Question:
Prove $\ell^\infty$ is complete.
I'm very close to proving this, but I am stuck on essentially the final step. My proof goes as follows:
Let $\{x^{(n)}\}_{n \in \Bbb N}$ be a Cauchy sequence in $\ell^\infty$ (note that each $x^{(n)}$ is itself a sequence in $\ell^\infty$). Let $\epsilon \gt 0$. Then $\exists N \in \Bbb N$ such that: $$d_\infty(x^{(n)}, x^{(m)}) = sup_k|x^{(n)}_k - x^{(m)}_k| \lt \epsilon, \forall n,m \ge N$$ In particular: $$|x^{(n)}_k - x^{(m)}_k| < \epsilon, \forall k \in N, \forall n,m \ge N$$ Then, for each $k \in \Bbb N$, $\{x_k^{(n)}\}_{k \in N}$ is a Cauchy sequence in $\Bbb R$.
Since $\Bbb R$ is complete, we have $\forall k \in \Bbb N$, as $n \rightarrow \infty$: $$x_k^{(n)} \rightarrow y_k$$ Consider the sequence $\{y_k\}_{k \in \Bbb N} = y$.
We show $\{x^{(n)}\}$ converges to y. Again, choose an $\epsilon \gt 0$. Then $\exists N \in \Bbb N$ such that $$|x^{(n)}_k - x^{(m)}_k| \lt \epsilon / 2, \forall k \in \Bbb N, \forall n,m \ge N$$ Taking $m \rightarrow \infty$, we get: $$|x^{(n)}_k - y_k| \le \epsilon / 2, \forall k \in \Bbb N, \forall n \ge N$$ Then: $$sup_k|x_k^{(n)} - y_k| \le \epsilon / 2, \forall n \ge \Bbb N$$ Thus: $$d_{\infty}(x^{(n)},y) \le \epsilon / 2 \lt \epsilon, \forall n \ge \Bbb N$$ Hence, we have that $x^{(n)} \rightarrow y$ as $n \rightarrow \infty$.
I now have to show that y is indeed in $\ell^{\infty}$, and this is where I'm stuck.
We must show that $sup_k|y_k| \lt \infty$. We've defined $y_k$ as the limit of the sequence $\{x_k^{(n)}\}_{n \in \Bbb N}, \forall k \in \Bbb N$ What is to say that the limits of each sequence in the sequences of sequences $\{x_k^{(n)}\}_{n \in \Bbb N}$ are not monotonically increasing (over k)? In this case, couldn't the sequence $y_k$ diverge?
Any hints or corrections of my proof are appreciated!
You already have the result.
Since $|w_k^\,| \le \|w\|_\infty$, we see that each element $x_k^n$ is Cauchy and hence converges to some limit value $x_k^n \to x_k^\,$.
You have shown that for any $\epsilon>0$ there is some $N$ such that for $n \ge N$ we have $\|x^n-x\|_\infty < \epsilon$. Choose $\epsilon =1$ and let $N$ be the corresponding index.
Then note that $|x_k^\,| \le |x_k^N-x_k^\,| + |x_k^N| \le 1 + \|x^N\|_\infty$ from which we get $\|x\|_\infty \le 1 + \|x^N\|_\infty$.