# Step and piecewise continuous linear function on $[0,1]$ are separable?

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.

• See tinyurl.com/y2o2t3ky for why $(E([0,1]),d_\infty)$ is not separable. – wjm Apr 1 '19 at 20:58
• @Cleric I got it. We take $Y \subset E[0,1]$ with $d_\infty (f_t, f_s) = 1, \forall t \neq s$. How Y is uncountable, then $(E[0,1], d_\infty)$ is not separable. – Thiago Alexandre Apr 1 '19 at 21:58
• Your argument for (b) is wrong. I think you misled yourself by writing $|t-s|=r$, because the distance between $f_t$ and $f_s$ varies with $t$ and $s$, and in particular can become arbitrarily small. To show a set is not separable by this argument, you must find a fixed constant $c$ and an uncountable set of elements for which any pair is more than distance $c$ apart. You have not done that, and in fact it is not possible; $E[0,1]$ is actually separable with respect to the $d_1$ metric. – Nate Eldredge Apr 1 '19 at 23:15
• @NateEldredge Thanks. I understand my wrong. So I need to find a subset dense and countable in $(E[0,1], d_1)$. – Thiago Alexandre Apr 1 '19 at 23:27
• @NateEldredge I found this result math.stackexchange.com/questions/195070/… – Thiago Alexandre Apr 1 '19 at 23:30

Let $$\mathscr P_n=\{\{0=x_0 let $$\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 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.
• $S$ is countable because it is taken on all countable partitions $P$ and $S$ is dense because given $\epsilon > 0$, $g \in E[0,1]$, there is a $f \in S$ such that $\int_{0}^{1}|f(x)-g(x)|\;dx < \epsilon$. This is because $g$ has a finite partition $P$ and there is a partition $P_n$ of rational ones that approximate $f$ of $g$ making the integral as small as it wants. Is it make sense? – Thiago Alexandre Apr 2 '19 at 0:50
• I got it. For any step function $f \in E[0,1]$ there is a partition $P$ that approaches by rational numbers for the partition of $f$ function. So, we have $g \in S$ such that $d_1(f,g) = |c_1 - q_1| + |c_2 - q_2| + \cdots + |c_{n+1}-q_{n+1}| < \epsilon$ because each parcel |c_i - q_i| we can get that is smaller than $\frac{\epsilon}{n+1}$. – Thiago Alexandre Apr 3 '19 at 14:42