Problem about superior limit could someone help with this problem?
Let be $x \in \ell_\infty$, prove that 
$$\inf \{ \| x-c \|_\infty : c \text{ is a sequence which converges to zero} \} = \limsup |x_n|.$$
Thanks.
 A: Let $c_0$ be the space of all sequences that converge to zero and let $$\lambda=\inf\{\|x-c\|_\infty|\ c\in c_0\}.$$ We will prove that for each $\epsilon>0,$ there exists $n_\epsilon\in \mathbb{Z^+}$ such that for each $n>n_\epsilon,|x_n|<\lambda+\epsilon$ and for each $n\in \mathbb{Z^+},$ there exists $m>n$ such that $|x_m|>\lambda-\epsilon.$
Let $\epsilon>0$. Then there exists $c\in c_0$ such that $\|x-c\|_\infty<\lambda+\dfrac{\epsilon}{2}$. Since $c\in c_0$ we can choose $n_\epsilon\in\mathbb{Z^+}$ such that for each $n>n_\epsilon,\ |c_n|<\dfrac{\epsilon}{2}$. Now let $n>n_\epsilon$. Then $|x_n|-|c_n|<|x_n-c_n|\leq\|x-c\|_\infty<\lambda+\dfrac{\epsilon}{2}$. Therefore for each $n>n_\epsilon,$ $|x_n|<\lambda+\dfrac{\epsilon}{2}+|c_n|<\lambda+\epsilon.$ 
Suppose there exists $\epsilon_0>0$ and some $N\in\mathbb{Z^+}$ such that for each $n>N,|x_n|<\lambda-\epsilon_0$. Put $c_n=x_n$ for each $1\leq n\leq N$ and $c_n=0$ for each $n>N$. Then the sequence $c'=(c_n)\in c_0$. Therefore $\lambda\leq\|x-c'\|_\infty=\sup_{n>N}|x_n|\leq \lambda-\epsilon_0$ which is a contradiction. Hence for each $\epsilon>0$ and for each $n\in\mathbb{Z^+},$ there exists $m>n$ such that $|x_m|>\lambda-\epsilon.$
Hence we conclude that $\lambda=\limsup|x_n|$.
Edit: Suppose $x_n$ is a bounded sequence. Let $E$ be the set of real numbers $x$ such that $x_{n_k}\to x$ for some subsequence $(x_{n_k})$ of $(x_n)$. Then $\limsup x_n=\sup E$. This is the definition of limsup from Rudin's Principles of Mathematical Analysis.
Now we'll prove that the statement "for each $\epsilon>0,$ there exists $n_\epsilon\in \mathbb{Z^+}$ such that for each $n>n_\epsilon,x_n<\lambda+\epsilon$ and for each $n\in \mathbb{Z^+},$ there exists $m>n$ such that $x_m>\lambda-\epsilon$ " implies $\lambda=\limsup x_n$.
Claim 1: $\lambda\in E$.
Proof: Let $\epsilon>0$. Choose $n_\epsilon\in\mathbb{Z^+}$ such that $x_n<\lambda+\epsilon$ for each $n>n_\epsilon$. Choose $n_1>n_\epsilon$ such that $x_{n_1}>\lambda-\epsilon.$ Next choose $n_2>n_1$ such that $x_{n_2}>\lambda-\epsilon.$ In this manner we choose an increasing sequence of integers $(n_r)$ such that $x_{n_r}>\lambda-\epsilon$. Since $n_r>n_\epsilon,\ x_{n_r}<\lambda+\epsilon$. So for each $r, \lambda-\epsilon<x_{n_r}<\lambda+\epsilon.$ Therefore $x_{n_r}\to \lambda$ and hence $\lambda\in E$.
Claim 2: For each $l\in E,\ l\leq \lambda$.
Proof: Let $l\in E$ and $\epsilon>0$. Then there exists some subsequence $(x_{n_r})$ of $(x_n)$ such that $x_{n_r}\to l$. Choose $R\in\mathbb{Z^+}$ such that $n_R>n_\epsilon.$ Then for each $r>R,\ x_{n_r}<\lambda+\epsilon$. Therefore $l\leq\lambda+\epsilon.$ Hence $l\leq\lambda$.
Now by claim 1 and 2 we have $\lambda=\limsup x_n$.
