Let $l^{\infty}(\mathbb{Z^+})$ be the set of all bounded complex functions on $\mathbb{Z^+}$. Then prove that $l^{\infty}(\mathbb{Z^+})$ is not separable.
My attempt: Suppose $E\subset l^{\infty}(\mathbb{Z^+})$ and $\overline E= l^{\infty}(\mathbb{Z^+})$
Each real number $t\in(0,1)$ can be expressed as $$t=\sum_{n=1}^{\infty}\frac{x(n)}{2^n}$$ with $x(n)=0$ or $x(n)=1$, $\forall n$
for each $t\in(0,1)$ we can find $h(t)=x$ in $l^{\infty}(\mathbb{Z^+})$ satisfying the above representation.Then we can find $g(x)$ in $E$ such that $${\|g(x)-x\|}_{\infty}<\frac{1}{4}$$
Let $f(t)=g(h(t))$, $0<t<1$
Then I have proved that $f:(0,1)\to E$ is one-to-one,this proves that $E$ is uncountable. Since $E$ was arbitrary dense set,this shows that $l^{\infty}(\mathbb{Z^+})$ cannot have dense subsets which are countable.Hence $l^{\infty}(\mathbb{Z^+})$ is not separable.
My problem: Is this my proof correct? Because we do this for the set of all real valued bounded functions on $\mathbb{Z^+}$, but here I have complex valued functions.You can also share some other idea to prove this. Thanks in advance.
