# The quotient norm on $\ell^{\infty}(\mathbb{N}) / c_0 (\mathbb{N})$ is given by $\limsup_{n \in \mathbb{N}} |x_n|$.

I try to show that the norm on the quotient space $\ell^{\infty}(\mathbb{N}) / c_0 (\mathbb{N})$ is given by $\limsup_{n \in \mathbb{N}} |x_n|$, where $x = (x_n)_{n \in \mathbb{N}} \in \ell^{\infty} (\mathbb{N})$. My attempt is the following:

By definition, the quotient norm $\| \cdot \|_{\ell^{\infty} / c_0}$ is defined as $$\| x + c_0 (\mathbb{N}) \|_{\ell^{\infty} / c_0} = d( x + c_0 (\mathbb{N}), 0 + c_0 (\mathbb{N}) ),$$ which can be re-written as $$\| x + c_0 (\mathbb{N}) \|_{\ell^{\infty} / c_0} = \inf \{ \| x + y \|_{\infty} \; | \; y \in c_0 (\mathbb{N}) \}.$$ Since $0 = (0,0,...) \in c_0 (\mathbb{N})$, it follows that $$\| x + c_0 (\mathbb{N}) \|_{\ell^{\infty} / c_0} \leq \| x \|_{\infty} = \sup_{n \in \mathbb{N}} |x_n|.$$

I thought about using Bolzano-Weiserstrass' Theorem at a certain point, but I am not sure how to obtain the wished equality from here.

Any help is appreciated. Thanks in advance.

I'll prove it showing both inequality's, first since $$y_n = (x_1,x_2,...,x_n,0,0,...)\in c_0$$ for all $$n\in\mathbb{N}$$ one has $$\| x + c_0 (\mathbb{N}) \|_{\ell^{\infty} / c_0} \leq \inf_n \| x-y_n\|_\infty = \inf_n \sup_{m\geq n} |x_n| = \limsup_n |x_n|.$$ On the other hand by definition, for all $$y=(y_1,y_2,y_3,...)\in c_0$$ there is an $$n\geq 1$$ such that $$y_m =0$$ for all $$m\geq n$$, hence $$\| x+y\|_\infty \geq \sup_{m\geq n}|x_n| \geq \limsup_n |x_n|$$
$$\| x + c_0 (\mathbb{N}) \|_{\ell^{\infty} / c_0} = \inf \{\|x+y\|_\infty\;|\; y\in c_0\} \geq \limsup_n |x_n|.$$
• I think maybe you are unclear on what $c_0$ is - it is the space of sequences which converge to zero. To define a norm on a quotient space, we require the subspace to be closed, so the space of finite sequences cannot be quotiented. – Jason Aug 30 '17 at 15:15
• It's not much different to do directly - first fix $\varepsilon>0$, and then instead of "$y_m=0$ for all $m\ge n$", write "$|y_m|<\varepsilon$ for all $m\ge n$". You then get $\|x+y\|_\infty\ge\limsup_n|x_n|-\varepsilon$, from which you can conclude. – Jason Aug 30 '17 at 17:54