Show that every dense subset of $L^{\infty}([0,1])$ is uncountable.

Question

I received this solution from a friend of mine, and I am unclear on part of his work:

Let $A = \{ \chi_{(0,t)} : t \in (0,1)\}$. Then for any $\chi_{(0,t_1)}, \chi_{(0,t_2)} \in A$ such that $t_1 \neq t_2$ $$\|\chi_{(0,t_1)} - \chi_{(0,t_2)}\|_{\infty} = 1$$ Without loss of generality, $t_1 < t_2$, then $\chi_{(0,t_1)} - \chi_{(0,t_2)} = 0,1,0$ on $(0,t_1),(t_1,t_2),(t_2,1)$ respectively.

Consider $\left\{ B_{\left(\chi_{(0,t)}, \frac{1}{3} \right)} \right\}$ which is an uncountable collection of disjoing balls. And given any dense set $S \subset L^{\infty}([0,1])$ has elements in each ball by definition of density. Thus $S$ is uncountable. QED

My question is in reference to the jump made in the 2nd line of the proof. How is it that $\chi_{(0,t_1)}, \chi_{(0,t_2)} \in A$ such that $t_1 \neq t_2$ implies $\|\chi_{(0,t_1)} - \chi_{(0,t_2)}\|_{\infty} = 1$?

Also, is this proof indeed correct? I believe if the step in question above is, then the rest should be as well. Thanks in advance.

Answer 1

As you correctly deduced, assuming $t_1 < t_2$ we have:

$$\chi_{(0,t_2)}(x) - \chi_{(0,t_1)}(x) = \begin{cases} 0, & \text{if $x \in [0,t_1\rangle$} \\ 1, & \text{if $x \in [t_1, t_2\rangle$} \\ 0, & \text{if $x \in [t_2,1]$} \\ \end{cases}$$

Hence:

$$\left\|\chi_{(0,t_2)} - \chi_{(0,t_1)}\right\|_\infty = \sup_{x\in[0,1]}\left|\chi_{(0,t_2)}(x) - \chi_{(0,t_1)}(x)\right| = 1$$

Answer 2

Recall the definition of the the $L^{\infty}$ norm/metric. For the second line, remember the $L^{\infty}$ distance is $1$ between the characteristic functions.

Answer 3

The proof is indeed correct, and is the standard way of demonstrating why $L^{\infty}([0,1])$ is not separable. That the distance between any two characteristic functions of the form $\chi_{(0,t_1)},\chi_{(0,t_2)}$ is $1$ when $t_1<t_2$ is very easy to verify: the only values that the difference can achieve are $\pm 1$ and zero, and when you choose $t_1<x<t_2$ you clearly get $1$. The rest of the proof is correct.