Show that a normed space is not separable

Question

Question:

Let $B[0,1]$ be the normed space of bounded real-valued functions on $[0,1]$ with the norm $\|f\| =\sup\left\{\left|f(x)\right| : x\in [0,1]\right\}$. Is this metric space separable?

My problem is that I have no real intuition about this metric space. I have a solution for the problem that uses the "characteristic equation" but do not know how I would come up with this solution. Any insight as well as different solutions would be appreciated.

The characteristic equation is $\mathcal{X}_x(s) = 1$ if $s=x$ , $0$ otherwise

Sorry, I'm not sure how to do a piecewise function.

Answer 1

The family $\{\chi_{[0,a]}\colon 0<a<1\}$ forms an uncountable discrete subset of $B[0,1]$. Every discrete subset of a separable metric space is countable, so $B[0,1]$ is non-separable.

Answer 2

For every $i \in [0,1]$ define $f_i:[0,1] \longrightarrow \mathbb{R}$ such that $f_i(x)=1$ if $x=i$ and $f_i(x)=0$ if $x \neq i$

Let $A=\{f_i| i \in [0,1]\} \subseteq B[0,1]$ the set of these functions,which is clearly uncountable.

We have that $||f_i-f_j||=sup_{x \in [0,1]}|f_i(x)-f_j(x)| = 1,\forall i \neq j$ in $[0,1]$

If you take the collection of disjoint open balls $S=\{B(f_i,\frac{1}{3})| i \in [0,1]\}$ then there does not exist a countable dense subset in $B[0,1]$

If it existed a countable dense subset $M \subseteq B[0,1]$, then from the definiton of denseness we would have that $M \cap B(f_i,\frac{1}{3}) \neq \emptyset, \forall i \in [0,1]$

$M=\{g_1,g_2.....\}$

For every $i \in [0,1]$ pick an element $g_i \in M$ where $g_i \in B(f_i,\frac{1}{3})$

You can find such element because we assumed that $M$ is dense.

Then take the function $\Phi: [0,1] \longrightarrow M$ such that $\Phi(i)=g_i$

This function is $1-1$ because of the disjointness of the open balls.

We proved that $cardinality([0,1]) \leqslant cardinality (M)$ thus $[0,1]$ is countable which is a contradiction.

Answer 3

For each $p \in [0,1]$ let $f_p$ be the function on $[0,1]$ that $f_p(p) = 1$ and $f_p(x) = 0$ for $x \neq p$. Then for $p \neq q$, we have $d(f_p, f_q) = ||f_p - f_q|| = 1$. So if $D$ is a dense set of functions, we can pick for each $p$, some $d_p$ such that $||f_p - d_p||< \frac{1}{2}$,as $B(f_p, \frac{1}{2}) \cap D \neq \emptyset$ by denseness of $D$.

Now, if $p \neq q$: $d_p \neq d_q$ otherwise if $d_p = d_q$:

$$||f_p - f_q|| \le ||f_p - d_p|| + ||d_p-d_q|| + ||d_q - f_q|| < 1$$ contradicting $||f_p-f_q||= 1$. So the map $[0,1] \to D, p \to d_p$ is injective and $D$ has at least size continuum so cannot be countable.