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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.

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  • $\begingroup$ See tinyurl.com/y2o2t3ky for why $(E([0,1]),d_\infty)$ is not separable. $\endgroup$ – wjm Apr 1 at 20:58
  • $\begingroup$ @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. $\endgroup$ – Thiago Alexandre Apr 1 at 21:58
  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Apr 1 at 23:15
  • $\begingroup$ @NateEldredge Thanks. I understand my wrong. So I need to find a subset dense and countable in $(E[0,1], d_1)$. $\endgroup$ – Thiago Alexandre Apr 1 at 23:27
  • $\begingroup$ @NateEldredge I found this result math.stackexchange.com/questions/195070/… $\endgroup$ – Thiago Alexandre Apr 1 at 23:30
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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.

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    $\begingroup$ $S$ 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? $\endgroup$ – Thiago Alexandre Apr 2 at 0:50
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    $\begingroup$ 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 $\frac{\epsilon}{n+1}$. $\endgroup$ – Thiago Alexandre Apr 3 at 14:42

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