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.

  • $\begingroup$ See tinyurl.com/y2o2t3ky for why $(E([0,1]),d_\infty)$ is not separable. $\endgroup$ – wjm Apr 1 '19 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 '19 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 '19 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 '19 at 23:27
  • $\begingroup$ @NateEldredge I found this result math.stackexchange.com/questions/195070/… $\endgroup$ – Thiago Alexandre Apr 1 '19 at 23:30

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.

  • 1
    $\begingroup$ $S$ 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 '19 at 0:50
  • 1
    $\begingroup$ 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 '19 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.