# How to decide completeness of $\ell^\infty$?

Let $\ell^\infty$ denote the set of all bounded sequences $x \colon = (\xi_j)_{j=1}^\infty$, $y \colon= (\eta_j)_{j=1}^\infty$ of complex numbers with the metric $d$ defined as follows: $$d(x,y) \colon= \sup_{j\in\mathbb{N}} |\xi_j - \eta_j|.$$

Then how to determine if $\ell^\infty$ is complete?

My work:

Let $(x_n)$, where $x_n \colon= (\xi_j^{(n)})$, be a Cauchy sequence in $\ell^\infty$. Then, given $\epsilon > 0$, there exists an integer $N$ such that $m$, $n > N$ implies that $$d(x_m, x_n) = \sup_{j\in\mathbb{N}} |\xi_j^{(m)} - \xi_j^{(n)}| < \epsilon.$$ So, for each $j \in \mathbb{N}$, we have $$|\xi_j^{(m)} - \xi_j^{(n)}| < \epsilon,$$ from which it follows that, for each $j\in \mathbb{N}$, the sequence $(\xi_j^{(n)})_{n=1}^\infty$ is a Cauchy sequence in the complete mertic space $\mathbb{C}$, the set of complex numbers under the usual metric, and hence this sequence is convergent; let $$\xi_j \colon= \lim_{n\to\infty} \xi_{j}^{(n)}$$ for each $j \in \mathbb{N}$.

Let $x \colon= (\xi_j)_{j=1}^\infty$. We now need to show that $x\in \ell^\infty$ and that $x_n \to x$ as $n \to \infty$. In order to show that $x \in \ell^\infty$, we show that $x$ is a bounded sequence.

Since $\xi_j^{(n)} \to \xi_j$ as $n \to \infty$, there exists $N_j$ such that $n > N_j$ implies that $$|\xi_j^{(n)} - \xi_j| < \epsilon.$$ And since, for example, $x_{N+1} = (\xi_j^{(N+1)} )_{j=1}^\infty$ is a sequence in $\ell^\infty$, it is a bounded sequence of complex numbers; so there is a non-negative real number $k_{N+1}$ such that $$|\xi_j^{(N+1)} | \leq k_{N+1}$$ for each $j \in \mathbb{N}$.

Hence, for each $j \in \mathbb{N}$, we have $$|\xi_j| = | \xi_j - \xi_j^{(N+1)} + \xi_j^{(N+1)} | \leq | \xi_j - \xi_j^{(N+1)} | + |\xi_j^{(N+1)} | < \epsilon + k_{N+1},$$ which shows that $x \in \ell^\infty$.

Am I right so far?

Now how to rigorously prove that $(x_n)$ converges to $x$?

My effort:

Fix $m \in \mathbb{N}$.

Then as, for each $j = 1, \ldots, m$, we have $$\lim_{n\to\infty} \xi_{j}^{(n)} = \xi_j,$$ so, given $\epsilon > 0$, we can find an integer $M_j$ such that $n > M_j$ implies that $$|\xi_j^{(n)} - \xi_j| < \frac{\epsilon}{2}.$$

Now let's take $M \colon= \max(M_1, \ldots, M_m)$.

Then $n > M$ implies $$|\xi_j^{(n)} - \xi_j| < \frac{\epsilon}{2}$$ for each $j= 1, \ldots, m$.

That is, $n > M$ implies that $$\sup \{|\xi_j^{(n)} - \xi_j| \colon j = 1, \ldots, m \} \leq \frac{\epsilon}{2}.$$

What next?

• Well isn't your conclusion what you want to show? I only skimmed through your work, but it looks ok. Maybe I will read it in detail tomorrow. – IAmNoOne Oct 30 '14 at 7:22
• Please do. In the last inequality, I would like to have the supremum over all $j \in \mathbb{N}$. – Saaqib Mahmood Oct 30 '14 at 7:42
• I believe there is a tiny flaw in the part where you try to prove that $x \in \ell^\infty$ as $N_j$ depends on $j$ and so you do not get a uniform bound for the $|\xi_j|$. I think you will have to use that you have a Cauchy-sequence to get a uniform bound. The same applies for the part where you want to show convergence. – Matthias Klupsch Oct 30 '14 at 8:05
• Matthias Klupsch, you see the $N$ doesn't depend on the $j$; the $N$ comes from the fact that $(x_n)$ is a Cauchy sequence. So I think we do have a uniform bound. – Saaqib Mahmood Oct 31 '14 at 4:09

## 1 Answer

For boundedness, I suggest you set $\epsilon=1$, that is $|\xi_j| < 1 + k_{N+1}$ for some $k_{N+1}\in \mathbb{R}$,so that you get a fixed upper bound for $|\xi_j|$.

For convergence,

Since $(x_n)$, where $x_n \colon= (\xi_j^{(n)})$ is a Cauchy sequence in $\ell^\infty$. Then, given $\epsilon > 0$, there exists an integer $N$ such that $\forall m,n > N$ implies that $$d(x_m, x_n) = \sup_{j\in\mathbb{N}} |\xi_j^{(m)} - \xi_j^{(n)}| < \frac{\epsilon}{2}.$$ So, for each $j \in \mathbb{N}$, we have $$|\xi_j^{(m)} - \xi_j^{(n)}| < \frac{\epsilon}{2},$$

Since $\xi_j^{(m)} \to \xi_j$ as $m \to \infty$, let $m \to \infty$, we have $\forall n > N$ implies that $$|\xi_j^{(n)} - \xi_j| \leq \frac{\epsilon}{2}\quad \forall j.$$ That is $$d(x_n, x) = \sup_{j\in\mathbb{N}} |\xi_j^{(n)} - \xi_j| \leq \frac{\epsilon}{2}<\epsilon$$.

Hence $x_n\to x$.