I'm working through Neal Carothers' Real Analysis and I'm stuck on trying to show that the set $B$ of bounded, real-valued functions over $[0,1]$ is not separable. The metric of this set is $$||f-g||=sup{|f(x)-g(x)|:x\in [0,1]}$$ Here is what I have so far:
Consider the set of all sequences whose terms are either $0$ or $1$. Associate with each sequence a bounded function whose value at the $i^{th}$ rational in $[0,1]$ is the $i^{th}$ term of the associated sequence (recall that the rationals are countable, so the set of rationals in $[0,1]$ is countable).
So the set of functions generated by these sequences, say $S$, is uncountable since the set of sequences is uncountable (by Cantor's Diagonal Argument), and $$||f-g||\ge 1$$ if $f,g\in S$, $f\neq g$ since $f$ and $g$ differ at a rational by $1$.
Now assume for contradiction that $B$ is separable, or that $A$ is a countable, dense subset of $B$. Here's where I'm stuck: I want to show there are functions $a\in A, f\in S, g\in S$ such that $$||f-a||<\dfrac{1}{2}, ||g-a||<\dfrac{1}{2}$$ which, by triangle inequality, would give me $||f-g||<1$, and I would have a contradiction.
By the definition of density, every neighbourhood of $h\in S$ contains infinitely many $a\in S$; however, I can't seem to show that two neighbourhoods of radius $\dfrac{1}{2}$ centered at different functions in $S$ share a common point that belongs to $A$. Any help at all would be much appreciated!
Best Answer
Suppose $A$ is the countable dense set. You want to find an injective map from $S$ into $A$, which leads to a contradiction.
Let $\epsilon=\frac{1}{3}$. But definition of dense, for each $s\in S$, $B(s,\epsilon)\cap A\neq \emptyset$. That is $\exists a\in A$, such that $||s-a||<\frac{1}{3}$ Hence we can associate each $s$ with one $a$.
Now we show it's injective, that is, it's not possible that there is $a\in A$, which is associated with $s_1,s_2\in S$, because you can use triangle inequality to derive a contradiction.