Given $E[0,1]$ be the set of all step functions on $[0,1]$ and $L[0,1]$ be the set of all piecewise linear continuous functions on $[0,1]$.
Then
(a) $(E [0,1], d_{\infty})$ is separable?
(b) $(E [0,1], d_{1})$ is separable?
(c) $(L [0,1], d_{\infty})$ is separable?
First all, I am using the following definitions:
1) $(X,d)$ is called a separable metric space if it contains a countable, dense subset.
2) $d_\infty: X \times X\rightarrow \mathbb{R}$ given by $d_{\infty}(f,g) = \sup_{x \in [0,1]}|f(x)-g(x)|$.
3) $d_1 : X \times X\rightarrow \mathbb{R}$ given by $d_{1}(f,g) = \int_{0}^{1}|f(x)-g(x)|\;dx$.
4) $E[0,1]$ is the set of all functions $f:[0,1] \rightarrow \mathbb{R}$ such that there are $0 = x_0 < x_1 < \cdots < x_n < x_{n+1}=1$, $n \geq 0$, where $f$ is constant in all open subinterval $(x_i, x_{i+1}), i = 0, \ldots, n$.
5) $L[0,1]$ is the set of all continuous functions $f:[0,1] \rightarrow \mathbb{R}$ such that there are $0 = x_0 < x_1 < \cdots < x_n < x_{n+1}=1$, $n \geq 0$, where $f(x) = f(x_i)+
\dfrac{(f(x_{i+1}) – f(x_i))}{x_{i+1}-x_i}(x-x_i)$ if $x \in [x_i, x_{i+1}]$, $0 \leq i \leq n$.
edit Problem solved.
Best Answer
Let $$\mathscr P_n=\{\{0=x_0<x_1<\cdots<x_n<x_{n+1}=1\}:x_i\in\mathbb Q\},$$ let $$\mathscr P=\bigcup_{n=0}^\infty\mathscr P_n,$$ let $$\mathscr Q_P=\{(x_i,x_{i+1})\}_{i=0}^n\text{ for every }P=\{0=x_0<x_1<\cdots<x_n<x_{n+1}=1\}\in\mathscr P,$$ and let $$S=\{f:[0,1]\to\mathbb Q:P\in\mathscr P\text{ and }f\text{ is constant on every }I\in\mathscr Q_P\}.$$ Can you show that $S$ is a countable, dense subset of $E[0,1]$ with respect to the $d_1$ metric?
Alternatively, you can show that $[0,1]$ is a separable measure space, so that $L^1[0,1]\supset E[0,1]$ is separable and thus $E[0,1]$ is separable.