of the sum. Hence, $$|x_n^i - x^i|^2 < \epsilon^2 \iff |x_n^i - x^i| < \epsilon,$$ since $|x_n^i - x^i|$ is non-negative and $\epsilon >0$. Therefore, for all $i \in \{1,\dots,k\}$, the sequence $(x_n^i)$ converges with limit $x^i$.
Secondly, suppose that for all $i \in \{1,\dots, k\}$, that the sequence $(x_n^i)$ converges with limit $z^i \in \mathbb{C}$, with $\mathbf{z} := (z^1,\dots , z^k)$. Let $\delta >0$, and $\gamma := \sqrt{\frac{\delta}{k}} >0$. Furthermore, since $(x_n^i)$ converges to $z^i$, there is a natural number $N_i$ such that for all $n \in \mathbb{N}: \ n \geq N_i$, we have $$|x_{n}^i - z^i| < \gamma.$$ Let $M := \max\{N_1,\dots,N_k\}$, and $m \in \mathbb{N}$ such that $m \geq M$. It then follows that for each $i \in \{1,\dots, k\}$, $$|x_m^i - z^i| < \gamma \iff |x_m^i - z^i|^2 < \gamma^2,$$ since $|x_m^i - z^i|$ is non-negative and $\gamma$ is strictly positive. Therefore, \begin{align} ||\mathbf{x}_m - \mathbf{z}|| &= \sum_{j=1}^k |x_m^j - z^j|^2 \\ &< k \gamma^2 \\ &= k \left| \frac{\delta}{k}\right| \\ &= \delta. \end{align} Therefore, the sequence $(\mathbf{x}_n)$ is convergent with limit $\mathbf{z}$. This concludes the proof of the theorem.
Like $\mathbb{C}^k$ is a finite dimensional vector space, all norms are equivalent, hence you can use this norm: $$ \|z_n-z\|_\infty = \max_i|z_n^i-z^i| $$ for which the theorem proof is much shorter.