I am trying to show that $\ell_{\infty}/c_0$ is not separable. I am aware of a more general proof which involves showing that if $X/M$ and $M$ are separable, then $X$ must be separable. However, I am trying show a more constructive proof. This is what I have done so far:
Define $B:=\{x=(x_1, x_2, \ldots) : x_i =0 \text{ or } x_i=1 \ \forall \ i \in \mathbb{N}\}$. This is the set of all sequences that have entries either $0$ or $1$. Let $x$, $y \in B$. Then, we claim that if $x_i \neq y_i$ for finitely many $i$'s, then $x$ and $y$ belong in the same coset of $\ell_{\infty}/c_0$. Notice that if $x_i \neq y_i$ for finitely many $i$'s, then there exists an $M \in \mathbb{N}$ such that for every $m > M$, $x_m-y_m=0$, implying that $x-y \in c_0$. On the other hand, we claim that if $x_i \neq y_i$ for infinitely many $i's$, then $\|x-y-c_0\|=1$. Notice that $x_i \neq y_i$ for infinitely many $i's$ implies that for every $M \in \mathbb{N}$ there exists $m > M$, such that $x_m \neq y_m$, implying $|x_m -y_m|=1$. Then, $\|{x-y}\|_{\ell_{\infty}/c_0}=\lim \sup_{n} |x_n -y_n|=1$. So in this scenario, $x$ and $y$ are distinct elements in $\ell_{\infty}/c_0$, that have distance $1$.
If $B/c_0$ were uncountable, things would fall nicely in place for we would have the following:
Now, if we consider the elements in $B/c_0$, we can take $r=1/2$, then $\{ \mathscr{B}_{r}(x)\}_{x \in B/ c_0}$ gives us a uncountable set of disjoint open balls in $\ell_{\infty}/c_0$. If a set $A$ were dense, its elements would have to belong in each of these balls, implying that the $A$ must be uncountable. Thus, $\ell_{\infty}/c_0$ is not separable.
However, the issue is that I have been stuck trying to show that the set of elements in $B/c_0$ is uncountable for days now. At this point I no longer know if this set is even uncountable anymore. Any help would be appreciated.
Best Answer
Note that if $x,y\in B$ belong to the same coset, then not only $x-y\in c_0$, but each component of $x-y$ can only be $-1,0,1$, so $x-y\in \bigcup_n\{-1,0,1\}^n\times \{0\}^\mathbb N=:c_0^B$. Now clearly $c_0^B$ is countable, so for each $x$ there exist at most countably many $y\in B$ that are in the same subset. But clearly $B$ corresponds to the power set of $\mathbb N$ and must thus be uncountable. Combining these two things you get that $B/c_0^B$ must be uncountable, for else:
If $B/c_0^B$ only contains at most countably many classes, then $$ B \subset \bigcup_{x\in B/c_0^B}x $$ would be an at most countable union of at most countable sets, thus at most countable.