Here $l^{\infty}$ is the set of bounded sequence
And a simple sequence is a sequence $(x_n)$ such that $\left \{ x_n: n\in\mathbb{N} \right \}$ is finite.
The norm that I'm using is the sup norm.
That is $||x_n||=\sup{(|x_n|:n\geq 1)}$
I know that any measurable function can be approximated by a sequence of simple functions (Done in measure theory).
So here I think we can consider sequence as a measurable function (Domain is $\mathbb{N}$) and thus we will be able to approximate through simple functions. and as the domain of those simple functions is $\mathbb{N}$ we have simple sequences. Am I correct?
Anyway I'm looking for a direct method to obtain the dense nature of bounded sequences ($l^{\infty}$).
That is to show that $\forall \epsilon>0, \forall (l_n)\in l^{\infty} \exists$ a simple sequence $(a_n)$ such that, $||(l_n-a_n)||<\epsilon$
Appreciate your help
Best Answer
Fix $(l_n) \in \ell^\infty$ and $\varepsilon > 0$. We want to show that there is a simple sequence $(a_n)$ such that $\|(l_n - a_n)\|_\infty \leq \varepsilon$.
Since $(l_n) \in \ell^\infty$, $L = \sup_n |l_n| < \infty$ and hence there is an $N \in \mathbb{N}$ such that $[-L,L] \subseteq [-N \varepsilon, N \varepsilon]$. Then, for each $n \in \mathbb{N}$ we know that $l_n \in [-N\varepsilon,N \varepsilon]$ and hence there is an integer $k_n \in [-N,N]$ such that $|k_n \varepsilon - l_n| < \varepsilon$.
To conclude, let $a_n = k_n \varepsilon$ and notice that $\|(l_n - a_n)\|_\infty \leq \varepsilon$ by construction and that $$|\{a_n: n \in \mathbb{N}\}| \leq [-N,N] \cap \mathbb{Z} = 2N < \infty$$ so that $(a_n)$ is a simple sequence.