Consider the normed space
$(l^\infty, \|\cdot\|) = \{u = \{u_n\}{n \in \mathbb{N}} \subset \mathbb{C} \text{ such that } \|u\| := \sup{n \in \mathbb{N}} |u_n|\}.$
I attempted to prove that $(l^\infty, \|\cdot\|)$ is complete as following, but encountered several problems towards the end. Please provide a solution.
Let $u^n = (u^n_1, u^n_2, \ldots)$ and assume that $\{u^n\}_{n \in \mathbb{N}}$ is a Cauchy sequence in $l^\infty$ , so we have
$\forall \varepsilon > 0, \exists N \in \mathbb{N} \text{ such that for all } m, n \geq N, \Rightarrow \|u^m - u^n\| < \varepsilon.$
Fix $k \in \mathbb{N}$ . Then, we have
$|u^m_k - u^n_k| \leq \sup_{k \in \mathbb{N}} |u^m_k - u^n_k| = \|u^m - u^n\| < \varepsilon$,
which means that $\{u^n_k\}_{n \in \mathbb{N}}$ is a Cauchy sequence in $K$ . Since $K = \mathbb{C}$ or $\mathbb{R}$ is complete, the limit $u_k = \lim_{n \to \infty} u^n_k \in K$ exists. Now, let $u = \{u_n\}_{n \in \mathbb{N}}$ and show that $u \in l^\infty $ and $ |u^n - u| \to 0 $ as $n \to \infty$ .
Showing that $u \in l^\infty$ is straightforward by using the boundedness of Cauchy sequences in $K$ , so I will omit it.
Next, I will show that $\|u^n - u\| \to 0 $ as $ n \to \infty$ , but here is where the problem arises. Since $\{u^n\}$ is a Cauchy sequence, we have
$|u^m - u^n| < \varepsilon $ for $m, n \geq N$. For any $k \in \mathbb{N}$ , the inequality $|u^m_k - u^n_k| < \sup_{k \in \mathbb{N}} |u^m_k - u^n_k| = |u^m - u^n|$ holds. Therefore, we obtain the following:
$\forall \varepsilon > 0, \forall k \in \mathbb{N}, \exists N > 0 \text{ such that for all } m, n \geq N, \Rightarrow |u^m_k - u^n_k| < \varepsilon.$
At this point, I tried to conclude that $\sup_{k \in \mathbb{N}} |u^m_k - u_k| < \varepsilon$ by letting $n \to \infty$ , using the fact that $ |u^m_k - u_k| < \varepsilon$ , but the uniformity over $k$ is lost when taking the limit as $n$ goes to infinity, making it difficult to justify taking the supremum at the end.
To clarify this issue, let’s rewrite it using the $\varepsilon$-$N$ method:
Since $u^n_k \to u_k$ as $n \to \infty$ , we have
$\forall \varepsilon > 0, \exists N_k \in \mathbb{N} \text{ such that for all } n \geq N_k, |u^n_k - u_k| < \varepsilon.$
Let $\tilde{N} = \max\{N, N_k\}$ . For $n \geq \tilde{N}$ ,
$|u^n_k - u_k| = |u^n_k - u^{\tilde{N}}_k + u^{\tilde{N}}_k - u_k| \leq |u^n_k - u^{\tilde{N}}_k| + |u^{\tilde{N}}_k - u_k| < 2\varepsilon.$
However, $\tilde{N}$ still depends on $k$ .
To resolve this problem, do I need to use a different approach? Or is there a way to address and justify my concerns? I would appreciate any advice on how to proceed.